PANIMALAR ENGINEERING COLLEGE

(A CHRISTIAN MINORITY INSTITUTION)

JAISAKTHI EDUCATIONAL TRUST

(ACCREDIATED BY NATIONAL BOARD OF ACCREDITATION)

BANGALORE TRUNK ROAD, VARADHARAJAPURAM,

NASARATHPET, POONAMALLEE,

CHENNAI – 600 123.

DEPARTMENT OF ELECTRONICS AND INSTRUMENTATION

ENGINEERING

IV SEMESTER EIE

QUESTION BANK

(2018-2019 EVEN SEMESTER)

i
 PANIMALAR ENGINEERING COLLEGE
Department of Electronics & Instrumentation Engineering

PROCESS OF DEFINING VISION AND MISSION

PROCESS OF DEFINING PROGRAM EDUCATIONAL OBJECTIVES

ii
 College Vision
To transform the budding engineers into academically excellent, highly intellectual and self
disciplined engineering graduates to mould them as good citizens with the spirit of integrity and
morality that would cater to the needs of our nation.

College Mission
To impart quality education with high standards of excellence in engineering and technology,
to provide an excellent infrastructure in a serene and conducive atmosphere that would motivate the
students in their pursuit of knowledge in the field of engineering and technology.

Department Vision
To produce technically competent Electronics and Instrumentation Engineers with ethical and
moral values, who can tackle the challenges facing the society and industry at national level and
global level.
Department Mission
M1: To provide fundamental knowledge and quality technical education through effective teaching-
learning process.

M2: To prepare graduates to meet the needs of industry through creative projects, soft skills and
industry interaction.

M3: To create passion for learning to pursue higher studies and research towards serving society with
moral and ethical standard.

Program Educational Objectives (PEOs)

PEO 1: To acquire a strong foundation in the mathematical and scientific concept and
applying them in core engineering.

PEO 2: To gain in-depth knowledge in the field of electronics and Instrumentation

Engineering which is necessary to formulate, analyze and solve engineering problems.

PEO 3: To train students in carrying out project works to synthesize the platforms and fields
of engineering.

PEO 4: To produce ethical engineers with adequate soft skill to exhibit professionalism in
multi-disciplinary field.

PEO 5: To create interest for lifelong learning to excel in higher studies and research.

iii
 Program Outcomes (POs)
PO 1: Engineering knowledge: Apply the knowledge of mathematics, science, engineering
fundamentals, and an engineering specialization to the solution of complex engineering
problems.

PO 2: Problem analysis: Identify, formulate, review research literature, and analyze

complex engineering problems reaching substantiated conclusions using first principles of
mathematics, natural sciences, and engineering sciences.

PO 3: Design/development of solutions: Design solutions for complex engineering

problems and design system components or processes that meet the specified needs with
appropriate consideration for the public health and safety, and the cultural, societal, and
environmental considerations.

PO 4: Conduct investigations of complex problems: Use research-based knowledge and

research methods including design of experiments, analysis and interpretation of data, and
synthesis of the information to provide valid conclusions.

PO 5: Modern tool usage: Create, select, and apply appropriate techniques, resources, and
modern engineering and IT tools including prediction and modeling to complex engineering
activities with an understanding of the limitations.

PO 6: The engineer and society: Apply reasoning informed by the contextual knowledge to
assess societal, health, safety, legal and cultural issues and the consequent responsibilities
relevant to the professional engineering practice.

PO 7: Environment and sustainability: Understand the impact of the professional

engineering solutions in societal and environmental contexts, and demonstrate the knowledge
of, and need for sustainable development.

PO 8: Ethics: Apply ethical principles and commit to professional ethics and responsibilities
and norms of the engineering practice.

PO 9: Individual and team work: Function effectively as an individual, and as a member or

leader in diverse teams, and in multidisciplinary settings.

PO 10: Communication: Communicate effectively on complex engineering activities with

the engineering community and with society at large, such as, being able to comprehend and
write effective reports and design documentation, make effective presentations, and give and
receive clear instructions.

iv
PO 11: Project management and finance: Demonstrate knowledge and understanding of
the engineering and management principles and apply these to one’s own work, as a member
and leader in a team, to manage projects and in multidisciplinary environments.

PO 12:Life-long learning: Recognize the need for, and have the preparation and ability to
engage in independent and life-long learning in the broadest context of technological change.

Program Specific Outcomes (PSOs)

PSO 1: Apply the fundamentals of mathematics, science and basic engineering knowledge to
analyze and implement electric and electronic circuits, transducers and measurement devices.

PSO 2: Apply appropriate techniques and tools to identify, formulate, design and analyze
engineering problems in the field of electronics, instrumentation and process control.

PSO 3: Understand ethical issues, environmental impact and acquire managerial skills to
communicate effectively in industry and society.

v
 CONTENT

SUBJECT
S.NO. SUBJECT NAME PAGE NO.
CODE

1 MA8491 Numerical Methods 1

2 EI8452 Industrial Instrumentation - I 63

3 EE8451 Linear Integrated Circuits And Applications 90

4 IC8451 Control Systems 117

5 EC8395 Communication Engineering 154

6 EI8451 Electrical Machines 173

vi
 NUMERICAL METHODS

MA8491-NUMERICAL METHODS

COURSE OUTCOMES

Course
Statement
Outcome

CO1 Solve Algebraic and Transcendental Equations.

CO2 Build a function using an appropriate Numerical Method

CO3 Evaluate a derivative at a value using an appropriate Numerical Method

CO4 Apply Numerical Integration for Mathematical Problems

CO5 Find Numerical Solution of Ordinary Differential Equations

CO6 Make use of Numerical Methods for Boundary Value Problems

CO - PO MAPPING

PO PO PO PO PO PO PO PO PO PO1 PO1 PSO PSO PSO

CO PO12
1 2 3 4 5 6 7 8 9 0 1 1 2 3

CO1 3 3 - - - - - - - - - 1 - - -

CO2 3 3 - - - - - - - - - 1 - - -

CO3 3 3 - - - - - - - - - 1 - - -

CO4 3 3 - - - - - - - - - 1 - - -

CO5 3 3 1 - - - - - - - - 1 - - -

CO6 3 3 1 - - - - - - - - 1 - - -

AVG 3 3 1 - - - - - - - - 1 - - -

3 – Substantially 2 – Moderate 1 – Slightly

1 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

UNIT-I
Solution of equations and eigen values
PART-A
1. State the order of convergence and the criterion for the convergence in Newton’s
method. [M/J12,N/D14,N/D15, M/J16,A/M17,N/D17,A/M18]

The order of convergence of Newton Raphson method is 2 (quadratic) and

convergence condition is f ( x) f ( x)   f ( x) .
2

2. What is Newton’s algorithm to solve the equations x2 = 12 ? [N/D 10]

Given : x 2  12
x 2 12  0
Let f x   x 2 12
Hence f x   2 x
By Newton Raphson method,
f xn 
x n 1  x n 
f x n 

 xn 
x n 12
2

2 xn
2 xn  xn  12 xn  12
2 2 2

 
2 xn 2 xn
3. Evaluate 15 using Newton-Raphson’s formula. [M/J14]
Let x  15
Hence x 2  15  0
Let f x   x 2  15 and f x   2 x
Now f 3  32  15   6 = -ve
f 4  4 2  15  1 = + ve
Therefore, the root lies between 3 and 4
By Newton Raphson method, we have
f xn 
x n 1  x n  Let x0  3.5
f x n 
f  x0  f 3.5
x 1  x0   3.5   3.89
f x0  f 3.5
f x1  f 3.89
x 2  x1   3.89   3.87
f x1  f 3.89

2 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

f x 2  f 3.87 
x 3  x2   3.87   3.87
f x2  f 3.87 
Therefore, the value of 15 is 3.87 .
4. Using Newton’s method, find the root between 0 and 1 of x3 = 6x – 4 [N/D11]
Given x 3  6 x  4
x3  6 x  4  0
Let f x   x 3  6 x  4
Hence f x   3 x 2  6
Let x0  0
f xn 
By Newton’s formula, x n 1  x n 
f x n 
f x0  4 2
x1  x0   0   0.67
f x0   6 3
f x1  0.2807
x2  x1   0.67   0.7303
f x1   4.6533
f x2  0.0076
x3  x 2   0.7303   0.732
f x2   4.3999
Hence the root is 0.73
5. Write the iterative formula for finding N where N is a real number, by Newton’s
method. [N/D 06,A/M 08,A/M15,N/D16,N/D16]
Let x  N
Hence x 2  N  0
Let f x   x 2  N
Hence f x   2 x

By Newton method, x n 1  xn 
f xn 
 xn 

xn  N
2
1  N
  xn  
f x n  2 xn 2  xn 
6. Derive Newton’s algorithm for finding the p th root of a number N , where N  0 .
Let x p  N  0 [N/D15]
Let f x   x p  N
Hence f x   p x p 1

By Newton method, x n 1  x n 
f xn  
x p N
 x n  n p 1 

p xn p  xn p  N
f x n  p xn p x n p 1
 p 1 xn p  N
 .
p x n p 1

3 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

7. Write Newton-Raphson formula to obtain the cube root of N. [N/D10]

1
Let x  N 3

Hence x 3  N  0
Let f x   x 3  N
Hence f x   3 x 2
f xn 
By Newton’s formula, x n 1  x n 
f x n 

xn 1  xn 
x n
3
N  
3xn  xn  N
3 3


2 xn  N
3

.
2 2 2
3 xn 3xn 3 xn
8. Find an iterative formula to find the reciprocal of a given number N. [M/J13]
1
Let x 
N
1 1
Hence N  , N   0
x x

1 1
Let f ( x )  N  and f ( x ) 
x x2
1
f  xn  N 
xn  1  xn  hence xn  1  xn 
, xn  1  2 xn  N xn2 x
f   xn  1
x2
9. Write down the order of convergence and the condition for convergence of fixed
point iteration method. [N/D12,A/M17]
What do you mean by the order of convergence of an iterative method for finding
the root of the equation f x   0 . [N/D13]
Order of convergence = 1
Given f x   0
Write f x  as x   x  provided  x   1 for all x in I .
10. Write sufficient condition for convergence of an iterative method for f x  0 ;
written as x  g x . [A/M10,M/J16,N/D16]
g x   1 ,  x I
11. Find the positive root of x2 + 5x – 3 = 0 using fixed point iteration starting with 0.6 as
first approximation. [N/D08]
Given : x  5 x  3
2

x x  5  3
3
x
x5

4 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Let  x   where  x   1 ,  x 0 ,1 

3
x5
By fixed point formula, xn 1   xn 
Let x0  0.6

x 1   x0    0.6 
3
 0.5357
0.6  5

x2   x1    0.5357  
3
 0.5419
0.5357  5

x3   x2    0.5419 
3
 0.5413
0.5419  5

x4   x3    0.5413 
3
 0.5413
0.5413  5
Hence the root is 0.5413 .

11. Solve e x  3 x  0 by the method of iteration. [N/D11]

Let us consider,
f x   e x  3 x
f ( 0 )   ve
f (1)   ve
Hence I = ( 0, 1)
Given e x  3 x  0
1 x
Which implies , x  e
3
ex
Let  x   where  x   1 ,  x 0 ,1 
3
By fixed point formula, xn 1   xn 
Let x0  1 ( After 10th iteration, α = 0.62 )
Hence the root is 0.62 .
12. Name the two methods to solve a system of linear simultaneous equations.
[M/J 12,N/D15 , M/J16]
Direct method : (1) Gauss Elimination method and (2) Gauss Jordan method.
Indirect method : (1) Gauss Seidel method and (2) Gauss Jacobi method.

13. What are the advantages of iterative methods over direct methods for solving a
system of linear equations? [N/D12]
Direct method Iterative method
i) We get exact solution i) Approximate solution
ii) Simple, take less time ii) Time consuming laborious.

5 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

14. Using Gauss elimination method, solve x + y = 2, 2x + 3y = 5. [M/J09,A/M11]

1 1 2  1 1 2
The augmented matrix is [A,B] =     R2  R2  2 R1
2 3 5 0 1 1
By back substitution, x  y  2 ----(1)
y 1
(1) becomes, x 1  2
x 1
Hence x  1 and y  1 .
15. Solve the system of equations x – 2y = 0, 2x + y = 5 by Gauss elimination method.
[M/J06]
1  2 0 1  2 0
The augmented matrix is A , B     R2  R2  2 R1 1
2 1 5 0 5 5
1  2 0 R
  R2  2
0 1 1 5
By back substitution x  2 y  0 ……………(1)
y 1
Equation (1) becomes x  2  0
x2
Hence, x  2 and y  1 .
16. Solve the system of equations 2x + 3y = 11, 4x – y = 1 by Gauss elimination
method. [A/M11]
2 3 11 2 3 11 
The augmented matrix is [A,B] =    0  7  21 R2  R2  2 R1
4  1 1  
By back substitution, 2 x  3 y  11 ----(1)
 7 y   21  y  3
(1) becomes, 2 x  9  11
2x  2  x 1
Hence x  1 and y  3 .
17. Using Gauss elimination method solve 5 x  4 y  15 , 3 x  7 y  12 . [M/J14]
The augmented matrix is

A , B  
5 4 15

3 7 12
5 4 15
  R2  5 R2  3 R1
0 23 12
15
By back substitution, we have 5 x  4 y  15 and 23 y  15 implies y 
23
 15  60 285 57
5 x  4    15 implies 5 x  15   implies x 
 23  23 23 23

6 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

57 15
Hence x  and y  .
23 23
18. Solve the equations x  2 y  1 and 3 x  2 y  7 by Gauss Elimination method.
The augmented matrix is [N/D13]

A , B  
1 2 1

3  2 7
1 2 1
  R2  R2  3R1
0  8 4 
1
By back substitution, we have x  2 y  1 and  8 y  4 implies y  
2
 1
x  2     1 implies x  1  1 implies x  2
 2
1
Hence x  2 and y   .
2
19. State the rate of convergence of Gauss Jacobi and Gauss Seidel method.[N/D17]
The rate of convergence in Gauss-seidel method is roughly two times that of Gauss-
Jacobi method.
20. Why Gauss-seidel method is better than Gauss-Jacobi method? [A/M18]
Since the current value of the unknowns at each stage of iteration are used in proceeding
to the next stage of iteration, the convergence in Gauss-seidel method will be more rapid
than in Gauss-Jacobi method.
21. Write down the condition for the convergence of Gauss-Seidel iteration scheme.
[M/J 07,A/M 08]
The absolute value of the leading diagonal element is greater than the sum of the
absolute values of the other elements in that row, which is called diagonally
dominant.
22. Which of the iterative methods for solving linear system of equations converge
faster? Why? [A/M15,N/D16]
In Gauss Seidel method the latest values of unknowns at each stage of iteration are
used in proceedings to the next stage of iteration. Hence the convergence in Gauss Seidel
method is faster than Gauss Jacobi method.
23. Write the procedure involved in Gauss Jordan method. [A/M15,N/D15]
In Gauss Jordan, the co-efficient matrix is reduced to a diagonal matrix with row
operations and we get the solution without using the back substitution method.
24. Compare Gaussian elimination and Gauss-Jordan methods in solving the linear
system[ A ] { X } = { B }. [N/D07]
Gauss Elimination Gauss Seidel
i.Direct method i.Indirect method
ii.Used to find inverse of ii. Used to solve system of
the matrix also. equations only
iii.Diagonally dominant condition iii. Diagonally dominant condition
is not insisted. is insisted.

7 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

1 3
25. Find the inverse of A    by Gauss-Jordan method. [N/D08,N/D14,M/J16]
2 7 

 A , I   
1 3 1 0

2 7 0 1 
1 3 1 0
   R2  R1  2 R2
0 1  2 1 
1 0 7  3
 
1  2 1 
 R1  R1  3 R2 
 I , A 1 
0
 7  3
Hence A 1    .
 2 1 
26. What is the use of Power method ? [M/J13]
Power method is used to find the dominant eigen value.
27. Write down the procedure to find the numerically smallest Eigen value of a matrix
by power method. [A/M10]
Procedure:
 Find A 1
 Find the largest eigen value  of A 1 and eigen vector v of A 1 using
power method.
1
 Smallest eigen value of A  and the corresponding eigen vector = v .

28. Define Eigen value and Eigen vector. [N/D07]
Let A be a square matrix of order n . We can find a column matrix X and a constant
 such that A X   X .
On expansion of A   I  0 , we get n th degree equation in  called the
characteristic equation. Its roots  i  i  1, 2 , ...... , n are called eigen values and the
corresponding to each eigen value,  A   I  X  0 will have a non-zero solution
X  x1 x2 ........ xn  which is known as the eigen vector.
T

29. To what kind of a matrix, can the Jacobi’s method be applied to obtain the
Eigen values of a matrix? [N/D 10]
Symmetric matrix.

PART-B

1. Solve for a positive root of the equation x 4  x 10  0 using Newton-Raphson method.
[A/M10]
2. Find the approximate root of x e  3 by Newton’s method correct to 3 decimal places.
x

[A/M11 , M/J16]

8 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

3. Find an iterative formula to find the reciprocal of a given number N and hence find the
1
value of . [N/D11]
19
4. Find the Newton’s iterative formula to calculate the reciprocal of N and hence find the
1
value of . [N/D12]
23
5. Using Newton’s method find the real root of x log 10 x  1.2 correct to five decimal
places. [A/M10,N/D13,A/M15,N/D16]
6. Find by Newton-Raphson method, the real root of e  2 x 1 0 correct to four decimal
x

places. [A/M11]
7. Using Newton’s method, find the root between 0 and 1 of x  6 x  4 correct to 2
3

decimals. [M/J12,A/M17]
8. Find by Newton’s Raphson method a positive root of the equation 3 x  cos x 1  0 .
[N/D14 , M/J16]
9. Using Newton Raphson method find the real root of f x   3 x  sin x   e x  0 by
choosing initial approximation x0  0.5 . [A/M15]
10. Find the smallest positive root of 3 x  1 sin x correct to three decimal places by
iterative method. [N/D10]
11. Solve e x  3 x  0 by the method of fixed point iteration. [M/J12,N/D15]
12. Find a real root of the equation cos x  3 x 1 correct to three decimal places using fixed
point iteration method. [A/M17]
3
13. Find the smallest positive root of x -2x-5=0 by the fixed point iteration method, correct
to three decimal places. [N/D17]
14. Find the root of 4 x  e x  0 that lies between 2 and 3 by Newton-Raphson method.
[N/D15]
4 1 2
 
15. By Gauss elimination method find the inverse of the given matrix A   2 3 1 .
1  2 2 
 
[A/M10]
16. Using Gauss elimination method, solve x  2 y  z   5 , x  y  6 z   12 and
3x y z  4. [N/D11]
17. Solve by Gauss elimination method 3 x  4 y  5 z  18 , 2 x  y  8 z  13 and
5 x  2 y  7 z  20 . [M/J12]
18. Solve the system of equations x  2 y  z  8 , 2 x  3 y  4 z  20 , 4 x  3 y  2 z  16
using Gauss elimination method. [N/D16]
19. Solve the following system of equations by Gauss Elimination method,
x  2 y  5 z   9 , 3x  y  2 z  5 , 2 x 3 y  z  3. [A/M17]

9 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

20. Apply Gauss – Jordan method to solve the following system of equations x  y  z  9 ,
2 x  3 y  4 z  13 , 3 x  4 y  5 z  40 . [A/M11]
21. Apply Gauss – Jordan method to find the solution of the following system
10 x  y  z  12 , 2 x 10 y  z  13 , x  y  5 z  7 . [N/D11, M/J16]
22. Solve the system of equations by Gauss-Jordan method; 5 x1  x2  9 ;
 x1  5 x2  x3  4 ;  x2  5 x3   6 . [M/J14]
23. Using Gauss-Jordan method to solve 2 x  y 3z  8,  x2 y z  4,
3x y4 z  0 . [N/D14,M/J16,N/D16]
0 1 2
 
24. Find the inverse of the matrix by Gauss – Jordan method  1 2 3  . [A/M10]
3 1 1
 
0 1 1 
 
25. Find the inverse of the matrix A   1 2 0  using Gauss-Jordan method.
 3 1  4 
 
[N/D11]
4 1 2
 
26. Find the inverse of A   2 3 1 by using Gauss-Jordan method.
1  2 2 
 
[N/D10,N/D15,N/D17]
1 1 3

27. Using Gauss-Jordan method, find the inverse of the matrix  1 3  3 .
 2  4  4
[M/J12,A/M17]
1 2 6 
28. Using Gauss-Jordan method, find the inverse of the matrix 2 5 15  [N/D12]
6 15 46
.
 3 1 1 
 
29. Find the inverse of the matrix   15 6  5  using Gauss-Jordan method. [N/D13]
 5  2 2 

2 2 6
30. Using Gauss-Jordan method, find the inverse of 2 6  6 [M/J14,A/M15,N/D15]

4  8 8 
1 1 1 
31. Find the inverse of A  4 3 1 by Gauss-Jordan method. [N/D16]
3 5 3 

10 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

1 2 1
32. Using Gauss-Jordan method, find the inverse of the matrix A  4 1 0  . [A/M17]
2 1 3 
4 1 2
 
33. Consider the system of equations of the form AX=B, where A   2 3 1
1  2 2 
 
x  7 
X=  y  and B=  3 .Find by using Gauss-Jordan method, i)A-1 and ii) the numerical
 
 z  7 
solution of the given system. [A/M18]

34. Using Gauss-Seidel iterative method to obtain the solution of the equations
9 x  y  2 z  9 , x 10 y  2 z  15 , 2 x  2 y 13 z  17 . [A/M10]
35. Solve by Gauss-Seidel method, the system of equations 20 x  y  2 z  17 ,
3 x  20 y  z  18 , 2 x  3 y  20 z  25 [N/D10,M/J16,M/J16,N/D16,N/D16,A/M18]
36. Solve the following system by Gauss-Seidel method, x  y  54 z  110 ,
27 x  6 y  z  85 , 6 x 15 y  2 z  72 . [A/M11]
37. Solve by Gauss-Seidel method, the following system, 28 x  4 y  z  32 ,
x  3 y 10 z  24 , 2 x 17 y  4 z  35 . [N/D11,N/D15,N/D17]
38. Apply Gauss-Seidel method to solve the equations 20 x  y  2 z  17 ,
3 x  20 y  z  18 , 2 x  3 y  20 z  25 . [M/J12,N/D14]
39. Solve the following system of equations by Gauss-Seidel method: 8 x  3 y  2 z  20 ;
4 x 11 y  z  33 ; 6 x  3 y 12 z  35 . [N/D10]
40. Solve the given system of equations by Gauss-Seidel method 8 x  y  z  18 ;
2 x 5 y  2 z  3 ; x  y 3 z   6 . [A/M11]
41. Solve the following system of equations using Gauss-Seidel method 10 x  2 y  z  9 ,
x 10 y  z   22 ,  2 x  3 y 10 z  22 . [N/D12,N/D13]
42. Using Gauss-Seidel method, solve the following system of linear equations
4 x  2 y  z  14 ; x  5 y  z  10 ; x  y  8 z  20 . [M/J14,A/M15,A/M17]
43. Solve the following system of equations, starting with the initial vector of 0 , 0 , 0 using
Gauss-Seidel method 6 x1  2 x2  x3  11 ,  2 x1  7 x2  2 x3  5 , x1  2 x2  5 x3   1 .
[A/M15]
44. Solve the following system by Gauss-Seidel iterative procedure : 10 x  5 y  2 z  3 ,
4 x 10 y  3 z   3 , x  6 y 10 z   3 . [N/D15]
45. Find the solution of the system of following equations by Gauss-Seidal method (Upto 4
iterations). x  2 y  5 z  12 , 5 x  2 y  z  6 , 2 x  6 y  3 z  5 . [A/M17]
46. Find the dominant Eigen value and the corresponding Eigenvector of the matrix

11 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

1 6 1
 
A  1 2 0 [A/M10]
 0 0 3
 
47. Find by power method, the largest Eigen value and the corresponding Eigenvector of a
1 3 1
 
matrix A   3 2 4  with initial vector 1 1 1T . [N/D10,A/M18]
 1 
 4 10 
48. Determine the largest eigen value and the corresponding eigen vector of the matrix
 1 3  1
 
 3 2 4  with 1 0 0 as the initial vector by power method.
T
[N/D13]
  1 4 10 
 
1 6 1
 
49. Find the largest Eigen value and corresponding eigen vector of a matrix  1 2 0  by
 0 0 3
 
T
using Power method starting with initial vector (1 0 0) . [N/D 11,N/D15,N/D17]

50. Determine the largest Eigen value and the corresponding Eigen vector of the matrix
 2 1 0 
1 2 1 . [M/J 12,A/M15]
 
 0 1 2 
 25 1 2 
 
51. Find numerically largest eigen value of A   1 3 0  and the corresponding eigen
 2 0  4
 
vector. [N/D10,N/D11,M/J14,N/D14,M/J16]
1 2
52. Find the dominant eigen value of A    by power method and hence find the
3 4
other eigenvalue also. [M/J12]
5 0 1 
53. Find all the eigen values of A  0  2 0 using power method.
1 0 5
[A/M15,N/D15,N/D16,N/D16,A/M17]
54. Using power method find the dominant eigen value and the corresponding eigen
 15  4  3 
vector for the given matrix A    10 1 2  6  . [M/J16]
 20 4  2

12 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

UNIT-II
Interpolation
PART-A

1. Define the terms interpolation and extrapolation. (OR) Distinguish between

interpolation and extrapolation. [M/J09,N/D17]
The process of computing the value of a function inside the given range is called
interpolation. The process of computing the value of a function outside the given range is
called extrapolation.
2. What do you understand by inverse interpolation? [N/D07]
It is the process of finding the values of x corresponding to a value of y, not present in
the table.
 2  E ex
3. Evaluate   e x . 2 x . [N/D09]
E  e
ex  exh  ex


 e x  e x eh  e x  eh  1 e x 

2 e x  e h  1 e x 
2

 2  x
     
  e  E 1 2 e x  E 1 e h 1 e x  e h 1 E 1 x  e h 1 e x  h
E
2 2 2
  
 2  x E e x exh
2

  e . 2 x  e h 1 e x  h . 
e2 x
 ex .
 
E  e e 1 e
h 2 x
e x
 
4. Name two interpolation methods for unequal intervals. [M/J12]
(i) Newton’s divided difference method.
(ii) Lagrange’s method.
5. Find the polynomial which takes the following values [N/D06]
x 0 1 2
y 1 2 1
Difference table

x y y 0 2 y 0
0 1
1
1 2 -2
-1
2 1

13 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Newton’s forward formula

u u 1 2 x  x0 x0
y  y 0  u y 0   y 0 where u   x
2! h 1
x x 1
y  1  x (1)   2   x 2  2 x 1 .
2
6. Write down Newton’s forward and backward difference formula. [N/D07,N/D16]
Newton’s forward interpolation formula :
u u 1 2 u u 1u  2 3 x  x0
y  y 0  u y 0   y0   y0  ....... where u 
2! 3! h
Newton’s backward interpolation formula :
u u 1 2 u u 1u  2 3 x  xn
y  y n  u y n   yn   y n  ....... where u  .
2! 3! h
7. Write the Newton’s forward difference interpolation formula.
[A/M08, N/D10,M/J13,N/D13,M/J16,M/J16,A/M17]
Newton’s forward interpolation formula :
u u 1 2 u u 1u  2 3 x  x0
y  y 0  u y 0   y0   y0  ....... u
2! 3! where h
8. A third degree polynomial passes through ( 0, -1 ) , ( 1 , 1 ) , ( 2 , 1 ) and ( 3 , - 2 )
Find its value at x = 4 ? [M/J09,N/D16]
Difference table
x y y 0 2 y 0 3 y 0
0 -1
2
1 1 -2
0 -1
2 1 -3
-3
3 -2

u u 1 2 u u 1u  2 3
Newton’s forward formula, y  y0  u y 0   y0   y0
2! 3!
x  x0 x0
where u   x
h 1
x x 1
y   1  x (2)   2  x x 1x  2  1
2 6
1

y   x 3  3 x 2  16 x  6
6

Hence y 4 
581
.
3
9. Find the interpolating polynomial for the given data: [A/M10]

x -1 0 1 2
f 4 2 2 4

14 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Difference table

x y y 0 2 y 0 3 y 0
-1 4
-2
0 2 2
0 0
1 2 2
2
2 4

Newton’s forward formula

u u 1 2 x  x0 x 1
y  y 0  u y 0   y 0 where u    x 1
2! h 1
y  4  x  1 (2) 
x 1x  2
2
y  x x2.
2

10. Fit a polynomial from the following data using Newton’s forward difference
interpolation formula [A/M11]
x 0 2 4 6
f 2 4 14 32

Difference table

x y y 0 2 y 0 3 y 0
0 2
2
2 4 8
10 0
4 14 8
18
6 32

Newton’s forward formula

u u 1 2 x  x0 x0 x
y  y 0  u y 0   y 0 where u   
2! h 2 2
 x  x  x  2  1 
y  2    (2)        8
2  2  2  2 
y  x2  x  2 .

11. Given y0  3 , y1  12 , y 2  81 , y3  200 , y 4  100 . Find 4 y 0 . [A/M15]

15 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Difference table
x y y 0 2 y 0 3 y 0 4 y 0
0 3
9
1 12 60
69 -10
2 81 50 -21
119 -31
3 200 19
-
4 100 100

Hence the value of 4 y0   21.

  f x 
12. Prove that  log  f  x   log  1 
f  x  
. [A/M11 ]

 f x   f x  h   f x 
R.H .S  log  
 f x  
 f x  h 
 log    log f x  h  log f x    log f x   L.H .S
 f x  

13. Find a polynomial for the following data by Newton’s backward difference
formula [M/J06]
x 0 1 2 3
f(x) -3 2 9 18

Difference table
0 -3
5
1 2 2
7 0
2 9 2
9
3 18
x y y n  2 yn  3 yn

Newton’s backward interpolation formula is

u u  1 2 x  xn x 3
y  y n  u y n   y n where u    x 3
2! h 1
y  18  x  39 
x  3x  2 2
2
y  x  4 x 3
2

16 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

14. State Newton’s backward difference formula. [N/D12,N/D14]

Newton’s backward interpolation formula :
u u 1 2 u u 1u  2 3 x  xn
y  y n  u y n   yn   y n  ....... where u  .
2! 3! h
15. When to use Newton’s forward interpolation and when to use Newton’s backward
interpolation? [N/D10,A/M18]
Newton forward formula is used for interpolating near x 0 (the beginning of the table)
and backward formula is used for interpolating near x n (end of the table).
16. Obtain the divided difference table for the following data [M/J06]

x 2 3 5
y 0 14 102

Divided difference table

2 0
14  0
 14
3 2
3 14 44 14
 10
102 14 52
 44
5 102 53

17. Obtain the divided difference table for the following data [N/D06]

x -1 0 2 3
f(x) -8 3 1 12

Divided difference table

-1 -8
38
 11
0 1
0 3 1 11
 4
1 3 2 1
 1 44
20 2
2 1 11  1 3 1
4
12 1 3 0
 11
3 2
3 12

18. If f  x   , find f a , b and f a , b , c  by using divided differences.

1
x2
[M/J07]

17 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Divided difference table

a 1
a2

a  b 
a2 b2
b 1 ab  bc  ca 
b2

b  c  a2 b2 c2
b2 c2
c 1
c2

19. Obtain a divided difference table for the following data: [A/M08]

x 5 7 11 13 17
y 150 392 1452 2366 5202

Divided difference table

5 150
121
7 392 24
265 1
11 1452 32 0
457 1
13 2366 42
709
17 5202

20. Form the divided difference table for [N/D08]

x -1 1 2 4
y -1 5 23 119

Divided difference table

-1 -1
3
1 5 5
18 1
2 23 10
48
4 119

18 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

21. Find the divided differences of f x   x 3  x  2 for the arguments 1, 3, 6, 11.

Given f x   x 3  x  2 [A/M10,A/M11,N/D16]
For x  1, 3 , 6 ,11 , f x   4 , 32 , 224 , 1344
Divided difference table
1 4
14
3 32 10
64 1
6 224 20
224
11 1344

22. Write the Newton’s divided difference formula. [N/D10]

y  y0  x  x0  y0  x  x0 x  x1   y0  x  x0 x  x1 x  x2   y0  ..........
2 3

23. What is the nature of nth divided difference of a polynomial of nth degree?
The nth divided differences of a nth degree polynomial are constants. [N/D17]
24. Obtain the Newton’s divided difference table for the following data [A/M11]

x 1 2 4 6
y -26 12 256 844

Divided difference table

1 -26
38
2 12 28
122 3
4 256 43
294
6 844

25. Form the divided difference table for the data  0 , 1 , 1 , 4  ,  3 , 40  and
 4 , 85  . [A/M 10,N/D15]
Divided difference table
0 1
3
1 4 5
18 1
3 40 9
45
4 85

19 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

26. Find the first and second divided differences with arguments a , b , c of the function

f x  .
1
[N/D10,M/J14,A/M18]
x
Divided difference table

a 1
a 1

ab
b 1 1
b 1 abc

c bc
1
c

3  1    1
27. Show that 
bcd  a  abcd
. [A/M15,N/D15]

Let f a  
1
a
Then
 1   1   a  b 
  
f b   f a   b   a   a b  b  a    1
f a , b     
ba ba ba a b b  a  ab
 1   1   a  c
      
f b , c   f a , b   b c   a b   abc 
f a , b , c    
ca ca ca


c  a   1
a b c c  a  a b c
 1   1  a d
    
f b , c , d   f a , b , c   b c d   a b c   abcd 
f a , b , c , d    
d a d a d a
 d  a  1
 
a b c d d  a  abcd

28. Construct the divided difference table for the following data [M/J12]

x 0 1 2 5
f(x ) 2 3 12 147

20 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Divided difference table

0 2
1
1 3 4
9 1
2 12 9
45
5 147

29. Find the divided differences of f x   x 3  x 2  3 x  8 for the arguments 0 , 1 , 4 , 5 .

[N/D13,M/J16]
For the arguments of x  0 , 1 , 4 , 5 , we have y  f x   8 , 11 , 68 , 123

Divided difference table

0 8
11  8
3
1 0
1 11 19  3
4
68 11 40
 19 94
4 1 1
4 68 50
55  19
9
123  68 5 1
 55
54
5 123

30. Find the second degree polynomial through the point 0 , 2 , 2 ,1 , 1, 0 using
Lagrange’s formula. [N/D14]
By Lagrange’s formula,
x  x1 x  x2  x  x0 x  x2  x  x0 x  x1 
y y0  y1  y
x0  x1 x0  x2  x1  x0 x1  x2  x2  x0 x2  x1  2
y
x  2x  1 2  x  0x  1 1  x  0x  2 0
0  20  1 2  02  1 1  01  2

1 2
2

x  3x2 
1 2
2
 1
   1

x  x  x2 3 x  2  x2  x  2 x2  4 x  2
2 2
 
y  x  2 x 1
2

31. Using Lagrange’s interpolation, find the polynomial through  0 , 0 , 1 , 1  and

 2 , 2 . [M/J 07]
Lagrange’s formula :
x  x1 x  x2  x  x0 x  x2  x  x0 x  x1 
y y0  y1  y
x0  x1 x0  x2  x1  x0 x1  x2  x2  x0 x2  x1  2

21 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

x x  2
y0 1  x x 1 2  x.
1 1 21
32. Using Lagrange’s formula, find the polynomial to the given data. [M/J13]
X : 0 1 3
Y : 5 6 50
x  x1 x  x2  x  x0 x  x2  x  x0 x  x1 
y y0  y1  y
x0  x1 x0  x2  x1  x0 x1  x2  x2  x0 x2  x1  2
1
 47 x 2  41 x  30 
 y
6
33. Given y  f  x  is the exact curve and y  Pn  x  is the interpolating polynomial
curve, write the error in polynomial interpolation for any x where x 0  x  x n and
x0  c  xn . [M/J12]
u u 1u  2 .......... u  r  r  1 x  x0
Error 
r 1!
 y 
xc
where u 
h
34. State Lagrange’s interpolation formula for unequal intervals. [N/D11]

( x  x1 )( x  x 2 )( x  x3 ).......( x  x n )
y  f ( x)  y0
( x 0  x1 )( x 0  x 2 )( x 0  x3 ).......( x 0  x n )
( x  x0 )( x  x 2 )( x  x3 ).......( x  x n )
 y1
( x1  x0 )( x1  x 2 )( x1  x3 ).......( x1  x n )
+…………..…….+
( x  x 0 )( x  x1 )( x  x 2 )( x  x3 ).......( x  x n 1 )
 yn
( x n  x 0 )( x n  x1 )( x n  x 2 )( x n  x3 ).......( x n  x n 1 )

35. Distinguish between Newton’s divided difference interpolation and Lagrange’s

interpolation. [A/M15]
Lagrange’s Interpolation formula can be used for equal and unequal intervals. But
Newton’s divided difference formula can be used only for unequal intervals. Lagrangian
method involves more arithmetic operations than does the divided difference method.
36. Define a cubic spline S  x  which is commonly used for interpolation.
[A/M10,N/D12,M/J14]

   
xi  x 3 M i 1  x  xi 1 3 M i  xi  x  yi 1  h M i 1 
2
S x  
1
6h h  6 
x  xi  1  h2 
 i
y  Mi  .
h  6 
37. For cubic splines, what are the ( n – 1 ) conditions required to evaluate the
unknowns. [M/J12,N/D15]

22 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

6
yi 1  2 yi  yi 1 , where i 1, 2 , 3, .........., n 1 .
M i 1  4 M i  M i  1 
h2
38. What are the advantages of cubic spline fitting. [A/M15]
Cubic spline provide better approximation to the behaviour of functions that have
abrupt local changes. Further splines perform better than higher order polynomial
approximation.

PART-B

1. Given the following table, find the number of students whose weight is between 60 and
70 lbs: [A/M10]
Weight(in lbs) x 0-40 40-60 60-80 80-100 100-120
No. of students 250 120 100 70 50

2. Find the value of tan 45 15 by using Newton’s forward difference interpolation
formula for [N/D10,A/M18]
x 45 46 47 48 49 50

tan x 1.00000 1.03553 1.07237 1.11061 1.15037 1.19175

3. The population of a town is as follows:

X year 1941 1951 1961 1971 1981 1991
Y population in
20 24 29 36 46 51
thousands

Estimate the population increase during the period 1946 to 1976. [N/D11]

4. Find the value of y when x  5 using Newton’s interpolation formula from the
following table: [N/D12,N/D17]
x 4 6 8 10
y 1 3 8 16

5. The table gives the distance in nautical miles of the visible horizon for the given heights
in feet above the earth’s surface [A/M15]
x=height 100 150 200 250 300 350 400
y=distance 10.63 13.03 15.04 16.81 18.42 19.9 21.27
Find the values of y when x  218 ft using Newton’s forward interpolation formula.

6. From the following table, estimate the number of students who obtained marks between
40 and 45 [A/M 10 ,N/D 11]
Marks 30-40 40-50 50-60 60-70 70-80
No. of students 31 42 51 35 31

23 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

7. Find the value of y at x  21 and x  28 from the following data, using Newton’s
interpolation formula: [N/D 10,M/J 12,N/D16]
x 20 23 26 29
y 0.3420 0.3907 0.4384 0.4848

8. From the following data, find  at x  43 and x  84

x 40 50 60 70 80 90
 184 204 226 250 276 304
Also express  in terms of x . [A/M15]

9. Using Newton’s forward interpolation formula, find the cubic polynomial which takes
the following values: [N/D14,N/D15]
x 0 1 2 3
f x  1 2 1 10

10. Find a polynomial of degree two for the data by Newton’s forward difference formula.
[A/M17]
x 0 1 2 3 4 5 6 7
f x  1 2 4 7 11 16 22 29

11. From the following table find f x  and hence f 6 using Newton’s interpolation
formula [M/J12]
x 1 2 7 8
f x  1 5 5 4

12. For the following data prepare the finite difference table and express y as a function of
x using Newton’s backward difference formula and hence find y when x  3.5 [A/M11]
x 0 1 2 3 4
y 7 10 13 22 43

13. From the following data, find  at x  84

x 40 50 60 70 80 90
 184 204 226 250 276 304
Also express  in terms of x . [M/J12,N/D15]

14. Estimate sin 38 from the data given below: [A/M15,M/J16]
x 0 10  20  30  40 
sin x 0 0.17365 0.34202 0.5 0.64279

15. Derive Newton’s backward difference formula by using operator method. [M/J12]

24 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

16. Given : sin 45  0.7071 , sin 50  0.7660 , sin 55  0.8192 , sin 60  0.8660 .
Find sin 52  by Newton’s interpolation formula. [N/D16]
th
17. The values of y are consecutive terms of a series of which 23.6 is the 6 term. Find the
first and tenth terms of the series: [N/D16]
x: 3 4 5 6 7 8 9
y: 4.8 8.4 14.5 23.6 36.2 52.8 73.9
18. The following table gives the values of density of saturated water for various
temperatures of saturated steam.
Temperature  C 100 150 200 250 300
Density hg/m 3 958 917 865 799 712
Find by interpolation, the density when the temperature is 275 . [N/D13,M/J16]

19. Find the function f x  from the following table using Newton’s divided difference
formula [A/M10]
x 0 1 2 4 5 7
f(x) 0 0 -12 0 600 7308

20. Given the tables

x 5 7 11 13 17
f(x) 150 392 1452 2366 5202

Evaluate f 9 using Newton’s divided difference formula. [A/M11,N/D16,N/D16]

21. Determine f x  as a polynomial in x for the following data, using Newton’s divided
difference formulae. Also find f 2 . [N/D11]
Find f 1 by using divided difference interpolation from the following data: [A/M17]

x -4 -1 0 2 5
f(x) 1245 33 5 9 1335

22. Using Newton’s divided difference formula, find f x  from the following data and
hence find f 4 . [N/D12,M/J16,N/D16]
x 0 1 2 5
f(x) 2 3 12 147

23. Find f 8 by using Newton’s divided difference formula [A/M10,N/D14,A/M17]

x 4 5 7 10 11 13
f(x) 48 100 294 900 1210 2028

25 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

24. Find the value of f 6 for the data f 1  1 , f 2  5 , f 7  5 and f 8  4 by using
Newton’s divided difference formula. [N/D11]
25. Find f 3 by Newton’s divided difference formula for the following data: [M/J14]
x -4 -1 0 2 5
y 1245 33 5 9 1335

26. Using Newton’s divided difference formula find f 3 from the data: [N/D15]

x 0 1 2 4 5
f x  1 14 15 5 6
27. Use Lagrange’s formula to find a polynomial which takes the values f 0  12 ,
f 1  0 , f 3  6 and f 4  12 . Hence find f 2 . [A/M10]
28. Use Lagrange’s interpolation formula to fit a polynomial to the given data f 1   8 ,
f 0  3 , f 2  1and f 3  2 . Hence find the value of f 1 . [N/D10,A/M18]
29. Find the expression of f x  using Lagrange’s formula for the following data : [A/M11]
x 0 1 4 5
f(x) 4 3 24 39

30. Find the value of x when y  20 using Lagrange’s formula from the following table
[A/M11]
x 1 2 3 4
y=f(x) 1 8 27 64

31. Using Lagrange’s interpolation, calculate the profit in the year 2000 from the following
data: [M/J12,N/D15]
year 1997 1999 2001 2002
Profit in lakhs of
43 65 159 248
Rs.

32. Use Lagrange’s method to find log 10 656 , given that log 10 654  2.8156 ,
log 10 658  2.8182 , log 10 659  2.8189 and log 10 661  2.8202 . [N/D12]

33. Using Lagrange’s formula for interpolation find y9.5 given: [N/D10,M/J12]
x 7 8 9 10
y 3 1 1 9

34. Find Lagrangian interpolating polynomial for the following data:

x 1 2 3 5
f(x) 0 7 26 124

26 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Hence find f 4 . [A/M11]

35. Find the polynomial f x  by using Lagrange’s formula and hence find f 3 for the
following values of x and y : [N/D11,N/D14,N/D17]
x 0 1 2 5
f(x) 2 3 12 147

36. Apply Lagrange’s formula to find y ( 27 ) to the data given below .

x : 14 17 31 35
y : 68.8 64 44 39.1 [ M / J 13 ]
37. Use Lagrange’s formula to find the value of y at x  26 from the following data:
x: 3 7 9 10
y : 168 120 72 63 [N/D13,M/J16]
38. Using Lagrange’s interpolation formula, find y2 from the following data:
y0  0 ; y1 1 ; y3  81; y4  256 ; y5  625 [M/J14]
39. Using Lagrange’s interpolation find the interpolated value for x  3 of the table.
[A/M15]
x 3.2 2.7 1.0 4.8
f x  22.0 17.8 14.2 38.3

40. Fit a Lagrange polynomial to the data:

x 1 2 3 5
y 0 1 26 124

and hence find y when x  3.5 [A/M15]

41. Find the Lagrange polynomial f x  satisfying the following data :
x : 1 3 5 7
y  f x  : 24 120 336 720
and hence find f 4 . [N/D15]
42. Using Lagrange’s formula find the value of log 10 323.5 for the given data: [M/J16]
x : 321.0 322.8 324.2 325.0
log 10 x : 2.50651 2.50893 2.51081 2.51188
43. If f 0  1 , f 1  2 , f 2  33 and f 3  244 , find a cubic spline approximation,
assuming M 0  M 3  0 . Also find f 2.5 . [A/M10]
44. Using Lagrange’s formula find the cubic polynomial which takes the following values.
x : 1 3 4 6
f(x) : 0 22 57 205 [A/M17]
45. Find the natural cubic spline to fit the data : [M/J16,N/D13]
x : 0 1 2
f x  : -1 3 29
Hence find f 0.5 and f 1.5 .

27 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

46. Find the natural cubic spline approximation for the function f x  defined by the
following data: [N/D10,N/D17A/M18]
x 0 1 2 3
f(x) 1 2 33 244
Hence find the value of f 2.5 and f 2.5 .

47. Find the cubic spline approximation for the function y  f x  from the following data,
 
given that y0  y3  0 . Hence find y 0.5 [A/M11,N/D12]
Hence find y 0.5 , y 0.5 and y 1.5 [A/M17]
x -1 0 1 2
f(x) -1 1 3 35

48. Find the cubic polynomial which takes the following values: [M/J12]
x 0 1 2 3
f(x) 1 2 1 10

49. The following values of x and y are given in table :

x 1 2 3 4
y 1 2 5 11

Find the cubic splines and evaluate y 1.5 and y 3 . [M/J12,A/M15,N/D15,A/M17]

50. Fit the cubic spline for the following table of values: [N/D11,M/J16]
x 1 2 3
- -
f(x) 16
6 1
Hence evaluate y1.5 and y 2

51. Fit the cubic splines for the following data: [M/J14]
x: 1 2 3 4 5
y: 1 0 1 0 1

52. From the following table :

x : 1 2 3
y : -8 -1 18
Compute y1.5 and y 1 , using cubic spline. [N/D15,N/D16]

53. Fit a cubic spline curve that passes through 0 ,1 , 1, 4 , 2 , 0 and 3 , 2 with the
natural end boundary conditions s  0  0 , s  3  0 . [N/D16]

28 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

UNIT-III
Numerical Differentiation and integration
PART-A
dy d2y
1. Write down the expressions for and at x  x 0 by Newton’s forward
dx dx 2
difference formula. [M/J06,M/J16,N/D17]
 dy  1  1 1 1 
    y 0  2 y 0  3 y 0  4 y 0  ............
 dx  x  x0 h  2 3 4 
d2y 1  11 4 
 2   2 2 y 0  3 y 0   y 0  ...............
 dx  x  x0 h  12 
dy d2y
2. Write down the expressions for and at x  x n by Newton’s backward
dx dx 2
difference formula. [N/D06,N/D10,A/M11,M/J14,M/J16]
 dy  1  1 1 1 
    y n   2 y n   3 y n   4 y n  ............
 dx  x  xn h  2 3 4 
d2y 1  11 4 
 2   2  2 y n   3 y n   y n  ...............
 dx  x  xn h  12 

3. Write the formula for y'(x) and y''(x) using Newton’s backward differences.
[A/M18]
 dy  1  2v  1 2 3v  6v  2 3
2
4v  18v  22v  6 4 3 2

    y n   yn   yn   y n  ............
 dx  h  2 6 24 
d y
2
1  6v  8v  11 4
2

 2   2  2 y n  (v  1) 3 y n   y n  ...............
 dx  h  12 
4. What is the order of the error in trapezoidal rule? [M/J08]
The order of the error in Trapezoidal rule is h 2 .
rd
1
5. State the local error term in Simpson’s rule. [N/D14]
3

The local error term in Simpson’s

1
rd
rule is E 
b  a  h 4 M where M is in the
3 180
interval a , b  .
6. Write the errors in Trapezoidal and Simpson’s rules of numerical integration?
[A/M15]

The error in Trapezoidal rule is E 

b  a  h M where M is in the interval a , b .
2

12

The error in Simpson’s

1
rd
rule is E 
b  a  h 4 M where M is in the interval a , b  .
3 180

29 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS


7. Evaluate  sin x dx by Trapezoidal rule by dividing ten equal parts. [M/J13]
0

ba  0 
Here n  10 , y  f x   sin x and h   
n 10 10
  3 2  3 7 4 9
x 
0 10 5 10 5 2 5 10 5 10

y
0 0.309 0.5878 0.809 0.9511 1 0.9511 0.809 0.5878 0.309 0

By Trapezoidal rule,


 sin x dx  2  y  y10   2  y1  y 2  y3  y 4  y5  y 6  y 7  y8  y9 
h
0
0

 
 
   0  0  2 0.309  0.5878  0.809  0.9511  1  0.9511  0.809  0.5878  0.309
10
2


 sin x dx  1.9835
0
2
dx
8. Taking h  0.5 , evaluate  1 x
1
2
using Trapezoidal rule. [M/J14,M/J16]

Given h  0.5 , y  f x  
1
1 x 2
x 1 1.5 2

1
y 0.5 0.3077 0.2
1 x2

By Trapezoidal rule, we have

2

 1 x
dx
2

h
 y0  y 2   2 y1   0.5 0.5  0.2  2 0.3077  0.32885 .
1
2 2
6
dx
9. Evaluate  1 x
0
2
using Trapezoidal rule. [N/D13]

ba 6 0
 1 , y  f x  
1
Let n  6 , then h  
n 6 1 x 2
x 0 1 2 3 4 5 6

1
y 1 0.5 0.2 0.1 0.0588 0.0385 0.027
1 x2

30 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

By Trapezoidal rule, we have

2

 1 x
dx
2

h
 y0  y6   2  y1  y 2  y3  y 4  y5 
1
2
2
dx
 1 x 2

1
1 0.027  2 0.5  0.2  0.1 0.0588  0.0385  1.4108 .
1
2
1
dx
10. Evaluate  1 x
0
using Trapezoidal rule. [N/D10]

ba 1
Let n  4 h    0.25
n 4
x 0 0.25 0.5 0.75 1
1
y 1 0.8 0.66 0.57 0.5
1 x

By Trapezoidal rule
I   y 0  y 4   2  y1  y 2  y3 
h
2
I 
0.25 1 0.5  2 0.8  0.66  0.57  0.695 .
2
11. Write down the Newton-cote’s formula for equidistant ordinates. [A/M11,M/J16]
xn
 n2 1  n3 n 2  2 1  n4  
 f x  dx  h  n y 0   y 0      y 0    n 3  n 2  3 y 0  ...............
x0  2 2  3 2  6 4  
1
dx
12. Evaluate  1 x
0
2
using Trapezoidal rule. [N/D12]

ba 1
Let n  4 h    0.25
n 4

x 0 0.25 0.5 0.75 1

1
y 1 0.94 0.8 0.64 0.5
1 x 2

By Trapezoidal rule
I   y 0  y 4   2  y1  y 2  y3 
h
2
I 
0.25 1 0.5  2 0.94  0.8  0.64  0.7825 .
2
1
1
13. Evaluate  x dx by Trapezoidal rule, dividing the range into 4 equal parts.
1
2
[N/D15]

31 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

1
1  
ba  2   0.125
Let n  4 h  
n 4

x 0.5 0.625 0.75 0.875 1

1
y 2 1.6 1.3333 1.1429 1
x

By Trapezoidal rule,
I   y 0  y 4   2  y1  y 2  y3 
h
2
I 
0.125 2 1  2 1.6 1.3333  1.1429  0.697 .
2
14. Compare Trapezoidal rule and Simpson’s 1/3 rule for evaluating numerical
integration. [N/D07,N/D16]
i) In Newton Cotes Quadrature formula, if we put n = 1 we get Trapezoidal rule
whereas if we put n = 2, we get Simpson’s 1/3rd rule.
ii) In Trapezoidal rule, the interpolating polynomial is linear whereas in Simpson’s
1/3rd rule, the interpolating polynomial is of degree 2.
iii) In Trapezoidal rule, there is no restriction on the number of intervals whereas in
Simpson’s 1/3rd rule, the number of intervals should be even.
15. What approximation is used in deriving Simpson’s rule of integration? [N/D10]
h 4 is the approximation used in deriving Simpson’s rule of integration.
16. Give the order of error in the Simpson’s 1/3 rd rule. [M/J12]
The order of error in the Simpson’s 1/3 rd rule is h .4

17. When do you apply Simpson’s 1/3 rd rule, and what is the order of the error in
Simpson’s 1/3 rule. [A/M11]
4
The number of subintervals n should be even and order of error is h .
18. State Simpson’s one-third rule. [N/D11,M/J13,A/M17]
Simpson’s one-third rule is given by
x x0 x1 x2 ……………………….. xn
y y0 y1 y2 ……………………….. yn

I 
h
 y0  y n   2  y0  y2  .......  4  y1  y3  .........
3
4
19. Evaluate  f  x  dx
1
from the table by Simpson’s 3/8 rule [M/J06]

x 1 2 3 4
f(x ) 1 8 27 64

32 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Here n  3

I 
3h
 y0  y3   2 0  3  y1  y 2 
8
I  1  64  3 8  27  
3 255
 63.75 .
8 4
20. State the formula of Simpson’s 3/8 th rule. [M/J07]
Simpson’s 3/8 th rule is given by
x x0 x1 x2 ……………………….. xn
y y0 y1 y2 ……………………….. yn

3h
I   y0  yn   2  y3  y6  .......  3  y1  y 2  y4  y5  .........
8
21. Write Simpson’s 1/3 rd and 3/8 th formulae. [N/D09]
Simpson’s one-third rule is given by
x x0 x1 x2 ……………………….. xn
y y0 y1 y2 ……………………….. yn

h
I   y0  y n   2  y0  y2  .......  4  y1  y3  .........
3
Simpson’s 3/8 th rule is given by
x x0 x1 x2 ……………………….. xn
y y0 y1 y2 ……………………….. yn

3h
I   y0  yn   2  y3  y6  .......  3  y1  y 2  y4  y5  .........
8
22. Under what condition , Simpson’s 3/8 rule can be applied and state the formula.
[M/J12,N/D15]
The number of subintervals n should be a multiple of 3 .

23. State the Romberg’s integration formula with h1 and h2 . Further, obtain the
h
formula when h1  h and h2  . [A/M10]
2
 I h  Ih 
 h  
I  h ,   Ih   2 
 2 2  3 
 
1
dx
24. Use two-point Gaussian quadrature formula to solve  1 x
1
2
.

[A/M10,M/J12,A/M17,A/M17,A/M18]

33 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

 1   1 
1 1
dx dt
11 x 2  11 t 2  f  3   f   3  where f t   1 t 2
1

 0.75  0.75  1.5

2

e
 x2
25. Evaluate dx by two point Gaussian quadrature formula. [N/D10,N/D15]
0

 a b   ba  ba 
Let x    t , dx    dt
 2   2   2 
x  1 t , dx  dt
1
I   e  1  t  dt
2

1

Let f t   e  1  t 
2

Gaussian two point quadrature formula is given by

 1   1 
I  f    f     0.0831  0.8364  0.9195
 3  3
  t   
1

4 
26. Using two point Gaussian quadrature formula evaluate I  sin   dt .
1
4 
[A/M15]

Here f t   sin
t 1
4
By two point Gaussian quadrature formula, we have
 1
 t       1   1 
4 
I  sin   dt   f     f  
1
4  4   3  3 
  1   1 
 sin    1  sin   1
4 4 3  4  3 

I  0.9454  0.3259  0.9985 .
4
2 x

27. Evaluate e
2
2
dx by Gauss two point formula. [N/D13]

Here a   2 and b  2
 ab   b  a  b  a
Let x    t dx    dt
 2   2   2 

x  2 t and dx  2 dt
2 x 1 1

e
2
2
dx   e  t 2 dt  2  e  t dt
1 1

Here f t   e t

34 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

By Gaussian two point formula, we have

2

x 1
  1   1 
e
2
2
dx  2  e  t dt  2  f  
1  
  f 
3

 3 
  1 1

 2 e 3
e 3
  2 0.5614  1.7813  4.6854
 
28. Write down two point Gaussian quadrature formula. For what class of
functions f(x) does it give exact answer. [A/M 11,N/D17]
 a b   ba 
Change the interval  a , b  into  1 , 1  by using x    t ,
 2   2 
ba 
dx    dt Gaussian two point quadrature formula is
 2  .
 1   1 
1
I   f t  dt 
1
f  

  f 
3  3
 . It is exact for polynomials up to degree 3.

1
29. Write down the three point Gaussian quadrature formula to evaluate  f  x  dx .
1

[N/D12,A/M15]
b 1

 f x  dx   f t  dt where the interval a , b is changed into  1 , 1 by the

a 1

 a b   ba 
transformation x    t
 2   2 
1
Then  f t  dt  A f t   A f t   A f t 
1
1 1 2 2 3 3 where A1  A3  0.5555 , A2  0.8888

and t1   0.7745 , t 2  0 , t 3  0.7745

1
30. State two point Gaussian quadrature formula to evaluate  f  x  dx .
1

[N/D06,N/D07,N/D15]

 1   1 
Gaussian two point quadrature formula is I  f     f   .
 3  3
1
1
31. Evaluate  1 x
1
4
dx using Gaussian quadrature with two points. [N/D08]

Let f x  
1
1 x 4
 1   1 
Gaussian two point formula is I  f     f    0.9  0.9  1.8 .
 3  3

35 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

PART-B

1. From the following table of values of x and y , obtain y x  and y x  for x 16
[N/D10,A/M18]
x 15 17 19 21 23 25
y 3.873 4.123 4.359 4.583 4.796 5

2. Find the first two derivatives of  x  3 at x  50 and x  56 , for the given table: [N/D11]
1

x 50 51 52 53 54 55 56

y  x  3.68 3.70 3.73 3.75 3.779 3.80 3.82

40 84 25 63 8 30 59

3. A slider in a machine moves along a fixed straight rod. Its distance x cm along the rod is
given below for various values of the time 't ' seconds. Find the velocity of the slider
when t 1.1 second. [M/J12]
1.
t 1.0 1.1 1.2 1.4 1.5 1.6
3
9.
7.98 8.40 8.78 9.45 9.75 10.0
x 12
9 3 1 1 0 31
9

4. Find f x  at x 1.5 and x  4.0 from the following data using Newton’s formulae for
differentiation. [N/D12,M/J16]
[OR] Find the first and second derivatives of the function tabulated below at x  1.5
[N/D13,M/J16]

x 1.5 2.0 2.5 3.0 3.5 4.0

y=f(x) 3.375 7.0 13.625 24.0 38.875 59.0

5. Find the value of sec 31 from the following data [N/D11,A/M15]
 31 32 33 34
tan  0.6008 0.6249 0.6494 0.6745

6. The population of a certain town is given below. Find the rate of growth of the population
in the year 1931 and 1945 [M/J12,N/D16]
Year x 1931 1941 1951 1961 1971
Population in thousand y 40.6 60.8 79.9 103.6 132.7

7. For the given data, find the first two derivatives at x  1.1 [M/J14,A/M15]

36 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

x 1.0 1.1 1.2 1.3 1.4 1.5 1.6

y 7.989 8.403 8.781 9.129 9.451 9.750 10.031
d2y
8. From the following table, obtain the value of at x  0.96 [N/D16]
dx 2
x 0.96 0.98 1.00 1.02 1.04
y 0.7825 0.7739 0.7651 0.7563 0.7473

9. Using backward difference, find y 2.2 and y 2.2 from the following table: [A/M17]
x 1.4 1.6 1.8 2.0 2.2
y 4.0552 4.9530 6.0496 7.3891 9.0250

10. Given the following data, find y 6 and the maximum value of y (if it exists)
[A/M10]
x 0 2 3 4 7 9
y 4 26 58 112 466 922

11. Find the first derivative of f x  at x  2 for the data f 1   21, f 1 15 , f 2 12
and f 3  3 using Newton’s divided difference formula. [N/D10,A/M18]
12. Find f 10 from the following data [A/M11,N/D15]
x 3 5 11 27 34
-
f(x) 23 899 17315 35606
13
13. Find the first and second derivatives of y with respect to x at x=10 from the following
data: x : 3 5 7 9 11 [N/D17]
y: 31 43 57 41 27
14. Find the values f 8 and f 9 from the following table, using divided difference
interpolation formula: [A/M11]
x 4 5 7 10 11
f(x) 48 100 294 900 1210

15. Find f 3 and f 3 for the following data : [N/D15]
x : 3.0 3.2 3.4 3.6 3.8 4.0
f x  : - 14 -10.032 -5.296 - 0.256 6.672 14
1
dx 1 1 1
16. Find the approximate value of I   1 x
0
using Trapezoidal rule with h  , ,
2 4 8
and then by Romberg’s method. [A/M11]
6
1
17. Evaluate I   1 x dx using Trapezoidal rule and check by direct integration. [M/J12]
0

37 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS


18. By dividing the range into ten equal parts evaluate  sin x dx
0
by Trapezoidal and

Simpson’s one-third rule. [N/D10]

6
ex
19. Evaluate I  0 1 x dx using Simpson’s 1/3 rule with h 1.
rd
[N/D11]

1 1
x2 1
20. Find the value of log 2 3 from 0 1 x 3 dx using Simpson’s 3 rule with h  0.25 .
[M/J16]
21. The velocity v (km/min) of a moped which starts from rest, is given at fixed intervals of
time t (min) as follows:
x : 0 2 4 6 8 10 12
y : 0 10 18 25 29 32 20
 Estimate approximately the distance covered in 12 minutes, by Simpson’s 1/3rd rule.
 Estimate the acceleration at t  2 seconds. [ A/M15]

22. The velocities of a car running on a straight road at intervals of 2 minutes are given
below: [N/D13]
Time (min) : 0 2 4 6 8 10 12
Velocity (km/hr) : 0 22 30 27 18 7 0
1
Using Simpson’s - rd rule find the distance covered by the car.
3
23. The velocity v of a particle at a distance s from a point on its path is given by the table
below:
s(ft) 0 10 20 30 40 50 60
v(ft/sec) 47 58 64 65 61 52 38

Estimate the time taken to travel 60 feet by Simpson’s 1/3 rd rule and Simpson’s 3/8 th
rule. [A/M10,N/D14,N/D16]
24. The following data gives the corresponding values for pressure (p) and specific
volume(v) of a superheated steam. Find the rate of change of pressure with respect to
volume when v =2:
v: 2 4 6 8 10
p: 105 42.7 25.3 16.7 13 [N/D17]
0.6

 e dx correct to three decimal places by

x
25. Using Simpson’s one-third rule, evaluate
2

step-size=0.1. [N/D17]
6
1
26. Evaluate I   1 x dx by
0
using (i) direct integration (ii) Trapezoidal rule (iii)

Simpson’s one-third rule (iv) Simpson’s three-eighth rule. [N/D11,A/M17]

38 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS


2
27. Compute  sin x dx using Simpson’s 3/8 rule. [N/D12,M/J16]
0
1.3
28. Taking h  0.05 , evaluate 
1
x dx using Trapezoidal rule and Simpson’s three-eighth

rule. [M/J14]
1
1
29. Use Romberg’s integration to evaluate  1 x
0
2
dx . [A/M10,A/M11]

1
1
30. Using Romberg’s rule evaluate  1 x dx correct
0
to three decimal places by taking

h  0.5 , 0.25 and 0.125 . [N/D10,A/M15,A/M18]

1
2
x
31. Evaluate  sin x dx correct to three decimal places using Romberg’s method.
0
[M/J14]

1
dx
32. Evaluate  1 x
0
and correct to 3 decimal places using Romberg’s method and hence

find the value of log e 2 . [N/D14,N/D15,N/D16,N/D16,A/M17]

1
33. The following table gives the value of y  . Take h  0.5, 0.25, 0.125 and use
1  x2
1
1
Romberg’s method to compute 1  x
0
2
dx . Hence deduce an approximate value of  .

[A/M17]
x 0 0.125 0.25 0.375 0.5 0.675 0.75 0.875 1
y 1 0.9846 0.9412 0.8767 0.8 0.7191 0.64 0.5664 0.5

34. Use the Romberg’s method to get an improved estimate of the integral from x  1.8 to
x  3.4 from the data in table with h  0.4 [A/M15]
x : 1.6 1.8 2.0 2.2 2.4 2.6
f x  : 4.953 6.050 7.389 9.025 11.023 13.464
x : 2.8 3 3.2 3.4 3.6 3.8
f x  : 16.445 20.056 24.533 29.964 36.598 44.701
2
1
35. Evaluate  1 x
1
3
dx using Gauss three point formula. [A/M11,N/D14,M/J16]

x2  2 x 1
2
36. Evaluate 
0 1   x 1 
2
dx by Gaussian three point formula. [M/J13,N/D15]

1
1
37. Evaluate I 
1
 1 t 2
dt by one-point, two-point and three-point Gaussian formula and

compare with the exact value. [A/M10]

39 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

3
dt
38. Apply Gaussian three point formula to evaluate  1 t .
2
[N/D10,A/M17]

 3 x 
1
39. Using Gaussian three-point formula evaluate 2
 5 x 4 dx . [M/J12]
1
1
sin x
40. Apply three point Gaussian quadrature formula to evaluate 
0
x
dx [N/D13]

1.5

e
 x2
41. Evaluate dx using three point Gaussian quadrature formula. [A/M15]
0.2

1
1
42. Evaluate  1  x 2 dx using Gauss three point formula. [N/D16]
0
5
43. Evaluate  log 1  x  dx by three points Gauss quadrature formula.
0
10 [A/M17]

1
1
44. Use Romberg’s method to compute  1 x
0
2
dx correct to 4 decimal places. Also

evaluate the same integral using three-point Gaussian quadrature formula. Comment on
the obtained values by comparing with the exact value of the integral which is equal to

. [M/J12,N/D15]
4
1.2 1.4
dx dy
45. Evaluate   x y
1 1
by trapezoidal formula by taking h  k  0.1 . [A/M10]

4 1
5

46. Evaluate    dx  dy by trapezoidal rule in x -direction with h 1 and Simpson’s
1 1
x y 
one-third rule in y -direction with k 1 . [N/D10]
1 1
1
47. Evaluate   x  y 1 dx dy by using Trapezoidal rule taking h  0.5 and k  0.25 .
0 0

[A/M11,N/D14,N/D15]
2 2
dx dy
48. Using Trapezoidal rule, evaluate  x
1 1
2
 y2
numerically with h  0.2 along x -

direction and k  0.25 along y -direction. [M/J12,A/M15]

2 4
1
49. Evaluate   x  y 
1 3
2
dx dy by taking h  k  0.5 using Trapezoidal and Simpson’s rule.

[N/D10,N/D15]
3 2

 e
x2 y
50. Evaluate dx dy by Trapezoidal rule with  x  0.5 and Simpson’s one-third
1 0

rule with  y  0.5 . [N/D11]

40 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

2.4 4.4
51. Evaluate   x y dx dy
2 4
by Trapezoidal rule taking h  k  0.1 [N/D13]


2 

52. Evaluate   cos x  y  dx dy by using Trapezoidal rule by taking h  k  4 .
0
[M/J16]
2

1.4 2.4
1
53. Evaluate   xy
dx dy using Trapezoidal rule by taking h  k  0.1 and verify with
1 2
actual integration. [N/D16]
 /2 /2
54. Evaluate   sin x  y  dx dy by using Trapezoidal rule. [N/D16]
0 0
2 2
55. Evaluate 
0 0
f ( x, y ) dx dy by Trapezoidal rule for the following data, correct to three

decimal places:
x 0 0.5 1 1.5 2
y
0 2 3 4 5 6
1 3 4 6 9 11
2 4 6 8 11 14
2.6 4.4
1
56. Apply Simpson’s rule to evaluate the integral I   xy
2 4
dx dy . [A/M10,A/M18]

1 1
dx dy
57. Numerically evaluate   1 x 2  y 2 by taking  x   y  0.25 using Simpson’s 1/3rd
0 0
rule. [A/M11]
2 1
1 1
58. Evaluate   4 x y dx dy using Simpson’s rule by taking h  4
0 0
and k  .
2
[N/D12,M/J16]
1 1

1 2 2
sin x y 
59. Taking h  k 
4
, evaluate 
0 0
1 x y
dx dy using Simpson’s rule. [M/J14]


22

60. Evaluate   sin x  y  dx dy by Simpson’s rule taking h  k 
4
. Compare with the
00
actual value. [A/M17]
1 1
1 dx dy
61. Using Simpson’s
3
rule, evaluate   1 x y
0 0
with h  k  0.25 . [A/M17]

41 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

UNIT-IV
Initial value problems
PART-A
1. Find y  0 from the following table [A/M15]
x : 0 1 2 3 4 5
y : 4 8 15 7 6 2
Difference table

x y y 0 2 y 0 3 y 0 4 y 0 5 y 0
0 4
4
1 8 3
7 -18
2 15 -15 40
-8 22 -72
3 7 7 -32
-1 -10
4 6 -3
-4
5 2

By Newton’s forward formula, we have

1  
y 0  y 0  2 y 0  3 y 0  4 y 0  5 y 0 
1 1 1 1
h  2 3 4 5 
 
y 0 
4  2 3  3  18  4 40  5  72   27.9 .
11 1 1 1
1
2. By Taylor’s series method, find y 1.1 given y   x  y , y  1  0 .
[N/D 06,M/J 09, A/M 10, A/M 11,M/J13,N/D16]
Given x0  1 , y 0  0 , h  0.1
y  x  y y 0  1
y   1 y  y 0  2
y   y  y 0  2
h2
y1  y 0  h y 0 

y0 
h3
y0  0  0.11 
0.1 2  0.1 2  0.1103 .
2 3

2! 3! 2 6
3. Write Taylor’s series formula to solve y   f  x , y  with y x 0   y 0 . [M/J07]
h2 h3
y n  1  y n  h y n  y n  y n  .................... .
2! 3!
 y 2  x , y0 1 .
dy
4. Find by Taylor’s series method, the value of y at x  0.1 from
dx
[A/M15]

42 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Here x0  0 , y0  1 , h  0.1

Also given y   y 2  x y 0  y 0  x0  1  0  1
2

 
y   2 y y   1 y0  2 y0 y0 1  211 1  3
y   2 y y   2 y  2
 2
y0  2 y0 y0  2 y0  213  21  6  2  8
2

By Taylor’s series method, we have

 h  h 
2 3
y0.1  y0  h y 0  y0  y 0  ..............
2! 3!

 1  0.11 
0.12 3  0.13 8  ......  1 0.1 0.005  0.0013  1.1063 .
2! 6
5. Find the Taylor series upto x term satisfying 2 y   y  x  1 , y0  1 . [N/D08]
3

Given x0  0 , y 0  1 and let h  0.1

y 
1
x  y 1 y 0  0
2
y   1  y 
1
y 0  0.5
2
y 
y    y0   0.25
2
h2
y1  y 0  h y 0  
y0 
h3
y0  1  0.10 
0.1 0.5  0.1  0.25  1.0024 .
2 3

2! 3! 2 6
6. State the merits and demerits of Taylor’s series method of solution.
[A/M10,M/J14,N/D15]
Merits: (i) It is easily derived for any order according to own interest.
(ii) The values of y for any x are easily obtained.
Demerits: This method suffers from the time consumed in calculating the
higher derivatives.
7. What is the major drawback of Taylor series method? [M/J12]
If it is possible to find the successive derivatives in easy manner then only Taylor
series method is powerful.
8. State Euler’s Formula [M/J13]
Given y’ = f ( x ) , y ( x0 ) = y0
Formula: yn + 1 = yn + h f ( xn , yn )

9. Find y 0.1 if  1  y , y 0  1 using Taylor series method.

dy
[N/D12,M/J16]
dx
Given x0  0 , y 0  1 , h  0.1
y   1 y y 0  2
y   y  y 0  2
y   y  y 0  2

43 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

h2
y1  y 0  h y 0  y 0 
h3
y0  1  0.12 
0.1 2  0.1 2  1.1103 .
2 3

2! 3! 2 6
10. Find y 1.1 , given  x  y , y 1  2 by Euler’s method.
dy
[M/J06,M/J13]
dx
Given x0  1 , y 0  2 , h  0.1 , y’ = x+y
Let f x , y   x  y , Hence f x0 , y0   x0  y0  3
Euler’s formula is y1  y0  h f x0 , y0 
y1  2  0.1 f 1, 2  2  0.1 3  2.3 .

11. Find y 1.1 , using Euler’s method from  x 2  y 2 , y 1  1 .

dy
[A/M08]
dx
Given x0  1 , y 0  1 , h  0.1 , y   x 2  y 2
Let f x , y   x 2  y 2
f  x0 , y 0   x0  y 0  2
2 2

Euler’s formula is y1  y0  h f x0 , y0 

y1  1 0.12  1.2 .
12. Write down the modified Euler’s formula for ODE? [M/J09]
The modified Euler’s formula is
 
y n  1  y n  h f  xn  , y n  f xn , y n  .
h h
 2 2 
13. Find y 0.2 when y    2 x y 2 , y0  1 and h  0.2 , by Euler’s method. [N/D09]
Given x0  0 , y 0  1 , h  0.2 , y    2 x y 2
Let f x , y    2 x y 2
f  x0 , y 0    2 x0 y 0  0
2

Euler’s formula is y1  y0  h f x0 , y0 

y1  1 0.20  1 .
14. Using Euler’s method find y 0.2 given y   x  y , y 0  1 . [A/M10,N/D16]
Given x0  0 , y 0  1 , h  0.2 , y   x  y
Let f x , y   x  y
f  x0 , y 0   x0  y 0  1
Euler’s formula is y1  y0  h f x0 , y0 
y1  1 0.21  1.2 .
15. Given y   x 2  y , y0  1by using Euler’s method find y 0.1 . [A/M11]
Given x0  0 , y 0  1 , h  0.1 , y   x 2  y
Let f x , y   x 2  y
f  x0 , y 0   x0  y 0  1
2

Euler’s formula is y1  y0  h f x0 , y0 

44 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

y1  1 0.11  1.1 .

 1  y , y 0  0 using Euler’s method for x  0.1 with h  0.1 .

dy
16. Solve
dx
Given x0  0 , y 0  0 , h  0.1 , y   1  y [N/D11,A/M17]
Let f x , y   1  y
f  x0 , y 0   1  y 0  1
Euler’s formula is y1  y0  h f x0 , y0 
y1  0  0.11  0.1 .

 f  x , y  with initial
dy
17. Give the modified Euler’s method to find y(x1) for solving
dx
condition y x 0   y 0 . [M/J12]
Given y   f x , y  , yx0   y0
Modified Euler’s formula is
 
y1  y 0  h f  x0  , y 0  f x0 , y 0  .
h h
 2 2 

18. Use Euler’s method to find y 0.2 and y 0.4 given  x  y , y 0  1 .
dy
dx
Given x0  0 , y 0  1 , h  0.2 , y   x  y [A/M 10,A/M15]
Let f x , y   x  y
f  x0 , y 0   x0  y 0  1
Euler’s formula is y1  y0  h f x0 , y0 
y1  1 0.21  1.2 .
Hence x1  0.2 , y1  1.2
f x1 , y1   x1  y1  1.4
y2  y1  h f x1 , y1   1.2  0.21.4  1.48

19. Find y 0.1 by using Euler’s method given that  x  y , y 0  1 .

dy
dx
Given x0  0 , y 0  1 , h  0.1 , y   x  y [N/D10,N/D14,N/D15,N/D15]
Let f x , y   x  y
f  x0 , y 0   x0  y 0  1
Euler’s formula is y1  y0  h f x0 , y0 
y1  1 0.11  1.1 .
20. Find y 0.2 for the equation y   y  e x , given that y0  1 by using Euler’s
method. [A/M11]
Given x0  0 , y 0  1 , h  0.2 , y   y  e x
Let f x , y   y  e x
f x0 , y0   y0  e x0  2

45 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Euler’s formula is y1  y0  h f x0 , y0 

y1  1 0.22  1.4 .

 f  x , y  with y x 0   y 0 .
dy
21. State Euler’s method to solve [N/D11]
dx
Given y   f x , y  , y  x0   y 0
y n  1  y n  h f xn , y n  .
22. Using Euler’s method find the solution of the initial value problem y   y  x 2  1 ,
y0  0.5 at x  0.2 taking h  0.2 . [N/D13]
Here h  0.2 , x0  0 , y 0  0.5
y   f x , y   y  x 2  1
By Euler’s formula, we have
y1  yx1   y0.2  y0  h f x0 , y0   y0  h y0  x0 1  2

 
 0.5  0.2 0.5  0 2 1  0.8 .
23. Using Euler’s method, find the solution of the initial value problem

 log x  y  , y0  2 at x  0.2 by assuming h  0.2 .

dy
[M/J12]
dx
Given x0  0 , y 0  2 , h  0.2 , y   log x  y 
Let f x , y   log x  y 
f x0 , y0   log x0  y0   log 2  0.3010
Euler’s formula is y1  y0  h f x0 , y0 
y1  2  0.20.3010  2.0602 .
24. State the advantages of RK-method over Taylor’s series method. [A/M15].
The RK-methods are designed to give greater accuracy and they possess the
advantage of requiring only the function values at some selected points on the sub
interval.

 f  x , y  with y x 0   y 0 .
dy
25. Write Runge-Kutta’s 4th order formula to solve
dx
Given y   f x , y  , yx0   y0 [M/J07,N/D12,M/J16]
Runge-Kutta’s 4th order formula is given by
k1  h f x0 , y0 
 h k 
k 2  h f  x0  , y 0  1 
 2 2
 h k 
k 3  h f  x0  , y 0  2 
 2 2
k 4  h f  x0  h , y 0  k 3 

Let  y 
1
k1  2 k 2  2 k3  k 4 
6

46 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

y1  y0   y0 .
26. What is Predictor – Corrector method? [N/D07]
Predictor – corrector methods are methods which require the values of y at
xn , xn 1, xn  2 ,.......... for computing the values of y at x n 1 . We first use a formula to
find the value of y at x n 1 and this is known as a predictor formula. The value of y so
got is improved or corrected by another formula is known as corrector formula.
27. Mention the multistep methods available for solving ordinary differential equation.
[N/D07]
The multistep methods available for solving ordinary differential equation are
 Milne’s predictor-corrector method
 Adam-Bashforth predictor-corrector method.
28. What are multi-step methods? How are they better than single step method?
[A/M18]
The multistep methods available for solving ordinary differential equation are
 Milne’s predictor-corrector method
 Adam-Bashforth predictor-corrector method.
Multi step methods are better than single step methods since it use information about
the solution at more than one point.
29. Write down the Milne’s predictor – corrector formula.
yn 1 , p  yn  3  4h
3

2 y n  2  y n 1  2 y n [N/D10,M/J14,N/D14,N/D16,A/M17]

y n  1 , c  y n 1  
h
3

y n 1  4 y n  y n  1

30. Define single step and multi step methods for the solution of the differential

 f  x , y  , y x 0   y 0 .
dy
equation [N/D11]
dx
In solving the differential equation y   f x , y  , if only one initial value yx0   y0 is
given then it is called single step method.
If there are four initial values y0 , y1 , y 2 , y3 then it is called multi step meyhod.
31. Give the error for Milne’s predictor formula. [M/J12]
5
14h 5
The truncation error in Milne’s predictor formula is y ( )
45
 h5 5
The truncation error in Milne’s corrector formula is y ( )
90

32. How many prior values are required in predictor-corrector formulae?

4 prior values are needed. [N/D17]

33. How many values are needed to use Milne’s predictor-corrector formula prior to
the required value? [M/J16]

47 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

Four values are needed to use Milne’s predictor-corrector formula prior to the required
value.
34. State Adams – Bashforth predictor and corrector formula.
[M/J 06,N/D 06,A/M 08,N/D 08,N/D 09,A/M 11,N/D13,N/D15,A/M17,A/M18]
yn 1 , p  yn 
h
24
 
55 y n  59 y n 1  37 y n  2  9 y n  3

yn 1 , c  yn 
h
24
 
9 y n  1  19 y n  5 y n 1  y n  2

35. Distinguish between single step methods and multi step methods. [A/M15,N/D16]
In solving difference equation y   f x , y  , if only one initial value yx0   y0 is
given then it is called single step method. If there are four initial values y0 , y1 , y 2 , y3 ,
then it is called multistep method.

PART-B

1. Use Taylor series method to find y 0.1 and y 0.2 given that  3 e x  2 y , y0  0 ,
dy
dx
correct to 4 decimal accuracy. [A/M10]
 x 2 y 1 , y0 1 , by Taylor’s series
dy
2. Find the value of y at x  0.1 , 0.2 given
dx
method up to four terms. [N/D10,N/D16,A/M17]
 x 2 y  1 , y0 1 . [N/D14]
dy
3. Using Taylor’s series method, find y at x  0 if
dx

 x  y given y1 1 , and find y 1.1 and y 1.2 by Taylor’s series method.
dy
4. Solve
dx
[N/D10]
5. Using Taylor’s series method, compute y 0.2 correct to 4 decimal places given

 1  x y and y0  0 .
dy
[A/M11]
dx
1
6. Using Taylor series method, find y 1.1 correct to four decimal places given
dy
 x y3
dx
and y1 1 . [M/J12]
7. Obtain y by Taylor’s series method, given that y   x y 1 , y0 1 , for x  0.1 and 0.2
correct to four decimal places. [N/D13]
dy
8. Using Taylor’s series method, find y at x 1.1 by solving the equation  x2  y2 ;
dx
y1  2 . Carryout the computations upto fourth order derivative. [M/J14,M/J16]
9. Solve y   x  y ; y0 1 by Taylor’s series method. Find the values of y at x  0.1 and
x  0.2 . [A/M15]

48 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

 1  x y with y0  2 . Find y0.1 , y0.2 and

dy
10. Using Taylor series method solve
dx
y0.3 . [N/D15]
11. Using Taylor’s series method, compute the value of y0.2 correct to 3 decimal places

 1 2 x y given that y0  0 .

dx
from [M/J16]
dy
dy
12. Find the values of y at x=0.1 given that  x 2  y , y(0)=1 by Taylor’s method.
dx
[N/D17,A/M18]
13. Given y    y and y0 1 , determine the values of y at x  0.01 , 0.02 , 0.03 by
Euler’s method. [M/J12]
14. Using Modified Euler’s method find y 0.1 and y 0.2 given  x 2  y 2 , y0 1 .
dy
dx
[N/D10,M/J16]

, y0 1 by modified Euler’s method to find y 0.1 with h  0.1 .

dy 2x
15. Solve  y
dx y
[N/D11]
yx
16. Solve y '  , y  0   1 at x = 0.1 by taking h = 0.02 by using Euler’s method.
yx
[ M/J 13 ]
17. Using Modified Euler’s method, find y 4.1 and y 4.2 if 5 x
 y 2  2  0 ; y4 1
dy
dx
[N/D12]
18. Apply modified Euler’s method to find y0.2 and y0.4 given y   x  y , y0 1
2 2

by taking h  0.2 . [N/D14]

19. Solve 1  x    y 2 , y0  1 by Modified Euler’s method by choosing h  0.1 , find

dy
dx
y0.1 and y0.2 . [N/D16]

 y  x 2  1 , y0  0.5 . Find y0.2 by modified Euler’s method. [A/M17]

dy
20. Given
dx
dy
21. Find the values of y at x=0.1 given that  x 2  y , y(0)=1 by modified Euler’s
dx
method. [A/M18]
dy y 2  x 2
22. Use Runge - Kutta method of fourth order to find y 0.2 , given  , y0 1 ,
dx y 2  x 2
taking h  0.2 . [A/M10]
dy y 2  x 2
23. Using R-K method of fourth order solve  2 with y0  1 at x  0.2 .
dx y  x2
[A/M15,N/D16]

49 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

24. Find y 0.8 correct to 4 decimal places by using Runge-Kutta method of 4th order if
y   y  x 2 , y0.6 1.7379 . [A/M10]
25. Find the value of y1.1 using Runge-Kutta method of 4 th
order for the given equation

 y 2  x y ; y1 1 .
dy
[M/J16]
dx
26. Apply the fourth order Runge-Kutta method to find y 0.2 given that y   x 3  y ,
y0  2 . [A/M11]
27. Using Runge-Kutta method of fourth order,find the value of y at x  0.2 , 0.4 , 0.6 given

 x 3  y , y0  2 . Also find the value of y at x  0.8 using Milne’s predictor

dy
that
dx
and corrector method. [M.J14,M/J16]
y 0.3 , using Runge-Kutta method of fourth order, given that
dy 2x y
28. Find  1 ,
dx 1 x 2
y0  0 , y0.1  0.1006 , y0.2  0.2052 and then find the value of y 0.4 using Milne
predictor-corrector method. [A/M11]
2
d y dy dy
29. Solve 2
x  y  0 , given y 1 ,  0 at x  0 by the fourth order Runge-
dx dx dx
Kutta method to find y 0.1 with step size  0.1 . [N/D11]
30. Using Runge-Kutta method of fourth order, find y 0.7  correct to 3 decimal places if
y   y  x 2 , y0.6 1.7379 . [M/J12]
31. Find the value of y 0.1 by Runge-kutta method of fourth order given y   x y   y  0 ,
y0 1 and y 0  0 . [N/D10,N/D11,N/D14,A/M17]
32. Consider the second order initial value problem y   2 y   2 y  e 2 t sin t with
y0   0.4 and y 0   0.6 using fourth order Runge kutta algorithm, find y 0.2 .
[M/J12]
33. Solve y   x y   y 2  0 using Runge-Kutta method for x  0.2 correct to 4 decimal
places. The given conditions are y0  1 ; y 0  0 . [N/D16]

34. Using Runge-Kutta method of order four, find y when x 1.2 in steps of 0.1 given that
y   x 2  y 2 and y1 1.5 [N/D13]
35. Employ the classical fourth order Runge-Kutta method to integrate y   4 e 0.8 t  0.5 y
from t  0 to t  1 using a stepsize of 1 with y0  2 . [A/M15]
36. Find y0.8 given that y   y  x 2 , y0.6  1.7379 by using Runge-Kutta method of
order four. Take h  0.1 . [N/D15]

50 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

37. Solve for y 0.1 and z 0.1 from the simultaneous differential equations
dy
 2 yz;
dx
 y  3 z ; y0  0 , z 0  0.5 using Runge-kutta method of the fourth order. [N/D12]
dz
dx

38. Use Milne’s predictor-corrector formula to find y 0.4 , given

dy

 
1 x 2 y 2
, y0 1 ,
dx 2
y0.1 1.06 , y0.2 1.12 and y0.3 1.21 . [A/M10]

 x y  y 2 , y0 1 , y0.1 1.1169 and y0.2 1.2774 , find (i) y 0.3 by

dy
39. Given
dx
Runge-kutta method of fourth order and (ii) y 0.4 by Milne’s method.
[N/D10,N/D17,A/M18]
40. Using Runge-kutta method of fourth order, find y for x  0.1 , 0.2 , 0.3 given that
y   x y  y 2 , y0 1 . Continue the solution at x  0.4 using Milne’s method. [A/M11]

41. Given that

dy 1
dx 2
 
 1  x 2 y 2 ; y0 1 , y0.1 1.06 , y0.2 1.12 and y0.3 1.21 ,

evaluate y 0.4 and y 0.5 by Milne’s predictor – corrector method. [N/D11]

42. Given that y   x y   y  0 , y0 1 , y 0  0 . Obtain y for x  0.1 , 0.2 and 0.3 by
Taylor’s series method and find the solution for y 0.4 by Milne’s method. [M/J12]

 1  y 2 ; y0.6  0.6841, y0.4  0.4228 , y0.2  0.2027 , y0  0 , find

dy
43. Given that
dx
y  0.2 using Milne’s method. [N/D12]
44. Determine the value of y 0.4 using Milne’s method given y   x y  y 2 , y0 1 . Use
Taylor’s series method to get the values of y 0.1 , y 0.2 and y 0.3 .
[A/M10,N/D15,N/D16,A/M17]
45. Using Milne’s method to find y 4.4 given 5 x  y 2  2  0 , with y4 1 ,
dy
dx
y4.1 1.0049 , y4.2 1.0097 , y4.3 1.0143 . [N/D10,N/D14,A/M15,N/D15]
46. Using Milne’s predictor-corrector method, find y 4.5 given 5 x y   y 2  2  0 , y4 1 ,
y4.1 1.0049 , y4.2 1.0097 , y4.3 1.0143 and y4.4 1.0187 . [N/D11]

47. Use Milne’s method to find y0.8 , given y   , y0  2 , y0.2  2.0933 ,
1
x y
y0.4  2.1755 , y0.6  2.2493 . [N/D13]

 x y  y 2 and y0 1 , y0.1 1.1169 , y0.2 1.2773 , y0.3  0.2267 ,

dy
48. Given
dx
evaluate y0.4 by Milne’s predictor corrector method. [A/M15]

 x  y 2 , y0  1 , y0.2  0.02 , y0.4  0.0795 and y0.6  0.0.1762 .

dy
49. Given
dx
Compute y0.8 using Milne’s method. [N/D16]

51 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

 x y  y 2 , y0 1 , y0.1 1.1169 , y0.2 1.2774 and y0.3 1.5041 . Use

dy
50. Given
dx
Adam’s method to estimate y 0.4 . [A/M10]
51. Solve y   x  y 2 , y0 1 to find y 0.4 by Adam’s method. Starting solutions required
are to be obtained using Taylor’s method using the value h  0.1 . [A/M11]

 x  y 2 , y0 1 to find y0.4 by Adam’s

dy
52. Solve the initial value problem
dx
Bashforth predictor corrector method and for starting solutions, use the information
below y0.1  0.9117 , y0.2  0.8494 . Compute y0.3 using Runge Kutta
method of fourth order. [A/M15]
53. Obtain y 0.8 given y   x  y , y0 1 with h  0.2 by Adam’s method. [A/M10]
54. Using Adam’s Bashforth method, find y4.4 given that 5 x y   y 2  2 , y4 1 ,
y4.1 1.0049 , y4.2 1.0097 , y4.3 1.0143 . [M/J14,M/J16]
55. Using Adam-Basforth predictor-corrector method, find y 4.5 given 5 x y   y 2  2  0 ,
y4 1, y4.1 1.0049 , y4.2 1.0097 , y4.3 1.0143 and y4.4 1.0187 . [N/D11]

56. Find y 0.4 given   x y  , y0 1 , y0.1 1.01 , y0.2 1.022 , y0.3 1.023
dy 1
dx 2
by Adam’s method. [M/J12,M/J16,N/D17]
 x 2 1  y  , y1  1 , y1.1  1.233 , y1.2  1.548 , y1.3  1.979 ,
dy
57. Given
dx
evaluate y1.4 by Adam’s-Bashforth method. [A/M15,N/D15]
58. Find y0.1 , y0.2 and y0.3 from y   x  y 2 , y0  1 by using Runge-Kutta method
of Fourth order and then find y0.4 by Adam’s method. [N/D15]

UNIT-V
Boundary value problems
PART-A

1. Define Boundary value problem. [N/D10]

If the conditions are prescribed at two or more points, then the problem is called as
boundary value problem.
2. Classify the partial differential equation u xx  2 u xy  4u yy  0 ; x , y  0 .
[A/M11N/D16]
Given A  1 , B  2 , C  4
B 2  4 A C  4  4 4  12  0
Therefore, the given pde is elliptic.
3. Classify the partial differential equation u xx  x u yy  0 . [N/D11]
Given A  1 , B  0 , C   x

52 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

B 2  4 AC  0  4 x  4 x
B 2  4 A C  0 when x  0 , implies parabolic.
B 2  4 A C  0 when x  0 , implies elliptic.
B 2  4 A C  0 when x  0 ,implies hyperbolic.
4. Classify the differential equation f xx  2 f xy  f yy  0 . [A/M10,M/J16]
Given A  1 , B   2 , C  1
B 2  4 AC  4  4  0
Therefore, the given pde is parabolic.
 2u  2u  2 u u u
5. Classify the following equation 4 4 2  2  0.
x 2
x y y x y
Here A  1 , B  4 , C  4 [A/M15,N/D16]
B 2  4 AC  4  4 14  16 16  0
2

Hence, the given equation is a parabolic equation.

6. Write down the standard five point formula to find the numerical solution of
Laplace equation. [A/M 10,N/D14,A/M15,N/D16,A/M17]
Standard five point formula is
1
 
u i , j  u i 1, j  u i  1, j  u i , j  1  u i , j  1 .
4
7. Solve  2 U  0 numerically for the following square mesh with boundary values as
shown in figure [A/M11]

Initially we assume a = 1, b = 3. Using Standard five point formula, we get

The value of a  1 and b2

53 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

8. State the diagonal five-point formula for solving Laplace equation.

[M/J12,N/D13,M/J16,N/D17]
Diagonal five point formula is
1
 
u i , j  u i 1, j  1  u i  1, j 1  u i 1, j 1  u i 1, j  1 .
4
9. What is the error for solving Laplace and Poisson’s equations by finite difference
method? [N/D10,A/M18]
2
Error for solving Laplace and Poisson equations is the order of h .
10. Write the difference scheme for solving the Poisson equation  2 u  f  x , y  .[M/J12]
u i  1, j  u i  1, j  u i , j  1  4 u i , j  h 2 f  i h , j h 
11. Write down the explicit finite difference method for solving one dimensional wave
equation. [A/M10,N/D15,N/D16]
 
u i , j 1  2 1 2 a 2 u i , j  2 a 2  u i 1, j  u i 1, j   u i , j 1
1
To get the simplest form, choose  
a
u i , j  1  u i  1, j  u i  1, j  u i , j  1 .
12. Write down one dimensional wave equation and its boundary conditions. [M/J07]
One dimensional wave equation is  2 u x x  ut t
Boundary conditions are :
 u t  x , 0  0
 u  x , 0  f 1
 u 0 , t   f 2
 u c , t   f 3
13. Write down Bender-Schmidt’s difference scheme in general form and using suitable
value of  , write the scheme in simplified form. [N/D12]
u i , j 1   u i 1, j  1 2  ui , j   u i 1, j
k k
where   2
 2 ( since a  1 )
ah h
1
To get the simplest form, choose   . Then
2
1
u i , j 1 
2

u i  1, j  u i 1, j 
14. State the implicit finite difference scheme for one dimensional heat equation.
[M/J06]
State Crank-Nicholson difference scheme for solving one dimensional heat
equation. [N/D10,N/D12,N/D15,A/M18]

 u i 1, j 1  u i 1, j 1   2  1u i , j 1  2  1u i , j   u i 1, j  u i 1, j 

To get the simplest form, choose   1

54 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

1
u i , j 1 
4

u i  1, j  1  u i 1, j  1  u i  1, j  u i 1, j 
15. State whether the Crank Nicholson’s scheme is an explicit or implicit scheme. Justify.
[M/J14]
The solution value at any point i , j 1 on the  j 1th level is dependent on the
 j 1th level and j th level values. Hence it is an implicit method.
16. Write down the explicit formula to solve the hyperbolic equation utt  9u xx
1 1 1 1
when Δx=0.25 and t  .Given a2=9, x  h  , t  k  ,therefore   .
16 4 16 4
 9
ui , j  1  2 1   ui , j 
9
 
u i  1, j  u i  1, j  u i , j 1
 16  16
 2u 2  u
2
17. Express 2  c in terms of difference approximation. [A/M15]
t x 2
Given c 2 u xx  utt  0 ...............1
u i 1 , j  2 u i , j  u i  1, j ui , j 1  2 u i , j  u i , j 1
Substituting u xx  and u tt  in 1 , we
h2 k2
have
 u i 1, j  2 u i , j  u i  1, j   u i , j 1  2 u i , j  u i , j  1 
c2   0
 h2   k2 
k c u i 1, j  2 k c u i , j  k c u i  1, j  h u i , j  1  2 h 2 u i , j  h 2 u i , j  1  0
2 2 2 2 2 2 2

2 2 2
k 2 k 2 k 2
  c u i 1, j  2   c u i , j    c u i  1, j  u i , j 1  2 u i , j  u i , j  1  0
h h h
2 ui , j  2 2 c 2 ui , j  2 c 2 ui 1, j  ui 1, j   u i , j 1  ui , j 1

 
ui , j 1  2 1  2 c 2 ui , j  2 c 2 ui 1, j  ui 1, j   ui , j 1 where  
k
h
.

18. Write the finite difference approximations of y  x  and y  x  . [A/M15,N/D15]

The finite difference approximation of y  x  and y  x  is given by
 yi  1  yi  1  yi  1  2 yi  yi 1
yi x   and yi x   , i  1, 2 , 3 , ........n  1.
2h h2
19. Using finite difference solve y   y  0 given y0  0 , y1 1 , n  2 . [N/D13]
1
Given y   y  0 , y 0  0 , y 2  1 , n h 1 implies 2 h 1 implies h 
2
Let yi  yi  0 ...........1
 yi 1  2 yi  yi  1
and h  in 1 , we have
1
Substituting yi  2
h 2
yi 1  2 yi  yi  1
2
 yi  0
1
 
2
55 II YEAR EIE QUESTION BANK PEC
 NUMERICAL METHODS

4 yi 1  8 yi  4 yi  1  yi  0
4 yi 1  9 yi  4 yi 1  0 ..................... 2
Substituting i  1 in 2  , we have
4 y0  9 y1  4 y 2  0
Given y 0  0 , y 2  1 , we have 4 0  9 y1  4 1  0 implies
 9 y1   4 implies
4
y1  .
9
1
x 0 1
2

4
y 0 1
9

20. Write down the finite difference scheme for the differential equation
d2y dy
2
3  2. [M/J06]
dx dx
Given y n  3 y n  2
 y n 1  2 y n  y n  1   y n  1  y n 1 
 2 3 2
 h   2h 
 
2  y n  1  2 y n  y n  1  3 h y n  1  3 h y n  1  2 2 h 2
2  3 h yn 1  2 yn  2  3 h yn 1  4 h 2 .
d2y
21. Obtain the finite difference scheme for the differential equation 2  y  5.
dx 2
Given 2 y n  y n  5 [N/D06,N/D07,M/J13,M/J16]
 y n 1  2 y n  y n  1 
2   yn  5
 h2 
 
2 y n 1  2 y n  y n  1  h 2 y n  5
 
2 y n 1  h 2  2 y n  2 y n 1  5h 2
d2y
22. State finite difference approximation for and state the order of truncation
dx 2
error. [A/M08]
y n 1  2 y n  y n  1
y n 
h2
23. Write the finite difference scheme for the second order differential equation
1
y   f , h  . [N/D09]
n

56 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

y n 1  2 y n  y n  1
y n 
h2
Given y   f x , y  hence y n  f xn , y n  and h 
1
n

Therefore, f xn , y n   n 2 y n 1  2 y n  y n 1 
dy d2y
24. State central finite difference expression for and . [A/M11,N/D14, N/D15]
dx dx 2
y n 1  y n 1 y n 1  2 y n  y n  1
y n  y n 
2h h2
25. Write down the difference equation to solve y   4 y   4 y  0 , y0  0 , y1  1 .
Given y n  4 y n  4 y n  0 [N/D11]
 y n 1  2 y n  y n  1   y n  1  y n 1 
   4    4 yn  0
 h2   2h 
ba 1
h 
n 4
Hence 2 y n 1  7 y n  6 y n 1  0 for n  1, 2 , 3 ,........

PART-B

1. Deduce the standard five point formula  2 u  0 . Hence solve it over the square region
by the boundary conditions as in the figure below: [A/M10,N/D15]

2. Solve the elliptic equation u xx  u yy  0 for the following square mesh with boundary
values as shown: [M/J12,N/D15]

57 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

 2u  2u
3. Solve the Laplace’s equation   0 at the interior points of the square region
x 2 y 2
given as below: [M/J16]
0 11.1 17.0 19.7 18.6

u1 u2 u3
0 21.9

u4 u5 u6
0 21.0

0 u7 u8 u9 17.0

0 8.7 12.1 12.8 9.0

4. Solve u x x  u y y  0 , for the following square mesh with boundary condition as shown
below. Iterate until the maximum difference between successive values at any grid point
is less than 0.001 . [A/M15]
1 2
A B

uuuuuuu u1
1 2

2 1

C D
2 1
5. Solve u xx  u yy  0 over the square mesh of side 4 units satisfying the following
boundary conditions [A/M 10,N/D 11,M/J 12,N/D14,N/D16 ,A/M18]
(i) u0 , y   0 for 0  y  4
(ii) u4 , y   12  y for 0  y  4
(iii) ux , 0  3 x for 0  x  4
(iv) ux , 4  x 2 for 0  x  4

6. Given the values of u  x , y  on the boundary of the square in fig. evaluate the function
u  x , y  satisfying the Laplace equation  2 u  0 at the pivotal points of this fig. by
Gauss Seidel method. [A/M15]
58 II YEAR EIE QUESTION BANK PEC
 NUMERICAL METHODS

1000 1000 1000 1000

u1 u2
2000 500

u3 u4
2000 0

1000 500 0 0
7. Solve  u  8 x y in the square region  2  x , y  2 with u  0 on the boundaries
2 2 2

after dividing the region into 16 subintervals of length 1 unit. [N/D10,M/J13,M/J16]

 
8. Solve the equation  2 u  10 x 2  y 2 10 over the square mesh with sides x  0 ,
y  0 , x  3 , y  3 with u  0 on the boundary with mesh length 1unit.
[A/M11,N/D12,M/J14,N/D16,N/D17]
9. Solve  u  8 x y for square mesh given u  0 on the four boundaries dividing the
2 2 2

square into 16 sub squares of length 1 unit. [N/D11]

10. Solve the Poisson equation u xx  u yy   81 x y , 0  x 1; 0  y 1 u0 , y   0 ,

u1, y   100 , ux , 0  0 , ux ,1  100 and h 

1
. [A/M17]
3
11. Solve the Poisson’s equation  2 u  8 x 2 y 2 for the square mesh of fig. with
ux , y   0 on the boundary and mesh length 1 . [A/M15]
Y

u1 u2 u1

u2 u3 u2

u1 u2 u1

12. By iteration method solve the elliptic equation u x x  u y y  0 over the square region of
side 4 , satisfying the boundary conditions. [N/D13]
i  u 0 , y   0 , 0  y  4
59 II YEAR EIE QUESTION BANK PEC
 NUMERICAL METHODS

ii  u 4 , y   8  2 y , 0  y  4
x2
iii  u x , 0  , 0  x  4
4
iv  u x , 4  x 2 , 0  x  4
Compute the values at the interior points correct to one decimal with h  k 1 .
13. Solve numerically 4 u xx  utt with the boundary conditions u0 , t   0 , u4 , t   0 and
the initial conditions ut x , 0   0 and ux , 0  x 4  x  , taking h 1 (for 4 time steps)
[N/D10,N/D15,M/J16,A/M17,A/M17]
14. Solve 25 u xx  utt  0 for u at the pivotal points, given u0 , t   u5 , t   0 , ut x , 0   0
and ux , 0  x 5  x  for one half period of vibration (taking h 1 ). [A/M11]
 2u  2u u
15. Solve  2 , 0  x 1 , t  0 given ux , 0  0 , x , 0  0 , u0 , t   0 and
t 2
x t
u1, t  100 sin  t . Compute ux , t  for four times steps with h  0.25 .[N/D10,A/M18]
 2u  2u
16. Solve the equation  2 , 0  x 1 , t  0 satisfying the conditions ux , 0  0 ,
x 2 t
u
x , 0  0 , u0 , t   0 and u1, t   1 sin  t . Compute ux , t  for 4 time- steps by
t 2
1
taking h  . [N/D12]
4
2 u 2 u
17. Solve the wave equation  , 0  x 1 , t  0 , u 0 , t   u 1, t   0 , t  0 ,
 x2 t2
 1 , 0  x  0.5 u
u x , 0   and x , 0  0 , using h  k  0.1 , find u for three time
1 , 0.5  x 1 t
steps. [M/J14]

18. Evaluate the Pivotal values of the equation ut t  16 u x x taking  x  1 upto t  1.25 .
The boundary conditions are u0 , t   u 5 , t   ut x , 0  0 and ux , 0  x 2 5  x  .
[A/M15,N/D16]
19. Solve yt t  y x x upto t  0.5 with a spacing of 0.1 subject to y0 , t   0  y 1, t  ,
yt x , 0  0 and yx , 0  10  x 1  x  . [A/M15]

20. Solve ut t  u x x , 0  x  2 , t  0 subject to u x , 0  0 , ut x , 0  100 2 x  x 2 ,
u 0 , t   0 , u 2 , t   0 , choosing h 
1
compute u for four time steps. [N/D13]
2

2 f f
21. Given 
t
 
, f 0 , t   0  f 5 , t  , f x , 0  x 2 25  x 2 , find f in the range
x 2

taking h  1 and upto 5 seconds. [A/M15]

60 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

 
22. Solve ut  u xx in 0  x  5 , t  0 given that u0 , t   0 , u5 , t   0 , ux , 0  x 2 25  x 2 .
Compute u up to t  2 with  x 1 , by using Bender-Schmidt formula. [N/D10]
 u u
, given u0 , t   0 , u5 , t   0 , ux , 0  x 2 25  x 2 , find u in the range
2
23. Solve 
x 2
t
taking h 1 up to 3 seconds using Bender-Schmidt recurrence equation. [A/M11]
 2 u u
24. Sol  , subject to u0 , t   u 1, t   0 , ux , 0  sin  x , 0  x 1 using
ve x
2
t
Bender-Schmidt method. [M/J12,N/D15]
2 u u
25. Using Bender-Schmidt’s method solve  given u 0 , t   0 , u 1, t   0 ,
 x2 t
u x , 0  sin  x , 0  x 1 and h  0.2 . Find the value of u upto t  0.1 . [M/J14,N/D16]
 2 u u
26. Solve  given u0 , t   0 , u4 , t   0 , ux , 0  x 4  x  assuming h  k 1 . Find
x 2 t
the value of u upto t=4 with Δx=Δt=1. [A/M11,N/D17]
27. Solve u xx  32 ut , h  0.25 for t  0 , 0  x 1 , u0 , t   0 , ux , 0  0 , u1, t   t . [M/J16]
k c2
28. Obtain the Crank-Nicholson finite difference method by taking    1 . Hence
h2
 2 u u
find ux , t  in the rod for two times steps for the heat equation  , given
x 2 t
ux , 0  sin  x  , u0 , t   0 , u1, t   0 . Take h  0.2 . [A/M10]
 u  2u
29. Solve the equation  subject to the condition ux , 0  sin  x , 0  x  1 ;
t  x2
u0 , t   u1, t   0 using Crank-Nicholson method. [A/M15]
u  2 u
Using Crank-Nicholson method, solve  2 subject to ux , 0  0 , u0 , t   0 and
t x
u1, t   t (i) taking h  0.5 and k  and (ii) taking h  and k  . [A/M10,N/D16]
1 1 1
8 4 8
 2 u u
30. Use Crank-Nicholson method to solve the equation  , satisfying the conditions
x 2 t
ux , 0  0 , u0 , t   0 and u1, t   200 t . Compute u for one time step, taking  x  0.25
and t  0.125 . [A/M11]
u  2 u
31. Solve  2 in 0  x  5 , t  0 given that ux , 0  20 , u0 , t   0 and u5 , t  100 .
t x
Compute u for one time step with h 1 by Crank-Nicholson method. [N/D11]
32. Using Crank-Nicholson implicit scheme, solve the heat equation
u xx  ut , t  0 , 0  x 1 subject to the conditions ux , 0  0 , u0 , t   0 and u1, t   t for
two time steps. [M/J12]

61 II YEAR EIE QUESTION BANK PEC

 NUMERICAL METHODS

u 2 u
33. Using Crank-Nicolson’s scheme, solve 16  , 0  x 1 , t  0 subject to
t  x2
u x , 0  0 , u 0 , t   0 , u 1, t  100 t . Compute u for one step in t direction taking
1
h . [N/D13,A/M17]
4
 2u  u
34. Solve by Crank-Nicholson’s method  for 0  x  1 , t  0 given that
 x2 t
u0 , t   0 , u1, t   0 and ux , 0  100 x 1 x  . Compute u for one time step with
1 1
h and k  . [N/D14,A/M17]
4 64
35. Find the values of the function ux , t  satisfying the differential equation ut  4 u xx and
x2
the boundary condition u0 , t   0  u8 , t  and u x , 0  4 x  at the point x  i ,
2
1
i  0 ,1, 2 , 3 , 4 , 5 , 6 , 7 , 8 , t  j , j  0 ,1, 2 , 3 , 4 , 5 . [N/D15]
8
 u  2u
, u 0 , t   0 , u4 , t   0 and u x , 0  16  x 2 . Find
x
36. Given that 
t  x 2
3
ui j ; i 1, 2, 3, 4 and j  1, 2 by using Crank-Nicholson method. [M/J16]
37. Solve the boundary value problem y   x y subject to the conditions y0  y 0 1 ,

y1 1 taking h  , by finite difference method.

1
[N/D10,A/M18]
3
38. Using the finite difference method, compute y0.5 , given y   64 y 10  0 ,
x  0 , 1 , y0  y1  0 , subdividing the interval into (i) 4 equal parts (ii) 2 equal
parts. [N/D11]
39. Solve the equation y   x  y with the boundary conditions y0  y1  0 .
[M/J12,N/D15]
40. Solve y   y  0 with the boundary conditions y0  0 and y1 1 . [N/D12]
41. Solve y   y  0 , with y0  0 , y1 1 using finite difference method with h  0.2 .
[A/M15,N/D15]
42. Solve y   y  x , x 0 ,1 given y0  y1  0 using finite differences by dividing the
interval into four equal parts. [M/J14,A/M17]
43. Write Liebmann’s iteration process. [M/J16]
44. Solve x y   y  0 , y1 1, y2  2 with h  0.25 using finite difference method.
[A/M17,N/D17]

62 II YEAR EIE QUESTION BANK PEC

 INDUSTRIAL INSTRUMENTATION – I

EI 8452 - INDUSTRIAL INSTRUMENTATION – I

COURSE OUTCOMES

Course
Statement
Outcome

CO1 Explain the principle of operation, construction and characteristics of force,

torque and speed transducers.
CO2 Explain the principle of operation, construction and characteristics of
acceleration, vibration and density transducers
CO3 Explain the principle of operation, construction and characteristics of
viscosity, humidity and moisture transducers
CO4 Explain the principle of operation, construction and characteristics of
temperature transducer.
CO5 Design of signal conditioning circuits and compensators for temperature
measuring instruments
CO6 Explain the principle of operation, construction and characteristics of pressure
transducer

CO - PO MAPPING

PO PO PO PO PO PO PO PO PO PO1 PO1 PSO PSO PSO

CO PO12
1 2 3 4 5 6 7 8 9 0 1 1 2 3

CO1 3 - - - - - - - - - - - 3 - -

CO2 3 - - - - - - - - - - - 3 - -

CO3 3 - - - - - - - - - - - 3 - -

CO4 3 - - - - - - - - - - - 3 - -

CO5 3 1 - - - - - - - - - - 3 1 -

CO6 3 1 - - - - - - - - - - 3 1 -

AVG 3 1 - - - - - - - - - - 3 1 -

3 – Substantially 2 – Moderate 1 - Slightly

63 II YEAR EIE QUESTION BANK PEC

 INDUSTRIAL INSTRUMENTATION – I

UNIT – I
Measurement of Force, Torque and Speed
PART-A

1. Define force. [R]

Force can be defined as a push or a pull that changes or tends to change the state of
rest or uniform motion of an object or changes the direction or shape of an object. It is
measured in ‗Newton‘ (N).
The basic formula for force is, F = m.a, where, ‗m‘ stands for the mass in kilograms
and ‗a‘ stands for acceleration in m/sec2 . The other unit of force is Pounds.
2. Define speed. [R]
Speed is the distance travelled per unit of time. It gives information about how fast an
object is moving. Speed is the scalar quantity. The SI units for speed are m / s (meters
per second).
The basic formula for speed is, r = d / t, where r is the rate, or speed, d is the distance
moved and t is the time it takes to complete the movement
Conversions for Unit of Speed
km/h mph knot ft/s
1 m/s = 3.6 2.236936 1.943844 3.280840
3. Define Torque? [R]
Torque is the tendency of a force to cause or change the rotational motion of a body.
It is a twist or turning force on an object. . It is a vector quantity. The SI units of torque
are newton-meters or N*m
The basic formula for torque is τ = F.d where F is force applied over the body and d is
the distance moved by it. Torque is also known as Moment, moment of force.
4. Which is the most frequently used speed measuring instrument? [R]
Tachometer is the most frequently used speed measuring instrument
5. What are the types of tachometer? [R]
 A.C. tacho generator
 D.C. tacho generator
6. What are the types of D.C. tacho generator? [R]
 Permanent magnet type
 Separately excited field type
7. What is the working principle of DC tachogenerator? [Nov-Dec 2016] [R]
D.C. Tachometer generator consists of a small armature which is coupled to the
machine whose speed is to be measured. This armature revolves in the field of a
permanent magnet. The emf generated is proportional to the product of flux and speed.
The polarity of output voltage indicates the direction of rotation. This emf is measured
with the help of a moving coil voltmeter having a uniform scale and calibrated directly in
terms of speed.

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8. What are the disadvantages of D.C. tacho generator? [R]

A.C. ripple is present in the output signal. The magnitude of the ripple is 2% of the
output D.C. level.
9. Name the types of rotor used in A.C. tacho generators. [R]
 Drag cup rotor
 Squirrel cage rotor
10. What are the advantages of Squirrel cage rotor? [R]
 Cheaper
 Occupies less space
11. Give some applications of drag cup tacho generator. [AP]
It is commonly used in the speedometers of motor vehicles and as a speed indicator
for aero engines.
12. Give the formula to calculate the shaft speed. [R]
Shaft speed = [disk speed * No. of openings in the disk] / No. of images
13. What is the other name for inline stationary torque sensor? [R]
Relative regular twist torque sensor
14. What are the types of torque transducer? [R]
 Inline rotating torque sensor
 Inline stationary torque sensor
 Optical torque sensor
 Proximity torque sensor
15. Write the relationship between torque and force. [U]
T = F × D; where T is the Torque, F is the Force and D is the perpendicular distance
between the axis of rotation of the line of action of the force
16. What are the advantages of optical torque sensor? [R]
Low cost, small physical size
17. What is meant by strobotron? [April-May 2017][R]
Strobotron is a gas-filled electron tube with a cold cathode used especially as a source
of stroboscopic light.
18. Describe the stroboscopic method of rotational speed measurement .
[Nov-Dec 2015][AP]
A mechanical disk type stroboscope consists essentially of a whirling disk attached to
motor whose speed can be varied and measured. A reference mark on the rotating shaft
on the shaft appears to be stationary. For this condition, the shaft speed equals that of
rotating disk, or some even multiple of this speed and is given by:

19. Mention the advantages and disadvantages of stroboscope. [Nov-Dec 2015] [R]
Advantages:
 Imposes no load on the shaft hence no power loss.
 non contact type hence, no attachments needed.
 convenient to use for spot checks on machinery speeds and laboratory work.

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Disadvantages:
 The variable frequency oscillator circuit cannot be stabilized to give a fixed
frequency hence less accurate than digital meters.
 cannot be used where ambient light is above a certain level.
 requires well defined lightening conditions for efficient operations.
 errors are caused due to slight variation in the frequency.
20. Write the formula for determining the actual speed in a stroboscope. [U]
Actual speed, fr = flfn [n-1] / [ fn - fl ], where
fn is the highest flashing speed
fl is the lowest flashing speed
n is the no. of flashing frequencies
21. Define load cell. Mention its types. [Nov-Dec 2016] [R]
A load cell is a transducer that is used to create an electrical signal whose magnitude
is directly proportional to the force being measured. The various load cell types include
hydraulic, pneumatic, and strain gauge.
22. Write down the range, accuracy and resolution in hydrostatic or hydraulic load cell.
 Range of force measurement = 0 to 30000 N [U]
 Accuracy= 0.1%
 Resolution= 0.02%
23. Why magneto elastic load cell is also called pressductor load cell? [R]
The degree of change has a direct relationship with the applied stress or force. So it is
called pressductor load cell.
24. What are the advantages of magneto elastic load cell? [R]
 Extremely robust transducer
 Produces relatively high output signal levels
 Overload ratings are as high as 15 times the rated loads
25. How force is measured using piezoelectric load cell? [Nov-Dec 2015] [U]
Piezoelectric load cells work on the principle of Piezoelectric Effect which is the
ability of certain materials to generate an electric charge in response to applied
mechanical stress. Piezoelectric load cells are useful for dynamic measurements of force.

The figure shows the centers of symmetry of the charges move apart on the application
of force and charge can be measured at the top and bottom of the crystal.
26. What is magneto elastic effect? [APRIL/MAY 2017][R]
The inverse magnetostrictive effect (also known as magnetoelastic effect or Villarian
effect) is the name given to the change of the magnetic susceptibility of a material when
subjected to a mechanical stress.
Magnetostriction is a property of ferromagnetic materials that causes them to change
their shape when subjected to a magnetic field. The effect was first identified in 1842 by

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James Joule when observing a sample of nickel. This effect can cause losses due to
frictional heating in susceptible ferromagnetic cores.
27. Write down the excitation voltage and accuracy of strain gauge load cell. [U]
 Excitation voltage may be a.c. or d.c in the range of 5 to 25 volts.
 Accuracy is +0.1% of the full scale output.

PART–B
1. Explain in detail the strain gauge type load cells and piezo-electric load cell.
[Apr-May2017][Nov-Dec 2016][U]
2. Derive the relationship between gauge factor and pistons ratio for a given strain gauge.
[Nov-Dec 2017] [U]
3. Explain the construction and operation of hydraulic load cell and pneumatic load cell.
[Apr-May 2015] [U]
4. Explain the construction and operation of magneto elastic type load cell.
[Nov-Dec 2015][Nov-Dec 2016][U]
5. Explain any two types of torque measurement with neat sketch. [Nov-Dec 2016] [U]
6. List the basic means of force measurement techniques. [Apr-May 2015] [R]
7. Explain speed measurement using capacitive tacho and drag cup type tacho.
[Nov-Dec 2015][U]
8. Describe the construction and working of differential pressure type densitometer.
[Nov-Dec 2017][AP]
9. Explain how calibration of pressure gauge is carried out using dead weight tester and
mentions the factors affecting the accuracy of dead weight. [Nov-Dec 2017] [U]

PART-C
1. In a LVDT pick up the output voltage is a linear function of the core position, the stoke
being 2cm from the null positions and the maximum output is 21V. Determine the change
in output voltage when sensing a displacement of 0.1cm. If the noise voltage is 50mV,
what is the maximum possible resolution? [Nov-Dec 2015] [AP]
2. What is meant by tacho generators? Explain about A.C and D.C Tacho generators.
[Apr-May2017] [Nov-Dec 2015] [U]
3. Illustrate with a neat diagram, the principle of operation, construction and working of
Stroboscope [Apr-May2017] [Nov-Dec 2016] [AP]
4. A piezoelectric type accelerometer has a sensitivity of 100Mv/g. The transducer is
subjected to a constant acceleration of 5g. Find the steady state output of the transducer
[Nov-Dec 2015] [AP]
5. A strain gauge has a gauge factor of 4. If the strain gauge is attached to a metal bar, that
stretches from 0.25m to 0.258 m when strained. What is the percentage of change in
resistance? If the unstained value of the gauge is 120 ohm, what is the resistance value of
the gauge after it is stained? [Nov-Dec 2017] [AP]

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UNIT – II
Measurement of Acceleration, Vibration and Density
PART-A

1. Define acceleration. [R]

Acceleration is the rate of change of velocity of an object with respect to time. An
object's acceleration is the net result of any and all forces acting on the object, as
described by Newton's Second Law. The SI unit for acceleration is meter per second
squared (m s−2). Acceleration is a vector quantity.
2. What are the advantages of using Baume scale? [Nov-Dec 2015] [R]
It is used for liquids both higher and heavier than water.
3. Write the abbreviation of API scale. What is API gravity Scale
[Apr-May 2015][Nov-Dec2015] [U]
The American Petroleum Institute gravity, or API gravity, is a measure of how
heavy or light a petroleum liquid is compared to water: if its API gravity is greater than
10, it is lighter and floats on water; if less than 10, it is heavier and sinks.
API gravity is mathematically a dimensionless quantity; it is referred to as being in
'degrees'. API gravity is graduated in degrees on a hydrometer instrument. API gravity
values of most petroleum liquids fall between 10 and 70 degrees.
4. Write the disadvantages of bridge type gas densitometer. [U]
The major disadvantage is that the variations in ambient temperature will introduce
errors.
5. Write the formula for determining the height in a pressure head type densitometer.
[U]
H = Span / Specific gravity maximum - Specific gravity minimum
6. Give some of the materials which are used for the manufacturing of a float. [R]
Pyrex, Plastic

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7. For what purpose accelerometers are used? [U]

 For the measurement of shock & vibration
 For gross measurement of acceleration of vehicles like aircraft, submarines etc.
8. What are the advantages of LVDT? [R]
 It is used for steady state and low frequency vibration measurements.
 Smaller mass, so, it is used for the measurement of vibrations of higher
frequencies.
9. How a piezo electric accelerometer works. [Apr-May 2015][R]
A piezoelectric accelerometer consists of a mass attached to a piezoelectric crystal
which is mounted on a case. When the accelerometer body is subjected to vibration, the
mass on the crystal remains undisturbed in space due to inertia. As a result, the mass
compresses and stretches the piezoelectric crystal. This force is proportional to
acceleration in accordance with Newton‘s second law, F = ma, and generates a charge.
The charge output is then converted into low impedance voltage output with the help of
electronics.
10. What are the features of piezo electric accelerometers? [R]
 Small in size and weight
 It can be used for vibration and shock measurements.
 High output impedance.
 Their response is poor at low frequencies
11. List some of the applications of strain gauge accelerometers. [R]
It is used for the measurement of acceleration and vibration in vehicles, aircrafts,
bridges, hoists, cranes & lifts.
12. Seismic instrument can be used as accelerometer &vibrant – True or false. [E]
True
13. What are the two modes of seismic instrument? [R]
 Displacement mode
 Acceleration mode
14. Define density. [Apr-May 2017][R]
It is defined as the mass per unit volume of a substance under fixed conditions.
15. Define specific gravity. [R]
It is defined as the ratio of density of one substance to the density of another reference
substance both obtained at same temperature & pressure.
16. What is the other name for specific gravity? [R]
Relative density
17. What are the units of density? [Apr-May 2017][R]
The SI unit for density is: Kg/m3 or gm / liter. Liquid water has a density of about
1 kg/dm3, solids and liquids have densities between 0.1 and 20 kg/dm3.
18. For what purpose mechanical type vibration instruments are used? [U]
It is used for the measurement of
 Motion between 2 points with respect to each other [relative motion]
 Motion with respect to earth [absolute motion ]

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19. Draw a neat sketch of gas densitometer. [Nov-Dec 2016][R]

PART-B

1. Write short notes on variable reluctance type accelerometer and strain gauge
accelerometer [Apr-May2017] [Nov-Dec 2016][Nov-Dec 2015][U]
2. Write a short note on Calibration of vibration pickup. [Nov-Dec 2015][U]
3. Explain the construction and operation of Float type, ultrasonic type densitometer.
[Nov-Dec 2016], and Hydrostatic type densitometer [Nov-Dec 2016][U]
4. Derive and explain the amplitude and phase response of seismic Accelerometer.
[Nov-Dec 2016][U]
5. Explain the construction and operation of Bridge type densitometer and Pressure head type
densitometer. [Apr-May 2015][U]
6. Derive the expression of output from drag cup rotor AC tachogenerator
[Apr-May 2015][Nov-Dec 2017][AP]
7. Explain the construction and operation of mechanical type vibrometers [U]
8. Describe the construction and working of differential pressure type densitometer.
[Nov-Dec 2017] [AP]
9. Describe the working of following gas densitometer with neat sketch a) Electromagnetic
suspension type b) Thermal conductivity densitometer. [Apr-May 2017][R]

PART-C

1. Illustrate with a neat diagram, the principle of operation, construction and working of
LVDT type Accelerometer and piezo electric type Accelerometer .
[Apr-May2017][Nov-Dec 2015/2017] [AP]
2. A seismic instrument has a natural frequency of 4 Hz and a damping ratio of 0.66. If the
system is excited by a frequency 6Hz, determine the error due to the proximity of excited
frequency with natural frequency of the instrument [Apr-May 2015] [AP]
3. Seismic instrument as an accelerometer and vibrometer. Justify. [Nov-Dec 2015] [E]
4. Analyse three stages of measuring system for seismic mass accelerometer with the help of
neat sketch. [Nov-Dec 2017] [AZ]

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UNIT-3
Measurement of Viscosity, Humidity & Moisture
PART-A

1. Define Viscosity. [R]

It is the property of the fluid which gives the resistance to the flow. It is a measure of
fluidity of the system.
2. What are the units of viscosity? [R]
Dynamic viscosity: [Pascal second or Ns/m2 or kg/ms] (SI) or poise (CGS)
Kinematic viscosity: m2 /s
3. Define Newtonian fluids. [R]
If the force-flow relation is linear then the fluid is Newtonian.
4. Define Non Newtonian fluids. [R]
If the force-flow relation is non- linear then the fluid is Newtonian.
5. Define Kinematic Viscosity. [R]
It is the ratio of absolute viscosity to the density of the fluid.
6. Define Specific Viscosity. [R]
It is the ratio of absolute viscosity of the fluid to the absolute viscosity of a standard
fluid at the same temperature. Specific Viscosity, µ s =µ / µh
7. Define Relative Viscosity. [R]
It is the ratio of absolute viscosity of the fluid at a given temperature to the absolute
viscosity of a standard fluid at 20°C.
8. Define Viscosity index. [R]
Viscosity index is a measure for the change of viscosity with temperature. It is used
in the automotive industry to characterize lubricating oil. It is an empirical number that
indicates the effect of change of temperature on viscosity of a fluid.
9. Define fluidity. [R]
-1 −1
It is the reciprocal of viscosity. Its unit is poise or cm·s·g
10. Define Humidity. [R]
Humidity is a term used to describe the amount of water vapor present in air.

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11. Define Absolute Humidity. [R]

Absolute humidity is the total mass of water vapor present in a given volume or mass
of air. It changes with change in temperature.
Absolute humidity is the mass of the water vapour (mH2O), divided by the volume of
the air and water vapour mixture (Vnet) , which can be expressed as: AH = (mH2O) / (Vnet)
12. Define Specific Humidity. [R]
Specific Humidity is defined as the ratio of the mass of water vapor to the mass
of dry air in a known volume of air.
13. Define Relative Humidity. [R]
Relative Humidity is the amount of water vapor in the air relative to what the air can
hold.
14. Define dew point. [R]
It is the temperature at which the water vapour starts to condense.
15. Define various units of Humidity. [R]
Vppm = parts per million / volume.
G/ kg = weight concentration
Relative humidity = in % Dew point in °C.
16. Define Hygrometer. [R]
It is used to measure the moisture content in air. It also used to measure humidity.
17. What is the basic principle of Hygrometer? [R]
It consists of mechanical device measuring the dimension change of humidity
sensitive materials like animal hair, animal membrane, paper etc.
18. Define Moisture. [R]
It is defined as the amount of water absorbed by solids, liquids and gases.
19. What are the various methods of measurement of moisture? [R]
 Based on the weight of the particle
 Based on the resistance and capacitance
20. How will you find the % moisture content on wet-weight basis? [R]
The formula to calculate the moisture content on a wet-weight basis is:
Moisture content (%) = [(W2 - W3) x 100)] / (W2-W1) where, W1 = weight of
container with lid; W2 = weight of container with lid and sample before drying; and W3
= weight of container with lid and sample after drying
21. What are the different types of viscometer? [R]
 Say bolt viscometer,
 Rotameter type
 Torque type and
 Consistency meters.
22. What is a Psychrometer? [R]
Psychrometer is a device that uses the bulb thermometers to measure humidity. It is
also used in air conditioning systems for maintaining humidity.
23. What are the different types of hygrometer? [R]
 Hair hygrometers
 Wire electrode hygrometers

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 Electrolysis type hygrometers

 Resistive type
 Capacitive type and Microwave reflector
24. Explain the principle of Saybolt viscometer. [U]
As the viscosity of the fluid varies, the flow rate and hence time taken to drain the
fluid through the capillary tube varies. The time indicates the viscosity and is denoted by
Saybolt number.
25. What is meant by consistency? [R]
It is the general term for viscosity and more often used in connection with Non-
Newtonian fluids.
26. Explain the principle of oscillating type consistency meters. [U]
When the inner cylinder is given an axial
sinusoidal motion through a mechanical drive
rod, the fluid in the annular space gets a
shearing force and the motion to the inner
cylinder will be transmitted to the outer
cylinder because of the metal bellows-the
magnitude of this transmission will depend on
the consistency of fluid flowing through the
instrument.

27. List out the limitations of psychrometer. [R]

Limitations of psychrometer:
 Psychrometers show impaired accuracy below 15% RH as the cooling of wet-bulb to
its full depression becomes difficult. Though few psychrometers work well below 5%
RH to 10% RH.
 Wet-bulb measurements at temperature below 0º C (32º F) are difficult to obtain with
a high degree of confidence.
 Psychrometers cannot be used in small, closed volumes.
 Dirt, oil or contamination on the wick, insufficient water flow, etc., tends to increase
the apparent wet-bulb temperature.
28. Explain about dry and wet bulb psychrometer. [U]
Psychrometer, is an instruments used for measuring the water vapor content or
relative humidity of the atmosphere. It consists of two identical
thermometers—the wet-bulb thermometer, its bulb is covered
with a jacket of tight-fitting muslin cloth that can be saturated
with distilled water; and the dry-bulb thermometer. When the
cloth is soaked and the thermometers are properly ventilated,
the wet-bulb temperature will be lower than the dry-bulb
temperature (actual air temperature) because of cooling due to
the evaporation of water from the cloth. The drier the air is, the
greater the evaporation and thus the more the wet-bulb
temperature is depressed. Psychrometric tables list various

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humidity variables, such as relative humidity, according to dry-bulb temperature and

wet-bulb depression at equilibrium. Ventilation is provided by whirling the thermometers
at the end of a chain (sling psychrometer) or by a suction fan (aspiration psychrometer).
Newer psychrometers use special electronic sensors.
29. Describe Dew cell. [U]
Dewcells are instruments used for determining the dew point. They consist of a
small heating element surrounded by a solution of lithium chloride. As the LiCl
absorbs moisture from the air, conduction across the heating element increases, current
in it increases, and heat increases, evaporating moisture from the salt solution. At a
certain temperature the amount of moisture absorbed by the salt solution equals the
amount evaporated.
Inside the dewcell core a thermistor composite changes electrical resistance with the
temperature created by the heating. A front end processor provides a reference voltage,
measures the output of the network, and calculates the dew point.
30. Show why viscosity measurement is important for industrial process. [C]
Measuring viscosity reveals the degree of a liquid or gaseous substance's resistance to
flow, usually in terms of its unwillingness to flow under conditions where it is deformed
by shear or extensional stress.
Measuring viscosity is an effective way of determining the state [properties of matter]
or fluidity of a liquid or gas. It plays an important role in the quality control and various
research and development stages in lab, process and research environments as well as a
wide range of industries and applications including Food, Chemical, Pharmaceutical,
Petrochemical, Cosmetics, Paint, Ink, Coatings, Oil and Automotives.
31. List out the factors which should be considered as possible source of error in
humidity measurements. [R]
 An electrical resistance thermometer is required for measuring the equilibrium
temperature; the usual sources of error for thermometry are present.
 The equilibrium temperature achieved is determined by the properties of the
solute, and significant amounts of contaminant will have an unpredictable effect.
 Variations in aspiration affect the heat exchange mechanisms and, thus, the
stability of operation of the instrument. A steady aspiration rate is required for a
stable operation.
32. List the moisture measurement application in industries [R]
 Animal Food manufacturing industry  Renewable Energy
 Chemicals industry  Snack Foods manufacturing industry
 Human Food manufacturing industry  Textiles industry
 Minerals industry  Tobacco industry
 Paper industry  Wood industry
33. Explain the working of capacitive type moisture measuring instrument? [R]
A capacitive humidity sensor is used to measure relative humidity with a usual range
of 5 to 95 % relHum. The sensor is a capacitor, made from two metal electrodes with a
porous dielectric substance between them. Water vapour is able to penetrate this layer,
changing the dielectric constant of the dielectric material thereby changing the total

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capacitance. Typical capacitance at 50% relHum is 100 to 500pF, with 0.2 to 0.5pF
change per 1% relHum.
As the sensor uses water vapour in the air, placing the sensor in the path of moving air
allows it to respond faster. Relative humidity is directly dependant on the temperature
and the sensor has a thermistor used for calibration.
34. How microwave type moisture measurement instruments work? [R]
In materials with bipolar molecular structure, (i.e. water), the electric field of
microwaves can induce oscillations whilst travelling through the medium. During
transmission, (where the substance to be analysed is placed between the micro wave
emitter and detector) the micro wave intensity arriving at the detector decreases with
increasing of moisture. The moisture content can then be calculated by taking into
account the absorption of the dry substance and some geometrical factors.
In reflection the procedure is equivalent, except that both micro wave emitter and
detector are mounted on the same side of the substance.
35. List the advantages of micro wave moisture determination technique. [R]
 MWs are not sensitive to pH and conductivity of the material to be measured.
 Total moisture (surface as well as inherent) is measured.
 Contactless measurement: No abrasion of the sensors, no sticking of material, no
interference with the process.
 The measurement gives instantaneous results ("real time measurement") and can be
utilised for process control.
36. List the materials whose moisture content is measured using micro wave moisture
determination technique. [R]
Some examples of material which can be measured are coal, sand, gravel, wood chips,
potatoes, pommes chips, tobacco, seed corn etc.
37. Limitations of micro wave moisture determination technique. [R]
Nearly all bulk goods can be measured by means of microwaves with few restrictions.
The water molecules are not able to oscillate, if they are captured in small areas like
capillaries which causes problem in the measurement. For the same reason crystal water
and ice do not respond to the measurement are the few limitations of microwave
moisture measurement technique.
38. How an IR type moisture measuring instrument works? [R]
A portion of the light (typically quartz halogen bulb), collimated and filtered into
specific wavelengths is directed onto the surface of the product whose moisture has to be
measured. The reflected light falls on the detector (usually lead sulfide). Specific
wavelengths of light are absorbed by water content (moisture) in the product. The
amplitude of the reflected wavelengths will be proportionate to the amount of water in
the product.
39. How a resistive type moisture measuring instrument works? [R]
The measuring electrode forming a part of the bridge circuit is permitted to contact
the sample. A regulated voltage applied to the electrodes maintains a small current flow
through the material. Changes in the moisture level of the sample result in a change of

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electrode current and bridge imbalances. The amount of imbalance is related to the
moisture content.
40. List the limitations of resistive type moisture measuring instrument. [R]
The measurement is strongly influenced by contact pressure between sample and
electrodes, operating temperature, packing density and particle size.
The resistance measured by the instrument is very high therefore good insulation is
required at the electrodes to prevent leakage currents from introducing an error into the
instrument.
41. How a NMR type moisture measuring instrument works? [R]
NMR is capable of distinguishing the different atoms in a sample, based on the
different absorption characteristics of the nucleus of these atoms of radio frequency
energy, when the atoms are placed in a constant magnetic field.
42. List the advantages and disadvantages of NMR type moisture measuring instrument. [R]
Advantages: Disadvantage:
 Speed ● high cost
 Accuracy
 Non-destructive testing

PART – B
1. Explain the construction and principle of operation of dry bulb psychrometer with neat
sketch. N/D-14, N/D-15, A/M-15, M/J-16 [U]

2. Summarize the principle of humidity measurement. Explain the working principle of any
one type of hygrometer with neat sketch. N/D-14, N/D-15, A/M-15, M/J-16 [U]
3. Explain how moisture is measured in granular materials and solid penetrable materials.
N/D-14, N/D-15, A/M-15, M/J-16 [U]
4. Discuss the principle of operation of different methods of moisture measurement. [C]
5. Generalize commercial type dew point meter and rotameter type viscometer in detail with
a neat sketch. N/D-14, N/D-15, A/M-15, M/J-16 [U]
6. Analyze consistency meters with neat sketch. N/D-14, A/M-15, M/J-16 [AZ]
7. Describe the moisture measurement using distillation method with a block diagram.
N/D-15 [U]
PART-C
1. A rotameter uses a cylindrical float of 3.5 cm height, 3.4 cm diameter and density of
3985 kg/cm3. The maximum inside diameter of the metering tube is 6cm. Determine the
maximum flow rate handling capacity in [m3/sec] of the rotameter if the fluid id water.
Assume Cd = 0.75. [E]
2. Determine the nominal flow velocity [in cm/sec] at the orifice [diameter: 30mm] kept in
a pipe of 60mm diameter. Reynolds number R is 105. Assume density of water= 1000
kg/m3 and kinematic viscosity [k] is 10-2 cm2/sec. [E]
3. The flow of cooling water is measured with the help of a horizontal venturimeter with
200 mm inlet and 100 mm throat. A U-tube manometer connected between the inlet and

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throat of the venturimeter shows a differential pressure of 235 mm of mercury is 13.6,

density of water is 1 g/cm3. [E]
3
4. A rotameter is constructed using a float of density 1900 kg/m . It is calibrated for
metering a gas of density 1.3 kg/m3 and has a scale ranging from 0.018 m3/min to 0.18
m3/min. Now, it is intended to use this meter for metering the same gas within a flow
range of 0.036 m3/min to 0.36 m3/min. What should be the density [ kg/m3 ] of the new
float?. Both the floats can be assumed to have the same volume and shape. [E]
5. A right angled V-notch is employed to measure the discharge of water. If the head H
above the still is measured as 0.25 m, estimate the discharge [in lt/sec] if cd = 0.62.
[width of notch L = 0.35 m, acceleration due to gravity g = 9.81 m/sec2, water density =
1000 kg/m3]. [E]
6. A doppler shift ultrasonic flow meter uses two piezoelectric crystal, each having a
natural frequency of 3.6 MHz. These crystals are used as transmitter and receiver. The
transmitter directs an ultrasonic wave into the pipe which makes an angle of 45 degree
with the direction of flow. Calculate the received frequency for a fluid velocity of 10
m/sec. Assume the velocity of sound in fluid to be 1000 m/sec. [E]
7. A turbine flow meter consists of four mild steel blades rotating at an angular velocity
given by the following relation ω = 50000.Q, where Q is the flow rate in m3/sec. Total
flux linked with the coil of the magnetic transducer is given by Φ = 4.25 + cosθ mWb,
where θ is the angle between the blade assembly and the transducer. Range of the flow
meter is 0.5 to 5 lt/s. Calculate the amplitude and frequency of the transducer output at
[i] maximum and
[ii] minimum flow rates. [E]

UNIT - 4
Temperature Measurement
PART- A
1. Define temperature. [R]
The temperature of a substance is a measure of hotness or coldness of that substance.

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2. What is the difference between temperature and heat? [AZ]

Temperature is defined as ―degree of heat‖. Heat is used to mean ―quantity of heat‖.
3. Define boiling point, freezing point and triple point. [Apr-May 2017][R]
The boiling point of a substance is the temperature at which the vapour pressure of
the liquid equals the pressure surrounding the liquid and the liquid changes into a vapour.
Freezing is a phase transition in which a liquid turns into a solid when
its temperature is lowered below its freezing point.
In thermodynamics, the triple point of a substance is the temperature and pressure at
which the three phases (gas, liquid, and solid) of that substance coexist
in thermodynamic equilibrium. For example, the triple point of mercury occurs at a
temperature of −38.83440 °C and a pressure of 0.2 mPa.
4. What are the temperature scales? [R]
 Lower fixed point or ice point.
 Upper fixed point or steam point.
5. Give few types of thermocouple with composition and temperature range.
[Apr-May 2017][R]

6. Give the relationship between Celsius scale, Fahrenheit scale and Kelvin scale? [U]
C/5 = [F-32]/9 = [K+273]/5
7. Write the relationship between Kelvin scale & Rankine scale? [U]
The Kelvin scale measures thermodynamic temperature with the kelvin being a
base unit of the International System [SI] of units, whereas Rankine is typically
associated with the imperial system or the US customary system of units;
They are related by a scale factor: 1 R = [5/9] K.
8. Define triple point. [R]
A particular temperature and pressure at which three different phases of one substance
can exit in equilibrium is known as ―triple point‖.
9. Mention the properties of good manometric liquid. [Nov-Dec 2015] [R]
 it should be of high density.
 it should be of low vapour pressure.
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 it should not evaporate.

 it should not be mix with water
10. Name the sources of error in filled system thermometer.
 Ambient temperature effect [Apr-May2017] [Nov-Dec 2016] [R]
 Head or elevation effect
 Barometric effect
 Immersion effect
 Radiation effect
11. Mention the advantages and disadvantages of filled system thermometer.
Advantages: [Nov-Dec 2015] [U]
 Simple and low cost
 Quite rugged construction and less chances of damage in handling
 Self operated, no power requires ad generate sufficient power for controller
mechanism.
 Good response, accuracy and sensitivity
 vapour actuated thermometer are most widely used as they are less costly and simple.
It has good speed of response.
 It can be used for remote indication.
Disadvantages:
 Lower accuracy, sensitivity and temperature span compared to electrical temperature
instruments.
 Compensation is required in case of variation in ambient temperature.
 The transient response is primary dependent on the bulb size and thermal properties of
the filled fluid.
 To get good accuracy, requires large size of bulb results in poor response.
 In case of damage, entire system has to be replaced.
 As filled system thermometers don‘t produce electrical signals, it is difficult to
implement them in process controls that rely on electrical and computerized control.
12. Write some methods of measurement of temperature? [U]
 Expansion Thermometer.
 Filled system Thermometer.
 Electrical Thermometer.
 Pyrometer.
13. What are the different types of filled system Thermometer? [R]
 Gas-filled Thermometer.
 Liquid-filled Thermometer.
 Mercury-filled Thermometer.
 Vapour-filled Thermometer.
14. How radiation error occurs? [U]
It occurs due to the temperature difference between bulb and other solid bodies in
filled system Thermometer.

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15. Define Seebeck effect? [Nov-Dec 2015] [R]

If two dissimilar metals are joined together to form a closed circuit, there will be two
junction where they meet each other. If one of these junctions is heated, then a current
flow in the circuit which can be detected by a galvanometer. The amount of current
depends on the difference in temperature between the two junctions and on the
characteristics of the two metals. This was observed by Seebeck & hence known as
Seebeck effect.
16. Which effect is used in thermocouple? [U]
Seebeck effect is used in thermocouple.
17. What are the various types of the thermocouples? [Nov-Dec 2016] [R]
Copper-Constantan, Iron-Constantan, Platinum-Rhodium, Chromal-Constantan
&Chromal-Alumel.
18. What is the purpose of protecting tube in a thermocouple? [U]
It is used to protect the thermocouple from harmful atmosphere, corrosive fluids and
also to prevent from mechanical damage.
19. List the advantages and disadvantages of Thermocouple. [U]
Advantages of Thermocouple
 Self powered  Inexpensive
 Simple  Wide variety
 Rugged  Wide temperature range
Disadvantages of Thermocouple
 Non-Linear  Least stable
 Low voltage  Least sensitive
 Reference required
20. Draw the characteristics curves for thermocouple, thermistor and RTD. [R]

21. Give some of the temperature instruments? [R]

 Resistance thermometer.
 Thermocouple.
 Thermistor.
22. Define Barometric effect. [R]
The effect due to change in atmospheric pressure is known as Barometric effect.
23. How Barometric error can be minimized? [AP]
It can be minimized by keeping the filled system at a pressure sufficiently larger than
the atmospheric pressure.
24. What are the advantages of Filled system Thermometer? [R]
 Low cost.
 Less maintenance requirement.

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 Rugged construction.
 Absence of need of electric power.
25. List some of the disadvantages of bimetallic thermometers. [U]
 Possibility of calibration change due to rough handling.
 Not very accurate
 Not suitable for measuring lower temperature as the metals and metallic alloys show
nearly same expansion or contraction in lower range of temperature.
 Availability of indication type only.
26. What are the methods used for thermometer calibration? [April-May 2015][R]
 Ice point method
 boiling point method
 triple point method
27. How an IC sensor measures temperature? [U]
Transistors are sensitive to temperature variations. It has been found that if two
identical transistors are operated at a constant ratio of collector current densities, then the
difference in their base-emitter voltages will be directly proportional to absolute
temperature.
Therefore as temperature decreases the base bias must be increased to maintain the
collector current constant. The base bias voltage is usually converted to current by a low
temperature coefficient thin film resistor. The current output units are usually set for a
one microampere output change per kelvin (Celsius), while the voltage output
configuration generates 10mV per degree kelvin
28. Define first law of thermocouple. [R]
The thermal emf of a thermocouple with junctions at T1 and T2 is totally unaffected
by temperature elsewhere in the circuit if two metals used are each homogenous.
29. List the types of thermocouple. [Apr-May 2015] [R]
 Nickel-alloy thermocouples : Type E, Type J, Type K, Type M, Type N, Type T
 Platinum/rhodium-alloy thermocouples: Type B, Type R, Type S
 Tungsten/rhenium-alloy thermocouples: Type C, Type D, Type G
30. What are the two types of signal conditioning? [R]
 Analog  Digital
31. What are the possible ways of signal conditioning? [R]
 Linearisation  signal transmission
 Conversion  Digital interface
32. Define heat radiation. [R]
When a body is heated, it emits thermal energy known as heat radiation.
33. Define Stefan Boltzmann law. [Nov-Dec 2016][R]
Stefan-Boltzmann law, statement that the total radiant heat energy emitted from a
surface is proportional to the fourth power of its absolute temperature. If E is the radiant
heat energy emitted from a unit area in one second and T is the absolute temperature [in
degrees Kelvin], then E= σT4, the Greek letter sigma [σ] representing the constant of
proportionality, called the Stefan-Boltzmann constant.
This constant has the value 5.670367 × 10−8 watt per metre2 per K4. The law applies

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only to blackbodies, theoretical surfaces that absorb all incident heat radiation.
34. Define pyrometry. [R]
Pyrometry is the technique for measuring the body‘s temperature by measuring it‘s
electromagnetic radiation.
35. What are the two types of pyrometer? [R]
 Radiation pyrometer
 Optical pyrometer.
36. Give the values of temperature coefficient of resistance in platinum and nickel? [R]
Pt - 0.004, Ni – 0.005
37. What is the advantage of optical pyrometer? [R]
It is used to measure high temperature.
38. How the calibration is adjusted in optical pyrometer? [E]
By adjusting the emissivity.
39. What are the errors in total radiation pyrometer? [R]
 They are sensitive to emittance errors.
 Sensitive to any obstructions in the line of sight between the pyrometer and the hot
body.
40. List some of the applications of total radiation pyrometer? [Apr-May2017][R]
 used for moving target.
 used in furnaces.
 used for the temperatures above the practical operating range of thermocouple.
41. What are the advantages of total radiation pyrometer? [R]
 High temperature measurement.
 Fast response speed.
 Moderate cost and high output.
42. State the radiation pyrometer principle. [Apr-May 2015][R]

The radiation pyrometer measures the heat emitted by a hot object. The radiation
pyrometers operate on the principle that the energy radiated from a hot body is a function
of its temperature. The energy radiated by the hot body whose temperature is measured is
focused by the lens to the detector. The detector is a thermocouple or bolometer
[bolometer is a thermal device that changes electrical resistance with temperature
change]. The detector output is given to a PMMC instrument, digital display or recorder.

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43. What are the advantages of selected radiation pyrometer? [R]

The transmission losses are minimised and accuracy is improved.
44. What are the types of two colour pyrometer? [R]
 Selected radiation pyrometer  Chopper broad- band pyrometer
 broad band pyrometer  Narrow band radiation pyrometer
45. What is the other name for two colour radiation pyrometer? [R]
Ratio pyrometer.
46. What are the advantages in ratio pyrometer? [R]
 Less accurate
 More cost about 50-100% more than other types of pyrometer.
47. What are the disadvantages of total radiation pyrometer? [R]
 Non linear scale.
 Emissivity of target material affecting measurement.
 Errors due to presence of gases and vapours.
48. What is thermograph? [R]
The principle of infrared thermography is based on the physical phenomenon that
any body of a temperature above absolute zero (-273.15 °C) emits electromagnetic
radiation. There is clear correlation between the surface of a body and the intensity and
spectral composition of its emitted radiation.
Thermographic cameras usually detect radiation in the long-infrared range of the
electromagnetic spectrum (roughly 9,000–14,000 nanometers or 9–14 µm) and produce
images of that radiation, called thermograms.
49. Give the formula for rate of radiation emitted per second [R]
4
E=KT
where, T-> Temperature;
K-> Constant

PART-B
1. Explain in detail about RTD. Explain the need and working of 3 lead and 4lead RTD.
[Nov-Dec 2015][U]
2. Write short notes on [U]
i. Bimetallic thermometer [Apr-May 2017] [Nov-Dec 2016/2017]
ii. Thermistor [Nov-Dec 2015]
iii. Different standards and units of temperature scale
3. Write short notes on [U]
i. Calibration thermometer
ii. Vapour pressure filled system thermometer
4. Explain in detail about different types of filled system thermometers with neat sketch.
Explain how they are calibrated. [Apr-May 2017][R]
5. What is the need for reference junction compensation in thermocouples? Describe the
various methods of practical thermocouple reference junction compensation.
[Nov-Dec 2017][R]

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6. Explain the construction and working of Fabrication of thermocouple. [U]

7. Explain the construction and working of Total radiation pyrometer
[Apr-May 2017] [Nov-Dec 2016] [U]
8. Explain the construction and working of optical pyrometer.
[Apr-May 2017] [Nov-Dec 2017][Nov-Dec 2016] [U]
9. Explain the construction and working of Ratio pyrometer
[Apr-May2015][Nov-Dec 2016] [U]
10. State the basic laws of thermocouple [Apr-May2017][Apr-May 2015][R]
11. What are the special techniques adopted for measuring high temperature using
thermocouple [Apr-May 2015] [R]
12. Explain the following [U]
 Thermowell
 Thermopile
 Seebeck effect and Peltier effect
 Signal conditioning circuit

PART-C
1. What are the Different sources of errors in filled system thermometers and explain their
compensation. [Nov-Dec 2017][Nov-Dec 2016][Apr-May 2015] [R]
2. Explain the construction and working of Special techniques for temperature
measurement. [U]
3. Explain the construction and working of Response of thermocouple and describe any
three references or cold junction compensation. [Apr-May 2017][U]
4. Explain the construction and working of Laws of thermocouple and Fiber optic method
to measure temperatures. [Nov-Dec 2017] [U]
5. Explain in details the Thermocouple signal conditioning circuit and its characteristics [U]
6. Explain how fiber optic temperature measurement is advantageous than other methods.
[Apr-May2015] [U]
7. Suggest a suitable measuring system to measure the inside temperature of nuclear
reactor. [Nov-Dec 2017][E]

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UNIT - 5
Pressure Measurement
PART - A
1. What are the main parts of an electrical pressure transducer? [R]
 Pressure sensing element such as a bellow, a diaphragm or a bourdon tube.
 Primary conversion element. e.g., resistance or voltage.
 Secondary conversion element.
2. Draw the diagram to show the range of different pressures. [R]

Figure 1 illustrates that relationship of absolute and gauge pressure

Figure 2 illustrates the relationship of absolute and vacuum pressures.

Figure 3 illustrates the relationship of differential pressure and absolute pressure

3. What is absolute pressure? [R]
Absolute pressure is measured relative to high vacuum (0 PSIA). It is referred to as
pounds per square inch (absolute) or PSIA. The electrical output of an absolute pressure
transducer is 0 VDC at 0 PSIA and full scale output (typically 5 VDC) at full scale
pressure (in PSIA).
4. What is gauge pressure? [R]

Gauge pressure is pressure measured relative to ambient atmospheric pressure

(approximately 14.7 PSIA). It is referred to as pounds per square inch (gauge) or PSIG.
The electrical output of a gauge pressure transducer is 0 VDC at 0 PSIG (14.7 PSIA) and
full scale output (typically 5 VDC) at full scale pressure (in PSIG).

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5. What is vacuum pressure? [R]

Vacuum pressure is measured relative to ambient atmospheric pressure. It is referred
to as pounds per square inch (vacuum) or PSIV. The electrical output of a vacuum
pressure transducer is 0 VDC at 0 PSIV (14.7 PSIA) and full scale output (typically 5
VDC) at full scale vacuum, 14.7 (0 PSIA).
6. What are the advantages of strain gauge pressure transducer? [R]
 Small & easy to install
 Good accuracy
 More stable
 Fast speed response
 Simple to maintain
7. What are the disadvantages of strain gauge pressure transducer? [R]
 High cost
 Requires constant voltage supply.
 Electrical readout is necessary.
 Temperature compensation is required.
8. What is the principle of operation of a piezoelectric pressure transducer?
[Nov-Dec 2016] [R]
The sensing material in a piezoresistive pressure sensor is a diaphragm formed on a
silicon substrate, which bends with applied pressure. A deformation occurs in the crystal
lattice of the diaphragm because of that bending. This deformation causes a change in the
band structure of the piezoresistors that are placed on the diaphragm, leading to a change
in the resistivity of the material. This change can be an increase or a decrease according
to the orientation of the resistors.
9. List the application of piezoresistive pressure transducer.
[Apr-May2017][Nov-Dec2016][U]
Application Areas
 Household Appliances: Washing machines, dishwashers, vacuum cleaners;
 Automotive Applications: Oil level, gas level, air pressure detection;
 Biomedical Applications: Blood pressure measurement, etc...
10. State the principle of thermal conductivity gauge. [Nov-Dec 2016] [R]
A hot wire, placed within an envelope, will transfer thermal energy from the wire to
any gas molecules that come into contact with it, and that energy will again transferred to
the walls of the envelope. With continual motion of the gas molecules, a thermal
equilibrium will be reached as long as the number of gas molecules [pressure] remains
constant. If the pressure changes, the temperature of the wire will change, this means that
the temperature of the wire can be used as an indication of the pressure within the
envelope.
11. What are the types of thermal conductivity gauges? [R]
 Pirani gauge
 Thermocouple gauge
12. What is the purpose of ionization gauge? [U]
Ionization gauge is used to measure the density of a gas.

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13. State the principle of McLeod gauge? [Apr-May2017] [R]

A McLeod gauge is a scientific instrument used to measure very low pressures, down
to 10−6 Torr. It was invented in 1874 by Herbert McLeod. McLeod gauges operates
under the principle by taking in a sample volume of gas from a vacuum chamber, then
compressing it by tilting and infilling with mercury. The pressure in this smaller volume
is then measured by a mercury manometer, and knowing the compression ratio (the ratio
of the initial and final volumes), the pressure of the original vacuum can be determined
by applying Boyle's law.
14. Write down the formula for calculating the pressure in a McLeod gauge? [U]
P = KHHo[1- KH]
K->Constant
H-> Difference in heights of the two mercury columns.
Ho->Height of the top of the closed capillary tube above the line marked on the tube.
15. What is the function of a dead weight tester? [Apr-May 2015] [R]
Dead weight tester is used to calibrate bourdon gauges .It is used as a measuring device
and also as a calibration method.
16. What are the disadvantages of using thermocouple gauge? [R]
 Easily damaged by organic vapours
 The filaments can be coated with a deposit of vapours which alters the way the
filament transfers heat.
17. Define thermal conductivity [R]
The ability of the material to carry heat by conduction is called as the thermal
conductivity.
18. What are the materials used in the construction of piezoelectric pressure transducer? [R]
These devices use piezoelectric characteristic of crystalline and ceramic materials
such as quartz.
19. What are the main parts in electrical pressure transducer? [R]
 Pressure sensing element such as a bellow, a diaphragm or a bourdon tube.
 Primary conversion element
 Secondary conversion element.
20. What are the advantages of capacitive pressure transducer? [R]
 It gives rapid response to changes in pressure
 It can withstand a lot of vibration
 It has a good frequency response and can measure both static & dynamic
changes.
21. What is a vacuum pressure? [R]
Pressures which are below the atmospheric pressure are called vacuum pressure.

22. What is the purpose of calibrating a pressure measuring instrument? [U]

It is used to adjust the output signal to a known range of pressure. It includes zero,
span and linearity adjustment.

23. Write down the applications of ionization gauges? [U]

It is used to measure low vacuum and ultra high vacuum pressure.
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24. List out the limitations of ionization gauge. [Apr-May 2015] [R]
 The lower range of hot cathode gauges is limited by the residual current.
Therefore, the measurement has an increasingly high uncertainty, if the pressure
is smaller than 3 to 4 times the x-ray limit.
 The life time of the hot cathode gauge is typically limited by the contamination
by the vacuum system or by the depletion of the filament coating layer.
25. Determine the range of pressure which can be measured by Pirani gauge &
thermocouple gauge [U]
-5
 In Pirani gauge-> Pressure range from 10 torr to 1 torr can be measured.
 In thermocouple gauge-> Pressure range from 10-4 torr can-be measured.
26. What are the disadvantages of LVDT pressure transducer? [R]
 Large core displacement are required for appreciable amount differential output.
 Temperature affects the performance of the transducer.
 They are sensitive to stray magnetic fields.
27. Write the principle of operation of an ionization gauge? [U]
Operation of ionization gauge follows Boyle's law
[i.e.,] at constant temperature, the ratio of pressure of two
gases is equal to the ratio of the two densities. A hot
filament emits electrons which are accelerated to go inside
a cylindrical wire cage. In the cage the electrons hit rest gas
molecules and ionize them. The ions are collected at the
wire in the middle of the cage and the current from that
wire is a measure for the pressure. The ion current is proportional to the pressure of the
rest gas.
28. What is the principle of operation of a capacitive pressure transducer? [U]
A linear change in capacitance with changes in the physical position of the moving
element may be used to provide an electrical indication of the element‘s position. It is
based on the principle of the capacitance equation of the parallel plate capacitor
C = ε0εrA/d farad
where, ε0 = 8.85 * e -12 farad / m
εr = Dielectric constant
A = Area of each plate
d->Distance between two plates.
From this equation, it is seen that capacitance increases if the effective area of the
plate is increased, and if the material has a high dielectric constant. The capacitance is
reduced if the spacing between the plates is increased. Transducers which make use of
these three methods of varying capacitance have been developed.

PART-B
1. Describe the methods of measurement of Low pressure using
i. Ionization type vacuum gauge [Apr-May 2017][R] [Apr-May 2015][AP]
ii. Thermal Conductivity gauge
2. Discuss the methods of pressure measurement using [R]

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i. Bourdon tube
ii. Bellows
iii. Diaphragm
3. State the Different Types of Manometer and explain the construction and working of any
three. [U]
4. With neat sketch describe the method of differential pressure using [AP]
i. capacitive transducer
ii. resonator pressure sensor [Apr-May 2015]
5. Describe working of capacitive type pressure gauge and potentiometric pressure gauge
[Nov-Dec 2015/2016] [AP]
6. Explain in detail about electrical method of pressure measurement using neat diagram.
[Apr-May 2017][R]

PART-C
1. Explain capacitive type pressure gauge and Strain gauge type pressure sensor with
diagram and application. [U]
2. Explain how calibration of pressure gauge is carried out using dead weight tester and
mention and what are the factors affect the accuracy of dead weight tester?
[Nov-Dec 2015/2016] [U]
3. Explain the construction and operation of McLeod Gauge and Resonator type pressure
transducer [Nov-Dec 2015][U]
4. How LVDT can be used for process pressure measurement? Explain. [Nov-Dec 2017] [U]

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EE8451- LINEAR INTEGRATED CIRCUITS AND

APPLICATIONS

COURSE OUTCOMES

Course
Statement
Outcome

CO1 Understand the manufacturing process involved in the fabrication of IC.

CO2 Analyze the DC & AC characteristics of operational amplifier.
CO3 Understand and acquire knowledge on the Applications of Op-amp
CO4
Describe the internal functional blocks of Timer IC and its applications.
CO5 Describe the internal functional blocks of PLL and Analog Multiplier IC and
its applications.
CO6 Describe the internal functional blocks of application ICs such as Voltage
regulator, Power amplifier, Function Generators.

CO - PO MAPPING

PO PO PO PO PO PO PO PO PO PO1 PO1 PSO PSO PSO

CO PO12
1 2 3 4 5 6 7 8 9 0 1 1 2 3

CO1 3 2 3 2

CO2 3 3 3 2

CO3 3 3 2 2 3 2
CO4 3 3 3 2
CO5 3
3 3 2
CO6 3 2 3 2

AVG 3 2.67 2 2 3 2

3 – Substantially 2 – Moderate 1 - Slightly

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UNIT I
IC Fabrication
PART – A

1. Define the term photolithography in IC fabrication. (A/M 18) [R]

Photolithography is a process used in microfabrication to pattern parts of a thin film
or the bulk of a substrate. It uses optical radiation to image the mask on a silicon wafer
using photoresist layers.

2. State the limitations of IC technology. (N/D 17) (N/D 16) (A/M 16) [U]
 Devices using ICs must generally be designed to work with discrete inductors
(coils) external to the ICs themselves.
 High-power operation necessitates a certain minimum physical mass and volume
because the components generate a lot of heat. Effective removal of this heat
requires hefty objects, such as heatsinks.
3. Distinguish between dry and wet etching. (N/D 17) (A/M 16) [AZ]

Dry etching Wet etching

 Gaseous mixture is used as the  Chemical reagents used are in the
chemical liquid reagent form.
 Smaller line openings ( 1µm)  Line opening are larger
 It produces straight walled It produces patterns with
etching undercutting.

4. State the advantages of CMOS circuits. (A/M 17) [U]

 High input impedance.
 The outputs actively drive both ways
 Good speed to power ratio compared to other logic types.
 Very little power consumption when held in a fixed state.
 CMOS gates are very simple.
5. What is lithography ? (A/M 17) [R]
Lithography is a process by which the pattern appearing on the mask is transferred to
the wafer. It involves two steps: the first step requires applying a few drops of photo
resist to the surface of the wafer & the second step is spinning the surface to get an even
coating of the photo resist across the surface of the wafer.

6. Mention different available IC package configurations. (N/D 15) [R]

 Single inline package (SIP)
 Dual inline package (DIP)
 Quadruple inline package (QIP)
 Zig-zag inline package (ZIP)

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7. Classify ICs on the basis of application, device used and chip complexity.
(N/D 15) [R]

8. What is ion implantation ? Give its advantages. (A/M 15) [R]

Ion implantation is a low-temperature process by which ions of one element are
accelerated into a solid target, thereby changing the physical, chemical, or electrical
properties of the target. Its advantages are :
 It is performed at low temperature. Therefore, previously diffused regions have a
lesser tendency for lateral spreading.
 In diffusion process, temperature has to be controlled over a large area inside the
oven, whereas in ion implantation process, accelerating potential & beam content are
dielectrically controlled from outside.
9. What is PV cell ? [R]
A solar cell or photovoltaic cell is an electrical device that converts the energy of light
directly into electricity by the photovoltaic effect. They are used as a photodetector detecting
light or other electromagnetic radiation near the visible range, or measuring light intensity.
10. Define an Integrated circuit. [R]
An integrated circuit(IC) is a miniature, low cost electronic circuit consisting of active
and passive components fabricated together on a single crystal of silicon. The active
components are transistors and diodes and passive components are resistors and capacitors.

11. Give the difference between monolithic and hybrid ICs [AZ]

Monolithic Integrated Circuits Hybrid Integrated Circuits

 In Monolithic circuits, all circuit  Hybrid Integrated circuits
components both active and passive separated component parts are
elements and their interconnections attached to a ceramic substrate and
are manufactured into or on top of a interconnected by means of either
single chip of silicon. metallization pattern or wire boards.
 It is used for more applications in  It is used for adopt less applications
Linear and digital IC.  Cost wise is slightly higher
 Cost wise is less. compared to Monolithic ICs.

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12. What are basic processes involved in fabricating ICs using planar technology? [R]
 Silicon wafer (substrate) preparation
 Epitaxial growth
 Oxidation
 Photolithography
 Diffusion
 Ion implantation
 Isolation technique
 Metallization
 Assembly processing & packaging
13. List out the steps used in the preparation of Si – wafers. [R]
 Crystal growth &doping
 Ingot trimming & grinding
 Ingot slicing
 Wafer policing & etching
 Wafer cleaning
14. Write the basic chemical reaction in the epitaxial growth process of pure silicon. [R]
The basic chemical reaction in the epitaxial growth process of pure silicon is the
hydrogen reduction of silicon tetrachloride.
1200oC
SiCl4 + 2H2 <-----------> Si + 4 HCl
15. What are the two important properties of SiO2? [R]
 SiO2 is an extremely hard protective coating & is unaffected by almost all reagents
except by hydrochloric acid. Thus it stands against any contamination.
 By selective etching of SiO2, diffusion of impurities through carefully defined
windows in the SiO2 can be accomplished to fabricate various components.
16. Explain the process of oxidation. [U]
The silicon wafers are stacked up in a quartz boat & then inserted into quartz furnace
tube. The Si wafers are raised to a high temperature in the range of 950 to 1150oC & at the
same time, exposed to a gas containing O2 or H2O or both. The chemical action is
Si + 2H2O -----------> Si O2+ 2H2
17. What is meant by molecular beam epitaxy (MBE)? [R]
In the molecular beam epitaxy, silicon along with dopants is evaporated. The
evaporated species are transported at a relatively high velocity in a vacuum to the substrate.
The relatively low vapour pressure of silicon & the dopants ensures condensation on a low
temperature substrate. Usually, silicon MBE is performed under ultra high vacuum (UHV)
condition of 10-8 to 10-10 Torr.
18. What are the advantages of Molecular Beam Epitaxy( MBE )? [U]
 It is a low temperature process, useful for VLSI. This minimizes out diffusion & auto
doping.
 It allows precise control of doping& permits complicated profiles to be generated.
 Linear doping profile desirable for varactor diode in FM, can be obtained with MBE.
 Wider choice of dopants can be used.
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19. What is oxidation induced defects in semi conductor? [R]

 Stacking faults
 Oxide isolation defects
Stacking faults:
Structural defects in the silicon lattice are called oxidation induced stacking faults.
The growth of stacking faults is a strong function of substrate orientation, conductivity type
& defect nuclei present. The stacking faults formation can be suppressed by the addition of
HCl.
Oxide isolation defects:
The stress along the edges of an oxidized area produces severe damage in the silicon.
Such defects result in increased leakage in nearby devices. High temperature (around 950oC )
will prevent stress induced defect formation.
20. What is bird’s beak? [R]
In local oxidation process, the oxidation of silicon proceeds slightly under the nitride
as well. Also, a large mismatch in the thermal expansion co-efficient of Si3N4 & Silicon
results in damage to the semi conductor during local oxidation.
This damage can be greatly reduced by growing a thin layer of SiO2 prior to
placement of the Si3N4 mask. Typically 100 to 200Ao is used for this purpose.
Unfortunately, this greatly enhances the penetration of oxide under the nitride masked
regions, resulting in oxide configurations called bird’s beak.
21. What are the different types of lithography? What is optical lithography? [R]
The different types of lithography are:
 Photolithography
 Electron beam lithography
 X ray beam lithography
 Ion beam lithography
Optical lithography:
Optical lithography comprises the formation images with visible or UV radiation in a
photo resist using contact, proximity or projection printing.
22. What are the two processes involved in photolithography? [R]
 Making a photographic mask
 Photo etching
The development of photographic mask involves the preparation of initial artwork and
its reduction, decomposition of initial artwork or layout into several mask layers. Photo
etching is used for the removal of SiO2 from desired regions so that the desired impurities
can be diffused.
23. What is meant by reactive plasma etching? [R]
The term reactive plasma is meant to describe a discharge in which ionization &
fragmentation of gases takes place& produce chemically active plasma species, frequently
oxidizers and reducing agents. Such plasmas are reactive both in the gas phase & with solid
surfaces exposed to them. When these interactions are used to form volatile products so that
material is removed or etching of material form surfaces that are not masked to form
lithographic patterns, the technique is known as reactive plasma etching.

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24. What are isotropic & anisotropic etchings processes? [R]

Isotropic etching is a wet etching process which involves undercutting. Isotropic
etching is a dry etching process which provides straight walled patterns.
25. Define diffusion. [R]
The process of introducing impurities into selected regions of a silicon wafer is called
diffusion. The rate at which various impurities diffuse into the silicon will be of the order of
1µm/hr at the temperature range of 900oC to 1100oC .The impurity atoms have the tendency
to move from regions of higher concentrations to lower concentrations.
26. What is dielectric isolation? [R]
In dielectric isolation, a layer of solid is dielectric such as SiO2 or ruby completely
surrounds each components thereby producing isolation, both electrical & physical. This
isolating dielectric layer is thick enough so that its associated capacitance is negligible. Also,
it is possible to fabricate both pnp & npn transistors within the same silicon substrate.
27. What is metallization? [R]
The process of producing a thin metal film layer that will serve to make
interconnection of the various components on the chip is called metallization.

PART – B

1. Describe the steps involved in the fabrication of monolithic IC transistors.

(A/M 18) [U]

2. Elaborate the fabrication of MOS ICs with suitable diagram. (A/M 18) [R]
3. Explain the basic process used in silicon planar technology with neat diagram.
(N/D 17) [U]

4. Write a note on classification of IC and IC packages. (N/D 17) (A/M 15) [R]
5. With neat illustrations explain the various steps involved in the IC fabrication process.
(A/M 17) [U]
6. With circuit diagram explain the steps involved in the fabrication of the circuit shown
below using IC technology. (A/M 17) [U]

7. Describe about epitaxial growth process. (N/D 16) (A/M 15) [U]
8. Explain in detail about the Photolithography process with neat diagram. (N/D 16) [U]
9. Write a note on masking and etching process in IC fabrication. (N/D 16) [R]
10. State the limitations of IC technology. (A/M 16) [U]

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11. Distinguish between dry etching and wet etching. (A/M 16) [AZ]
12. Why IC 741 is not used for high frequency applications. (A/M 16) [U]
13. Explain the various steps involved in fabrication of a typical transistor into monolithic
ICs. (N/D 15) [U]
14. What is thin and thick film technology ? Explain various methods used for deposition of
thin film technology. (N/D 15) [R]
15. Briefly explain the various process involved in fabrication monolithic IC which
integrates diode, capacitance and FET. (A/M 15) [U]

PART – C

1. What are the new trends in Integrated Circuit technologies and explain about its scope
for future technologies. (A/M 18) (N/D 17) (N/D 16) [U]
2. Write a note on recent fabrication technologies of FET for industrial applications.
(A/M 18) [R]
3. Explain in detail the recent fabrication methods of diode and capacitance for industrial
applications. (N/D 17) [U]
4. Write a note on recent fabrication methods of FET for industrial applications.
(N/D 16) [U]

UNIT II
Characteristics of OP-AMP
PART – A

1. What is the drawback of IC 741 ? (N/D 17) [U]

 Most OP-AMP are designed to for lower power operation.
 For high output desire then the OP-AMP specifically designed for that purpose
must be seen.
 Most commercial OP-AMP shuts off when the load resistance is below the
specific level.

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2. Compare the ideal and practical op-amp characteristics. (A/M 18) [AZ]

Ideal OPAMP Practical OPAMP

 Infinite voltage gain, so that it can  Voltage gain is not infinite, but typically
amplify input signals of any amplitude. 10^5 to 10^8, so it is not able to amplify
input signals smaller than 100 uV.
 Infinite input resistance, so that almost  Input resistance is typically 10^6 to 10^12
any signal source can drive it and there ohm, so still it draws some current and not
is no loading of preceding stage. all source can drive it.
 Zero output resistance, so that output  Output resistance is typically 75 ohm for
can drive an infinite number of other standard Op-Amps, so it has limit to
devices. deliver current to output devices.
 Zero output voltage when input voltage  It is not able to give zero at output when
is zero. input is zero, due to mismatching of input
transistors.
 Infinite bandwidth, so that any  Op-Amp has its own Gain-Bandwidth
frequency signal can be amplified product, so input frequency should not
without attenuation. exceed from that particular frequency
range.
 Infinite CMRR, so that the output  CMRR is typically 90 dB, so still it gives
common-mode noise voltage is zero. output voltage even if both input terminals
are shorted.
 Infinite slew rate, so that output voltage  Slew rate is typically 0.5 to 90 V/uS, so
changes occur simultaneously with output cannot be change simultaneously
input voltage changes. with input, there is some delay.

3. How an op-amp can be used as a voltage follower ? (A/M 18) [U]

Voltage Follower is a special type of the non-inverting amplifier circuit. Here the input
signal is connected directly to the non-inverting input of the amplifier and thus the
output signal is not inverted resulting in the output voltage being equal to the input
voltage, Vout = Vin. This then makes the voltage follower circuit ideal as a “Unity Gain
Buffer circuit” because of its isolation properties.

4. Write the advantages of ICs over discrete circuits. (N/D 16) (A/M 15) [U]
 Minimization & hence increased equipment density.
 Cost reduction due to batch processing.
 Increased system reliability
 Improved functional performance.
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 Matched devices.
 Increased operating speeds
 Reduction in power consumption
5. Draw the circuit diagram of a symmetrical emitter coupled differential amplifier.
(A/M 17) [R]

6. Write some applications of operational amplifier. (N/D 16) [U]

 Adder & Subtractor
 Voltage to current converter & current to voltage converters
 Instrumentation amplifier,
 Power amplifier
 Rectifier
 Peak detector
 Clipper and clamper
 Sample and hold circuit
 Log amplifier & anti –log amplifier
 Multiplier
7. Why IC 741 is not used for high frequency applications. (A/M 16) [U]
IC741 has a low slew rate because of the predominance of capacitance present in the
circuit at higher frequencies. As frequency increases the output gets distorted due to
limited slew rate.
8. What are the ideal characteristics of an op-amp ? (N/D 15) [R]
 Open loop gain infinite
 Input impedance infinite
 Output impedance low
 Bandwidth infinite
 Zero offset, ie,Vo=0 when V1=V2=0
9. What is integrator ? (N/D 16) (A/M 16) [R]
The output voltage of integrator is proportional to the input voltage integrated over time.

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10. What do you mean by input offset current and offset voltage ? (A/M 15) [R]
The input terminals conduct a small value of dc current to bias the input transistors.
Since the input transistors cannot be made identical, there exists a difference in bias
currents. The difference between the bias currents at the input terminals of the op-amp is
called as input offset current. A small voltage applied to the input terminals to make the
output voltage as zero when the two input terminals are grounded is called input offset
voltage.
11. Define CMRR. (A/M 15) [R]
The relative sensitivity of an op-amp to a difference signal as compared to a common –
mode signal is called the common –mode rejection ratio. It is expressed in decibels.

CMRR= AD/AC
12. What is OPAMP? [R]
An operational amplifier is a direct coupled high gain amplifier consisting of one or
more differential amplifiers, followed by a level translator and an output stage. It is a
versatile device that can be used to amplify ac as well as dc input signals & designed for
computing mathematical functions such as addition, subtraction, multiplication,
integration & differentiation.
13. What are the different kinds of packages of IC741? [R]
 Metal can (TO) package
 Dual- in- line package
 Flat package or flat pack
14. What are the assumptions made from ideal opamp characteristics? [R]
 The current drawn by either of the input terminals (non-inverting/inverting) is
negligible.
 The potential difference between the inverting & non- inverting input terminals is
zero.

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15. Mention some of the linear applications of op – amps: [U]

Adder, subtractor, voltage to current converter, current to voltage converters,
instrumentation amplifier, analog computation, power amplifier, etc are some of the linear
op-amp circuits.
16. Mention some of the non – linear applications of op-amps:- [U]
Rectifier, peak detector, clipper, clamper, sample and hold circuit, log amplifier, anti
–log amplifier, multiplier are some of the non – linear op-amp circuits.
17. What are the areas of application of non- linear op- amp circuits? [U]
 Industrial instrumentation,
 Communication
 Signal processing
18. What happens when common terminal of V+ and V- sources is not grounded? [U]
If the common point of the two supplies is not grounded, twice the supply voltage will
get applied and it may damage the op-amp.
19. In practical op-amps, what is the effect of high frequency on its performance? [U]
The open- loop gain of op-amp decreases at higher frequencies due to the presence of
parasitic capacitance. The closed- loop gain increases at higher frequencies and leads to
instability.
20. What is the need for frequency compensation in practical op-amps? [U]
Frequency compensation is needed when large bandwidth and lower closed loop gain
is desired. Compensating networks are used to control the phase shift and hence to improve
the stability.
21. Mention the frequency compensation methods. [R]
 Dominant-pole compensation
 Pole- zero compensation.
22. What are the merits and demerits of Dominant-pole compensation? [U]
 Noise immunity of the system is improved.
 Open-loop bandwidth is reduced.
23. Define slew rate. [R]
The slew rate is defined as the maximum rate of change of output voltage caused by
a step input voltage. An ideal slew rate is infinite which means that op-amp’s output voltage
should change instantaneously in response to input step voltage.
24. What causes slew rate? [U]
There is a capacitor with- in or outside of an op-amp to prevent oscillation. It is this capacitor
which prevents the output voltage from responding immediately to a fast changing input.
25. Define thermal drift. [R]
The bias current, offset current & offset voltage change with temperature. A circuit
carefully nulled at 25oC may not remain so when the temperature raises to 35oC.This is
called thermal drift. Often, offset current drift is expressed in nA/ oC and offset voltage drift in
mV/oC.

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26. Define supply voltage rejection ratio (SVRR) [R]

The change in OPAMP’s input offset voltage due to variations in supply voltage is
called the supply voltage rejection ratio. It is also called Power Supply Rejection Ratio
(PSRR) or Power Supply Sensitivity (PSS).

PART – B

1. Explain the working principle of emitter coupled differential amplifier. (A/M 18) [U]
2. How common mode rejection ratio can be increased using constant current source ?
(A/M 18) [U]

3. Draw the inverting amplifier circuit of an op-amp in closed loop configuration. Obtain
the expression for the closed loop gain. (A/M 18) [R]
4. For a non-inverting amplifier using an op-amp assume R1 = 470 Ω and R2 = 4.7 kΩ.
Calculate the closed loop voltage gain of the amplifier. (A/M 18) [A]
5. Explain the following terms in an op-amp : (N/D 17) [U]
i. Bias current
ii. Thermal drift
iii. Input offset voltage and current
iv. Virtual ground
6. Draw the circuit of a symmetrical emitter coupled differential amplifier and derive for
CMRR. (N/D 17) [R]
7. Determine the output voltage for the following circuits : (A/M 17) [A]

8. With diagram explain the working principle of V/I converter. (A/M 17) [U]
9. Write a note on stability criterion and compensation techniques applicable to opamp
circuit. (A/M 17) [R]
10. Discuss in detail about the DC and AC characteristics of op-amp. (N/D 16) [R]
11. Explain the differential amplifier using op-amp. (N/D 16) [U]
12. Discuss the frequency response characteristics and compensation of an operational
amplifier. (A/M 16, 15) (N/D 15) [R]
13. Explain the application of op-amp as differentiator. (A/M 16) [U]
14. Find V0 for the given circuit. (A/M 16) [A]

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15. What is slew rate ? List the causes of the slew rate and explain its significance in
applications. (N/D 15) [R]
16. Draw and explain the operation of a current to voltage converter. (N/D 15) [R]
17. What are the limitations of an ordinary op-amp differentiator ? Draw the circuit of a
practical differentiator that will eliminate these limitations. (N/D 15) [U]
18. Design an op-amp circuit to give an output voltage V0 = 4V1 – 3V2 + 5V3 – V4 where V1
,V2 ,V3, V4 are inputs. (A/M 15) [C]
19. Explain voltage to current converter using operational amplifier. Also explain the
application of op-amp as integrator. (A/M 15) [U]
20. What is slew rate and how it can be improved ? (A/M 15) [R]
21. Draw the circuit diagram of op-amp differentiator, integrator and derive an expression
for the output in terms of the input. [R]
22. Explain in detail about voltage series feedback amplifier. [U]
23. Derive the gain of inverting and non-inverting. [A]
24. Explain and derive the condition for DC-characteristics of an operational amplifier. [U]

PART – C

1. Sketch the implementation of an instrumentation amplifier using three op-amps. Explain

the principle of operation and its applications. (A/M 17) [U]

2. Determine Vo and Io in the circuit shown below. [AZ]

3. Obtain the closed loop voltage gain of the circuit shown below. [AZ]

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UNIT III
Applications of OP-AMP
PART – A

1. What is the use of sample and hold circuit ? (A/M 18) (N/D 17, 16) (A/M 16) [U]
A sample and hold circuit is one which samples an input signal and holds on to its last
sampled value until the input is sampled again. This circuit is mainly used in digital
interfacing, analog to digital systems, and pulse code modulation systems.

2. Enlist the applications of comparators. (A/M 18) [U]

 Zero crossing detectors
 Window detector
 Time marker generator
 Phase detector
3. Write down applications of clipper and clamper. (N/D 17) [U]
Clippers find several applications, such as

 The excessive noise spikes above a certain level can be limited or clipped in FM
transmitters by using the series clippers.
 For the generation of new waveforms or shaping the existing waveform.
 The typical application of diode clipper is for the protection of transistor from
transients

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Clampers can be used in applications like

 These are frequently used in test equipment, sonar and radar systems.
 For the protection of the amplifiers from large errant signals clampers are used.
 Clampers can be used as voltage doublers or voltage multipliers.
4. Which is the fastest ADC ? State reason. (A/M 17) [U]
The flash ADC is the fastest type ADC. A flash ADC uses comparators one per voltage
step and a string of resistors. A 4-bit ADC will have 16 comparators, an 8-bit ADC will
have 256 comparators. The conversion speed of the flash ADC is the sum of the
comparator delays and the logic delay.
5. What is Zero crossing detector ? (A/M 15) [R]
A zero crossing detector is a type of voltage comparator, used to detect a sine waveform
transition from positive and negative. It changes the o/p between +Vsat & –Vsat when
the i/p crosses zero reference voltage.
6. Draw the circuit diagram of a zero cross detector with input and output waveforms.
(A/M 17) [R]

7. Draw the circuit of a log amplifier using two op-amps. (N/D 15) [R]

8. What is the need for an instrumentation amplifier? [R]

In a number of industrial and consumer applications, the measurement of physical
quantities is usually done with the help of transducers. The output of transducer has to be

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amplified So that it can drive the indicator or display system. This function is performed by
an instrumentation amplifier.
9. List the features of instrumentation amplifier. [R]
 High gain accuracy
 High CMRR
 High gain stability with low temperature co-efficient
 Low dc offset
 Low output impedance
10. What is a comparator? [R]
A comparator is a circuit which compares a signal voltage applied at one input of an
op-amp with a known reference voltage at the other input. It is an open loop op - amp with
output + Vsat.
11. What is a Schmitt trigger? [R]
Schmitt trigger is a regenerative comparator. It converts sinusoidal input into a square
wave output. The output of Schmitt trigger swings between upper and lower threshold
voltages, which are the reference voltages of the input waveform.
12. What is a multivibrator? [R]
Multivibrators are a group of regenerative circuits that are used extensively in timing
applications. It is a wave shaping circuit which gives symmetric or asymmetric square output.
It has two states stable or quasi- stable depending on the type of multivibrator.
13. What do you mean by monostable multivibrator? [U]
Monostable multivibrator is one which generates a single pulse of specified duration
in response to each external trigger signal. It has only one stable state. Application of a
trigger causes a change to the quasi-stable state. An external trigger signal generated due to
charging and discharging of the capacitor produces the transition to the original stable state.
14. What is an astable multivibrator? [R]
Astable multivibrator is a free running oscillator having two quasi-stable states. Thus,
there is an oscillation between these two states and no external signals are required to
produce the change in state.
15. What is a bistable multivibrator? [R]
Bistable multivibrator is one that maintains a given output voltage level unless an
external trigger is applied. Application of an external trigger signal causes a change of state,
and this output level is maintained indefinitely until an second trigger is applied. Thus, it
requires two external triggers before it returns to its initial state
16. List the broad classification of ADCs. [R]
 Direct type ADC.
 Integrating type ADC.
17. List out the direct type ADCs. [R]
 Flash (comparator) type converter
 Counter type converter
 Tracking or servo converter
 Successive approximation type converter

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18. List out some integrating type converters. [R]

 Charge balancing ADC
 Dual slope ADC
19. What is integrating type converter? [R]
An ADC converter that perform conversion in an indirect manner by first changing
the analog I/P signal to a linear function of time or frequency and then to a digital code is
known as integrating type A/D converter.
20. Explain in brief the principle of operation of successive Approximation ADC. [U]
The circuit of successive approximation ADC consists of a successive approximation
register (SAR), to find the required value of each bit by trial & error. With the arrival of
START command, SAR sets the MSB bit to 1. The O/P is converted into an analog signal &
it is compared with I/P signal. This O/P is low or High. This process continues until all bits
are checked.
21. What are the main advantages of integrating type ADCs? [U]
 The integrating type of ADC’s do not need a sample/Hold circuit at the input.
 It is possible to transmit frequency even in noisy environment or in an isolated form.
22. Define conversion time. [R]
It is defined as the total time required to convert an analog signal into its digital
output. It depends on the conversion technique used & the propagation delay of circuit
components. The conversion time of a successive approximation type ADC is given by
T(n+1) Where T---clock period, c---conversion time, n----no. of bits
23. Define resolution of a data converter. [R]
The resolution of a converter is the smallest change in voltage which may be
produced at the output or input of the converter.
 Resolution (in volts)= VFS/2n-1=1 LSB increment.
The resolution of an ADC is defined as the smallest change in analog input for a one-
bit change at the output.
24. Explain the application of instrumentation amplifier for transducer bridge [U]

25. What is meant by linearity? [R]

The linearity of an ADC/DAC is an important measure of its accuracy & tells us how
close the converter output is to its ideal transfer characteristics. The linearity error is usually
expressed as a fraction of LSB increment or percentage of full-scale voltage. A good
converter exhibits a linearity error of less than ±½LSB.
.
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PART – B

1. Design a weinbridge oscillator for a frequency of 5 kHz. Assume C = 0.01μF.

(A/M 18) [A]
2. Explain the operation of a triangular waveform generator using op-amp. (A/M 18) [U]
3. Discuss the operation of successive approximation type A/D converter.
(A/M 18) (A/M 16) [R]
4. What are the advantages of continuous type A/D converter over the counter type A/D
converter ? (A/M 18) [U]
5. With neat diagram, explain the working principle of (N/D 17) [U]
 R-2R ladder type DAC
 Weighted resistor DAC
6. Draw and explain the circuit of a second order butterworth low pass filter and derive its
transfer function. (N/D 17) [U]
7. With diagram explain the following applications of op-amp : (A/M 17) [U]
 Clippers and clampers
 Triangular waveform generator
8. Explain the working principle of R-2R ladder type D/A converter.
(A/M 17) (A/M 15) [U]
9. Design a second order butterworth low pass filter with cutoff frequency 2 KHz.
(A/M 17) [C]

10. Write a note on logarithmic and antilog amplifier using op-amp. (N/D 16) [R]
11. Explain the working of SAR type and flash type A/D converter. (N/D 16) [U]
12. Design a Schmitt trigger using op-amp. (A/M 16) (A/M 15) [A]
13. Draw the instrumentation amplifier using 3 op-amp and derive its output voltage
equation. (A/M 16) [U]
14. Explain the first order low pass butterworth filter with a neat diagram. Derive its
frequency response and plot the same. (A/M 16) [U]
15. Design a second order butterworth low pass filter having upper cut-off frequency of 1
kHz. (N/D 15) [C]
16. Explain how to measure the phase difference between two signals. (N/D 15) [U]
17. Draw a sample and hold circuit and explain its operation. (N/D 15) [R]
18. Design a circuit of a clipper which will clip the input signal below a reference voltage.
(N/D 15) [C]

19. Discuss the second order high pass filter with its frequency response and design the
circuit with the cut-off frequency of 5 kHz. (A/M 15) [C]
20. Explain the working of Instrumentation amplifier. (A/M 15) [U]
21. Explain the working of any one of sinusoidal oscillators. [U]
22. Explain how a comparator can be used as a zero crossing detector. [U]
23. Draw the circuit of a first order and second order butter worth active low pass filter and
derive its transfer functions. [A]

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PART – C

1. Explain the operation of Dual slope ADC with neat illustrations? Also prove that this ADC
is free from drifts? [U]
2. With neat schematic representations explain the operation of the following circuits.
i. Positive Peak follower.
ii. Active positive clamper to clamp the input signal above ground state by 5 V. [U]

UNIT IV
Special ICs
PART – A

1. Define the terms lock range and capture range with respect to PLL.
(A/M 18) (A/M 16) (N/D 15) [R]
When PLL is in lock, it can trap freq changes in the incoming signal. The range of
frequencies over which the PLL can maintain lock with the incoming signal is called as
lock range. The range of frequencies over which the PLL can acquire lock with the input
signal is called as capture range.
2. Mention the applications of analog multipliers. (A/M 18) (N/D 17) (A/M 17) [U]
 Ring modulator
 Product detector
 Frequency mixer
 Squelch
 Analog signal processing
 Automatic gain control
 True RMS converter
3. Define PULL time of PLL. (N/D 17) [R]
The total time taken by the PLL to establish lock is called pull- in time.

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4. Define PLL. (N/D 16) [R]

A PLL is a basically a closed loop system designed to lock output frequency and phase to
the frequency and phase of an input signal.

5. Draw the circuit diagram of a PLL circuit used as an AM modulator. (A/M 17)[R]

6. What is analog multiplier IC ? Where it is used ? (A/M 16) [R]

The analog multiplier ICs are mainly used for :

 Voltage squarer
 Frequency doubler
 Voltage divider
 Square rooter
 Phase angle detector
 Rectifier
7. Draw the functional block diagram of 555 timer IC. (N/D 16) [R]

8. Define duty cycle in astable multivibrator using IC 555. (A/M 15) [R]
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The ratio of high output and low output period is given by a mathematical parameter
called duty cycle. It is defined as the ratio of ON Time to total time. It is given by :
TON
Duty cycle =
TON  TOFF

9. List the applications of PLL. (A/M 15) [U]

Phase-locked loops are widely used :

 For synchronization purpose

 In space communications for coherent demodulation and threshold extension
 Bit and symbol synchronization
 To demodulate frequency-modulated signals.
10. What are the applications of 555 Timer? [U]
 Astable multivibrator
 Monostable multivibrator
 Missing pulse detector
 Linear ramp generator
 Frequency divider
 Pulse width modulation
 FSK generator
 Pulse position modulator
 Schmitt trigger
11. List the applications of 555 timers in monostable mode of operation: [U]
 Missing pulse detector
 Linear ramp generator
 Frequency divider
 Pulse width modulation.
12. List the applications of 555 timers in Astable mode of operation: [U]
 FSK generator
 Pulse-position modulator
13. Define 555 IC? [R]
The 555 timer is an integrated circuit specifically designed to perform signal
generation and timing functions.
14. List the basic blocks of IC 555 timer? [R]
 A relaxation oscillator
 RS flip flop
 Two comparator
 Discharge transistor.
15. List the features of 555 Timer? [R]
 It has two basic operating modes: monostable and astable
 It is available in three packages. 8 pin metal can, 8 pin dip, 14 pin dip.
 It has very high temperature stability.
16. Define AD633 IC. [R]
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The AD633 is a functionally complete, four-quadrant, analog multiplier. It includes high

impedance, differential X and Y inputs, and a high impedance summing input (Z).

17. List the features of 566 VCO. [R]

 Wide supply voltage range (10-24V)
 Very linear modulation characteristics
 High temperature stability
18. List the applications of 565 PLL. [U]
 Frequency multiplier
 Frequency synthesizer
 FM detector

PART – B

1. Explain the functional block diagram of NE 561 phase locked loop. (A/M 18) [U]
2. Narrate the process of FSK demodulation using PLL. (A/M 18) [R]
3. Describe the working principle of the variable trans-conductance analog multiplier.
(A/M 18) [U]

4. Briefly explain the difference between the two operating modes of 555 timer.
(N/D 17) [U]

5. List the important feature of 555 timer. (N/D 17) [R]

6. Write a note on analog multipliers and VCO. (N/D 17) [R]
7. Briefly explain the functional block diagram of NE 565 PLL – IC to operate as a
frequency divider. (A/M 17) [U]
8. Explain the functional block diagram of 555 timer IC. (A/M 17) (N/D 15) [U]
9. Design a monostable multivibrator with pulse duration of 1 msec using 555 timer IC.
(A/M 17) [C]
10. With the help of schematic diagram, explain the operation of IC 566 VCO and derive its
output frequency. (N/D 16) (A/M 16) [U]
11. What is PLL ? How frequency multiplication is done in PLL ? (N/D 16) [U]
12. Design and draw the waveform of a 1 kHz square wave generator using 555 timer for
duty cycle of 50 %. (A/M 16) [C]
13. Explain the operation of astable operation of IC 555 with necessary waveform.
(A/M 16) (A/M 15) [U]

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14. Discuss the operation of a FSK generator using 555 timer. (N/D 15) [R]
15. Draw the block diagram of a VCO and explain its operation. (N/D 15) [R]
16. In the astable multivibrator using 555 timer, R A = 2.2 kΩ, RB = 6.8 kΩ and C=0.01μF.
Calculate tHIGH , tLOW, free running frequency and duty cycle. (A/M 15) [C]
17. Explain the working of a voltage controlled oscillator. (A/M 15) [U]
18. Explain how frequency multiplication is done using PLL. (A/M 15) [U]
19. Draw the block diagram of an Astable multivibrator using 555timer and derive an
expression for its frequency of oscillation. [A]
20. Draw the block diagram of monostable multivibrator using 555timer and derive an
expression for its frequency of oscillation. [A]

PART – C

1. Explain with functional block diagram the operation of 566 Voltage Controlled Oscillator.
Determine the maximum and minimum output frequencies in the circuit shown below. [U]

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UNIT V
Application ICs
PART – A

1. What is an isolation amplifier ? (A/M 18) (N/D 17) [R]

An isolation amplifier is an amplifier that offers electrical isolation between its input and
output terminals.
2. List the features of opto-coupler ICs (A/M 18) (N/D 17) [R]
Features of opto-coupler IC MOC3020 are :
 400 V Photo-TRIAC Driver Output
 High isolation – 500 Vpeak
 Output driver designed for 220 Vac
 Standard 6-terminal plastic DIP
3. What is SMPS ? (N/D 16) [R]
An Switched Mode Power Supply (SMPS) is an electronic power supply that
incorporates a switching regulator to convert electrical power efficiently. It transfers
power from a DC or AC source to DC loads, such as a personal computer.
4. Give some examples of monolithic IC voltage regulators. (A/M 16) [U]
 78XX series fixed output, positive voltage regulators
 79XX series fixed output, negative voltage regulators
 723 general purpose regulator.
5. Define line regulation and load regulation. (N/D 15) [R]
 Line regulation is defined as the percentage change in the output voltage for a change
in the input voltage
 Load regulation is defined as the change in output voltage for a change in load
current. It is expressed in mill volts or as a percentage of the output voltage.
6. What is a voltage regulator? [R]
A voltage regulator is an electronic circuit that provides a stable dc voltage
independent of the load current, temperature, and ac line voltage variations.
7. Give the classification of voltage regulators: [R]
 Series / Linear regulators
 Switching regulators.
8. What is a linear voltage regulator? [R]
Series or linear regulator uses a power transistor connected in series between the unregulated
dc input and the load and it conducts in the linear region .The output voltage is controlled by
the continuous voltage drop taking place across the series pass transistor.
9. What is a switching regulator? [R]
Switching regulators are those which operate the power transistor as a high frequency on/off
switch, so that the power transistor does not conduct current continuously. This gives
improved efficiency over series regulators.

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10. What are the advantages of IC voltage regulators? [U]

 low cost
 high reliability
 reduction in size
 excellent performance
11. What is the purpose of having input and output capacitors in three terminal IC
regulators? [U]
A capacitor connected between the input terminal and ground cancels the inductive
effects due to long distribution leads. The output capacitor improves the transient
response.
12. What is meant by current limiting? [R]
Current limiting refers to the ability of a regulator to prevent the load current from
increasing above a preset value.
13. Give the drawbacks of linear regulators [U]
 The input step down transformer is bulky and expensive because of low line
frequency.
 Because of low line frequency, large values of filter capacitors are required to
decrease the ripple.
 Efficiency is reduced due to the continuous power dissipation by the transistor as it
operates in the linear region.
14. What is the advantage of switching regulators? [U]
 Greater efficiency is achieved as the power transistor is made to operate as low
impedance switch. Power transmitted across the transistor is in discrete pulses rather
than as a steady current flow.
 By using suitable switching loss reduction technique, the switching frequency can be
increased so as to reduce the size and weight of the inductors and capacitors.
15. Mention the advantages of opto-couplers: [U]
 Better isolation between the two stages.
 Impedance problem between the stages is eliminated.
 Wide frequency response.
 Easily interfaced with digital circuit.
 Compact and light weight.
 Problems such as noise, transients, contact bounce are eliminated.
16. What is an isolation amplifier? [R]
An isolation amplifier is an amplifier that offers electrical isolation between its input
and output terminals.
17. What are the features of isolation amplifier? [R]
 Easy to use
 Ultra low leakage
 18 pin DIP package
18. What is LM380? [R]

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It is a power amplifier produced by national semiconductor. It is capable of delivering

2.5 W min, to 8 ohm load.
19. Define AD623 IC ? [R]
The AD623 is an integrated, single- or dual-supply instrumentation amplifier that
delivers rail-to-rail output swing using supply voltages from 3 V to 12 V. With no external
resistor, the AD623 is configured for unity gain (G = 1), and with an external resistor, the
AD623 can be programmed for gains of up to 1000.

PART – B

1. Explain the working principle of basic linear voltage regulator using op-amp.
(A/M 18) [U]

2. Explain the operation of a monolithic IC Class A audio power amplifier LM 380.

(A/M 18) [U]

3. Write a detailed note on switching regulators. (A/M 18) (A/M 16) [R]
4. Briefly explain the working principle of switch mode power supply with necessary
circuit diagrams and waveforms. (N/D 17) [U]
5. Write short notes on the following : (N/D 17, 16) (A/M 17, 16) [R]
 LM 380 power amplifier
 ICL 8038 function generator.
6. With necessary diagram and waveforms explain the working principle of switched mode
power supply. (A/M 17) [U]
7. What do you mean by the fixed voltage and variable voltage regulator. List its various
applications. (N/D 16) [R]
8. Draw and explain the functional diagram of 723 IC regulator. (N/D 15) [R]
9. Explain fold back characteristics of 723 IC regulator. (N/D 15) [U]
10. Draw the circuit diagram of a LM 380 power audio amplifier and explain its operation.
(N/D 15) (A/M 15) [U]
11. What are the applications of LM 380 power amplifier ? (N/D 15) [U]
12. Explain the working of series voltage regulator. (A/M 15) [R]
13. Explain the working principle of IC 8038 function generator. (A/M 15) [R]
14. What is the principle of switch-mode power supplies ? Discuss its advantages and
disadvantages. (A/M 15) [R]
15. Explain i) Oscillation amplifier. ii) Voltage regulator [U]

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PART – C

1. Using 7805 design a current source to deliver a 0.2 A current to a 22 ohm 10 W load.
(A/M 17) [C]

2. Design an adjustable voltage regulator (5V to 15V) using a 723 voltage regulator. [C]

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IC 8451- CONTROL SYSTEM

COURSE OUTCOMES

Course
Statement
Outcome

Analyze transfer function model for physical system and control system
CO1
components.

CO2 Analyze the time response of system and steady state error.

CO3 Analyze the open loop and closed loop frequency response of the systems.

Analyze the system stability using Routh criterion, Root Locus &Nyquist
CO4
criterion.

Design Lead, Lag & Lead-Lag compensators and P, PI, PID controllers to
CO5 meet the desired specifications, which is required in the process control
Industry.

CO6 Describe the State Space representation of physical systems.

CO - PO MAPPING

PO PO PO PO PO PO PO PO PO PO1 PO1 PO1 PSO PSO PSO

CO 1 2 3 4 5 6 7 8 9 0 1 2 1 2 3

CO1 3 3 1 - 1 - - - - 1 - - 3 1 -

CO2 3 3 1 - 1 - - - - - - - 3 1 -

CO3 3 3 3 1 3 - - - - 1 - - 3 3 -

CO4 3 3 2 - 2 - - - - 1 - - 3 2 -

CO5 3 3 2 1 2 - - - - 1 - - 3 2 -

CO6 3 3 2 - - - - - - - - - 3 2 -

AVG 3 3 2 1 1.8 - - - - 1 - - 3 1.8 -

3 – Substantially 2 – Moderate 1 - Slightly

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UNIT I
Systems and their representation
PART A
1. What is a control system? (R)
A system is a collection of components, connected together in a sequence to perform
a certain task. In a system if the output quantity is controlled by varying the input quantity,
then the system is called control system.
The output quantity is called the controlled variable or response and the input quantity
is called command signal or excitation.
2. What is an open-loop control system? (R)
The control system in which the output is not fed back to the input side for correction is
termed as open loop control system. Here, the output quantity doesn’t have any effect on the
input quantity.
3. What is a closed loop control system? (R)
The control system in which the output is fed back to the input side so as to maintain the
desired output value is termed as closed loop control system. Here, any change in the output
quantity will automatically change the input quantity.
4. What is feedback? What type of feedback is preferred for control system? (R)
The feedback is a control action in which the output is sampled and a proportional signal is
given to input for automatic correction of any changes in the desired output. Negative
feedback is usually preferred for control system.
5. Why is negative feedback preferred in control system? (U)
Negative feedback is preferred in control system as it has the following features:
 Results in a stable system
 Rejects the disturbance signals
 Low sensitivity to parameter variations
6. Distinguish between open loop and closed loop system. (E)

Open loop system Closed loop system

 Inaccurate and unreliable  Accurate and reliable
 Simple and economical  Complex and costlier
 Changes in output due to external  Changes in output due to external
disturbances are not corrected disturbances are corrected
automatically. automatically.
 Usually stable  Usually unstable. Great care needs to
be taken to make it stable

7. Define linear system. (R)

A system is said to be linear if it obeys the principle of superposition and homogeneity. A
system is linear if the response of a system to a weighted sum of signals is equal to the

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corresponding weighted sum of the responses of the system to each of the individual input
signals.
8. What is a time invariant system? (R)
A system is said to be time invariant if its input output characteristics do not change
with time. i.e., if there is a delay or advance in the input to the system then the output also
should change by same amount.
Let be the response of the system to input , if the input
becomes , then the output should also be for the system to be time
invariant.
9. Define transfer function. (R)
The transfer function of a system is defined as the ratio of Laplace transform of output to the
Laplace transform of input with zero initial conditions. It can also be defined as the Laplace
transform of the impulse response of system with zero initial conditions.

10. What are basic elements used for modeling mechanical translational system? (R)
The model of mechanical translational system can be obtained by using three basic elements
mass, spring and dashpot.
11. Write the force balance equation of an ideal mass element. (AP)
Let a force f be applied to an ideal mass M. The mass will offer an opposing force
proportional to the acceleration.

M
f

12. Write the force balance equation of an ideal dashpot. (AP)

Let a force f be applied to an ideal dashpot, with viscous friction coefficient B. The dashpot
will offer an opposing force proportional to the velocity.
x

f f x1 x2

B B

13. Write the force balance equation of an ideal spring. (AP)

Let a force f be applied to an ideal spring, with spring constant K. The spring will offer an
opposing force proportional to the displacement.

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x X1 X2
f f

K K

14. What are the basic elements used for modeling mechanical rotational system ? (R)
The model of mechanical rotational system can be obtained by using three basic elements
mass with moment of inertia J, dashpot with rotational frictional coefficient B and the
torsional spring with stiffness K.
15. Write the torque balance equation of an ideal rotational mass element. (AP)
Let a torque T be applied to an ideal mass with moment of inertia J. The mass will offer an
opposing torque proportional to the angular acceleration.

T θ

16. Write the torque balance equation of an ideal rotational dashpot. (AP)
Let a torque T be applied to a rotational dashpot with friction coefficient B. The dashpot will
offer an opposing torque proportional to the angular velocity.

T θ B B
T θ1 θ2

17. Write the torque balance equation of an ideal rotational spring. (AP)

K K
T θ T θ1 θ2

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18. Write the analogous electrical elements in force-voltage analogy for the elements of
mechanical translational system. (E)
Mechanical translational System Electrical voltage System
 Force  Voltage
 Velocity  Current
 Displacement  Charge
 Frictional coefficient  Resistance
 Mass  Inductance
 Stiffness constant  Inverse of capacitance
 Newton’s second law  Kirchhoff’s voltage law

19. Write the analogous electrical elements in force-current analogy for the elements of
mechanical translational system. (E)
Mechanical translational System Electrical current System
 Force  Current
 Velocity  Voltage
 Displacement  Flux
 Frictional coefficient  Conductance
 Mass  Capacitance
 Stiffness constant  Inverse of inductance
 Newton’s second law  Kirchhoff’s current law

20. Write the analogous electrical elements in torque-voltage analogy for the elements
of mechanical rotational system. (E)
Mechanical rotational System Electrical voltage System
 Torque  Voltage
 Angular Velocity  Current
 Angular Displacement  Charge
 Frictional coefficient  Resistance
 Moment of inertia  Inductance
 Stiffness constant  Inverse of capacitance
 Newton’s second law  Kirchhoff’s voltage law

21. Write the analogous electrical elements in torque-current analogy for the elements
of mechanical rotational system. (E)
Mechanical rotational System Electrical current System
 Torque  Current
 Angular Velocity  Voltage
 Angular Displacement  Flux
 Frictional coefficient  Conductance
 Moment of inertia  Capacitance
 Stiffness constant  Inverse of inductance
 Newton’s second law  Kirchhoff’s current law

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22. What is block diagram? What are its basic components? (R)
A block diagram of a system is the pictorial representation of the functions performed by
each component of the system and shows the flow of signals. The basic elements of the block
diagram are block, branching point and summing point.
23. Questions on rules for block diagram reduction. (refer to class notes) (R)
24. What is a signal flow graph? (R)
A signal flow graph is a diagram that represents a set of simultaneous linear algebraic
equations. It is used to represent the control system graphically. It is easy to simplify and find
the transfer function of system represented by signal flow graph.
25. Define loop and non-touching loops. (R)
A loop is a closed path in a signal flow graph. Non-touching loops are those loops that don’t
have a common node.

26. Write the Mason’s gain formula. (R)

According to the Mason’s gain formula, the overall gain of the system is given by:

27. What are the basic elements of the thermal system? (R)
The basic elements of the thermal system are thermal resistance and thermal capacitance.
28. Define thermal resistance. (R)
The thermal resistance for heat transfer between two substances is defined as the ratio of
change in temperature and change in heat flow rate.

29. Define thermal capacitance. (R)

Thermal capacitance is defined as the ratio of change in heat stored and change in
temperature.

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30. Mention the electrical analogous of simple thermal system. (R)

The electrical analogous of simple first order thermal system is an R-C parallel circuit.
31. What is a synchro? (R)
A synchro is a device used to convert an angular motion to an electrical signal or vice versa.
It works on the principle of a rotating transformer (induction motor)
32. What is a synchro pair? (R)
A synchro pair is a system formed by the interconnection of the devices synchro transmitter
and synchro control transformer. A synchro pair is used to either transmit an angular motion
from one place to another or employed to produce an error voltage proportional to the
difference between two angular motions.
33. What are the differences between synchro transmitter and control transformer? (E)
Synchro transmitter Control transformer
 Rotor is dumb bell shaped  Rotor is cylindrical in shape
 Rotor winding is excited by an A.C  Induced emf in rotor is used as output
voltage signal (error signal)
34. What is electrical zero of a synchro? (R)
The electrical zero of a synchro transmitter is a position of rotor at which one of the coil to
coil voltages is zero. Any angular motion of rotor is measured with respect to the electrical
zero position of the rotor.
35. What is null position in synchro? (R)
The null position of a synchro control transformer in a servo system is that position of its
rotor for which the output voltage on the rotor winding is zero, with the transmitter in its
electric zero position.
36. What are the applications of synchro? (R)
The applications of synchros are:
 Error detector in position control system
 Conversion of angular displacement to proportional electric signal
 Transmission of angular motion from one place to another
37. What is servomotor? (R)
The motors used in automatic control systems or in servomechanisms are called servomotors.
They are used to convert electric signal to angular motion.
38. Compare AC and DC servomotors. (E)
DC servomotor AC servomotor
 Higher power output  Relatively lesser power output than a
 Linear characteristics DC servomotor of same size.
 Fast response due to low electrical  Nonlinear characteristic
and mechanical time constants  Slow response due to higher values
 Suitable for large power applications of time constants
 Suitable for low power applications.
39. What is the difference between AC servomotor and two phase induction motor?
(May/Jun-12) (E)
The AC servomotor has low value of X/R to achieve linear speed-torque characteristics.
But the conventional induction motors have large values of X/R for higher efficiency.

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The AC servomotor has low inertia rotor compared to the induction motor.

PART B
1. Derive the transfer function of armature controlled DC motor. (Nov/Dec-15) (E) (7)
2. Derive the transfer function of field controlled DC motor. [AP] (7)
3. Write short notes on electrical analogy of mechanical systems. [AP] (7)
4. Derive the transfer function of a simple thermal system. [AP] (7)
5. Construct the block diagram of armature controlled DC motor. [C] (7)
6. Construct the block diagram of field controlled DC motor. [C] (7)
7. What is a synchro? How is it used as synchro transmitter and control transformer?
[AZ] (May/June-16) (7)
8. Write short notes on D.C servomotors. (Nov/Dec-14) [AP] (7)
9. Write short notes on A.C servomotors. (Nov/Dec-14) [AP] (7)
10. Design the transfer function and draw the F-V and F-I analogous circuits for the
mechanical system shown in figure. (Nov/Dec-14) [C] (7)
X1 X2
K1 K2

f(t)
M1 M2
B12

B1 B2

11. Simplify the block diagram and obtain the shown in figure(May/June-16)(AZ)(13)

H2

-
R(s) + + + C(s)
G1 G2 G3

- - +

H1

G4

12. Solve the mathematical model of mechanical system shown in figure.

May/June-16,May/June- 17 [AP] , Apr/May-18 (13)

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f1
k1

X1
M1

k2 f2

X2
M2

13. Build the T.F of the electrical network shown in figure [AP] (13)
R1

C1

R2

ei eo

C2

14. Demonstrate mason’s gain formula find the T.F C(s) / R(s) for the SFG shown below.
[AP]MAY/JUN-17
G7
G6

R(s) G1 G2 G3 G4 G5 1 C(s)

-H1

-H2

PART C
1. Derive the transfer function of armature controlled DC motor. Evaluate the close loop
transfer function when input R(s) is at station II. [E] (15)

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2. Draw a signal flow graph and evaluate closed loop transfer function of a system using
Mason’s gain formula. [C] (15)

3. A voltage of V1(s) is applied to a RLC series circuit. A voltage V2(s) is taken across R.
Find the transfer function of the above network. [E] (15)

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UNIT II
Time Response
PART A

1. What is time response? [R]

The time response is the output of the closed loop system as a function of time. It is
denoted by C(t). It is given by inverse Laplace of the product of the input and transfer
function of the system.
The closed loop transfer function,

Response in S domain,

Response in time domain,

2. What is transient and steady state response? [R]

The transient response is the response of the system when the input changes from one state to
another. The response of the system as time tends to infinity is called steady state response.
3. What is the importance of test signals? [R]
 The test signals can be easily generated in test laboratories and the characteristics of
test signals resembles the characteristics of actual input signals.
 They can be used to predetermine the performance of the system.
 Response of the system to test signals can be considered as a measure of suitability
for practical applications.
4. Name the test signals used in control system. [U]
The commonly used test signals in control system are impulse, step, ramp, acceleration and
sinusoidal signals.
5. Define step signal. [R]
The step signal is one whose values changes from 0 to A and remains constant at A for
. Unit step signal is one whose value changes from 0 to 1 and remains constant at 1 for
.
The mathematical representation of step signal is:
A

6. Define ramp signal. [R]

A ramp signal is a signal whose value increases linearly with time from an initial value of
zero at . It is mathematically represented as:
r(t)

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7. Define parabolic signal. [R]

It is a signal in which the instantaneous value varies as square of the time from an initial
value of zero at .
It is mathematically represented as:

8. What is an impulse signal? [R]

A signal which is available for very short duration is called impulse signal. Ideal impulse
signal is a unit impulse signal which is defined as a signal having zero values at all times
except at . At , the magnitude becomes infinity. It is denoted by and is
mathematically expressed as:

9. What is weighting function? [R]

The impulse response of system is called weighting function. It is given by inverse Laplace
transform of system transfer function.
10. Define pole. [R]
The poles of a transfer function are the values of S for which the transfer function becomes
infinity. Poles of a transfer function can be obtained by equating the denominator to zero.
11. Define zero. [R]
The zeros of a transfer function are the values of S for which the transfer function becomes
zero. zeros of a transfer function can be obtained by equating the numerator to zero.
12. What is the order of a system? [R]
The order of the system is given by order of the differential equation governing the system. It
is also given by the maximum power of S in the denominator polynomial of transfer function.
The number of poles of transfer function is the order of system.
13. Define damping ratio. [R]
The damping ratio is defined as the ratio of the actual damping to critical damping.
14. Give the expression for damping ratio of mechanical and electrical system. [AZ]
The damping ratio of second order mechanical translational system,

The damping ratio of second order mechanical rotational system,

The damping ratio of second order electrical system,

15. Explain the classification of system based on value of damping. [U]

Depending on the value of damping, system can be classified as:
 Un-damped system,
 Under-damped system,
 Critically damped system,
 Over-damped system,

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16. What will be nature of response of second order system for different kinds of
damping? [U]
The response of second order system for various cases of damping is as follows:
 For un-damped system, the response is oscillatory
 For under-damped system, the response is damped oscillatory
 For critically damped system, the response is exponentially rising
 For over-damped system, response is exponentially rising, but rise time will be large
17. Sketch the response of second order under-damped system. [AP]
c(t)

0.5

0
t
18. What is damped frequency of oscillation? [R]
In under-damped system, the response is damped oscillatory. The frequency of damped
oscillations is given by
19. Give the expressions for natural frequency of oscillations of electrical and
mechanical systems. [R]
 The natural frequency of oscillation of second order mechanical translational

system,

 The natural frequency of oscillation of second order mechanical rotational

system,

 The natural frequency of oscillation of second order electrical system,

20. List the time domain specifications. [R]
The time domain specifications are:
 Delay time
 Rise time
 Peak time
 Maximum peak overshoot and
 Settling time
21. Define delay time. [R]
Delay time is defined as the time taken by the response to reach 50 % of the final value, for
the very first time.

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22. Define rise time. [R]

 For under-damped system, it is the time taken by the response to rise from 0 to 100 %.
 For over-damped system, it is time taken by the response to rise from 10% to 90%.
 For critically damped system, it is time taken by the response to rise from 5% to 95%.
 Rise time is calculated by the formula: , where ,

23. Define peak time. [R]

It is the time taken for the response to reach the peak value or the maximum value for the
very first time. It is given by the expression: .

24. Define peak overshoot. [R]

It is defined as the ratio of the maximum peak value measured from final value to the final
value. Let the final value be and the maximum value is , then the maximum peak

overshoot is given by:

25. Define settling time. [R]
It is defined as the time taken by the response to reach and stay within a specified error and
the error is usually specified as % of final value. The usual tolerable error is 2% to 5% of the
final value. Mathematically settling time is given by:
, for 2% tolerance error

, for 5% tolerance error

26. What is type number of system? What is its significance? [R]

The type number of a system is the number of poles of transfer function at origin. The type
number of the system decides the steady state error.
27. Distinguish between the order and type of system.(Nov/Dec-14) (Nov/Dec-11) [U]
 Type number is specified for loop transfer function only but order can be
specified for any transfer function.
 Type specifies the number of poles of system at origin but order specifies the
total number of poles of system.
28. What is steady state error? [R]
The steady state error is the value of error signal e(t) when t tends to infinity. The steady state
error is an indication of the system accuracy. These errors arise from the nature of inputs,
type of system and from nonlinearity of system components.
29. What are static error constants? [R]
The static error constants are associated with steady state error in a particular type of system
and for a standard input. The three types of static error constants are position, velocity and
acceleration error constants.

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30. Define positional error constant. [R]

The positional error constant is defined as . The steady state error in
type 0 system when the input is unit step is given by .

31. Define velocity error constant. [R]

The velocity error constant is defined as . The steady state error in
type 1 system when the input is unit ramp is given by .
32. Define acceleration error constant. [R]
The acceleration error constant is defined as . The steady state
error in type 2 system when the input is unit parabolic is given by .
33. What are generalized error coefficients? [R]
They are the coefficients of generalized error series. The generalized error series is given by:

The coefficients are called generalized error coefficients or dynamic

error coefficients. The nth coefficient is given by: , where

34. Give the relation between generalized and static error coefficients. [AP]
The following equations give the relation between generalized and static error coefficients:

35. Mention two advantages of generalized error constants over the static error
constants. [R]
 Generalized error series gives error signal as a function of time
 Using generalized error constants, the steady state error for any type of input can
be determined, while the static error constants can be applied only for standard
inputs.
36. What is the effect on system performance when a proportional controller is
introduced in the system? [AP]
 Improves the steady state tracking accuracy
 Improves the disturbance signal rejection
 Improves the relative stability of the system
 Increases the loop gain and hence reduces the sensitivity of system to parameter
variations.
37. What is the drawback of proportional control? [R]
The drawback of proportional control is that it produces a constant steady state error.
(offset)
38. What is the advantage and drawback of integral control action? [R]
The advantage in integral action is that it eliminates the steady state error, but the
drawback is that it can sometimes make the system unstable.

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39. What is reset rate? [R]

The reset rate is the reciprocal of integral time or reset time. The rest rate is the number of
times per minute that the proportional part of the control action is duplicated and is measured
in terms of repeats/minute.
40. What is the effect of PI controller on system performance? [R]
As the PI controller increases the order of system by one, the steady state error reduces, but
the system becomes less stable than the original system.
41. What is the effect of PD controller on the system performance? [R]
The PD controller increases the damping ratio of the system and hence the peak overshoot is
reduced.
42. Why derivative controller is not used separately in control applications? [U]
The derivative controller produces a control action based on the rate of change of error signal
and it does not produce corrective action for any constant error. Hence derivative controller is
not used separately in control applications.
43. Write the transfer functions of P, PI, PD and PID controllers. [R]
The transfer function of P controller is

The transfer function of PI controller is

The transfer function of PD controller is

The transfer function of PID controller is

Where,

44. What is root locus? [R]

The path taken by a root of characteristic equation when open loop gain K is varied from 0 to
is called root locus.
45. How is the gain at a point on root locus determined? [U]
The gain K at a point on root locus is given by:

46. How to determine the root locus on real axis? [U]

To find the root locus on real axis, choose a test point on real axis. If the total number of
poles and zeros on the real axis to the right of this point is odd number then the test point lies
on the root locus. If it is even then the test point does not lie on the root locus.
47. What are asymptotes? How to find the angle of asymptotes? [R]
Asymptotes are straight lines which are parallel to root locus going to infinity and meet the
root locus at infinity. The angle of asymptotes is given by:

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48. What is centroid? How is it calculated? [R]

The meeting point of asymptotes with real axis is called centroid. The centroid is given by:

49. What is breakaway point and break-in point? How to determine them? [R]
 At breakaway point, the root locus breaks from the real axis to enter into the complex
plane.
 At break-in point the root locus enters the real axis from the complex plane.
 To determine the break away or break-in point, form an equation for gain K from the
characteristic equation and differentiate the equation of K w.r.to S. Then find the roots
of equation dK/dS = 0. The roots are the break away or break-in points, provided for
this value of root, the gain K should be positive and real.
50. Sketch the step response of P and PI controllers. [AP]
C(t) PI control action

Proportional control action

51. Sketch the ramp responses of P, PD and PID controllers. [AP]

C(t)
PID control action

PD control action
Proportional control action

52. How to find the crossing point of root locus in the imaginary axis? [U]
The crossing point can be determined by two methods:
 By Routh-Hurwitz criterion
 By letting in the characteristic equation, separate the real and imaginary parts.
These two equations are equated to zero. Solve the two equations for . The value of
gives the point where the root locus crosses imaginary axis and the value of K is the gain
corresponding to the crossing point.

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53. What is dominant pole? [R]

The dominant pole is a pair of complex conjugate pole which decides transient response of
the system. In higher order systems, the dominant poles are very close to origin and all other
poles of the system are widely separated and so they have less effect on transient response of
the system.
54. How is the root locus modified when a zero is added to open loop transfer function?
Or What is the effect of addition of zero to open loop transfer function on the
system? [AP]
The addition of a zero to open loop transfer function will pull the root locus to the left, which
makes the system more stable and also leads to the reduction of settling time.
55. How is the root locus modified when a pole is added to open loop transfer function?
Or What is the effect of addition of pole to open loop transfer function on the
system? [AP]
The addition of a pole to open loop transfer function will pull the root locus to the right,
which reduces the stability of the system and increases the settling time.

PART B
1. Determine the response of first order system for unit step input. (AZ) (8)
2. Determine the response of second order system for unit step input.
(AZ)(Nov/Dec-15) (Nov/Dec-14)(May/June-17)(13)
3. Derive the time domain specifications. (May/June-16) (AZ) (13)
4. The system shown in figure below uses a rate feedback controller. Determine the
tachometer constant Kt so as to obtain the damping ratio as 0.5. Calculate the
corresponding tp,ts,Mp and d . (AP)(13)

5. The open loop transfer function of a system with unity feedback gain is given as
G(s) = . Sketch the root locus. (AP) (8)

6. For a system with G(s) = , Calculate the generalized error co-efficients and the steady
state error. Assume r(t) = 6 +5t. May/June-17 (AP) (8)
7. A single loop negative feedback system has open loop transfer function
G(s) = . Sketch root locus. May/June-17 (AP) (13)

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8. Explain briefly the PID controller action with block diagram and obtain its transfer
function model (Nov/Dec-15) (May/Jun-12) (R)
9. Explain the rules to construct root locus of a system. (Nov/Dec-15) (R)
10. Determine steady state errors of type 0, type 1, type 2 systems for unit step, unit ramp
and unit parabolic inputs. (May/Jun-12) (R)
11. Discuss the effect of P, PI, PD, PID Controllers. (Nov/Dec-11) (R)

PART C

1. For a system whoseG(s)= . Find steady state error when it’s subjected
to the input , r(t) = 1+ 2t + 1.5 t2 (AZ) (15)
2. A closed loop servo system is represented by differential equation
d2c/dt2 + 8 dc/dt = 64 c
Where c is the displacement of output shaft
r is the displacement of input shaft
e=r–c
3. Determine undamped natural frequency, damping ratio and % maximum over shoot for a
unit step input. May/June-17(AZ) (15)
4. For a given system, the time domain specification is as follows.
Tr = 0.6046 s Tp = 0.907 s % Mp = 16.3 Ts = 1.53 ( 5% error)
Find the transfer function of the system. (AZ) (15)
5. For a unity Feedback control system, the open loop transfer function is given by
G(S) = 10(s+2)/s2(S+1). Find (i) Find the position, velocity and acceleration error
coefficient. (ii) Also find the steady state error when the input is R(s) = (3/S)-
(2/S2)+(1/3S2) (May/June-16) (AZ)
6. With a neat diagram explain the effect of PD controller in detail. (May/June-16) (C)
7. Obtain the expression for dynamic error co-efficient of the following system
G(s) = 10/ S(1+S) (Nov/Dec-14) (C)
8. A unity feedback control system has an open loop transfer function,
G(S)=K/S(S2+4S+12). Sketch the Root Locus. Apr/May-18

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UNIT III
Frequency Response
PART A

1. What is frequency response? (R)

The frequency response is a steady state output of the system when the input is sinusoidal
signal.
2. What are the advantages of frequency response analysis? (R)
 The absolute and relative stability of system can be estimated form the knowledge of
open loop frequency response
 The practical testing of system can be easily carried out with available sinusoidal
signal generators
 The transfer function of complicated functions can be determined experimentally by
frequency response tests
 The design and parameter adjustment can be carried out more easily
 The corrective measure for noise disturbance and parameter variation can be easily
carried
 This method can be extended to certain nonlinear systems as well
3. What are frequency domain specifications? (R)
The frequency domain specifications are used to indicate the performance of the system in
frequency domain. Some frequency domain specifications are:
 Resonant peak
 Resonant frequency
 Bandwidth
 Cut-off rate
 Gain margin and
 Phase margin
4. Define resonant peak. (R)
The maximum value of the magnitude of closed loop transfer function is called resonant
peak. It is given by:

5. What is resonant frequency? (R)

The frequency at which the maximum value of magnitude of the closed loop transfer function
occurs is called resonant frequency. It is given by:
6. Define bandwidth. (R)
The bandwidth is the range of frequencies for which the system gain is more than –3dB. It is

given by the expression:

7. What is cut-off rate? (R)
The slope of log-magnitude curve near the cut-off frequency is called cut-off rate.

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8. Define gain margin. (R)

The gain margin is defined as the reciprocal of the magnitude of open loop transfer
function, at phase cross-over frequency, .
Gain margin,

When expressed in decibels, it is given by:

Gain margin in dB

9. Define phase margin. (R)

The phase margin, γ is that amount of additional phase lag at the gain cross-over frequency
required to bring the system to the verge of instability. It is given by , where
is the phase of at the gain cross over frequency.
Phase margin, ,
Where
10. What is phase and gain cross-over frequency? (R)
The gain cross over frequency is the frequency at which the magnitude of the open loop
transfer function is unity.
The phase cross over frequency is the frequency at which the phase of open loop transfer
function is .
11. What is Bode plot? (R)
The Bode plot is a frequency response plot of the transfer function of a system. It consists of
two plots- magnitude plot and phase plot.
12. Define corner frequency. (R)
The magnitude plot can be approximated by asymptotic straight lines. The frequencies
corresponding to the meeting point of asymptotes are called corner frequency. The slope of
the magnitude curve changes at every corner frequency.
13. What are the advantages of Bode plot? (R)
 The magnitudes are expressed in dB and so a simple procedure is available to add
magnitude of each term one by one
 The approximate Bode plot can be quickly sketched and the corrections can be made
at corner frequencies to get the exact plot.
 The frequency domain specifications can be easily obtained
 The Bode plot can be used to analyze both open and closed loop systems
14. What is polar plot? (R)
The polar plot of a sinusoidal transfer function is a plot of the magnitude of
versus the phase angle/argument of on polar or rectangular co-ordinates as is
varied from zero to infinity.
15. What is a minimum phase system? (R)
The minimum phase systems are systems with minimum phase transfer functions. In
minimum phase transfer functions, all poles and zeros will lie on left half of S plane.

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16. What is an all-pass system? (R)

The all pass systems are systems with all pass transfer functions. In all pass transfer
functions, the magnitude is unity at all frequencies and the transfer functions will have anti-
symmetric pole zero pattern (i.e., for every pole in the left half of S plane, there is a zero in
the mirror image position w.r.to the imaginary axis)
17. What is non-minimum phase transfer function? (R)
A transfer function which has one or more zeros in the right half of S plane is known as non-
minimum phase transfer function.
18. What is Nichols plot? (Nov/Dec-11) (R)
The Nichols plot is a frequency response plot of the open loop transfer function of the
system. It is a graph between magnitude of in dB and the phase of in degrees,
plotted on an ordinary graph sheet.
19. How is the start and end of polar plot determined for minimum phase systems?(AP)
The type number of system determines the quadrant in which the polar plot starts, and the
order of the system determines the quadrant in which the polar plot ends.
Type 3

3rd order 4th order

Type 2 Type 0

2nd order
1st order

Type 1

Start of polar plot End of polar plot

20. What are M and N circles? (R)

The magnitude, M of closed loop transfer function with unity feedback will be in the form of
a circle in complex plane for each constant value of M. The family of these circles is called
M circles.
Let , where is the phase of closed loop transfer function with unity feedback.
For each constant value of N, a circle can be drawn in the complex plane. The family of these
circles is called N circles.
21. How is closed loop frequency response determined from open loop frequency
response using the M and N circles? (U)
The locus or the polar plot of the open loop system is sketched on the standard M and
N circles chart. The meeting of M circle with locus gives the magnitude of closed loop
system, the frequency being same as that of open loop system. The meeting point of
locus with N circle gives the value of phase of closed loop system.
22. What is Nichols chart? (R)
The Nichols chart consists of M and N contours superimposed on ordinary graph. Along each
M contour, the magnitude of closed loop system, M will be a constant. Along each N contour,
the phase of closed loop system will be constant. The ordinary graph consists of magnitude in
dB, marked on the y-axis and the phase in degrees marked on x-axis. The Nichols chart is
used to find the closed loop frequency response from the open loop frequency response.

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23. How is closed loop frequency response determined from open loop frequency
response using the Nichols chart? (AP)
The locus or the Nichols plot is sketched on standard Nichols Chart. The meeting
point of M contour with locus gives the magnitude of closed loop system and the
meeting point with N circle gives the argument/phase of the closed loop system.
24. What are the advantages of Nichols Chart? (R)
 It is used to find closed loop frequency response from open loop frequency response.
 The frequency domain specifications can be determined from Nichols chart.
 The gain of the system can be adjusted to satisfy the given specifications.

PART B

1. Derive the frequency domain specifications. [AZ] (13)

2. For the given system, G (s) = , sketch the Bode plot.
(May/June-16) May/June- 17 [AP] (13)
3. For the following system, sketch the polar plot. G(s) H(s) =
May/June-16, May/June-17[AP] (13)
4. Explain the M and N circles in detail. Explain how they are used to determine the closed
loop response from open loop response. [AP] (13)
5. Write short notes on the correlation between time domain and frequency domain
specifications. (Nov/Dec-15) [AZ] (13)

PART C

1. G(s) = K e – 0.2 s, find K so that system is stable with gain margin of 2 dB. [AZ](15)
2. The open loop transfer function of a system is given by G(s) = K/[ s (s2+ s +4) ]
Using polar plot, determine the value of K, so that the phase margin is 50 o. What
is the corresponding value of gain margin. (15)[AZ]

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UNIT IV
Stability and Compensator Design
PART A

1. Define BIBO stability. (R)

A linear relaxed system is said to have BIBO stability if every bounded (finite) input results
in a bounded (finite) output.
2. What is the requirement for BIBO stability? (R)
The requirement for BIBO stability is that , where is the impulse
response of the system.
3. What is the characteristic equation? (R)
The denominator polynomial of c(s)/R(s) is the characteristic equation of the system.
4. How are the roots of characteristic equation related to stability? (AP)
If the roots of characteristic equation has real part, then the response of the system is not
bounded. Hence the system will be unstable. If the roots of the characteristic equation have
negative real parts, then the response is bounded for all bounded input, hence resulting in
stable system.
5. What is the necessary condition for stability? (R)
The necessary condition for stability is that all the coefficients of the characteristic equation
be positive.
6. What is the relation between stability and coefficient of characteristic polynomial?
(R)
If the coefficients of characteristic polynomial are negative or zero, then some of the roots lie
on right half of S plane. Hence the system is unstable. If the coefficients of characteristic
polynomial are positive and if no coefficient is zero then there is a possibility of the system to
be stable provided all the roots are lying on left half of S plane.
7. What will be nature of impulse response when the roots of the characteristic
equation are on the imaginary axis? [AZ]
If the roots of the characteristic equation lie on the imaginary axis, the nature of impulse
response is oscillatory.
8. What will be nature of impulse response when the roots of the characteristic
equation are lying on the right half of S plane? [AZ]
When the roots are on the real axis on the right half of S plane, the response is exponentially
rising. When the roots are complex conjugate and are on the right half of S plane, then the
response is oscillatory with exponentially increasing amplitude.
9. State the principle of argument. [R]
The principle of argument states that “let F(s) be an analytic function and if an arbitrary
closed contour in clockwise direction is chosen in the S plane so that F(s) is analytic at every
point of the contour. Then the corresponding F(S) plane contour mapped in the F(s) plane
will encircle the origin N times in the anticlockwise direction, where N is the difference
between number of poles and zeros of F(S) that are encircled by the chosen closed contour in
the S plane.

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10. What is the necessary and sufficient condition for stability? [R]
The necessary and sufficient condition for stability is that all of the elements in the
first column of the Routh array should be positive.
11. State Routh stability criteria. [R]
Routh criteria states that the necessary and sufficient condition for stability is that all of the
elements in the first column of the Routh array be positive. If this condition is not met, the
system is unstable and the number of sign changes in the elements of the first column of
Routh array corresponds to the number of roots of characteristic equation in the right half of
S plane.
12. What is the need for auxiliary polynomial? [R]
In the construction of Routh array a row of all zeros indicates the existence of an even
polynomial as a factor of the given characteristic equation. In an even polynomial the
exponents of S are even integers or zero only. This even polynomial factor is called auxiliary
polynomial. The coefficients of auxiliary polynomial are given by the elements of the row
just above the row of all zeros.
13. What is limitedly stable system? [R]
For a bounded input signal, if the output has constant amplitude oscillations then the system
may be stable or unstable under some limited constraints. Such a system is called limitedly
stable system.
14. State Nyquist Stability criterion? [R]
If G(S)H(S) contour in the G(S)H(S) plane corresponding to Nyquist contour in S plane
encircles the point in the anticlockwise direction as many times as the number of
right half S plane poles of G(S)H(S), then the closed loop system is stable.
15. What is compensation? [R]
Compensation is the design procedure in which the system behavior is altered to meet the
design specifications, by introducing additional devices called compensators.
16. What is compensator? What are the different types of compensators? [U]
A device inserted into the system for the purpose of satisfying the specifications is called
compensator. The different types of compensator are: lag compensator, lead compensator and
lag-lead compensator.
17. Why is compensation necessary in control system? [AZ]
In feedback control system, compensation is necessary for two reasons:
 When the system is unstable, then compensation is necessary to stabilize the system
and also to meet the desired performance.
 When the system is stable, compensation is required to obtain the desired
performance.
18. What is series compensation? [R]
The series compensation is a design procedure in which a compensator is introduced in series
with plant to alter the system behavior and to provide satisfactory performance.

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19. What is parallel or feedback compensation? [R]

The feedback compensation is a design procedure in which a compensator is introduced in
the feedback path so as to meet the desired specifications. It is also called parallel
compensation.

20. What are the factors to be considered for choosing series or shunt compensation?
The choice between series, shunt compensation depends on the following: [U]
 Nature of signals in the system
 Power levels at various points
 Components available
 Designer’s experience
 Economic considerations
21. What is a lag compensator? Give an example. [U]
A compensator having the characteristics of a lag network is called lag compensator. If a
sinusoidal signal is applied to a lag compensator, then in steady state the output will have a
phase lag w.r.to the input. R-C network shown below is a good example of electrical lag
network.

22. What is a lead compensator? Give an example. [U]

A compensator having the characteristics of a lead network is called lead compensator. If
sinusoidal signal is applied to a lead compensator, then in steady state the output will have a
phase lead w.r.to the input. R-C network shown below is a good example of electrical lead
network.

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23. When is lag/lead/lag-lead compensation employed? [AZ]

 Lag compensation is employed for a stable system for improvement in steady state
performance.
 Lead compensation is employed for stable/unstable system for improvement in
transient state performance.
 Lag-lead compensation is employed for stable/unstable system for improvement in
both steady state and transient state performance.
24. What is a lag-lead compensator? Give an example. [AZ]
A compensator having the characteristics of a lag-lead network is called lag-lead
compensator. If sinusoidal signal is applied to a lag-lead compensator, then in steady state the
output will have both phase lag and phase lead w.r.to the input, but in different frequency
regions. R-C network shown below is a good example of electrical lag-lead network.

25. What is the effect of adding a pole to open loop transfer function of a system?
(Nov/Dec-15) [E]
The addition of pole to open loop transfer function of a system will reduce the steady state
error. The closer the pole to origin, lesser will be the steady state error. Thus the steady state
performance of the system is improved.
Also the addition of pole increases the order of the system, making it less stable than the
original system.

26. What is the effect of adding a zero to open loop transfer function of a system?[AZ]
The addition of zero to the open loop transfer function of a system improves the transient
response. The addition of a zero reduces the rise time. If the zero is introduced close to origin,
then the peak overshoot will be charger. If the zero is introduced far away from the origin in
left half of s plane then the effect of zero on the transient response will be negligible.

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27. Give the transfer function of lag compensator and draw its pole-zero plot. [AZ]
The transfer function of lag compensator is: , where . The

pole of a lag compensator is nearer to origin. The pole-zero plot of lag compensator is

28. What are the characteristics of lag compensation? [R]

The characteristics of lag compensation are:
 Improves steady state performance
 Reduces the bandwidth
 Increases the rise time
 If the pole introduced by compensator is not cancelled by zero, then it increases the
order of system by one.
 If lag compensation is applied to stable system, then steady state performance is
improved
29. Draw the bode plot of lag compensator. [AP]
Let be the sinusoidal transfer function of lag compensator.

30. What is the relation between in a lag compensator? [AZ]

The relation is given by:

31. Give the transfer function of lead compensator and draw its pole-zero plot.
(May/Jun-12) (Apr/May-11) [AP]
The transfer function of lead compensator is: , where . The

zero of a lead compensator is nearer to origin. The pole-zero plot of lead compensator is
shown in fig below:

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32. What are the characteristics of lead compensation? [U]

The characteristics of lead compensation are:
 Increases the bandwidth
 Improves the speed of response
 Reduces the peak overshoot
 If the pole introduced by compensator is not cancelled by zero, then it increases the
order of system by one.
33. Draw the bode plot of lead compensator. [AP]
Let be the sinusoidal transfer function of lead compensator.

34. What is the relation between in a lead compensator? [AZ]

The relation is given by:

35. Give the transfer function of lag-lead compensator and draw its pole-zero plot.[AP]

The transfer function of lag-lead compensator is: ,

Where . The pole-zero plot of lag-lead compensator is shown in fig

below:

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36. What are the characteristics of lag-lead compensation? [U]

Lag-lead compensator has the characteristics of both lag and lead compensators. The
characteristics are:
 Improves the steady state performance
 Decreases the bandwidth
 Improves speed of response
 Reduces peak overshoot
 If the pole introduced by compensator is not cancelled by zero, then it increases the
order of system by one.
 This compensator is applied when improvements in both steady state and transient
response are required
37. Draw the bode plot of lag-lead compensator. [AP]
Let be the sinusoidal transfer function of lag-lead compensator.

PART B
1. Obtain Routh array for the system whose characteristic polynomial equation is S6 +2S5 +
8S4 + 12S3 + 20S2 + 16S + 16 = 0 . Check the stability. May/June-16, 17 [AP](8)
1
2. Construct a Nyquist plot for the system, G(s) = [AP](13)
s(1  s)(1  2s)
3. Explain the design of a lag compensator using Bode plot. (Nov/Dec-15)[C](13)
4. Explain the design of a lead compensator using Bode plot. (Nov/Dec-15[C](13)
5. Explain the design of a lag – lead compensator using Bode plot. (13)
1
6. Consider a unity feedback system , G (s) = , Design a lag compensator so
( s  1)( 0.5s  1)
that phase margin is at least 500 and steady state error to a unit step input is 0.1. [C](13)
1
7. For a system, G(s) = , H(s) =1, Design a cascade lead compensator so that the
s ( s  1)
phase margin is at least 450 and steady state error for a unit ramp input is <0.1.
May/June- 17 [AP][C](13)
1
8. Consider the following system with transfer function G(s) = . Design a
s(1  0.1s)(1  0.2s)
lag-lead compensator such that the phase margin is at least 45 0 at gain cross over
frequency around 10 rad/sec and the velocity error constant Kv is 30. [C](13)

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PART C

1. A feedback system has open loop transfer function, G(s) = K e -5/ [ s (s2+ 5s +9) ].
Determine the maximum value of K for stability of closed loop system. (AP)(15)
2. Find the stability of the given system using Nyquist stability criterion , G(s) = K /
[ s ( s2+ 6 s + 10 ) ]. May/June- 17 [AP] (AP) (15)
3. Consider a unity feedback control system whose transfer function G (s) = K / [ s
(s+2) (s+8) ].Design a lag-lead compensator so that Kv = 80s-1 and dominant
closed loop poles are located at - . [C](15)

UNIT V
State Variable Analysis
PART- A

1. What are the advantages of state space analysis? [U]

 They are applicable for linear, nonlinear, time variant, time invariant and MIMO
(Multiple Input Multiple Output) systems.
 They can be performed with initial conditions.
 The variables used to represent the system can be any variables in the system.
 Using this, the internal state of the system at any time instant can be predicted.
2. What are the drawbacks in transfer function model analysis? [U]
 It is defined under zero initial conditions only.
 It is applicable only to LTI (Linear Time Invariant) systems.
 It is restricted only to SISO (Single Input Single Output) systems
 This analysis cannot provide information regarding the internal state of the system
3. What is state and state variable? [R]
 State is the condition of the system at any time instant ‘t’.

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 State Variables are a set of variables which describe the state of the system at any
time instant.
4. Write the state model of nth order system. [R]
The state model of a system is a combination of state equation and output equation. The state
model of a nth order system with m inputs and p outputs is given by:

5. What is state space? [R]

The set of all possible values which the state vector X(t) can have at any time t forms the
state space of the system.
6. The state model of a LTI system is given by: [AP]

Obtain the expression for transfer function of the system.

Applying Laplace Transform to the state equation with zero initial conditions, we get,

Applying Laplace Transform to the output equation with zero initial conditions, we get,

Substituting (1) in above equation, we get,

7. What is state diagram? [R]

The pictorial representation of the state model of a system is called the state diagram. The
state diagram can be represented either using block diagram or using signal flow graph.
8. Draw the block diagram representation of state model. [C]
U(t X(t Y(t)
B ∫ C

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9. Draw the signal flow graph representation of state model. [C]

10. What are the basic elements used to construct the state diagram? [R]
The basic elements used to construct the state diagram are scalar, adder and integrator.
11. A system is characterized by the differential equation, [AP]

Determine its transfer function.

Taking Laplace Transform on both sides with zero initial conditions, we get,

12. The transfer function of a system is given by . Determine the

differential equation governing the system. [AP]

Given,

Applying inverse Laplace transform on both sides, we get,

13. Write the canonical form of state model of nth order system. [R]
The canonical form of state model is:

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14. What is the advantage and disadvantage in canonical form of state model? [R]
 The advantage of canonical form is that the state equations are independent of each
other.
 The disadvantage is that the canonical variables are not physical variables and so they
are not available for measurement and control.
15. What is state transition matrix and how is it related to state of a system? [R]
The matrix exponential is called state transition matrix. It is used to find the state of the
system at any time instant t from the knowledge of the state at time t0.
When the input is zero,
16. Write the properties of state transition matrix. [U]



17. Write the solution of homogeneous state equations. [R]
The solution of homogeneous state equation is:

Where,

18. Write the solution of non – homogeneous state equations. [R]

The solution of non – homogeneous state equation is:

Where,

19. What is resolvant matrix? [R]

The Laplace transform of state transition matrix is called resolvant matrix.
It is given by

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20. Define controllability. [R]

A system is said to be completely state controllable if it is possible to transfer the system
from any initial state to any other desired state in a finite time by a control vector
U(t).
21. What is the need for controllability test? [R]
The controllability test is necessary to find the usefulness of a state variable. If the state
variables are controllable, then by controlling the state variables, the desired outputs of the
system are achieved.
22. State the condition for controllability by Kalman’s method. [U]
th
For a n order system described by the state equation , the composite matrix
is given by
The system is completely state controllable if the rank of composite matrix is n.
Or in other words if , then the
system is completely state controllable.

23. What is the advantage and disadvantage in Kalman’s test of controllability? [R]
 The advantage is that the calculations are simpler.
 The disadvantage is that we can’t find the state variable which is uncontrollable.
24. Define observability. [R]
A system is said to be completely observable if every state X(t) can be completely identified
by measurements of the output Y(t) over a finite time interval.
25. What is the need for observability test? [U]
The observability test is necessary to find whether the state variables are measurable or not.
If the state variables are measurable, then the state of the system can be determined by
practical measurements of the state variables.
26. State the condition for observability by Kalman’s method. [U]
For a nth order system described by the state equation
, the composite matrix is given by

The system is completely observable if the rank of composite matrix is n.

Or in other words if,
, then the system is
completely observable.

27. What is the advantage and disadvantage in Kalman’s test of observability? [U]
 The advantage is that the calculations are simpler.
 The disadvantage is that the non – observable state variables cannot be determined.

28. State the duality between controllability and observability. [U]

The principle of duality states that a system is completely state controllable if and only if it’s
dual is completely observable or vice versa (i.e. if the system is observable then its dual is
controllable)

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PART B
1. Find the state model of electrical networks. [AP] (8)
R1

C1

R2

ei eo

C2

2. Find the state model of the given mechanical system. [AP] (8)

X1 X2
K1 K2

f(t)
M1 M2
B12

B1 B2

3. Determine the state model of armature controlled DC motor. [AZ] (8)

4. Determine the state model of field controlled DC motor. [AZ] (8)
5. Find the state model for the system whose transfer function is =

May/June- 17 [AP]
At
6. Find the solution of state equation, e [AP] (8)
0 1
A=  
 2  3
7. Consider a system with state space model given below
0 1 0 0
X=  0 0 1  x + 4 0 u ; y = [ 2 -4 0] x + ( 0 ) u
 
 1  2  3 4
8. Verify that the system is observable and controllable. (May/June-16) [AZ] (13)
9. Derive the transfer functions of the lag-lead compensator. (Nov/Dec-15) [AZ]
10. With a neat block diagram, derive the state model and its equations of a Linear
multiinput-multi-output system. (Nov/Dec-15) [AZ]
11. Explain the frequency response of lead compensator and write step to design a lead
compensator. (May/June-12) (Nov/Dec-11) [AZ]

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11. With a neat block diagram, derive the state model and its equations of a Linear
multiinput- multi-output system. (May/June-16) [AZ]
12. Explain about the effect of state feedback. (Nov/Dec-15) [AZ]

PART C
d 2x dx
2 dy
1. A system described by the following differential equation dx + 3 +2 y = x(t) is
initially at rest. For input x(t) = 2 u(t). Find the output. May/June-17[AP] (13)
2. Find the state model of the given electrical system [C] (13)

3. Consider a linear system described by the following transfer function, Y(S)/U(S) =

10/S(S+1)(S+2). Design a feedback controller with a state feedback so that the closed
loop poles are placed at -2, 1+j1,1-j1. Apr/May-18 (C)
4. Write the state equation for system shown in the figure, in which x1, x2 & x3 constitute
the state vectors. Determine whether the system is completely controllable and
observable

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EC8395 - COMMUNICATION ENGINEERING

COURSE OUTCOMES

Course
Statement
Outcome

CO1 Apply the Amplitude and Frequency modulation in Analog communication.

CO2 Apply the pulse modulation schemes in digital communication.

CO3 Apply the shift keying techniques in digital communication.

Analyze the coding efficiency using Shannon-fano and Huffman theorem for
CO4
data compression.

CO5 Analyze the errors using cyclic, convolution and viterbi error control codes.

Explain about the new modulation technique such as spread spectrum

CO6
modulation and multiple access techniques.

CO - PO MAPPING

PO PO PO PO PO PO PO PO PO PO1 PO1 PSO PSO PSO

CO PO12
1 2 3 4 5 6 7 8 9 0 1 1 2 3

CO1 3 3 3 3

CO2 3 1 3 1

CO3 3 2 3 2

CO4 3 3 3 3

CO5 3 3 1 3 3

CO6 3 1 3 1

AVG 3 2 1 3 2

3 – Substantially 2 – Moderate 1 - Slightly

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UNIT 1

Analog Modulation

PART – A
1. Define amplitude Modulation. [R]
Amplitude Modulation is the process of changing the amplitude of a high frequency
carrier signal in proportion with the instantaneous value of the modulating signal.
2. Define Modulation index and percent modulation for an AM wave. [R]
Modulation index is a term used to describe the amount of amplitude change present
in an AM waveform .It is also called as coefficient of modulation. Mathematically
modulation index ism = Em/ Ec
Where m = Modulation coefficient
Em = Peak change in the amplitude of the output waveform voltage.
Ec = Peak amplitude of the un modulated carrier voltage.
Percent modulation gives the percentage change in the amplitude of the output wave when
the carrier is acted on by a modulating signal.
3. Draw the frequency spectrum of AM. [May/June 2016][R]

4. Compare AM and FM [Nov/Dec 2017] [R]

Amplitude Modulation (AM) Frequency Modulation (FM)

The radio wave is called as a carrier wave The radio wave is called a carrier wave,
and the frequency and phase remain same but the amplitude and phase remain same.
Has poor sound quality, but can transmit Has higher bandwidth with better sound
longer distance. quality
The frequency range of AM radio varies The frequency range of FM is 88 to 108
from 535 to 1705 KHz MHz in higher spectrum
More susceptible to noise Less susceptible to noise.

6. Distinguish between low level and high level modulation. [AZ]

In low level modulation, modulation takes place prior to the output element of the
final stage of the transmitter. It requires less power to achieve a high percentage of
modulation.
In high level modulators, the modulation takes place in the final element of the final stage
where the carrier signal is at its maximum amplitude and thus ,requires a much higher
amplitude modulating signal to achieve a reasonable percent modulation.

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7. Define image frequency. [R]

An image frequency is any frequency other than the selected radio frequency carrier
that , if allowed to enter a receiver and mix with the local oscillator ,will produce a cross
product frequency that is equal to the intermediate frequency.
8. Define image frequency rejection ratio. [R]
The image frequency rejection ratio is the measure of the ability of preselector to
reject the image frequency.
9. Define Heterodyning. [R]
Heterodyne means to mix two frequencies together in a nonlinear device or to
translate one frequency to another using nonlinear mixing.
10. What are the disadvantages of conventional (or) double side band full carrier
system? [R]
In conventional AM, carrier power constitutes two thirds or more of the total
transmitted power. This is a major drawback because the carrier contains no information;
the sidebands contain the information . Second, conventional AM systems utilize twice as
much bandwidth as needed with single sideband systems.
11. Define Single sideband suppressed carrier AM. [R]
AM Single sideband suppressed carrier is a form of amplitude modulation in which
the carrier is totally suppressed and one of the sidebands removed.
12. Define AM Vestigial sideband. [R]
AM vestigial sideband is a form of amplitude modulation in which the carrier and
one complete sideband are transmitted, but only part of the second sideband is transmitted.
13. What are the advantages of single sideband transmission? [May/June 2016] [R]
The advantages of SSBSC are
 Power conservation: Normally, with single side band transmission, only one
sideband is transmitted and the carrier is suppressed. So less power is required
to produce essentially the same quality signal.
 Noise reduction: Because a single side band system utilizes half as much
bandwidth as conventional AM, the thermal noise power is reduced to half that
of a double side band system.
14. What are the disadvantages of single side band transmission? [May/June 2016] [R]
1. Complex r e c e i v e r s : Single s i d e b a n d s y s t e m s r e q u i r e m o r e c o m p l e x
a n d expensive receivers than conventional AM transmission.
2. Tuning Difficulties: Single side band receivers require more complex and precise
tuning than conventional AM receivers.
15. Define direct frequency modulation. [R]
In direct frequency modulation , frequency of a constant amplitude carrier signal is
directly proportional to the amplitude of the modulating signal at a rate equal to the
frequency of the modulating signal.
16. Define indirect frequency Modulation. [R]
In indirect frequency modulation ,phase of a constant amplitude carrier directly
proportional to the amplitude of the modulating signal at a rate equal to the frequency of the
modulating signal.

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17. Define instantaneous frequency deviation. [R]

The instantaneous frequency deviation is the instantaneous change in the frequency
of the carrier and is defined as the first derivative of the instantaneous phase deviation.
18. Define frequency deviation. [R]
Frequency deviation is the change in frequency that occurs in the carrier when it is
acted on by a modulating signal frequency. Frequency deviation is typically given as a peak
frequency shift in Hertz. The peak to peak frequency deviation is sometimes called
carrier swing. The peak frequency deviation is simply the product of the deviation
sensitivity and the peak modulating signal voltage.
19. State Carson’s rule. [R]
Carson rule states that the bandwidth required to transmit an angle modulated wave
as twice the sum of the peak frequency deviation and the highest modulating signal
frequency. Mathematically Carson’s rule is B=2(∆f+fm) Hz.
20. Define Deviation ratio. [R]
Deviation ratio is the worst case modulation index and is equal to the maximum
peak frequency deviation divided by the maximum modulating signal frequency.
Mathematically ,the deviation ratio is
DR= ∆f (max) fm(max)
21. Define Quantization error. [May/Jun 2018] [R]
The difference between an input value and its quantized value (such as round-off error)
is referred to as quantization error. A device or algorithmic function that performs
quantization is called a quantizer.
22. Compare Narrow band FM and Wideband FM. [May/Jun 2018] [AZ]
NBFM WBFM
Maximum modulating frequency is Maximum modulating frequency is
3Khz and Maximum frequency 30Khz to 50 Khz and Maximum
deviation is 75Khz frequency deviation is 75Khz
Modulation index is less than 1 Modulation index is greater than 1
Bandwidth is 2fm Bandwidth 15 times NBFM
Less suppressing of noise Noise is more suppressed
Use: Entertainment and broadcasting Use: Mobile communication

PART-B
1. Write the methods to generate AM waves. [May/Jun 2016] [U]
2. Discuss in details about the working of a SSB transmitting and receiver.
[May/Jun 2018] [U]
3. Describe an expression to show that for every modulating frequency component, the
amplitude modulated wave contains,2 sideband frequencies in addition to the carrier and
also that the amplitude of sideband components is equal to mVc/2 where m and Vc are
modulation Index and amplitude of unmodulated carrier. Draw the spectrum of AM for a
signal frequency component. [May/Jun 2017][AP]
4. What are the advantages of producing FM from PM over direct FM [May/Jun17/18] [R]

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5. Determine an expression to illustrate that when a modulating signal with only one
frequency component to frequency modulates a carrier. Also show the resulting
FM signal has got infinite sideband frequencies. [AZ]
6. Explain demodulation of AM using envelope detector. [U]
7. Determine an expression of AM wave and its power relations. [May/Jun 17][AP]
8. Explain modulation and demodulation of AM wave. [U]
9. Explain the indirect method of FM generation. [Nov/Dec ’17;May/Jun’16,17] [U]
10. Draw the block diagram for generation and demodulation of a VSB signal and explain the
principle of operation. [U]
11. Explain in details about FM modulation. [May/Jun 2018] [U]
12. Compare Narrow band FM and Wideband FM. [May/Jun 2015/2016][AZ]

UNIT 2
Pulse Modulation
PART – A

1. Statement of sampling theorem. [R]

A band limited signal of finite energy, which has no frequency components higher
than W hertz, is completely decribed by specifying the values of the signal at instants of
time separated by 1/2W
A band limited signal of finite energy, which has no frequency components higher than
W hertz, may be completely recoverd from the knowledge of its samples taken at the rate
of 2W samples per second
2. List out any two advantages and limitations of DPCM [R]
 Advantages:
 DPCM requires less bandwidth compared to PCM
 Its signal to noise ratio is better than DM and PCM
 Limitations:
 Implementation of DPCM is complex compared to PCM
 Slope overload distortion and quantization noise is present in DPCM
 DPCM requires high sampling frequency

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3. What are the advantages of digital transmission? [R]

 Digital signals are better suited to processing and multiplexing than analog
signals.
 Digital transmission systems are more noise resistant than the analog
transmission systems.
 Digital systems are better suited to evaluate error performance.
4. What are the disadvantages of digital transmission? [R]
The t r a n s m i s s i o n o f d i g i t a l l y e n c o d e d analog signals requires
significantly more bandwidth than simply transmitting the original analog signal.
Analog signal must be converted to digital codes prior to transmission and converted
back to analog form at the receiver, thus necessitating additional encoding and
decoding circuitry.
5. Define pulse code modulation. [R]
In pulse code modulation, analog signal is sampled and converted to fixed length,
serial binary number for transmission. The binary number varies according to the
amplitude of the analog signal.
6. What is the purpose of the sample and hold circuit? [R]
The sample and hold circuit periodically samples the analog input signal and converts
those samples to a multilevel PAM signal.
7. What is the Nyquist sampling rate and Nyquist interval. [R]
Nyquist sampling rate states that, the minimum sampling rate is equal to twice the
highest audio input frequency.
Nyquist interval: It is the time interval between any two adjacent samples when sampling
rate is Nyquist rate.
Nyquist rate: 2W hertz
Nyquist interval: 1/2W seconds
8. Define aliasing and list out different ways to avoid aliasing? [May/Jun2018] [R]
When the high frequency interferes with low frequency and appears as low frequency,
then the phenomenon is called aliasing.
Alaising can be avoided by two methods
 Sampling rate fs greater than equal to 2W.
 Strictly band limit the signal to W.
9. Define quantization. [R]
Quantization is a process of approximation or rounding off. Assigning PCM codes
to absolute magnitudes is called quantizing.
10. Define companding. [R]
Companding is the process of compressing, then expanding. With companded
systems, the higher amplitude analog signals are compressed prior to transmission, then
expanded at the receiver.
11. Define slope overload. How it is reduced. [R]
The slope of the analog signal is greater than the delta modulator can maintain, and
is called slope overload. Slope overload is reduced by increasing the clock frequency
and by increasing the magnitude of the minimum step size.

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12. Define granular noise. How it is reduced. [R]

When the original input signal has relatively constant amplitude, the reconstructed
signal has variations that were not present in the original signal. This is called
granular noise. Granular noise can be reduced by decreasing the step size.
13. Define adaptive delta modulation. [R]
Adaptive delta modulation is a delta modulation system where the step size of the AC
is automatically varied depending on the amplitude characteristics of the analog input
signal.
14. In what situation multiplexing is used ? [U]
Multiplexing is used in situations where the transmitting media is having bandwidth, but
the signals have lower bandwidth. Hence there is a possibility os sending number of
signals simultaneously. In this situation Multiplexing can be used to achieve the
following goals:
 To send large number of signals simultaneously.
 To reduce the cost of transmission
 To make effective use of the available bandwidth.
15. Distinguish between the two basic multiplexing techniques? [AZ]
TDM FDM
Fixed time slot is allotted to each Fixed frequency band is allotted to each
source source
Full bandwidth of the channel is Frequency band is available continuously
available to the source without interruption
Suitable for pulse modulation and Suitable for analog modulation and CW
digital signals signals
Time of the channel is utilized Bandwidth of the channel is utilized
effectively effectively

16. Why guard bands are used in FDM? [U]

In FDM , a number of signals are sent simultaneously on the same medium by allocating
separate frequency band or channel to each signal. Guard bands are used to avoid
interference between two successive channels.
17. Why sync pulse is required in TDM? [U]
In TDM, in each frame time slots are pre assigned and are fixed for each input sources.
In order to identify the beginning of each frame, a sync pulse is added at the beginning of
every frame.

PART-B

1. Write a note on Delta modulation. [Nov/Dec ’17;May/Jun’16,17] [R]

2. Describe in details about the operation of PSK and MSK with neat diagram.
[May/Jun2018] [U]
3. Write a detail note on PWM,PAM,PPM. [May/Jun 207][R]
4. Determine an expression for average output noise power in delta modulation. [AP]
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5. Explain DPCM with neat block diagram. [R]

6. Explain DPCM. How DPCM is better than PCM, justify [AZ]
7. Briefly explain the different step size control in ADM. [May/Jun 2017] [U]
8. Discuss the performance of DM, ADM, DPCM, PCM. [U]

UNIT – 3
Digital Modulation and Transmission
PART – A

1. Define bit rate. [R]

In digital modulation, the rate of change at the input to the modulator is called
the bit rate (fb) and has the unit of bits per second (bps).
2. Define Baud rate [R]
The rate of change at the output of the modulator is called baudrate, that means
number of bytes can be transmitted per second.
3. Define QAM. s [R]
Quadrature amplitude modulation is a form of digital modulation where the digital
information is contained in both the amplitude and phase of the transmitted carrier.
4. Define peak frequency deviation for FSK. [R]
Peak frequency deviation (∆f) is the difference between the carrier rest frequency
and either the mark or space frequency and either the mark or space frequency.
(∆f)=│fm-fs│
2
5. What is meant by Frequency Shift Keying (FSK)? [R]
Frequency Shift Keying is the relatively simple, low performance type of digital
modulation. Binary FSK is a form of constant amplitude angle modulation similar to
conventional frequency modulation except that the modulating signal is a binary signal
that varies between two discrete voltage levels rather than a continuously changing analog
waveform.
6. What do you meant by M-ary encoding? [R]
M-ary is a term derived from the word binary. M is simply a digit that represents
the number of conditions or combinations possible for a given number of binary variables.
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7. Analyze, why FSK is referred over ASK. [May/Jun2018][AZ]

Parameters ASK FSK

Variable Amplitude Frequency
characteristics
Noise immunity low High
Error probability High Low
Performance in Poor Better than ASK
presence of noise
Bit rate Suitable upto 100 bits/sec Suitable upto about 1200
bits/sec

FSK is less susceptible to errors than ASK – receiver looks for specific frequency
changes over a number of intervals, so voltage (noise) spikes can be ignored and also
popular in modern systems.
8. What does QPSK mean? [R]
Quaternary Phase Shift Keying (QPSK), or Quadrature PSK as it is sometimes
called, is another form of angle modulated, constant amplitude digital modulation. QPSK
is an M-ary encoding technique where M=4.
9. Define Bandwidth efficiency. [R]
It is defined as the ratio of the transmission bit rate to the minimum bandwidth
required for a particular modulation scheme. BW efficiency = transmission rate (bps) /
minimum BW (Hz) bits/cycle.
10. What is meant by DPSK? [R]
Differential Phase Shift Keying (DPSK) is an alternative form of digital
modulation where the binary input information is contained in the difference between
two successive signaling elements rather than the absolute phase.
11. What is meant by Probability of error & Bit Error Rate? [R]
Probability of error P (e) & Bit Error Rate (BER) are often used interchangeably,
although in practice they do have slightly different meanings. P (e) is a theoretical
expectation of the bit error rate for a given system. BER is an empirical record of a
systems actual bit error performance.
12. Compare the bandwidth efficiency of BPSK and QPSK modulated signals. [AZ]
The bandwidth efficiency of BPSK is 1 bit per cycle, where as that of QPSK is 2
bits per cycle. The bandwidth efficiency of QPSK is more because it encodes the signal
with 4 different phase shifts. Therefore it combines two successive bits.
13. Define QPSK. [R]
QPSK is Quadrature phase –shift keying. In QPSK the phase of the carrier takes
on one of the four equally spaced values Such as π/4, 3π/4, 5π/4 and 7π/4.
14. State any two advantages of MSK. [Apr/May 2017] [U]
 The sidebands of PSK modulated spectrum is minimized by this modulation
technique. Hence sideband power is reduced.
 The MSK or GMSK spectrum is less affected by noise and hence leads to good SNR.
This helps in achieving very stable and long distance communication.

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Due to this fact, the GMSK modulation technique is being employed in GSM
technology.
15. Draw Constellation diagram of QPSK [Nov/Dec 2017] [R]

16. What are the three broad types of synchronization? [R]

 Carrier synchronization
 Symbol & Bit synchronization
 Frame synchronization.
17. What is carrier synchronization? [R]
The carrier synchronization is required in coherent detection methods to generate
a coherent reference at the receiver. In this method the data bearing signal is modulated
on the carrier in such a way that the power spectrum of the modulated carrier signal
contains a discrete component at the carrier frequency.

PART – B
1. Determine the expression for QPSK. [May/Jun’16, 17][AP]

2. Compare MSK and QPSK. What is the bandwidth requirement of MSK? [AZ]
3. Compare BFSK and BPSK. [Nov/Dec 2017][AZ]
4. Discuss the details about GMSK with neat diagram. [May/Jun2018] [U]
5. Explain BPSK system (transmitter and receiver). [U]
6. With neat diagram explain FSK transmitter and FSK receiver. [U]

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UNIT – 4
Information Theory and Coding
PART – A
1. Define data communication codes. [R]
Data communication codes are prescribed bit sequences used for encoding
characters and symbols.
2. Define error detection. [R]
Error detection is simply the process of monitoring the received data and
determining when a transmission has occurred.
3. What is vertical redundancy checking? [R]
A group of messages are transmitted as a matrix. Each row of the matrix represent
one message. One additional row is appended to this matrix. The elements of this
additional row are selected such that number of 1’s in each column are either even or
odd. It is called vertical redundancy check or parity row. At the receiver, the elements of
this purity row are checked to detect the errors in received message matrix.
4. Define entropy. [R]
The average information per message is called entropy. It is represented in bits
per message. It is mathematically given as
Entropy, M
H= ∑ pk log2[1/pk]
K=1
5. What is meant by line coding? [R]
Line coding consists of representing the digital signal to be transported, by
a waveform that is optimally tuned for the specific properties of the physical channel
(and of the receiving equipment). The pattern of voltage, current or photons used to
represent the digital data on a transmission link is called line encoding. The common
types of line encoding are unipolar, polar, bipolar, and Manchester encoding.
6. What is information rate? [R]
The average number of bits of information per second is called information rate. It is
given as, R= rH
Here R is the information rate
R is the rate at which messages are generated.
H is the entropy or average information
7. Explain the significance of AMI code [U]
In this format successive 1’s are represented by pulses with alternate polarity and 0’s
are represented by no pulses Because of alternate polarity of pulses, AMI coded
waveform have no DC component. The ambiguities due to transmission sign inversion
are eliminated
8. State the difference between source coding and error control coding. [AZ]
Error control code: Encoder calculate extra control bits from the information that we
wish to transmit, and to transmit those control bits together with the information.

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Source code: The statement that the output of any information source
having entropy H units per symbol can be encoded into an alphabet having N symbols in
such a way that the source symbols are represented by code words having a weighted
average length not less than H/logN .
9. List the type of characters used in data communication mode. [May/Jun2018][R]
 ASCII : this is ANSI’s 7-bit American Standard Code for Information
Interchange
 EBCDIC : this is IBM’s 8-bit Extended Binary Coded Decimal Interchange
Code.
10. Give the significance of AMI code. [May/Jun2018][U]
 Eliminates DC build up on cable.
 Reduces bandwidth compared to polar.
 Provides error detecting; every bit error results in bipolar violation
 Guarantees transitions for timing recovery with long runs of ones.
11. What is mean by channel capacity theorem. [R]
The channel capacity of the discrete memory less channel is given as maximum average
mutual information. The maximization is taken with respect to input probabilities P(xi)
C = B log2(1+S/N) bits/sec
Here B is channel bandwidth.
12. What is meant by linear code? [R]
A code is linear if modulo-2 sum of any two code vectors produces another code
vector. This means any code vector can be expressed as linear combination of other code
vectors.
13. What are the error detection and correction capabilities of Hamming codes? [R]
The minimum distance (dmin) of Hamming codes is „3‟. Hence it can be used to
detect double errors or correct single errors. Hamming codes are basically linear block
codes with dmin = 3
14. What is meant by cyclic code? [R]
Cyclic codes are the subclass of linear block codes. They have the properly that a
cyclic shift of one codeword produces another code word. For example consider the
codeword.
X = (xn-1,xn-2,……x1,x0)
Let us shift above code vector to left cyclically,
X‟ = (xn-2,xn-3,…… x0, x1,xn-1) Above code vector is also a valid code vector.
15. State any four properties of entropy. [R]
 Entropy is zero if the event is sure or it is impossible
H = 0 if pk = 0 or 1
 When pk = 1/M for all the „M‟ symbols, then the symbols are equally likely for
such source entropy is given as H = log2M
 Upper bound on entropy is given as,
Hmax = log2M

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16. Define bandwidth efficiency. [R]

The ratio of channel capacity to bandwidth is called bandwidth efficiency.
i.e, Bandwidth efficiency = Channel capacity (C ) /Bandwidth (B)
17. Find the Hamming distance between the following code words C1 = (1000111) and
C0 = (0001011). [AP]
Hamming distance d(C1, C2) between a pair of code vector is defined as the number of
locations in which their respective elements differ.
C1 =1 0 0 0 1 1 1
C2 =0 0 0 1 0 1 1
The Hamming distance is 3
18. List out the source coding techniques. [U]
The various source coding techniques are
 Prefix coding
 Shannon-fano coding
 Huffman coding
19. What is convolution code? [R]
In convolution code the encoder accepts the message bits in serially rather than in larger
blocks and generates n-bit code word. The resultant bits are generated using modulo-2
addition and hence called convolution code.
20. Differentiate block code from convolution code. [AZ]
Block codes Convolutions code
The encoder accepts a k-bit The encoder accepts the message bits in
message block and generates a n-bit serially rather than in larger blocks and
code word. Thus, code words are generates n-bit code word. The resultant
produced on a block by block basis. bits are generated using modulo-2
addition and hence called convolution
code.

21. What is viterbi algorithm? [R]

Viterbi decoding scheme performs maximum likelihood decoding and it reduces the
computational load by taking advantage in code trellis. Decoding is done with
algorithm.
Metric: Metric is the discrepancy between the received and the decoding signal at
particular node.
Survivor path: Survivor path is the path of decoded signal with minimum metric.

PART - B
1. Analyze Source coding theorem, give the advantage and disadvantage of channel
coding in detail. [AZ]
2. Explain in detail Huffman coding algorithm and compare this with the other types of
coding. [Nov/Dec 2017] [AZ]
3. Discover the properties of entropy and with suitable example, explain the entropy of
binary memory less source. [AZ]
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4. What is entropy? Explain the important properties of entropy. [R]

5. What do you mean by binary symmetric channel? Derive channel capacity formula
for symmetric channel. [Nov/Dec 2017][AP]
6. A (15,5) cyclic code has a generator polynomial as g(x)=1+x+x2+ x5+ x8+ x10. Draw
the diagram of the encoder for this code. [AP]
7. Describe the concept of source coding theorem with example. [May/June 2018][AP]
8. Discrete memory less source has five symbols m0, m1, m2, m3,m4 with probability
0.4, 0.2, 0.2, 0.1, 0.1 respectively. Construct the code word and entropy using Huffman
coding and analyze the coding efficiency [Nov/Dec’17;May/Jun’16,17][AP]
9. State and prove Shannon’s noiseless coding theorem. [May/Jun2018] [U]
10. Prove the channel capacity theorem. [U]

UNIT -5
Spread Spectrum and Multiple Access
PART – A

1. Classification of spread spectrum systems. [R]

It can be classified as Direct sequence Spread Spectrum and Frequency Hopping
Spread Spectrum.
2. What is meant by wide band system? [R]
1. The main feature of wide band systems is that either complete spectrum is available
(e.g. CDMA, TDMA) or a considerable portion of it is used by each user (e.g.
TDMA+FDMA).
2. The advantage of wideband systems is that the transmission bandwidth always
exceeds the coherence bandwidth for which the signal experiences only selective fading.
That is, only a small fraction of the frequencies composing the
signal is affected by fading.
3. Signal can be distorted and therefore equalization is needed but unlikely a total
signal fade occurs.

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3. What is meant by narrow band system? [R]

1. Channel system: generally total spectrum is divided into a number of relatively narrow
radio channels (e.g. FDMA). If all the channels are being used, call blocking occurs.
Unused bandwidth in each channel cannot be used by other users.
2. Transmission experiences non selective fading. This means that when fading occurs,
whole of the information (i.e. the whole channel) is affected.
4. Write the types of Multiple Access Techniques? [U]
 Frequency Division Multiple Access (FDMA)
 Time Division Multiple Access (FDMA)
 Code Division Multiple Access (FDMA)
 Space Division Multiple Access (FDMA)
They all are grouped as narrow band and wide band system.
5. What is the difference between Multiple Access & Multiplexing? [R]
Both are very different terms.
1. Multiplexing is the process of transmitting several messages simultaneously on the
same circuit or channel. On the other hand Multiple Access are techniques that have been
developed in the satellite industry which allow satellite spectrum and power to be shared
efficiently among multiple users.
2. In multiple access, more than one simple signal can thus be transmitted as part of a
single complex signal and separated out at the receiving end. This is not possible in
multiplexing.
6. What is CDMA? [R]
CDMA is Code Division Multiple Access Technique.
1. For code-division multiple access, each transmitter is assigned a different pseudo -
noise (PN) sequence.
2. If possible, orthogonal sequences should be used.
3. The PN sequence for the transmitter is only given to the receiver that is to operate with
the transmitter.
4. The receiver will then only receive the correct signals and ignore all others.
7. Define spread spectrum techniques [R]
In spread spectrum, the signal occupies a bandwidth in excess of the minimum
necessary to send the information the band spread is accomplished by means of a code
that is independent of the data, and a synchronized reception with the code at the receiver
is used for despreading and subsequent.
8. Define time division multiplexing. [R]
Multiple channels are transmitted over a single channel with help of time division
multiplexing. Each channel is given a fixed time slot. Full bandwidth of the channel is
available for transmission during that slot. Normally TDM works effectively for pulse
modulation. The pulses from various channels are time multiplexed over one period.
9. What is the need for multiplexing? [R]
For pulsed signal, the channel is unused between the two pulses. Hence pulses from
other sources can be transmitted over the same channel to improve utilization of channel.
It is called Time division multiplexing.

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The transmitted signal does not utilize full bandwidth of the channel. Hence total
channel bandwidth is unused. Therefore signals from different sources can
simultaneously occupy the different frequency slots in the channel. It is called frequency
Division multiplexing.
10. Mention the advantages of CDMA system [R]
Maximum utilization of the channel takes place Synchronization is not necessary
11. Mention the application of multiple access technique in wired communication. [R]
 TV transmission
 Broad cast networks
 Local area networks
 Telephone networks
12. Define near – far problem. [May/June 2018] [R]
When the strength of unwanted signal is strong because its source is located close to t
he receiver and strength of the desired signal is weak because its transmitter is located far
from the receiver, the near-far problem arises. Because of this problem the desired signal
is suppressed. Near far problem can be avoided because of orthogonal codes.
13. Give the working principle of TDMA. [May/June 2018][U]
Time division multiple access (TDMA) is a channel access method (CAM) used to
facilitate channel sharing without interference. TDMA allows multiple stations to share
and use the same transmission channel by dividing signals into different time slots. Users
transmit in rapid succession, and each one uses its own time slot. Thus, multiple stations
(like mobiles) may share the same frequency channel but only use part of its capacity.

PART-B
1. Discuss in detail about CDMA techniques and compare the performance with FDMA
and TDMA. [U]
2. With a neat block diagram explain FDMA. Discuss the application in communication
[May/Jun 2017, 2018][AZ]
3. Describe the time division multiple access. [May/June 2018] [U]
4. Explain the application of CDMA in wireless communication
[May/Jun’16;Nov/Dec’17] [U]
5. Compare the performance of CDMA with FDMA and TDMA. [AZ]
6. Discuss about SDMA technique and its applications in wired and wireless
communication. [May/Jun’16,17][AZ]
7. Explain the model of spread spectrum and its applications. [May/Jun16;Nov/Dec17] [U]

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PART – C

1. The total power content of an AM signal is 1000 W. Determine the power being
transmitted at the carrier frequency and each of the bands when the % modulation is
100%. [AZ]
2. A complete modulation waveform consisting of a sine wave of amplitude 3V and
frequency 1 KHz plus a cosine wave of amplitude 5 V and frequency 3 KHz amplitude
modulates a 500 KHz and 10 V peak carrier voltages. Plot the spectrum of modulated
wave and determine the average power when the modulated wave is fed into 50Ω load.
[AP]
3. The carrier frequency of a broadcast signal is 100 MHz The maximum frequency
deviation is 75KHz. If the higher audio frequency modulating the carrier is limited to
15KHz. What is the approximate bandwidth? [AP]
4. The maximum deviation allowed in an FM broadcast system is 75 KHz. If the
modulating signal is of 10 KHz find the bandwidth of FM signal. Find the bandwidth
when the modulating frequency is doubled. [AZ]
5. Apply the Shannon–fano coding and Huffmann coding procedure for the following
message and also find the efficiency of the coding. [AP]
Symbols A B C D E F G
Probabilities 0.4 0.2 0.12 0.08 0.08 0.08 0.04

6. Supposed that binary PSK is used for transmitting information over an AWGN with a power
1 1
spectral density of N0 =10 -10 W/Hz. The transmitted signal energy is  b  A 2 T , where T
2 2
is the bit interval and A is the signal amplitude. Determine the signal amplitude required to
-6
achieve an error probability of 10 when the data rate is [AP]
a) 10 kbits/s.
b) 100 kbits/s
c) 1 Mbit/s

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7. Describe the FDMA, TDMA, CDMA, SDMA in the application of wireless

communication. [AZ]
8. Consider the two 8-point QAM signal constellations shown in Figure P5.20. The
minimum distance between adjacent points is 2A. Determine the average transmitted
power for each constellation, assuming that the signal points are equally probable. Which
constellation is more power-efficient? [E]
Figure P5.20

9. A speech signal is sampled at a rate of 8 kHz, logarithmically compressed and encoded

into a PCM format using 8 bits per sample. The PCM data is transmitted through an

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AWGN baseband channel via M-level PAM. Determine the bandwidth required for
transmission when a) M=4. b) M=8. c) M=16. [AP]

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EI8451 - ELECTRICAL MACHINES

COURSE OUTCOMES

Course
Statement
Outcome

CO1
Explain the Construction and principles of operations of DC machines
CO2 Describe the Construction and operation of single phase and three phase
Transformers
CO3
Describe the concepts of synchronous machine
CO4
Illustrate the basic operation of three phase Induction machines
CO5
Explain the basic operation of single phase Induction machines
CO6
Analyze the Special Electrical Machines and their applications

CO - PO MAPPING

PO PO PO PO PO PO PO PO PO PO1 PO1 PSO PSO PSO

CO PO12
1 2 3 4 5 6 7 8 9 0 1 1 2 3

CO1
3 3 2 - 1 - - - - 1 - - 3 3 -
CO2
3 3 2 - 1 - - - - - - - 3 3 -
CO3
3 3 2 - 3 - - - - 1 - - 3 3 -
CO4
3 3 1 - 2 - - - - 1 - - 3 3 -
CO5
3 3 1 - 2 - - - - 1 - - 3 3 -
CO6
3 3 2 - - - - - - - - - 3 3 -
AVG 3 3 1.6 - 1.8 - - - - 1 - - 3 3 -

3 – Substantially 2 – Moderate 1 – Slightly

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UNIT I
D.C. Machines
PART - A

1. How will you find the direction of emf using Fleming’s right hand rule? [R]
The thumb, forefinger & middle finger of right hand are held so that these fingers are
mutually perpendicular to each other, then
Forefinger - field Thumb - motion Middle - current.
2. How will you find the direction of force produced using Fleming’s left hand rule? [R]
The thumb, forefinger & middle finger of left hand are held so that these fingers are mutually
perpendicular to each other, then forefinger-field thumb-motion middle-current.
3. Write down the emf equation for d.c.generator? [U]
E=(ФNZ/60)(P/A) Volts.
p--->no. of poles Z--->Total no. of conductors
Ф--->flux per pole N--->speed in rpm.
4. Why the armature core in d.c machines is constructed with laminated steel sheets
instead of solid steel sheets? [A]
Lamination highly reduces the eddy current loss and steel sheets provide low reluctance path
to magnetic field.
5. Why commutator is employed in d.c.machines? [A]
Conduct electricity between rotating armature and fixed brushes, convert alternating emf into
unidirectional emf(mechanical rectifier).
6. Distinguish between shunt and series field coil construction? [A]
Shunt field coils are wound with wires of small section and have more no. of turns. Series
field coils are wound with wires of larger cross section and have less no of turns.
7. How are armature windings are classified based on placement of coil inside the
armature slots? [U]
Single and double layer winding.
8. How will you change the direction of rotation of d.c.motor? [A]
Either the field direction or direction of current through armature conductor is reversed.
9. What is back emf in d.c. motor? [Nov/ Dec 2015][R]
As the motor armature rotates, the system of conductor come across alternate north and south
pole magnetic fields causing an emf induced in the conductors. The direction of the
emfinduced in the conductor is in opposite to current. As this emf always opposes the flow of
current in motor operation it is called as back emf.
10. What is the function of no-voltage release coil in d.c. motor starter? [R & U]
As long as the supply voltage is on healthy condition the current through the NVR coil
produce enough magnetic force of attraction and retain the starter handle in ON position
against spring force. When the supply voltage fails or becomes lower than a prescribed value
then electromagnet may not have enough force to retain so handle will come back to OFF
position due to spring force automatically.

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11. Enumerate the factors on which speed of a d.c.motor depends? [AP]

N= (V-IaRa)/Ф so speed depends on air gap flux, resistance of armature, voltage applied to
armature.
12. Under what circumstances does a dc shunt generator fails to generate?
[Apr/ May 2015][A]
Absence of residual flux, initial flux setup by field may be opposite in direction to residual
flux, shunt field circuit resistance may be higher than its critical field resistance, load circuit
resistance may be less than its critical load resistance.
13. Define critical field resistance of dc shunt generator? [Nov/ Dec 2015][R]
Critical field resistance is defined as the resistance of the field circuit which will cause the
shunt generator just to build up its emf at a specified field.
14. Why is the emf not zero when the field current is reduced to zero in dc generator?
[A]
Even after the field current is reduced to zero, the machine is left out with some flux as
residue so emf is available due to residual flux.
15. On what occasion dc generator may not have residual flux? [A]
The generator may be put for its operation after its construction, in previous operation, the
generator would have been fully demagnetized.
16. State the conditions to be fulfilled for a dc shunt generator to build back emf? [U]
The generator should have residual flux, the field winding should be connected in such a
manner that the flux setup by field in same direction as residual flux, the field resistance
should be less than critical field resistance, load circuit resistance should be above critical
resistance.
17. Define armature reaction in dc machines? [Apr/ May 2017][R]
The interaction between the main flux and armature flux cause disturbance called as armature
reaction.
18. What are two unwanted effects of armature reactions? [R]
Cross magnetizing effect & demagnetizing effect.
19. What is the function of carbon brush used in dc generators? [R]
The function of the carbon brush is to collect current from commutator and supply to external
load circuit and to load.
20. Explain briefly why series motor cannot be started without load.
[May / Jun 2017][E]
If a DC series motor is started without load, its I a will be less which inturn its I f also
small.(For series motor If = Ia). As Ф  If and N  1/Ф, its speed becomes dangerously high.
Hence a series motor should never be started without some mechanical load on it.

PART - B

1. Describe with a neat diagram the principle of operation and constructional features of
DC machines. [Nov/Sec 2015] [Apr/May 2017] [E](13)
2. What are the different types of DC generators? Explain them with their schematic
diagrams. [R][U](10)

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3. Derive the emf equation of DC generator. [May/June 2016][C](10)

4. Discuss in detail the principle of operation of DC generator. [U](6)
5. With neat diagram explain in detail [U]
(i) Armature Reaction (8)
(ii) Commutation in DC machines (8)
6. Explain the principle of operation of DC motor. [U](8)
7. Derive the torque equation of a DC motor. [Nov/Sec 2015][AP](13)
8. Explain the different methods of speed control of DC shunt and series motor with neat
circuit diagrams. [Apr/May 2015] [Apr/May 2017][R][U](6,5)

9. Discuss in detail about the N-Ia, T- Iaand N-T characteristics for a DC series, DC shunt
and DC compound Generator &motor. [Nov/Sec 2015][May/June 2016][R][U](13,10)

PART – C

1. Explain in detail about construction and working principle of DC machine.

[Apr/May 2017][E](15)
2. calculate the emf generated by a 6 pole d.c. generator having 480 conductor and driven at
a speed of 1200rpm. The flux per pole is 0.012wb.Ans: Eg (lab wdg) =115.2V, Eg (wave
wdg) =345.6V. [AP][E](15)
3. A 4 pole generator with wave wound armature has 51 slots each having 24 conductors.
The flux/pole is 0.01weber.at what speed must the armature rotate to give an induced emf
of 250V. What will be voltage developed if the winding is lap connected and the armature
rotates at the same speed. Ans: i) N=612.74rpm, ii)Eg=125V. [AP][E](15)
4. A 10KW, 220 V, DC 6 pole shunt motor runs at 1000 rpm. Delivering full load. The a
534 la connected conductors. Full load core loss is 0.64 KW. The total brush drop is 1
Determine the flux per ole neglecting shunt current. [AP](15)

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UNIT II
Transformers
PART – A

1. Mention the difference between core and shell type transformers? [A]
In core type, the windings surrounded the core considerably and in shell type the core
surround the windings i.e winding is placed inside the core.
2. What is the purpose of laminating the core in a transformer? [A]
To reduce the eddy current loss in the core of the transformer.
3. Give the emf equation of a transformer and define each term? [AP]
Emf induced in primary coil E1= 4.44fФmN1 volts
Emf induced in secondary coil E2 =4.44 fФmN2 volts
f--->freq of AC input
Фm---->maximum value of flux in the core
N1,N2--->Number of primary & secondary turns.
4. Does transformer draw any current when secondary is open? [Apr/May 2017][A]
Yes, it(primary) will draw the current from the main supply in order to magnetize the core
and to supply for iron and copper losses on no load. There will not be any current in the
secondary since secondary is open.
5. Define voltage regulation of a transformer. [Apr/May 2015][R]
When a transformer is loaded with a constant primary voltage, the secondary voltage
decreases for lagging PF load, and increases for leading PF load because of its internal
resistance and leakage reactance. The change in secondary terminal voltage from no load to
full load expressed as a percentage of no load or full load voltage is termed as Regulation.
% regulation down = (V2noload-V2F.L)*100/V2noload
% regulation up = (V2noload-V2F.L)*100/V2F.L
6. Full load copper loss in a transformer is 1600W, what will be the loss at half load?[E]
If x is the ratio of actual load to full load, then copper loss = x2(F.L. copper loss)
hereWc = (0.5)2 * 1600=400W.
7. Define all day efficiency of a transformer? [R]
It is computed on the basis of energy consumed during a certain period, usually a day of 24
hrs. All day efficiency=output in kWh for 24 hrs / input in kWh for 24 hrs.
8. Why transformers are rated in kVA? [U]
Copper loss of a transformer depends on current & iron loss on voltage. Hence total losses
depend on Volt-Ampere and not on PF. That is why the rating of transformers are in kVA and
not in kW.
9. What are the typical uses of auto transformer? [AP]
1. To give small boost to a distribution cable to correct for the voltage drop.
2. As induction motor starter.
10. What are the applications of step-up & step-down transformer? [AP]
Step-up transformers are used in generating stations. Normally the generated voltage will be
either 11kV. This voltage(11kV) is stepped up to 110kV or 220kV or 400kV and transmitted
through transmission lines(simply called as sending end voltage). Step-down transformers are
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used in receiving stations. The voltage is stepped down to 11kV or 22KV and again stepped
down to 3phase 400V by means of a distribution transformer and made available at consumer
premises. The transformers used at generating stations are called Power transformers.
11. Explain on the material used for core construction? [R]
The core is constructed by sheet steel laminations assembled to provide a continuous
magnetic path with minimum of air gap included. The steel used is of high silicon content
sometimes heat treated to produce a high permeability and a low hysteresis loss at the usual
operating flux densities. The eddy current loss is minimized by laminating the core, the
laminations being used from each other by light coat of core-plate vanish or by oxide layer on
the surface. The thickness of lamination varies from 0.35mm for a frequency of 50Hz and
0.5mm for a frequency of 25Hz.
12. How does change in frequency affect the operation of a given transformer? [A]
With a change in frequency, iron and copper loss, regulation, efficiency & heating varies so
the operation of transformer is highly affected.
13. What is the angle by which no-load current will lag the ideal applied voltage? [U]
In an ideal transformer, there are no copper & core loss i.e loss free core. The no load current
is only magnetizing current therefore the no load current lags behind by angle 900. However
the winding possess resistance and leakage reactance and therefore the no load current lags
the applied voltage slightly less than 900.
14. List the arrangement of stepped core arrangement in a transformer? [R]
1. To reduce the space effectively
2. To obtain reduce length of mean turn of the winding
3. To reduce I2R loss.
15. What is the function of transformer oil in a transformer? [U]
1. It provides good insulation 2. Cooling.
16. Can the voltage regulation go negative? If so under what condition? [A]
Yes, if the load has leading PF.
17. Distinguish power transformers & distribution transformers? [Apr/May 2017][A]
Power transformers have very high rating in the order of MVA. They are used in generating
and receiving stations. Sophisticated controls are required. Voltage ranges will be very high.
Distribution transformers are used in receiving side. Voltage levels will be medium. Power
ranging will be small in order of kVA. Complicated controls are not needed.
18. What is the purpose of providing Taps in transformer and where these are
provided? [A]
In order to attain the required voltage, taps are provided, normally at high voltages side(low
current).
19. What are factors on which hysteresis loss depend? [U]
It depend on magnetic flux density, frequency & volume of the material.
20. What is core loss? What is its significance in electric machines? [A]
When a magnetic material undergoes cyclic magnetization, two kinds of power losses occur
on it. Hysteresis and eddy current losses are called as core loss. It is important in determining
heating, temperature rise, rating & efficiency of transformers, machines & other A.C run
magnetic devices.

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21. What is eddy current loss? [Nov/Sec 2015][R]

When a magnetic core carries a time varying flux, voltages are induced in all possible path
enclosing flux. Resulting is the production of circulating flux in core. These circulating
current do no useful work are known as eddy current and have power loss known as eddy
current loss.
22. How hysteresis and eddy current losses are minimized? [R]
Hysteresis loss can be minimized by selecting materials for core such as silicon steel & steel
alloys with low hysteresis co-efficient and electrical resistivity. Eddy current losses are
minimized by laminating the core.
23. Why are breathers used in transformers? [R]
Breathers are used to entrap the atmospheric moisture and thereby not allowing it to pass on
to the transformer oil. Also to permit the oil inside the tank to expand and contract as its
temperature increases and decreases.

PART – B

1. Discuss the working principle and derive the emf equation of a transformer. [R][A](8)
2. Draw the phasor diagram and explain the operation of practical transformer on-load.
[May/June 2016][R](10)
3. Explain the constructional details of different types of 1-phase transformer with neat
diagrams. [U]
4. Explain the constructional details of different types of 3-phase transformer with neat
diagrams. [Nov/Sec 2015][AP][E] (16)
5. A 250/500V transformer gave the following test results: [AP][E](16)
S.C. Test with low voltage winding short circuited: 20V,12A,100W
O.C. Test : 250V, 1A, 80W on low voltage side
Determine the circuit constants and draw the equivalent circuit. Also determine the efficiency
when the output is 10A at 500V and 0.8pf lagging.
6. Explain the construction and working of current and potential transformer.
7. Discuss the application of current and potential transformer.
8. Draw the phasor diagram of the transformer under (i) Resistive load (ii) capacitive load
(iii) Inductive load [R][U](8)
9. Prove that flux remains constant in the core even under the loaded condition of
transformer. [AP]

PART – C

1. Explain various types of three-phase transformer connections. [R](15)

2. Drive the condition for maximum efficiency in the transformer [Apr/May 2017] [AP] (15)

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3. A 20 KVA,1Φ Transformer has 200 turns in the Pry & 40 turns in the Secondary. The pry
is connected to 1000V,50Hz supply. Determine i) The Secondary voltage on open circuit ii)
The current flow through the two windings on full load iii) The maximum value of flux.
[E](15)
4. A 30KVA, 2000/200V,1Φ,50Hz transformer has a pry resistance of 3.5Ω & reactance
4.5Ω. The Secondary resistance & reactance are 0.015Ω & 0.02Ω respectively. Find
equivalent resistance, reactance and impedance i)referred to primary ii) referred to
Secondary. [E](15)

UNIT III
Synchronous Machines
PART - A

1. What do you mean by the salient-pole type rotor? [R]

Salient - pole type rotor means a low and moderate speed rotor having large diameter and
small axial length with projected poles coming out of the rotor frame the outer surface of
which almost follows the inner cylindrical surface of the stator frame.
2. Define voltage regulation of an alternator. [R]
The voltage regulation of an alternator is defined as the increase in terminal voltage when full
load is thrown off, assuming field current and speed remaining the same.
Percentage regulation = E0 - V x 100
V
E0 = No load terminal voltage
V = Full load rated terminal voltage.
3. What are the advantages of having rotating field system? [Apr/May 2017][U]
1. Better insulation 2. Ease of current collection
3. Increased armature tooth strength. 4. More rigid construction
5. Reduced armature leakage reactance. 6. Lesser number of sliprings.
7. Lesser rotor weight &intertia 8. Improved ventilation & heat dissipation.

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4. Why is EMF method called Pessimistic method? [U]

The value of voltage regulation obtained by EMF method is always more than the actual
value, therefore it is called Pessimistic method.
5. Why is MMF method called Optimistic method? [U]
Compared to EMF method, MMF method involves more no of complex calculation steps.
Further the OCC is referred twice and SCC is referred once while predetermining the voltage
regulation for each load condition. Reference of OCC takes core saturation effect. As this
method require more effort, final result is very close to actual value, hence this method is
called as optimistic method.(or)
The value of voltage regulation obtained by MMF method is less than the actual value,
therefore it is called Optimistic method.
6. Compare salient pole rotor &smooth cylindrical rotor. [A]
Salient Pole Rotor Cylindrical Rotor
1 Large diameter and short axial 1 . Small diameter and long axial length, length
2. Used for low speed alternators 2. Used for high - speed turboalternators
3.Has projecting poles 3. No projecting poles
4. Needs damper windings 4. Does not need damper windings.
5.Windage loss is more 5. Windage loss is less
7. How is the armature winding in alternators different from those used in dcmachines?
[A]
The armature winding of the alternator is placed in the stator, but the in case of dc machines,
armature winding is placed in rotor.
8. What are the methods by using zero p.f. lagging curve can be obtained? [U]
Zero power factor characteristic of an alternator gives the variation of terminal voltage with
field current, when the alternator is delivering its full rated current to a zero power factor
(lagging) load. This characteristic is obtained by running the machine at synchronous speed
and connecting a purely inductive 3phase load to its terminals. The load is varied in steps and
at each step the field current is adjusted, so that the armature current is equal to its rated
value.
9. What are squirrel-cage windings of alternators? How and why are they used? [A]
Damper windings are squirrel cage windings of the alternators. This winding is placed in
rotor pole shoes.
10. What is hunting? How is hunting minimized? [U]
When a synchronous motor is used for driving a fluctuating load, the rotor starts oscillating
about its new position of equilibrium corresponding to the new load. This is called hunting or
phase swinging.Toprevent hunting, dampers (or) damping grids are employed. Damper
windings are short circuited,copper bars are embedded in the faces of the field poles of the
motor.
11. When is a synchronous motor said to receive 100% excitation? [A]
When Eb = V, synchronous motor receive 100% excitation.
12. What is a synchronous capacitor? [R]
An over excited synchronous motor, running without any mechanical load, used specifically
for power factor correction is known as synchronous capacitor.

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13. When is a synchronous motor said to be under - excited? What will be the p.f at this
condition? [U]
Excitation emfE b less than supply voltage Eb< V, Lagging power factor.
14. What are the inherent disadvantages of synchronous motor? [R]
i) Higher cost
ii) Necessity of a dc excitation source
iii) Greater initial cost
iv) Highmaintanence cost
15. Mention four applications of synchronous motor? [AP]
i) Power factor correction
ii) Constant speed, constant load drives
iii) Voltage regulation of transmission lines.
16. What is the role of synchronous motor in a transmission line? How? [AP]
Synchronous motor acts as a voltage regulator in a transmission line. When line voltage
decreases due to inductive load, motor excitation is increased thereby increasing its power
factor which compensates for the line voltage drop. When the line voltage increases due to
line capacitive effect, synchronous motor excitation is decreased, thereby making its power
factor lagging which helps to the maintain the transmission line voltage at its normal value.
17. Enlist the advantages and disadvantages of synchronous motor.[May/June 2016][R]
Advantages of Synchronous Motors:
1. The speed is constant and independent of load.
2. These motors usually operate at higher efficiencies.
3. Electro magnetic power varies linearly with the voltage.
4. These motors can be constructed with wider air gaps than induction motors,
which make them better mechanically.
5. An Over excited synchronous motor having a leading power factor can be
operated in parallel with induction motors.
18. What are the disadvantages of Synchronous Motor? [R]
1. It cannot be started under load.
2. It requires dc excitation which must be supplied from external source.
3. It has a tendency to hunt.
4. It cannot be used for variable speed jobs as there is no possiblity of speed
adjustment
5. Collector rings and brushes are required.
19. Define pullout torque in synchronous motor. [R]
The maximum torque which the motor can develop without pulling out of step or
synchronism is called the pull out torque.
20. What is synchronous condenser? [R]
Synchronous motor is operating at an over excited condition is called synchronous condenser.
The synchronous condensers having leading power factor are widely used for improving
power factor of those power systems which employ a large number of induction motors and
other lagging power factor loads.

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22. Define pull in torque in synchronous motor. [R]

It pertains to the ability of the machine to pull into synchronism when changing from
induction to synchronous motor operation.
23. What is meant by V curves of synchronous motor? [R]
The V-curves show the relation that exists between the armature curren: and field current for
different constant power input.
24. Give the expression for the gross mechanical power developed bysynchronous
motor. [R]
Pm = Eb V sin α
Xs
Eb = excitation emf,
V = Supply voltage,
Xs = synchronous reactance,
α = load angle.
25. Name the important characteristics of a synchronous motor not found in an
induction motor. [R]
The essential features of synchronous machine are f) The rotor speed is synchronous with
stator rotating field, ii) The power factor can be easily varied by varying its field current, iii)
It is used for constant speed operation.
26. What is the common starting method used for synchronous motor? [R]
i) Starting with the help of damper winding.
ii) Starting with the help of separate small induction motor.
iii) Starting by using an ac motor coupled to the synchronous, motor
27. Why does the synchronous motor always run at synchronous speed? [U]
A synchronous motor always runs at synchronous speed because of the magnetic locking
between the stator and rotor pole
28. What are the advantages of salient type pole construction used in synchronous
machines? [May/June 2016][R]
They allow better ventilation, the pole faces are so shaped radial air gap length increases from
pole center to pole tips so flux distortion in air gap is sinusoidal so emf is also sinusoidal.
29. Which type of sy.generators are used in hydroelectric plants and why?
[Apr/May 2017][AP]
As the speed of operation is low, for hydro turbines used in hydroelectric plants, salient pole
type sy.generator is used because it allows better ventilation also better than smooth
cylindrical type rotor.
30. Why are alternators rated in KVA and not in KW? [A]
2
As load increases I R loss also increases, as the current is directly related to apparent power
delivered by generator, the alternator has only their apparent power in VA/KVA/MVA as
their power rating.

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PART - B

1. Derive the emf equation and explain the principle of an alternator.[Apr/May 2017][AP](8)
2. Explain the effect of excitation on armature current and power factor and hence draw the
‘V’ curves and inverted ‘V’ curves. [A][AP] (16)
3. What is the principle of synchronous motor? Explain. [R](8)
4. What are the methods employed in starting of synchronous motor?
[Apr/May 2017][Apr/May 2015][U] (8)
5. Explain the synchronous impedance method of determination of voltage regulation of an
alternator. [U](16)
6. Briefly explain the application of synchronous motor in various classes of services. [U]
7. Derive the equation of induced emf for an alternator and draw the vector diagram of a
loaded alternator. [Nov/Sec 2015] [AP](16)
8. Derive an expression for the power developed in an synchronous motor. [R][AP](8)

PART – C

1. Draw and explain the vector diagram when the alternator is loaded with
(i) Resistive (ii) capacitive and (iii) Inductive [Apr/May 2015][R][U](15)
2.A 4 pole, 50Hz, star connected alternator has a flux/pole of 0.12wb. It has 4 slots/ph,
conductor/slot being 4. If the winding coil span is 150 degree, find the line Emf. [E](15)
3. Write the advantage of stationary armature and rotating field in alternator. [U](15)

UNIT IV
Induction Machines
PART- A

1.What are the 2 types of 3phase induction motor? [R]

Squirrel cage and slip ring induction motor.

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2. Write two extra features of slip ring induction motor? [U]

Rotor has 3phase winding. extra resistance can be added in rotor circuit for improving PF
with the help of three slip rings.
3. Why an induction motor is called as rotating transformer? [A]
The rotor receives same electrical power in exactly the same way as the secondary of a two
winding transformer receiving its power from primary. That is why induction motor is called
as rotating transformer.
4. Why an induction motor never runs at its synchronous speed? [Apr/May 2017][A]
If it runs at sy.speed then there would be no relative speed between the two, hence no
rotoremf, so no rotor current, then no rotor torque to maintain rotation.
5. What are slip rings? [R]
The slip rings are made of copper alloys and are fixed around the shaft insulating it. Through
these slip rings and brushes rotor winding can be connected to external circuit.
6. What are the advantages of cage motor? [R]
Since the rotor have low resistance, the copper loss is low and efficiency is very high. On
account of simple construction of rotor it is mechanically robust, initial cost is less,
maintenance cost is less, simple starting arrangement.
7. Give the condition for maximum torque for 3phase induction motor, when it is
running? [U]
The rotor resistance and reactance should be same for max. torque i.e. R 2=sX2.
8. List out the method for speed control of 3phase cage type induction motor. [U]
 By changing supply frequency
 By changing no of poles
 By operating two motors in cascade.
9. How do you reverse the direction of rotation of 3 phase? [Nov/Sec 2015][A]
The direction of 3 phase can be reversed by interchanging any 2 terminals of 3 phase
winding.
10. What are the different speed control methods of squirrel cage induction motors? [U]
a) Supply frequency control
b) Supply voltage control.
c) Controlling number of stator poles.
d) Adding rheostats in stator.
11. What are the different types of starters used for induction motors? [R]
a)Stator resistance starter
b) Autotransformer starter.
c) Star delta starter.
d) Rotor resistance starter.
e) Direct to line starter.
12. How the stator poles can be changed? [A]
a)consequent poles method.
b)multiple stator winding method.
c)pole amplitude modulation method.

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13. What is the effect of change in input voltage on starting torque of induction motor?
[A]
There is no change in starting torque due to change in input voltage. the motor reacts by
drawing more current at lower speeds to keep the same torque.
14. What are the characteristics of double squirrel cage motor, compared to a squirrel
cage motor? [A]
(i) High starting torque
(ii) Excellent running performance
15. What is meant by Slip power recovery Scheme? [U]
This slip power can be returned to the supply source and can be used to supply an additional
motor which is mechanically coupled to the main motor. This type of drive is known as a slip
power recovery system and improves the overall efficiency of the system.
16. What are the various methods of speed control of 3 phase induction motor? [U]
(i) stator voltage control
(ii) stator frequency control
(iii) V/f method
(iv) pole changing method
17.What is the function of slip ring in 3 phase induction motor? [A]
Slip rings are used to connect external stationary circuit to internal 2 M rotating circuit.
18.Under what condition the slip in an induction motor is ? [Nov/Sec 2015][A]
a) Negative
b) Greater than 1
a)When rotor is running at a speed above the synchronous speed slip is negative.
b)When motor is rotated in opposite direction to that of rotating field slip is greater than 1
19.What are the 2 fundamental characteristics of a rotating magnetic field? [U]
a)The resultant of three alternating fluxes separated from each other by 120 degree
has constant amplitude of 1.5 .
b)The resultant always keeps on rotating with a certain speed in space.
20. What is induction generator? [R]
When the slip of the induction motor is negative the induction motor that runs as a generator
is called induction generator.
21. What are the purposes that could be served by external resistors connected in the
rotor Circuit of phase wound induction motor? [U]
a) Increasing starting torque.
b) For speed control
c) Limiting starting current.
22. What are the merits of inner and outer cage of double cage induction motor? [R]
 Merits of inner cage—
a)leakage reactance is high.
b)resistance is small.
 Merits of outer cage—
a)has high starting torque.

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b)resistance is high.
23. Define Synchronous speed in a 3 phase I.M? [R]
The speed at which the revolving flux rotates is called synchronous speed Ns and is given by
Ns =120f / P
Where f – Supply Frequency
P- Number of poles on the stator
24. What are the losses in induction motor? [R]
a)constant losses
b)variable losses.
25. What is crawling in I.M? [R]
The tendency of the motor to run stably at speeds as low as one seventh of its synchronous
speed with a low pitched howling sound is called crawling
26. What are the applications of 3 phase I.M motors? [AP]
a)driving fans b)blowers.
c)lathes d)lifts---etc.

PART – B

1. Draw and explain the slip-torque characteristics of 3-phase induction motor.

[Apr/May 2015] [May/June 2016][R][U](8)
2. Derive the equation for torque under running conditions in a 3-phase induction motor.
[Nov/Sec 2015] [Apr/May 2017][AP] (16,5)
3. With a neat circuit diagram explain the starting of slip-ring induction motor.
[Apr/May 2017][Nov/Sec 2015][U](15,16)
4. Explain in detail the production of rotating magnetic field with neat phasor diagram and
hence explain the principle of operation of induction motor. [E](8)
5. What are the various methods of speed control of induction motor? Explain in detail.
[Apr/May 2015] [May/June 2016][R](16)
6. Explain with a neat diagram the principle of an induction motor.
[Apr/May 2015] [May/June 2016][U](16)
7. Compare squirrel cage of induction motor and slip ring of induction motor based on
construction, principle and characteristics. [Apr/May 2017][A](8)
8. Derive the condition for maximum running torque in 3phase induction motor. [R][U](8)
9. Derive the equivalent circuit of a 3 phase induction motor. [AP](8)

PART – C

1. The name plate details of a 1-phase , 4-pole induction motor gives the following data
output=410w,supply voltage=230v,frequency=50hertz,input current=3.2A,power factor=
0.8, speed=1410 rpm. calculate;
(1) The efficiency of the motor, and
(2) the slip of the motor when delivering the rated output. [E](15)

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2. Explain in detail about construction and working principle of 3 phase induction motor
[Apr/May 2015] [May/June 2016][U](15)
3. Explain the following starters used for starti induction motors:
(i) auto-transformer starter.
(ii) star-Delta starter [Apr/May 2017][Nov/Sec 2015][U](15)

UNIT - V
Special Machines
PART – A

1. Name the two winding of single phase induction motor? [R]

Running and starting winding.
2. What are methods available for making single phase induction motor a self starting?
By slitting the single phase, by providing shading coil in the poles. [U]
3. What is the function of capacitor in single phase induction motor? [A]
To make phase difference between starting and running winding, to improve PF and to get
more torque.
4. State any 4 use of single phase induction motor? [AP]
Fans, wet grinders, vacuum cleaner, small pumps, compressors, drills.
5. What kind of motors used in ceiling fan and wet grinders? [AP]
Ceiling fan - Capacitor start and capacitor run single phase induction motor,wet grinders-
Capacitor start capacitor run single phase induction motor.
6. What is the application of shaded pole induction motor? [AP]
Because of its small starting torque, it is generally used for small toys, instruments,hair driers,
ventilators, etc.
7. In which direction a shaded pole motor runs? [A]
The rotor starts rotation in the direction from unshaded part to the shaded part.
8. Why single phase induction motors have low PF? [A]
The current through the running winding lags behind the supply voltage by large angle so
only single phase induction motors have low PF.
9. Explain why single phase induction motor is not a self starting one?
[May/June 2016][A]
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When motor fed supply from single phase, its stator winding produces an alternating flux,
which doesn’t develops any torque.
10. Differentiate between “capacitor start” & “Capacitor start capacitor run” single
phase induction motor? [Apr/May 2017][E]
Capacitor start – capacitor is connected series with starting winding, but it will be
disconnected from supply when motor pick up its speed.
Capacitor start capacitor run - starting winding and capacitor will not be disconnected from
supply even though motor pickup its speed.
11. What are the principal advantages of rotating field type construction? [R]
Relatively small amount of power required for field system can easily supplied to rotating
system using slip rings and brushes, more space is available in the stator part of the machine
to provide more insulation, it is easy to provide cooling system, stationary system of
conductors can easily be braced to prevent deformation.
12. What is the function of capacitor in a single phase induction motor? [R]
Capacitor is used to improve the power factor of the motor. Due to the capacitor connected in
series with the auxiliary winding, the capacitive circuit draws a leading current which
increases the split phase angle between two currents I m and Ist .
13. What is the use of shading coil in the shaded pole motor? [AP]
In shaded pole motors the necessary phase –splitting is produced by induction. These motors
have salient poles on stator and a squirrel cage type rotor . The poles are shaded ie each pole
carries a copper band one of its unequally divided part is called shading band. When single
phase ac supply is given to the stator winding due to shading provided to the poles a rotating
magnetic field is generated .
14 .Why capacitor –start induction motors advantageous? [A]
In capacitor start induction motors capacitor is connected in series with the auxiliary winding
.when speed of the motor approaches to 75 to80%of the synchronous speed the starting
winding gets disconnected due to the operation of the centrifugal switch . the capacitor
remains in the circuit only at start . the starting torque is proportional to phase angle _ and
hence such motors produce very high starting torque.
15 . List out 4 applications of shaded pole induction motor? [AP]
shaded pole motors have very low starting torque , low power factor and low efficiency. The
motors are commonly used for small fans , toy motors ,advertising displays , film projectors ,
record players ,gramophones ,hair dryers , photocopying machines etc
16. What are the drawbacks of the presence of the backward rotating field in a single
phase induction motor? [U]
Due to cutting of flux, emf gets induced in the rotor which circulates rotor current. The rotor
current produces rotor flux This flux interacts with forward component Φf to produce a
torque in one particular direction say anticlockwise direction, while rotor flux interacts with
backward component Φb to produce a torque in the clockwise direction. so if anti clock wise
torque is positive then clockwise torque is negative thus net torque experienced by the rotor is
zero at start
17. Why is hysteresis motor free from mechanical and magnetic vibrations? [A]
The stator of hysteresis motor carries main and auxiliary windings to produce rotating agnetic
field or of shaded pole type also. The rotor is smooth cylindrical type made up of hard
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magnetic material .The torque in this motor is constant at all speeds it runs at synchronous
speed. there is not relative motion between stator and rotor field so the torque due to eddy
current vanishes .only hysteresis torque is present which keeps rotor running at synchronous
speeds .the high retentivity ensures continuous magnetic locking between stator and rotor .
Hence it is free from magnetic vibrations
18 . What types of motor is used in computer drives and wet grinders? [AP]
For computer drives permanent magnet dc motors are used while in wet grinders universal
motor may be used.
19. What are the specific characteristic features of the repulsion motor? [U]
Repulsion motors give excellent performance characteristics. A very high starting torque of
about 300 to350% of full load can be obtained with starting currents of about 3 to 4 times the
full load current. Thus it has got very good operating characteristics. The speed of the motor
changes with load with compensated type of repulsion motor the motor runs with improved
power factor as the quadrature drop in the field winding is neutralised. Also the leakage
between armature and field is reduced which gives better regulation.
20. Discuss characteristics of single phase series motor. [E]
 To reduce the eddy current losses, yoke and pole core construction is laminated.
 The power factor can be improved by reducing the number of turns. But this reduces
the field flux. But this reduction in flux increases the speed and reducing the torque.
To keep the torque same it is necessary to increase the armature turns proportionately.
This increases the armature inductance.
21. What are the demerits of repulsion motor? [U]
 very expensive
 speed changes with load
 on no load speed is very high causing sparking at brushes
 low power factor on no load
22. List four applications of reluctance motors? [AP]
This motor is used in signalling devices, control apparatus, automatic regulators, recording
instruments, clocks and all kinds of timing devices, teleprinters, gramophones .
23. What is a universal motor? [R]
There are small capacity series motors which can be operated on dc supply or single phase ac
supply of same voltage with similar characteristics called universal motors. The construction
of this motor is similar to that of ac series motor.
24. List the disadvantages of a switched reluctance motor. [R]
1. Stator phase winding should be capable of carrying magnetizing current.
2. For high speed operation developed torque has undesirable ripples is a result develops
undesirable noises or acoustic noises.3. It requires a position sensors.
25. Why rotor position sensor is essential for the operation of switched reluctance
motor? [A]
It is normally necessary to use a rotor position for commutation and speed feed back. The
turning ON and OFF operation of the various devices of power semiconductor switching
circuit are influenced by signals obtained from rotor position sensor.

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26. What are the advantages of Switched reluctance motor? [R]

1. Construction is simple and robust.
2. There is no permanent magnet.
3. Rotor carries no windings, no slip rings, no brushes, less maintenance
4. Power semiconductor switching circuitry is simpler.
5. It is the self starting machine.
6. It is possible to get very high speed.
27. What are the applications of SRM. [AP]
1. Washing Machines.
2. Vacum cleaners.
3. Fans
4. Future auto mobile applications.
5. Robotics control applications.
28. What is the Phase winding? [R]
Stator poles carrying field coils. The field coils of opposite poles are connected in series such
that mmf’s are additive and they are called Phase windings of SRM.
29. What are the difference between SRM and Stepper motor. [E]
1. In SRM is designed for continous rotation. SRM requires a rotor position sensor.
2. In Stepper motor is designed to rotate in step by step rotation. It does not require rotor
position sensor.
30. Define Chopping mode of operation of SRM. [R]
In this mode, also called low speed mode, each phase winding gets excited for a period which
is sufficiently long.
31. Define Single pulse mode of operation of SRM. [R]
In single pulse mode, also called high speed mode, the current rise is within limits during the
small time interval of each phase excitation.
32. State the principle of operation of switched reluctance motor. [Nov/Sec 2015][R]
The SRM develops an electromagnetic torque due to variable reluctance principle. When air
gap is minimum, the reluctance will be minimum. Hence inductance will be maximum, so the
rate of change of inductance is zero. When the reluctance varies, there will be a change in
inductance so when a particular stator winding of SRM is excited, the rotor pole comes in
alignment with that stator pole and thus the rotor rotates.
33. What is the need for shaft position sensor for SRM? [A]
1. For commutation the turning on and turning off of various semiconductor devices in the
switching circuitry is influenced by the signals obtained from the rotor position sensor.
2. For speed control of the motor ,it is necessary to use the rotor position sensor.
34. Differentiate SRM and SyRM. [A]

SRM SyRM.
In order to have self starting The motor has the same number of
capability and bidirectional poles on stator and rotor.
control, the rotor of a SRM has
lesser poles than the stator.

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35. What are the advantages of brushless dc motor drives? [R]

1. Regenerative braking is possible.
2. Speed can be easily controllable.
3. It is possible to have very high speeds.
4. There is no field winding so that field copper loss is neglected.
36. What are the disadvantages of brushless dc motor drives? [R]
1. Motor field cannot be controlled.
2. It requires a rotor position sensor.
3. It requires a Power semiconductor switching circuit.
4. Power rating is restricted because of the maximum available size of Permanentmagnets.
37. List the various PM materials. [R]
Alnico, Rare earth magnet, Ceramic magnet, NdFeB magnet.
38. Mention the some applications of PMBL DC motor. [AP]
Power alternators, Automotive applications, Computer and robotics applications,
Textile and glass industries.
39. State the principle of operation of PM brushless DC motor. [R]
When d.c supply is given to the motor, the armature winding draws a current. This current
sets up an mmf which is perpendicular to the main mmf set up by the permanent magnet
field. Hence a force is experienced by the armature conductors according to Fleming’s left
hand rule. As it is in the stator, a reactive force develops a torque in the rotor. If this
developed torque is more than the load torque and frictional
torque, the motor starts rotating.
40. How are PMBLDC motor and PMSM different? [A]
PMBLDC Motor
1. Rectangular distribution of magnetic flux in the airgap.
2. Rectangular current waveforms.
3. Concentrated stator winding.
PMSM
1. Sinusoidal or quasi –sinusoidal distribution of magnetic flux in the air gap.
2. Sinusoidal or quasi-sinusoidal current waveforms.
3. Quasi-sinusoidal distribution of stator conductors.
41. When does a PM synchronous motor operate as a SyRM. [A]
If the cage winding is induced in the rotor and the magnets are left out or demagnetized, a
PM SyRM operates as a SyRM.
42. What are the types of single phase induction motor? [U]
a) Split type induction motor.
b) Capacitor start induction motor.
c) Shaded pole induction motor.
d) Capacitor start capacitor run induction motor ---etc.
43. Define step angle.
Step angle is defined as the angle through which the stepper motor shaft rotates for each
command pulse. It is denoted as β.

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44. What is stepper motor? [R]

A stepper motor is a digital actuator whose input is in the form of programmed
energization of the stator windings and whose output is in the form of discrete angular
rotation.
45.Define resolution. [R]
It is defined as the no.of steps required to complete one revolution of the shaft.
Resolution (Z) = no. of steps /revolution
46.Mention some applications of stepper motor [AP]
 Floppy disc drives
 Qurtz watch
 Camera shutter operation
 Dot matrix and line printers
 Small tool application
 Robotics
47.What are the advantages and disadvantages of stepper motor? [R&U]
Advantages:
a) it can be driven in open loop without feedback
b) it is mechanically simple
c) it requires little or no maintenance.
Disadvantages:
a) low efficiency
b) fixed step angle
c) limited power output
48. Write down the formula for motor speed of stepper motor. [R&U]

49.What are the different types of stepper motor? [R]

 Variable reluctance stepper motor
 Permanent magnet stepper motor

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 Hybrid stepper motor

50. What is meant by micro stepping in stepper motor? [U]
The method of modulating currents through the stator windings so as to obtain rotation of
stator magnetic field through a small angle to get very small step angle is known as micro
stepping.
Micro stepping means, the step angle of the VR stepper motor is very small. It is also
called mini - stepping. It can be achieved by two phases simultaneously as in 2-phase on
mode but with the two currents deliberately made unequal.
51. Differential between VR, PM and hybrid stepper motor. [A]

S.No. VR Stepper motor PM Stepper Motor Hybrid Stepper Motor

1. Low rotor inertia high inertia High inertia
2. Less weight More weight More weight
3. No detente torque Provides detente
Provides detente
available windings torque torque with
de-energized windings de-energized
4. Rotor is no permanent Rotor is permanent Rotor is permanent
magnet magnet magnet
5. Rotor is a salient pole Rotor is a Rotor is a salient
type cylindrical type pole type

PART - B

1. Describe the construction, working principle and applications of shaded pole single phase
induction motor with neat diagrams. [R][U] [AP] (16)
2. Explain the construction, working principle, characteristics and applications of Universal
motor with relevant diagrams. [May/June 2016], [Apr/May 2015][R][U][AP] (8,5)
3. With a neat diagram explain in detail the following: [R][U] (10)
(i) Capacitor start and Capacitor run induction motor
(ii) Split phase induction motor
4. With a neat diagram explain in detail the following: [R][U] (16)
(i) Repulsion motor
(ii) Hysteresis motor [May/June 2016],[Apr/May 2015] (8,5)
5. Explain with phasor diagram the double revolving field theory.
[Apr/May 2017] [Apr/May 2015][U](13,16)
6. Explain the construction, working principle, characteristics and applications of Brushless
DC motor with relevant diagrams. [Apr/May 2017][R][U][AP] (13)
7. Explain the following: [R][U] (16)
(i) Cross field theory in 1phase induction motor
(ii) Permanent magnet brushless DC motor

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8. Explain the construction and performance characteristics and applications of switched

reluctance motor with suitable sketches. [May/June 2016][R][U][AP] (16)

PART – C

1. Explain with phasor diagram the double revolving field theory. [Apr/May 2017] [U](15)
2. Explain the construction, working principle, characteristics and applications of Brushless
DC motor with relevant diagrams. [U] (15)
3.Explain the construction and various modes of excitation of VR stepper motor. [R]

195 II YEAR EIE QUESTION BANK PEC