A 4-bit priority encoder has inputs and in descending order of priority. The two-bit output is generated as 00,01, 10, and 11 corresponding to inputs , and , respectively. The Boolean expression of the output bit is _____

Given 4-bit priority encoder has inputs and in descending order of priority. The Boolean expression of the output bit

A. $\overline{D_{3}} \; \overline{D_{2}}$

C. $D_{3} \overline{D_{2}}+\overline{D_{3}} D_{1}$

D. $\overline{D_{3}} \; \overline{D_{1}}$

Select the Boolean function(s) equivalent to , where , and are Boolean variables, and + denotes logical OR operation.

A. x + z + xy = x(1 + y) + z = x + z B. (x + y) (x + z) = x + xz + xy + yz = x(1 + y + z) + yz = x + yz C. x + xy + yz = x(1+y) + yz = x + yz D. x + xz + xy = x (1 + z + y) = x

A function F(A,B,C) defined by three Boolean variables A, B and C when expressed as sum of products is given by where, are complements of the respective variable. The product of sums (POS) form of the function F is

C. $F=(A + B + \bar{C}) \cdot (A + \bar{B} + \bar{C} ) \cdot (\bar{A} + B + \bar{C}) \cdot$ $(\bar{A}+\bar{B}+C) \cdot (\bar{A}+\bar{B}+\bar{C})$

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. $F=(\bar{A} + \bar{B} + C) \cdot (\bar{A} + B + C) \cdot$ $(A + B + \bar{C}) \cdot (A+B+C)$

Which one of the following gives the simplified sum of products expression for the Boolean function , where , are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?

Given Boolean function is, It can be minimized by using K-map as given below.

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. $\bar{A}BC+\bar{A}\bar{C}+A\bar{B}C$

The Boolean expression 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. $(X+\bar{Y}+\bar{Z})(\bar{X}+Y+Z)(\bar{X}+\bar{Y}+Z)(X+Y+Z)$

A function of Boolean variables X, Y and Z is expressed in terms of the min-terms as Which one of the product of sums given below is equal to the function F(X,Y,Z)?

B. $(X+Y+Z)\cdot (X+ \bar{Y}+\bar{Z})\cdot (\bar{X}+Y+Z)$

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. $(X+Y+\bar{Z})\cdot (\bar{X}+Y+Z)\cdot (\bar{X}+Y+\bar{Z})$ $\cdot (\bar{X}+\bar{Y}+Z)\cdot (\bar{X}+\bar{Y}+\bar{Z})$

In the sum of products function f (X, Y, Z) = (2, 3, 4, 5), the prime implicants are

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. $\bar{X}Y\bar{Z},\bar{X}YZ,X\bar{Y}\bar{Z},X\bar{Y}Z$

The Boolean expression can be minimized to

A. $Y=\bar{A}\bar{B}\bar{C}D+\bar{A}B\bar{C}+A\bar{C}D$

B. $Y=\bar{A}\bar{B}\bar{C}D+BC\bar{D}+A\bar{B}\bar{C}D$

C. $Y=\bar{A}BC\bar{D}+\bar{B}\bar{C}D+A\bar{B}\bar{C}D$

D. $Y=\bar{A}BC\bar{D}+\bar{B}\bar{C}D+AB\bar{C}\bar{D}$

The Boolean expression AC+ B is equivalent to

Writing the SOP form of the above expression and filling the K-map, we get, Now, writing the SOP for the given options, we get, (A) (B) (C) (D) Now, comparing it which the expression given in the question, we find option (D) satisfies the given condition. Alternate method

B. $\bar{B}C+AC+B\bar{C}+\bar{A}C\bar{B}$

If the function W,X,Y and Z are as follows , , , Then,

K Map(1)= K.map (4) From map (2) & (4)

Boolean Algebra – GATE ECE Digital Circuits Practice Questions & PYQs

Q1GATE 2025MCQ

A 3-input majority logic gate has inputs ${X}, {Y}$ and $Z$ . The output $F$ of the gate is logic '1' if two or more of the inputs are logic '1'. The output $F$ is logic '0' if two or more of the inputs are logic '0'. Which one of the following options is a Boolean expression of the output $F$ ?

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