Boolean Algebra

Question 1
A function F(A,B,C) defined by three Boolean variables A, B and C when expressed as sum of products is given by

F=\bar{A}\cdot \bar{B} \cdot \bar{C} + \bar{A}\cdot B \cdot \bar{C} + A\cdot \bar{B} \cdot \bar{C}

where,\bar{A},\bar{B} \;and \; \bar{C} are complements of the respective variable. The product of sums (POS) form of the function F is
A
F=(A+B+C)\cdot (A+\tilde{B}+C)\cdot (\bar{A}+B+C)
B
F=(\bar{A}+\bar{B}+\bar{C})\cdot (\bar{A}+B+\bar{C})\cdot (A+\bar{B}+\bar{C})
C
F=(A + B + \bar{C}) \cdot (A + \bar{B} + \bar{C} ) \cdot (\bar{A} + B + \bar{C}) \cdot (\bar{A}+\bar{B}+C) \cdot (\bar{A}+\bar{B}+\bar{C})
D
F=(\bar{A} + \bar{B} + C) \cdot (\bar{A} + B + C) \cdot (A + B + \bar{C}) \cdot (A+B+C)
GATE EC 2018   Digital Circuits
Question 1 Explanation: 
\begin{aligned} F(A, B, C, D) &=\bar{A} \bar{B} \bar{C}+\bar{A} B \bar{C}+A \bar{B} \bar{C} \\ &=\Sigma m(0,2,4)=\Pi M(1,3,5,6,7) \\ =&(A+B+\bar{C})(A+\bar{B}+\bar{C})(\bar{A}+B+\bar{C}) \\ &(\bar{A}+\bar{B}+C)(\bar{A}+\bar{B}+\bar{C}) & \end{aligned}
Question 2
Which one of the following gives the simplified sum of products expression for the Boolean function F=m_{0}+m_{2}+m_{3}+m_{5}, where m_{0},m_{2},m_{3},m_{5}, are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?
A
\bar{A}B+\bar{A}\bar{B}\bar{C}+A\bar{B}C
B
\bar{A}\bar{C}+\bar{A}B+A\bar{B}C
C
\bar{A}\bar{C}+A\bar{B}+A\bar{B}C
D
\bar{A}BC+\bar{A}\bar{C}+A\bar{B}C
GATE EC 2017-SET-1   Digital Circuits
Question 2 Explanation: 
Given Boolean function is,
F=m_{0}+m_{2}+m_{3}+m_{5}
It can be minimized by using K-map as given below.


F=\bar{A} \bar{C}+\bar{A} B+A \bar{B} C
Question 3
Following is the K-map of a Boolean function of five variables P, Q, R, S and X. The minimum sum-of-product (SOP) expression for the function is
A
\bar{P}\bar{Q}S\bar{X}+P\bar{Q}S\bar{X}+Q\bar{R}\bar{S}X+QR\bar{S}X
B
\bar{Q}S\bar{X}+Q\bar{S}X
C
\bar{Q}SX+Q\bar{S}\bar{X}
D
\bar{Q}S+Q\bar{S}
GATE EC 2016-SET-3   Digital Circuits
Question 3 Explanation: 


\therefore minimum sum of product expression of the function is
=\bar{Q}S\bar{X}+Q\bar{S}X
Question 4
A function of Boolean variables X, Y and Z is expressed in terms of the min-terms as
F(X,Y,Z)=\sum (1,2,5,6,7)
Which one of the product of sums given below is equal to the function F(X,Y,Z)?
A
(\bar{X}+\bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z)\cdot (X+\bar{Y}+\bar{Z})
B
(X+Y+Z)\cdot (X+ \bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z)
C
(\bar{X}+\bar{Y}+Z)\cdot (\bar{X}+Y+\bar{Z})\cdot (X+\bar{Y}+Z) \cdot (X+Y+\bar{Z})\cdot (X+Y+Z)
D
(X+Y+\bar{Z})\cdot (\bar{X}+Y+Z)\cdot (\bar{X}+Y+\bar{Z}) \cdot (\bar{X}+\bar{Y}+Z)\cdot (\bar{X}+\bar{Y}+\bar{Z})
GATE EC 2015-SET-2   Digital Circuits
Question 4 Explanation: 
\begin{aligned} F(X, Y, Z) &=\Sigma(1,2,5,6,7) \\ &=\pi[0,3,4] \\ =(X+Y+Z)&(X+\bar{Y}+\bar{Z})(\bar{X}+Y+Z) \end{aligned}
Question 5
The Boolean expression f(X,Y,Z)=\bar{X}Y\bar{Z}+X\bar{Y}\bar{Z}+XY\bar{Z}+XYZ converted into the canonical product of sum (POS) form is
A
(X+Y+Z)(X+Y+\bar{Z})(X+\bar{Y}+\bar{Z})(\bar{X}+Y+\bar{Z})
B
(X+\bar{Y}+Z)(\bar{X}+Y+\bar{Z})(\bar{X}+\bar{Y}+Z)(\bar{X}+\bar{Y}+\bar{Z})
C
(X+Y+Z)(\bar{X}+Y+\bar{Z})(X+\bar{Y}+Z)(\bar{X}+\bar{Y}+\bar{Z})
D
(X+\bar{Y}+\bar{Z})(\bar{X}+Y+Z)(\bar{X}+\bar{Y}+Z)(X+Y+Z)
GATE EC 2015-SET-1   Digital Circuits
Question 5 Explanation: 
F(X, Y, Z)=\bar{X} Y \bar{Z}+X \overline{Y Z}+X Y \bar{Z}+X Y Z


F=(X+Y+Z)(X+Y+\bar{Z})(X+\bar{Y}+\bar{Z})(\bar{X}+Y+\bar{Z})
Question 6
For an n-variable Boolean function, the maximum number of prime implicants is
A
2(n-1)
B
n/2
C
2^{n}
D
2^{n-1}
GATE EC 2014-SET-2   Digital Circuits
Question 6 Explanation: 
For a variable Boolean function the maximum number of prime implicants is 2^{(n- l)}.
Question 7
Consider the Boolean function, F(w,x,y,z)=wy+xy+\bar{w}xyz+\bar{w}\bar{x}y+xz+\bar{x}\bar{y}\bar{z}. Which one of the following is the complete set of essential prime implicants ?
A
w,y,xz,\bar{x}\bar{z}
B
w,y,xz
C
y,\bar{x}\bar{y}\bar{z}
D
y,xz,\bar{x}\bar{z}
GATE EC 2014-SET-1   Digital Circuits
Question 7 Explanation: 
\begin{array}{l} f(w, x, y, z) \\ =w+x y+\bar{w} x y z+\bar{w} \bar{x} y+x z+\bar{x} \bar{y} \bar{z} \end{array}


y, x z, \bar{x} \bar{z}
From the above K-map, it can be seen that, the set of essential prime-implicants is
y, x z, \bar{x} \bar{z}
Question 8
The Boolean expression (X+Y)(X+\bar{Y})+\overline{(X\bar{Y}+\bar{X})} simplifies to
A
X
B
Y
C
XY
D
X+Y
GATE EC 2014-SET-1   Digital Circuits
Question 8 Explanation: 
\begin{aligned} \text{Let}\quad F &=(X+Y)(X+\bar{Y})+(\overline{X \bar{Y}+\bar{X}}) \\ F &=X+X \bar{Y}+X Y+(\overline{X \bar{Y}} \cdot \bar{X}) \\ &=X+(\bar{X}+\eta) \cdot X \\ &=X+X Y \\ &=X \end{aligned}
Question 9
In the sum of products function f (X, Y, Z) =\sum(2, 3, 4, 5), the prime implicants are
A
\bar{X}Y,X\bar{Y}
B
\bar{X}Y,X\bar{Y}\bar{Z},X\bar{Y}Z
C
\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}
D
\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}\bar{Z},X\bar{Y}Z
GATE EC 2012   Digital Circuits
Question 9 Explanation: 
f(X,Y,Z)=\sum(2,3,4,5)


f(X, Y, Z)=X \bar{Y}+\bar{X} Y
So prime implicants are X\bar{Y} and \bar{X}Y .
Question 10
If X=1 in logic equation
[X+Z\{\bar{Y}+(\bar{Z}+X\bar{Y})\}]\{\bar{X}+\bar{Z}(X+Y)\}=1, then
A
Y=Z
B
Y=\bar{Z}
C
Z=1
D
Z=0
GATE EC 2009   Digital Circuits
Question 10 Explanation: 
[X+Z\{\bar{Y}+(\bar{Z}+X\bar{Y})\}]\{\bar{X}+\bar{Z}(X+Y)\}=1
\begin{aligned} \Rightarrow Z(1+Y)&=1\\ \Rightarrow \quad\bar{Z}&=1\\ \Rightarrow \quad Z&=0 \end{aligned}
There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.