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+13 votes
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Consider the following circuit:

     

The flip-flops are positive edge triggered D FFs. Each state is designated as a two-bit string $Q_0Q_1$. Let the initial state be 00. The state transition sequence is

asked in Digital Logic by Veteran (59.6k points)
recategorized by | 1.8k views
0
After first state (0,0) q0 complement =1 and q1complement=1
next value for q0 is complement so its 1 which is also value for xor and another input is 1 which is value of q1complement so result of xor for (1,1) is 0 there for q1 is 0 and state becomes(1,0) how are you people getting 1,1
+2

(D) is the correct choice!

2 Answers

+20 votes
Best answer

Clearly $Q_{0}$ alternates in every clk cycle as $Q_{0}$$'$ is fed as input and it is D flipflop.

$Q_{1}$ becomes $1$ if its prev value and current $Q_{0}$ differs (EXOR). 

So, the sequence of transitions will be:  $00 -> 11 -> 01 -> 10 -> 00$, (D) choice. 

answered by Veteran (359k points)
edited by
+7
sir ,answer should be (D) ..

Q0next = D0= Q0'

Q1next = D1= Q0XORQ1'=Q0 XNOR Q1.

Q0Q1 = 00-->11--->01-->10-->00
0
Yes. I was taking OR Gate. Corrected now. Thanks..
+1
My pleasure ...sir :)
+1
sequence is 00->11->10->01
0
but i got 00--11--10--01.
0

your statement : "Q1 becomes 1 if its prev value and current Q0 differs (EXOR)." is wrong i guess as q1 becomes 1 if complement of its prev value and current Qo differs  

0
how to know about the sequence
0

 @Arjun sir,

$Q_1$ becomes $1$ if its prev value and current $Q_0$ differs (EXOR).   

 Shouldn't it be $Q_1$ becomes $1$ if the complement of its previous value and previous $Q_0$ differs (EXOR). ?

0 votes
answer - D
answered by Loyal (9k points)
edited by


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