GATE CSE 2011 | Question: 50

Question

Consider the following circuit involving three D-type flip-flops used in a certain type of counter configuration.

If at some instance prior to the occurrence of the clock edge, $P, Q$ and $R$ have a value $0$, $1$ and $0$ respectively, what shall be the value of $PQR$ after the clock edge?

• A. $000$
• B. $001$
• C. $010$
• D. $011$

Comments

FF numbered top to down as 0 ,1,  2
--------------------------------------------------
D0 =  Q2
D1 =  (Q2 + Q0) '
D2 =  Q2'.Q1

P     Q     R
----------------
0     0      0
0     1      0
0     1      1  ---------> (Ans)
1     0      0 
0     0      0

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Watch the Detailed Video Solution by clicking the button below..!

Watch Detailed Video Solution    

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Nothing new to learn here.

We have to follow basic process as discussed in the lecture, first write input equations and then simply write truth table and using current state and input find next state.

Same clock is given to all imples Synchronus.

Answer 1

Answer - D

As in D-flip-flop, next output is $Q^+ = D$

\[
P_{i+1} = R_i
\]
\[
Q_{i+1} = \overline{(P_i + R_i)}
\]
\[
R_{i+1} = \overline{R_i} \, Q_i
\]

\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\textbf{CLOCK} & D_1 = R & D_2 = {(P+R)'} & D_3 = Q\,{R}'\, & P & Q & R \\
\hline
1 & 0 & 1 & 0 & 0 & 1 & 0 \\
\hline
2 & 0 & 1 & 1 & 0 & 1 & 1 \\
\hline
3 & 1 & 0 & 0 & 1 & 0 & 0 \\
\hline
4 & 0 & 0 & 0 & 0 & 0 & 0 \\
\hline
5 & 0 & 1 & 0 & 0 & 1 & 0 \\
\hline
\end{array}
\]

Counter will go through states: $\boxed{\text{010} \to \text{011}} \to \text{100} \to \text{000} \to \text{010}$

So, value of $PQR$ after the clock edge will be $011$

Comments

Here $D_3$ is wrongly written.

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I think there is an error at $D_3$ at $CLOCK =1$.

Given, $P,Q$ and $R$ have a value $0, 1$ and $0$ respectively.

$D_{1}= R = 0$

$D_{2}=\overline{(P+R)} = 1$

$D_{3} = Q\overline{R} = 1$

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This question has 2 sub parts and in both the subparts the 2^nd subpart has been answered. There is no answer provided for the 1^st question (which should have been over here)

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@vaibhavkedia968 you can see other answers also if you didn’t understood this one

and in best answer also the question is answered correctly

Answer 2

D_P = Q_R

D_Q = ( Q_P + Q_R )’

D_R = Q_Q (Q_R)’

Characteristic Table:-

Input			Previous State			Next State
DP	DQ	DR	QP	QQ	QR	QPN	QQN	QRN
0	1	0	0	0	0	0	1	0
1	0	0	0	0	1	1	0	0
0	1	1	0	1	0	0	1	1
1	0	0	0	1	1	1	0	0
0	0	0	1	0	0	0	0	0
1	0	0	1	0	1	1	0	0
0	0	1	1	1	0	0	0	1
1	0	0	1	1	1	1	0	0

Thus the value of PQR after 010 is 011. Option D.

Answer 3

As simple as possible

Comments

is there any video link of this question??

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why did you try to write the D section, I could not think of other reason than to just confuse people just having the previous and next state is enough isn't? good explanation thought 👍

Answer 4

option d) 011

Answer 5

SIMPLE BUT GOOD QUESTION