1) Define binary logic?
Binary logic consists of binary variables and logical operations. The variables are
designated by the alphabets such as A, B, C, x, y, z, etc., with each variable having only
two
distinct values: 1 and 0. There are three basic logic operations: AND, OR, and NOT.
2) Convert (634) 8 to binary
634
110 011 100
Ans = 110011100
3) Convert (9B2 - 1A) H to its decimal equivalent.
N = 9 x 162 + B x 161 + 2 x 160 + 1 x 16-1 + A (10) x 16-2
= 2304 + 176 + 2 + 0.0625 + 0.039
= 2482.110
4) State the different classification of binary codes?
1. Weighted codes
2. Non - weighted codes
3. Reflective codes
4. Sequential codes
5. Alphanumeric codes
6. Error Detecting and correcting codes.
5) Convert 0.640625 decimal numbers to its octal equivalent.
0.640625 x 8 = 5.125
0.125 x 8 = 1.0
= 0.640 625 10 = (0.51) 8
6) Convert 0.1289062 decimal number to its hex equivalent
Staff Name:S.Yuvaraj Department:ECE
0.1289062 x 16 = 2.0625
0.0625 x 16 = 1.0
= 0.2116
7) Convert 22.64 to hexadecimal number.
16 22 -6
16 1 -1
0.64 x 16 = 10.24
0.24 x 16 = 3.84
0.84 x 16 = 13.44
.44 x 16 = 7.04
Ans = (16. A 3 D 7) 16
8) State the steps involved in Gray to binary conversion?
The MSB of the binary number is the same as the MSB of the gray code number. So
write it down. To obtain the next binary digit, perform an exclusive OR operation between
the bit
just written down and the next gray code bit. Write down the result.
9) Convert gray code 101011 into its binary equivalent.
Gray Code: 1 0 1 0 1 1
Binary Code: 1 1 0 0 1 0
10) Substract (0 1 0 1) 2 from (1 0 1 1) 2
1010
0101
Answer = 0 1 1 0
11) Add (1 0 1 0) 2 and (0 0 1 1) 2
1010
0011
Answer = (1 1 0 1) 2
12) Using 10’s complement subtract 72532 - 3250
M = 72532
10’s complement of N = + 96750
-----------
Sum = 169282
Staff Name:S.Yuvaraj Department:ECE
Discard end carry
Answer = 69282
13) Find 2’S complement of (1 0 1 0 0 0 1 1) 2
0 1 0 1 1 1 0 0 1 - 1’s Complement
+00000001
0 1 0 1 1 1 0 1 0 - 2’s complement.
14) Substract 1 1 1 0 0 1 2 from 1 0 1 0 1 1 2 using 2’s complement method
101011
+ 0 0 0 1 1 1 - 2’s comp. of 1 1 1 0 0 1
1 1 0 0 1 0 in 2’s complement form
Answer (0 0 1 1 1 0 )2
15) Find the excess -3 code and 9’s complement of the number 40310
403
010000000011
001100110011+
0 1 1 1 0 0 1 1 0 1 1 0 ----- excess 3 code
9’s complement 1 0 0 0 1 1 0 0 1 0 0 1
16) What is meant by bit?
A binary digit is called bit
17) Define byte?
Group of 8 bits.
18) List the different number systems?
i) Decimal Number system
ii) Binary Number system
iii) Octal Number system
iv) Hexadecimal Number system
19) State the abbreviations of ASCII and EBCDIC code?
ASCII-American Standard Code for Information Interchange.
EBCDIC-Extended Binary Coded Decimal Information Code.
20) What are the different types of number complements?
i) r’s Complement
ii) (r-1)’s Complement.
Staff Name:S.Yuvaraj Department:ECE
21) Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction
(a) X -Y and (b) Y - X using 2's complements.
a) X = 1010100
2's complement of Y = 0111101
--------------
Sum = 10010001
Discard end carry
Answer: X - Y = 0010001
b) Y = 1000011
2's complement of X = + 0101100
---------------
Sum = 1101111
There is no end carry, The MSB BIT IS 1.
Answer is Y-X = -(2's complement of 1101111) = - 0010001
22) Given the two binary numbers X = 1010100 and Y = 1000011, perform the subtraction
(a) X -Y and (b) Y - X using 1's complements.
a) X - Y = 1010100 - 1000011
X = 1010100
1's complement of Y = + 0111100
--------------
Sum = 10010000
End -around carry = + 1
--------------
Answer: X - Y = 0010001
b) Y - X = 1000011 - 1010100
Y = 1000011
1's complement of X = + 0101011
------------
Sum = + 1101110
There is no end carry.
Therefore the answer is Y - X = -(1's complement of 1101110) = -0010001
23) Write the names of basic logical operators.
Staff Name:S.Yuvaraj Department:ECE
1. NOT / INVERT
2. AND
3. OR
24) What are basic properties of Boolean algebra?
The basic properties of Boolean algebra are commutative property, associative property
and distributive property.
25) State the associative property of boolean algebra.
The associative property of Boolean algebra states that the OR ing of several variables
results in the same regardless of the grouping of the variables. The associative property
is stated
as follows:
A+ (B+C) = (A+B) +C
26) State the commutative property of Boolean algebra.
The commutative property states that the order in which the variables are OR ed makes
no difference. The commutative property is:
A+B=B+A
27) State the distributive property of Boolean algebra.
The distributive property states that AND ing several variables and OR ing the result with
a single variable is equivalent to OR ing the single variable with each of the the several
variables
and then AND ing the sums. The distributive property is:
A+BC= (A+B) (A+C)
28) State the absorption law of Boolean algebra.
The absorption law of Boolean algebra is given by X+XY=X, X(X+Y) =X.
29) Simplify the following using De Morgan's theorem [((AB)'C)'' D]'
[((AB)'C)'' D]' = ((AB)'C)'' + D' [(AB)' = A' + B']
= (AB)' C + D'
= (A' + B' )C + D'
30) State De Morgan's theorem.
De Morgan suggested two theorems that form important part of Boolean algebra.
They are,
1) The complement of a product is equal to the sum of the complements.
(AB)' = A' + B'
Staff Name:S.Yuvaraj Department:ECE
2) The complement of a sum term is equal to the product of the complements.
(A + B)' = A'B'
31) Reduce A.A'C
A.A'C = 0.C [A.A' = 1]
=0
31) Reduce A(A + B)
A(A + B) = AA + AB
= A(1 + B) [1 + B = 1]
= A.
32) Reduce A'B'C' + A'BC' + A'BC
A'B'C' + A'BC' + A'BC = A'C'(B' + B) + A'B'C
= A'C' + A'BC [A + A' = 1]
= A'(C' + BC)
= A'(C' + B) [A + A'B = A + B]
33) Reduce AB + (AC)' + AB'C(AB + C)
AB + (AC)' + AB'C(AB + C) = AB + (AC)' + AAB'BC + AB'CC
= AB + (AC)' + AB'CC [A.A' = 0]
= AB + (AC)' + AB'C [A.A = 1]
= AB + A' + C' =AB'C [(AB)' = A' + B']
= A' + B + C' + AB'C [A + AB' = A + B]
= A' + B'C + B + C' [A + A'B = A + B]
= A' + B + C' + B'C
=A' + B + C' + B'
=A' + C' + 1
= 1 [A + 1 =1]
34) Simplify the following expression Y = (A + B)(A + C' )(B' + C' )
Y = (A + B)(A + C' )(B' + C' )
= (AA' + AC +A'B +BC )(B' + C') [A.A' = 0]
= (AC + A'B + BC)(B' + C' )
= AB'C + ACC' + A'BB' + A'BC' + BB'C + BCC'
= AB'C + A'BC'
Staff Name:S.Yuvaraj Department:ECE
35) Show that (X + Y' + XY)( X + Y')(X'Y) = 0
(X + Y' + XY)( X + Y')(X'Y) = (X + Y' + X)(X + Y' )(X' + Y) [A + A'B = A + B]
= (X + Y' )(X + Y' )(X'Y) [A + A = 1]
= (X + Y' )(X'Y) [A.A = 1]
= X.X' + Y'.X'.Y
= 0 [A.A' = 0]
36) Prove that ABC + ABC' + AB'C + A'BC = AB + AC + BC
ABC + ABC' + AB'C + A'BC=AB(C + C') + AB'C + A'BC
=AB + AB'C + A'BC
=A(B + B'C) + A'BC
=A(B + C) + A'BC
=AB + AC + A'BC
=B(A + C) + AC
=AB + BC + AC
=AB + AC +BC ...Proved
37) Convert the given expression in canonical SOP form Y = AC + AB + BC
Y = AC + AB + BC
=AC(B + B' ) + AB(C + C' ) + (A + A')BC
=ABC + ABC' + AB'C + AB'C' + ABC + ABC' + ABC
=ABC + ABC' +AB'C + AB'C' [A + A =1]
38) Define duality property.
Duality property states that every algebraic expression deducible from the postulates of
Boolean algebra remains valid if the operators and identity elements are interchanged. If
the dual
of an algebraic expression is desired, we simply interchange OR and AND operators and
replace
1's by 0's and 0's by 1's.
39) Find the complement of the functions F1 = x'yz' + x'y'z and F2 = x(y'z' + yz). By
applying De-Morgan's theorem.
F1' = (x'yz' + x'y'z)' = (x'yz')'(x'y'z)' = (x + y' + z)(x + y +z')
F2' = [x(y'z' + yz)]' = x' + (y'z' + yz)'
= x' + (y'z')'(yz)'
= x' + (y + z)(y' + z')
Staff Name:S.Yuvaraj Department:ECE
40) Simplify the following expression
Y = (A + B) (A = C) (B + C)
= (A A + A C + A B + B C) (B + C)
= (A C + A B + B C) (B + C)
=ABC+ACC+ABB+ABC+BBC+BCC
=ABC
41) What are the methods adopted to reduce Boolean function?
i) Karnaug map
ii) Tabular method or Quine Mc-Cluskey method
iii) Variable entered map technique.
42) State the limitations of karnaugh map.
i) Generally it is limited to six variable map (i.e) more then six variable involving
expression are not reduced.
ii) The map method is restricted in its capability since they are useful for
simplifying only Boolean expression represented in standard form.
43) What is a karnaugh map?
A karnaugh map or k map is a pictorial form of truth table, in which the map diagram
is made up of squares, with each squares representing one minterm of the function.
44) Find the minterms of the logical expression Y = A'B'C' + A'B'C + A'BC + ABC'
Y = A'B'C' + A'B'C + A'BC + ABC'
=m0 + m1 +m3 +m6
=_m(0, 1, 3, 6)
45) Write the maxterms corresponding to the logical expression
Y = (A + B + C' )(A + B' + C')(A' + B' + C)
= (A + B + C' )(A + B' + C')(A' + B' + C)
=M1.M3.M6
=_ M(1,3,6)
46) What are called don’t care conditions?
In some logic circuits certain input conditions never occur, therefore the corresponding
output never appears. In such cases the output level is not defined, it can be either high
or low.
These output levels are indicated by ‘X’ or‘d’ in the truth tables and are called don’t care
conditions or incompletely specified functions.
Staff Name:S.Yuvaraj Department:ECE
47) What is a prime implicant?
A prime implicant is a product term obtained by combining the maximum possible
number of adjacent squares in the map.
48) What is an essential implicant?
If a min term is covered by only one prime implicant, the prime implicant is said to be
essential
