Digital Logic & Number representation

Since there are more than one outputs and number of outputs is less than inputs, it is a Priority encoder V=1 when input is valid and for priority encoder it checks first high bit encountered. Except all are having at least one bit high and ‘x’ represents the “don’t care” as we have found a high bit already. So answer is (A).

By Definition of XNOR, So Option-A is correct. Also by Definition of XOR, Option-B is So Option-B is also correct. Option-C is Option-C is also correct. Option-D x'⊕y' = x''y' + x'y'' = xy' + x'y = x⊕y ≠ x⊙y Therefore option (D) is false. This explanation is provided by Chirag Manwani.

The value of f(X, Y) is same as X for all input pairs. Also sum of product form of expression we get,

= XY’+XY
= X(Y’+Y)
= X *1
= X 

We see from truth table –

Column x = f(x,y) 
So , 
f(x,y)=x 

Option (A) is correct.

There are two prime implicants in the following K-Map- Prime Implicant higlighted in Green = Prime Implicant higlighted in Orange = So the Boolean expression is- Therefore option (B) is correct. This explanation is provided by Chirag Manwani.

There are two prime implicants in the following K-Map- Prime Implicant higlighted in Green = Prime Implicant higlighted in Orange = So the Boolean expression is- Therefore option (B) is correct. This explanation is provided by Chirag Manwani.

All options except D produce XOR as described below :

See following : (P+Q'+R').(P+Q'+R).(P+Q+R') = From the K-map, POS form is : P + Q'.R'

P' = R Q' = (P + R)' R' = QR' Given that (P, Q, R) = (0, 1, 0), next state P', Q', R' = 0, 1, 1 ----------------------------------------------------------------------------------------------- D flip flop truth table

D	Q(t+1)
0	0
1	1

Initially (p,q,r) =(0,1,0) D for p=R D for q=NOT(p xor r) D for r= (not)r.q So Q(t+1) for(p,q,r) p=>r=0 so p=0 q=> NOT(p xor r) => 1      so q=1 r=>(not)r.q => 1         so r=1 (p, q, r) = (0, 1, 1) Alternative approach - Truth table of a D Flip-Flop- By looking at the circuit diagram, it is clear that the boolean expressions of P, Q, and R are- Here the subscript t refers to the current clock cycle, and the subscript (t+1) refers to the next clock cycle. This explanation is provided by Chirag Manwani.

There are four distinct states, 000 → 010 → 011 → 100 (→ 000) so the answer is B

K-map,

= PQ + QR’ + PR’ 
= PQ(R+R’) + (P+P’)QR’ + P(Q+Q’)R’
= PQR + PQR’ +PQR’ +P’QR’ + PQR’ + PQ’R’ 
= PQR(m7) + PQR'(m6)+P’QR'(m2) +PQ’R'(m4) 
= m2 + m4 + m6 + m7 

Option (A) is correct.