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17536
MODEL ANSWER
WINTER– 18 EXAMINATION
Subject Title: CONTROL SYSTEM & PLC Subject Code: 17536
Important Instructions to examiners:
1) The answers should be examined by key words and not as word-to-word as given in the model answer
scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors should not be given more Importance (Not
applicable for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in the figure. The
figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent
figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values
may vary and there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgment on part of examiner of relevant answer
based on candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.

Q. No. Sub Answer Markin

Q.N. g
Scheme
Q.1 Attempt any THREE : 12-Total
Marks
a) Draw the block diagram of DC servo system. 4M

Ans: 4M

b) Classify the different modes of process control actions. 4M

Ans: Discontinuous control actions: on-off controller 4M

Continuous control actions: Proportional, Integral and Derivative
Combinational control actions: Proportional Integral (PI), Proportional Derivative (PD) ½
and Proportional Integral and Derivative (PID) marks
for each
modes

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Or

c) Name any four I/P and O/P devices each used with PLC. 4M

Ans:
Input devices used with PLC are push button, limit switch, pressure switch, Thermocouple, 2M(any
Level switch, RTD 4)

Output devices used with PLC are Motor, display, solenoid, heater, lamps, relays, buzzer,
stepper motor, alarm 2M(any
4)

d) Draw the block diagram of process control system. 4M

Ans: 4M

OR

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B) Attempt any ONE : 6M

a) Draw the block diagram of AC output module. Draw its wiring diagram. 6M

Ans:
3M for
block
diagram

3M for
relevant
wiring
diagram

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b) Derive the transfer function of the following block diagram in fig. 1: 6M

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Ans:
2M

2M

2M

Q2 Attempt any TWO : 16 -

Total
Marks
a) For a unity feedback system, the TF is given by C(s) 25 8M
_____ = -------------
R(s) S2 + 6S+25
Find i) rise time ii) peak time iii) peak overshoot iv) settling time.

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Ans: 𝑪(𝑺) 𝑮(𝑺) 𝑤 2 25 2M

Comparing with standard equation 𝑹(𝑺) = 𝟏+𝑮(𝑺)𝑯(𝑺) = 𝑺𝟐 +𝟐𝜉 𝑤𝑛 2
= 𝑺𝟐 +𝟔𝑆+25
𝑛 𝑆+𝑤𝑛

𝑤𝑛 2 = 25, 𝑤𝑛 = 5

2𝜉 𝑤𝑛= 6, 𝜉 = 0.6
2
𝑤𝑑 = 𝑤𝑛 √1 − 𝜉 = 4 𝑟𝑎𝑑/𝑠𝑒𝑐
1.5M
𝜋−𝜃
1) Rise Time: 𝑇𝑟 = 𝑤𝑑
√(1 − 𝜉 2 )
𝜃 = ta𝑛−1 ( ) = 0.92
𝜉
𝜋 − 0.92
𝑻𝒓 = = 𝟎. 𝟓𝟓𝒔𝒆𝒄
4 1.5M
𝜋 𝜋
2) Peak Time: 𝑇𝑝 = = = 𝟎. 𝟕𝟖𝟓 𝒔𝒆𝒄
𝑤𝑑 4
1.5M
3) % Peak overshoot: % 𝑴𝒑 = 𝒆− π ξ/√(1− ξ2) × 𝟏𝟎𝟎% = 𝟗. 𝟒𝟕%
4 1.5M
4) Settling time: 𝑇𝑠 = 𝜉 = 𝟏. 𝟑𝟑 𝒔𝒆𝒄
𝑤𝑛

b) An unity feedback system has G(S), 8M

𝟏𝟎 (𝐒 + 𝟏)
𝐆(𝐒) = 𝟐
𝐒 (𝐒 + 𝟐) (𝐒 + 𝟏𝟎)

Find,
i) Type of system
ii) Error coefficient kp, kv & ka
iii) Steady state error for i/p, r(t) = 1 + 4t +t2/2

Ans: i) Type of system is 2 1M

ii) Error coefficient 𝑲𝑷 = 𝐥𝐢𝐦 𝑮(𝑺)𝑯(𝑺) = ∞ 3M
𝒔→𝟎

𝑲𝑽 = 𝐥𝐢𝐦 𝐒 𝑮(𝑺)𝑯(𝑺) = ∞
𝒔→𝟎

𝟏𝟎
𝑲𝒂 = 𝐥𝐢𝐦 𝐒 𝟐 𝑮(𝑺)𝑯(𝑺) = = 𝟎. 𝟓
𝒔→𝟎 𝟐 × 𝟏𝟎

iii) Steady state error for i/p, r(t) = 1 + 4t +t2/2 4M

𝟏 𝟒 𝟏 𝑺𝟐 + 𝟒𝑺 + 𝟏
𝑹(𝑺) = + 𝟐 + 𝟑 =
𝑺 𝑺 𝑺 𝑺𝟑
𝑺𝟐 +𝟒𝑺+𝟏
𝑺 𝑹(𝑺) 𝑺𝟐
𝒆𝒔𝒔 = 𝐥𝐢𝐦 = 𝐥𝐢𝐦 𝟏𝟎 (𝐒+𝟏)
𝒔→𝟎 𝟏 + 𝑮(𝑺)𝑯(𝑺) 𝒔→𝟎
𝟏 + 𝐒𝟐 (𝐒+𝟐) (𝐒+𝟏𝟎)

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𝑺𝟐 +𝟒𝑺+𝟏
𝑺𝟐 (𝑺𝟐 +𝟒𝑺+𝟏)(𝐒+𝟐) (𝐒+𝟏𝟎) 𝟏×𝟐×𝟏𝟎
𝒆𝒔𝒔 = 𝐥𝐢𝐦 𝐒 𝟐 (𝐒+𝟐)(𝐒+𝟏𝟎)+𝟏𝟎 (𝐒+𝟏)
= 𝐥𝐢𝐦 = =𝟐
𝒔→𝟎 𝒔→𝟎 𝐒𝟐 (𝐒+𝟐)(𝐒+𝟏𝟎)+𝟏𝟎 (𝐒+𝟏) 𝟏𝟎
𝐒 𝟐 (𝐒+𝟐) (𝐒+𝟏𝟎)
𝒆𝒔𝒔 = 𝟐

c) Draw block diagram of PLC. Write function of each block. 8M

Ans:

4M

OR

Explanation:
The Basic PLC structure consists of –
1) Input Module:
Input Module works as an interface between the CPU and the real world input
devices attached to input module .The devices connected to input module are called
4M
as input devices. It accepts the incoming signal and converts this signal in the form
which is compatible with CPU.

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2) Output Module:
Output module works as an interface or link between the CPU and the real world
devices attached to the output module. The main function of output module is to
take control signal from CPU and based on signal received from CPU it changes the
status of output devices.
3) Central processing unit:
The CPU is the main part of any PLC .The CPU solves the user program logic, by
using real time input status from input module and updates the status of outputs
through output module.
4) Power supply:
Power supply is the part of PLC which is used to supply required amount of power
to CPU, input module and output module.
5) Programming device:
The programming device is used for communication between user and PLC. The
programming device help the user to enter and modify the required program into the
PLC memory and troubleshoot PLC ladder logic program.
6) Memory:
PLC memory is divided in two part, Data memory and User memory. Data memory
is used to store data associated with instruction address and user memory is used to
store user’s application program.

Q. 3 Attempt any FOUR : 16 -

Total
Marks
a) Derive the transfer function of given network. Refer fig. – 2 4M

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Ans: 4 marks
for
transfer
function

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b) Describe the role of PLC automation. 4M

Ans:  To reduce human efforts. 1 mark
 To get maximum efficiency from machine and control them with human logic each
 To reduce complex circuitry of entire system role (
 To eliminate the high costs associated with inflexible, relay-controlled systems. any
 Replacing Human Operators (Dangerous Environments & Beyond Human four )
Capabilities)
 Higher productivity.
 Superior quality of end product.
 Efficient usage of energy and raw materials
 Improved safety in working conditions.

c) Define transfer function. Derive the expression of T.F. of closed loop system. 4M
Ans: Transfer Function is defined as the ratio of Laplace transform of output variable to that of Definiti
input variable under the zero initial condition. on 1
Marks,
The transfer function for a general closed loop control Diagra
m 1 M,
derivati
on 02 M

R(s) – Laplace of reference i/p R(t).

C(s) – Laplace of controlled o/p C (t).
E(s) – Laplace of error signal e (t).
B(s) – Laplace of feedback signal b (t).
G(s) – Equivalent forward path transfer fn .
H(s) – Equivalent feedback path transfer function.
Referring to this Fig.
E(s) = R(s) ± B(s)-----------(1)
B(s) = C(s) H(s) -----------(2)
C(s) = E(s) G(s) ------------(3)
B(s) = C(s) H(s) and substituting in equation (1)
E(s) = R(s) ± C(s) H(s)
E(s) = C(s) / G( s)
C(s) / G(s) = R(s) ± C(s) H(s)
C(s) = R(s) G(s) ± C(s) G(s) H(s)
Hence, C(s) [1 ± G(s) H(s) ] = R(s) G(s)
C(s) / R(s) = G(s) / [1± G(s) H(s)]

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d) For system, characteristics equation is s4 + 22s3+10s2+s+k =0. Find K. 4M

Ans: 03 marks
for
rouths
array

01 mark
For K

e) Define the term scanning cycle, speed of execution in PLC. 4M

Ans: Scanning cycle 2
The processor controls the operating cycle or processor scan. This operating cycle marks
consists of a series of operations performed Sequentially and repeatedly. each
Input scan: During input scan, the CPU scans (examines) the external input devices. The term
ON or OFF input states are stored in the input status table.
Program scan: During the program scan, the processor scans the instructions in the
control program, uses the input status from the input status file and determines if an
output will or will not be energized
Output status: During output scan, the processor writes ON or OFF status, one word at a
time, to the associated output module.
The operating cycle typically takes 1 to 25 millisecond. The input and output scan are
normally short, relative to the time required for the program scan.
Speed of execution
“Speed at PLC scans memory and executes instructions is speed of execution”
 A period between one I/O update and the next is termed as “One scan”
 Time taken by PLC to update one of I/O terminal is Scan time
 Scan time is generally measured in msec.
 Speed of execution depends on speed of CPU, length of ladder diagram, types of
instruction used in program.
 Slower the CPU speed or longer the ladder diagram, takes more time to execute
the program

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Q. 4 A) Attempt any THREE : 12-Total

Marks
a) Explain with diagram sinking and sourcing concept in DC input modules. 4M
Ans: Sinking Input Module: 1 M for
diagram
&1M
for
explana
tion

Figure, Shows sinking i/p module where current from positive terminal of DC power
1 M for
supply flows first from i/p device to i/p module and then to common terminal. So here,
diagram
input module is sinking current from input device so it is sinking input module.
&1M
Sourcing input module:
for
The interface diagram of PLC input module as sourcing is shown in figure. In operation,
explana
PLC input module as sourcing, current from power supply first flows from input module
tion
to load and then to common terminal so the input module acts as source of current.

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b) Define . Give its S-plane representation. 4M

Ans: Definition :
Poles : The values of ‘s’ which makes the transfer function infinity after substitution in 1M
the denominator of a transfer function are called poles of the transfer function.
Zeros : The values of ‘s’ which makes the transfer function zero after substitution in the
numerator of a transfer function are called zeros of the transfer function. 1M

System stability is determined from the location of closed loop poles in the s- plane.
If the poles are located on the left half of s-plane, then such system is absolutely stable.
If the poles are located on the right half of s-plane, then such system is unstable.
If one or more pairs of non- repeated roots are located on the jw axis, then the system is
critically stable or marginally stable

2M

c) Compare proportional and integral controller on the basis of equation, 4M

advantages, response to error and application.

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Ans:

1M for
each
comparis
on.

Application In inner loop cascade control Robotic arm movement

d) Write any four logical instructions in PLC 4M
Ans: 1 mark
1. AND instruction: each
instructi
It performs the logical AND operation between two operands. on. Any
four
instructi
on (
descript
ion is
not
expecte
d)

2. OR instruction:

It performs the logical OR operation between two operands.

If one of the bit or both the bits of two operands are 1 then output bit is 1 otherwise 0.

3. XOR instruction:

It performs the logical EX-OR operation between two operands.

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If odd number of inputs are 1 then output of EX-OR is 1 otherwise 0

4. NOT instruction:

It has single source and perform logical NOT operation and store result in destination
memory.
Output is complement of input.
NOT instruction reverses all of the bits in the source word.

B) Attempt any ONE : 6 -Total

Marks
a) Draw block diagram of AC input module. 6M
Ans: 6 marks

b) Describe PID control action w.r.t. equation and response to error. 6M

Ans: PID controller is the proportional-integral-derivative controller. A PID controller 3 marks
calculates an "error" value as the difference between a measured process variable and a each
desired set point. The controller attempts to minimize the error by adjusting the process
control inputs.
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Equation: Pout= KPEp+ KP KI ∫t Ep(t)d(t)+KPKDd/dt Ep + P0

KP = proportional gain
Ep = error percentage
KI = integral gain
KD = derivative gain
P0 = controller output at time t = 0
When an error is introduced to a PID controller, the controller’s response is a
combination of the proportional, integral, and derivative actions. Assume the error is
due to a slowly increasing measured variable. As the error increases, the proportional
action of the PID controller produces an output that is proportional to the error signal.
The integral action of the controller produces an output whose rate of change is
determined by the magnitude of the error. In this case, as the error continues to increase
at a steady rate, the integral output continues to increase its rate of change. The
derivative action of the controller produces an output whose magnitude is determined by
the rate of change. When combined, these actions produce an output. The output
produced responds immediately to the error with a signal that is proportional to the
magnitude of the error and that will continue to increase as long as the error remains
increasing.
The proportional action of the controller stabilizes the process. The integral action
combined with the proportional action causes the measured variable to return to the set
point. The derivative action combined with the proportional action reduces the initial
overshoot and cyclic period.

Q.5 Attempt any TWO : 16 -

Total
Marks
a) List and explain the timer instructions of PLC. 8M
Ans: Depending on the time delay and operation there are two types of timers PLC timer
- (i) ON delay timer
(ii) OFF delay timer

ON delay timer 4M
This instruction counts time interval when conditions preceding it in the rung are true.
Produces an output when accumulated reaches the preset value.
Use Ton instruction to turn an output on or off after the timer has been on for a preset
time interval. The Ton instruction begins to count time base intervals when the rung
conditions become true.
The accumulated value is reset when the rung condition go false regardless of whether
the timer has timed out Instruction parameter- Timer TON is 3 word element.

Status bit explanation- Timer done bit (bit13)-DN is set when the accumulated value is
equal to or greater than the preset value. It is reset when rung condition become false.

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ii) Timer enable bit (bit 14)-EN is set when rung condition are true. It is reset when rung
condition become false.
iii) Timer timing bit (bit15)-TT is set when rung conditions are true & the accumulated
value is less than the preset value. It is reset when the rung conditions go false or when
the done bit is set.

OFF delay timer 4M

This instruction counts time interval when conditions preceding it in the rung are false.
Produces low output when accumulated value reaches the preset value.
Use Toff instruction to turn an output on or off after the timer has been off for a preset
timer has been off for a preset time intervals. The Toff instruction begins to count time
base intervals when the rung makes a true to false to transition.
As long as rung conditions remains false the timer increments its accumulated value
each scan until it reaches the preset value. The accumulated value is reset when the rung
conditions go true regardless of whether the timer has timed out. Instruction parameter-
Timer TOFF is 3 word element.

Status bit explanation-

i) Timer done bit (bit13)-DN is reset when the accumulated value is equal to or greater
than the preset value it is set when rung condition are true.
(ii) Timer enable bit (bit 14)-EN is set when rung condition are true. It is reset when
rung condition become false.
iii) Timer timing bit (bit1S)-TT is set when rung conditions are false & the accumulated
value is less than the preset value. It is reset when the rung conditions go true or when
the done bit is reset.

b) Consider the system with characteristic equation. 8M

S4+6 S3+26 S2+56 S+80 =0.
Determine stability of the system using Rout’s Criteria.

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Ans: 6M

2M

c) A unity feedback system has 40 (S+2) 8M

G(S) = --------------
S (S+1) (S+4)
Determine all error coefficients.

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Ans:

2M

2M

2M

2M

Q.6 Attempt any FOUR: 16-

Total
Marks
a) Draw the ladder diagram for 4M
i) AND gate
ii) NAND gate

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Ans: AND Gate for I1 & I2 Inputs 2M

I1 I2 Q1
0 0 0
0 1 0
1 0 0
1 1 1

NAND Gate for I1 & I2 Inputs 2M

I1 I2 Q1
0 0 1
0 1 1
1 0 1
1 1 0

b) State the Routh’s stability criteria. 4M

Ans: The necessary & sufficient condition for system to be stable is “All the terms in the first 4M
column of routh array must have same sign. There should not be any sign change in the
first column of Routh‟s array”.
If there are any sign changes existing then,
(1) System is unstable
(2) The number of sign changes equals the number of roots lying in the right half of the
S- plane.
c) List any two rules of block diagram reduction technique. 4M
Ans: i) Combining a block in cascade: When two or more blocks are connected in series, their Any
overall transfer function is the product of individual block transfer function. Two
Rules
2M for
each

ii) Combining two blocks in parallel: When two or more blocks are connected in parallel, their
overall transfer function is the addition or difference of individual transfer function
iii) Shifting a take-off point after a block: To shift take off point after a block, we shall add a
block with transfer function 1/G in series with signal having taking off from that point.

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iii) Shifting a take off point after a block: To shift take off point after a block, we shall add a
block with transfer function 1/G in series with signal having taking off from that point.

iv) Shifting a take off point before a block: To shift take off point before a block, we shall add a
block with transfer function G in series with signal having taking off from the take off point

v) Eliminating Feedback Loop:

vi) Interchanging Summing Points: The order of summing points can be interchanged, if two or
more summing points are in series and output remains the same

vii) Moving Take off point before a summing point: To shift a take off point before summing
point, add a summing point in series with take off point.

viii) Moving Take off point after a summing point: To shift a take off
point after summing point, one more summing point is added in series
with take off point.

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ix) Moving summing point after a block: To shift summing point after a block, another
block having transfer function G is added before the summing point.
x) Moving summing point before a block: To shift summing point before a block, another
block having transfer function 1/G is added before the summing point.
d) Explain on – off controller with neat diagram. 4M
Ans: This is the most elementary controller mode which has only two fixed position ON and OFF 2M
. positions are commonly used two positions in most of the control systems.
In this mode, when the error signed e(t) greater than the set point r(t), the output minimizes
When the error signal is less than the set point the output maximizes.
Fig (1) shows an iron which is an example of the ON/OFF control action, in this system
there are only two stages of the output i.e. either the heater coil turn ON or OFF.
In this the real time temperature is compared with the set point and error signal is
generated by the controller to activate the relay, which ON/OFF the coil supply.

2M for
any
relevan
t
exampl
e

e) Compare open loop and closed loop system 4M

Ans: Sr. Open Loop Control System Close Loop Control System 1 mark
No for
1 It is simple and economical It is complex and costlier each
2 It is easier to construct, as it It is not easy to construct, as it requires point (
requires less number of more number of components Any 04
components points)
3 It consumes less power It consumes more power
4 It is more stable It is less stable
5 It does not require feedback path It requires feedback path element
element
6 It has poor accuracy It has better accuracy
7 It does not give automatic It gives automatic correction for external
correction for external disturbances
disturbances
8 It is more sensitive to noise It is less sensitive to noise
9 It is dependent on operating It is not dependent on operating
condition conditions
10 Its operation is degraded if non Its operation is not independent on

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linearity is present conditions

11 It has slow response It has fast response
12 It has high bandwidth It has low bandwidth

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