Unit-4

Question Bank

Q1) Convert 268 to binary?

Sol: Step 1 – Converting octal to Decimal

Octal Number − 268 = Decimal Number − 2210

Step 2 − Converting Decimal to Binary

Decimal Number − 2210 = Binary Number − 101102

Octal Number − 268 = Binary Number − 101102

Q2) Find binary equivalent of 258?

Sol: Its Binary Equivalent −

Octal Number − 258 = Binary Number − 101012

Q3) Find hexadecimal equivalent of 101012

Sol: Its hexadecimal Equivalent −

Binary Number − 101012 = Hexadecimal Number − 1516

Q4) Find binary equivalent of 1516?

Sol: Its Binary Equivalent −

Hexadecimal Number − 1516 = Binary Number − 101012

Q5) Calculate the decimal equivalent of 111102?

Sol: Calculating Decimal Equivalent −

Binary Number − 111102 = Decimal Number − 3010

Q6) Find the complement of the Boolean function,

f = p’q + pq’.

Solution:

Using DeMorgan’s theorem, (x + y)’ = x’.y’ we get

⇒ f’ = (p’q)’.(pq’)’

Then by second law, (x.y)’ = x’ + y’ we get

⇒ f’ = {(p’)’ + q’}.{p’ + (q’)’}

Then by using, (x’)’=x  we get

⇒ f’ = {p + q’}.{p’ + q}

⇒ f’ = pp’ + pq + p’q’ + qq’

Using x.x’=0 we get

⇒ f = 0 + pq + p’q’ + 0

⇒ f = pq + p’q’

Therefore, the complement of Boolean function, p’q + pq’ is pq + p’q’.

Q7) Simplify  f(X,Y,Z)=∏M(0,1,2,4)f(X,Y,Z)=∏M(0,1,2,4)using K-map.

Therefore, the simplified Boolean function is

f = (X + Y).(Y + Z).(Z + X)

Q8) Simplify: F(P,Q,R,S)=∑(0,2,5,7,8,10,13,15)

F = P’Q’R’S’ + PQ’R’S’ + P’Q’RS’ +PQ’RS’ + QS

F = P’Q’S’ + PQ’S’ + QS

F = Q’S’ +QS

Q9) Simplify: F(A,B,C)=π(0,3,6,7)

F = A’BC +ABC +A’B’C’ +ABC’

F = BC + C’ ( A’B’ + AB )

Q10)Simplify, f(W,X,Y,Z)=∑m(2,6,8,9,10,11,14,15) and f(W,X,Y,Z)=∑m(2,6,8,9,10,11,14,15) using Quine-McClukey tabular method.

Solution:

Therefore, the prime implicants are YZ’, WX’ & WY.

           The prime implicant table is shown below.

The reduced prime implicant table is shown below.

                  Hence, f(W,X,Y,Z) = YZ’ + WX’ + WY