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
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
F=(A+B+C)\cdot (A+\tilde{B}+C)\cdot (\bar{A}+B+C) | |
F=(\bar{A}+\bar{B}+\bar{C})\cdot (\bar{A}+B+\bar{C})\cdot (A+\bar{B}+\bar{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}) | |
F=(\bar{A} + \bar{B} + C) \cdot (\bar{A} + B + C) \cdot (A + B + \bar{C}) \cdot (A+B+C) |
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?
\bar{A}B+\bar{A}\bar{B}\bar{C}+A\bar{B}C | |
\bar{A}\bar{C}+\bar{A}B+A\bar{B}C | |
\bar{A}\bar{C}+A\bar{B}+A\bar{B}C | |
\bar{A}BC+\bar{A}\bar{C}+A\bar{B}C |
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
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
\bar{P}\bar{Q}S\bar{X}+P\bar{Q}S\bar{X}+Q\bar{R}\bar{S}X+QR\bar{S}X | |
\bar{Q}S\bar{X}+Q\bar{S}X | |
\bar{Q}SX+Q\bar{S}\bar{X} | |
\bar{Q}S+Q\bar{S} |
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)?
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)?
(\bar{X}+\bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z)\cdot (X+\bar{Y}+\bar{Z}) | |
(X+Y+Z)\cdot (X+ \bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z) | |
(\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) | |
(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}) |
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
(X+Y+Z)(X+Y+\bar{Z})(X+\bar{Y}+\bar{Z})(\bar{X}+Y+\bar{Z}) | |
(X+\bar{Y}+Z)(\bar{X}+Y+\bar{Z})(\bar{X}+\bar{Y}+Z)(\bar{X}+\bar{Y}+\bar{Z}) | |
(X+Y+Z)(\bar{X}+Y+\bar{Z})(X+\bar{Y}+Z)(\bar{X}+\bar{Y}+\bar{Z}) | |
(X+\bar{Y}+\bar{Z})(\bar{X}+Y+Z)(\bar{X}+\bar{Y}+Z)(X+Y+Z) |
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})
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
2(n-1) | |
n/2 | |
2^{n} | |
2^{n-1} |
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 ?
w,y,xz,\bar{x}\bar{z} | |
w,y,xz | |
y,\bar{x}\bar{y}\bar{z} | |
y,xz,\bar{x}\bar{z} |
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}
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
X | |
Y | |
XY | |
X+Y |
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
\bar{X}Y,X\bar{Y} | |
\bar{X}Y,X\bar{Y}\bar{Z},X\bar{Y}Z | |
\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y} | |
\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}\bar{Z},X\bar{Y}Z |
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 .
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
[X+Z\{\bar{Y}+(\bar{Z}+X\bar{Y})\}]\{\bar{X}+\bar{Z}(X+Y)\}=1, then
Y=Z | |
Y=\bar{Z} | |
Z=1 | |
Z=0 |
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}
\begin{aligned} \Rightarrow Z(1+Y)&=1\\ \Rightarrow \quad\bar{Z}&=1\\ \Rightarrow \quad Z&=0 \end{aligned}
There are 10 questions to complete.