GATE1992-04-c

Question

Design a $3$-bit counter using D-flip flops such that not more than one flip-flop changes state between any two consecutive states.

Comments

Can we use Johnson Counter here??

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Yes.

Answer 1

State diagram will be as  (remember concept of $\textsf{GRAY}$ code) 

State table and $3$-bit synchronous counter with D FFs, will be as 

$\begin{array}{c|c|c}
\text{Present State}&\text{Next State}&\text{FF Inputs}\\\hline
ABC&\bar A \bar B \bar C & D_AD_BD_C\\\hline
001&011&011\\000&001&001\\010&110&110\\011&010&010\\100&000&000\\101&100&100\\110&111&111\\
\end{array}$

 

Comments

why from 000 state to 001 state we can also go from 000 to 100 or 010??

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@Arjun Sir @Deepak Poonia Sir why are we only going to only one state , many more states are possible. For example from 000 we can go to any state except 111 which holds true acording to the question.

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You can create any Counting/State Sequence here as long as any two consecutive states differ in exactly one bit position.

For eg, we can create Counter whose state sequence is:

$000 \rightarrow 010 \rightarrow 011 \rightarrow 001 \rightarrow  101 \rightarrow 111 \rightarrow 110 \rightarrow 100 \rightarrow 000 \dots  $

This state sequence also obeys all the given conditions. The Counter for this sequence is another correct answer.

Similarly, more sequences are possible.

Answer 2

We count in Gray Code :

Answer 3

	Present	state				Next	state
Q2	Q1	Q0	D2	D1	D0	Q2	Q1	Q0
0	0	0	0	0	1	0	0	1
0	0	1	0	1	1	0	1	1
0	1	1	0	1	0	0	1	0
0	1	0	1	1	0	1	1	0
1	1	0	1	1	1	1	1	1
1	1	1	1	0	1	1	0	1
1	0	1	1	0	0	1	0	0
1	0	0	0	0	0	0	0	0

D2=Q1Q0'+Q2Q0

D1=Q2'Q1+Q1Q0'

D0=Q2'Q1'+Q2Q1

Comments

I am getting

D2= Q1Q0' + Q2Q0

D1= Q2'Q0 + Q1Q0'

D0= Q2'Q1' + Q2Q1. Could you please check?

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how to get Do,D1,D2 coloum pllzzz explain

Answer 4

Consider $Q_{A},Q_{B},Q_{C}$ as the $3$ bits, $Q_{A}$ being MSB. The trend taken is that of the $3$-bit gray code. Solve for $D_{a}, D_{b}, D_{c}$  each as a function of $Q_{A},Q_{B},Q_{C}.$ $$\begin{array}{lll|ll}\hline  Q_{A} & Q_{B} & Q_{C} & Q_{A_{n}} = D_{A} & Q_{B_{n}}= D_{B} & Q_{C_{n}}= D_{C}   \\\hline  0&0&0 & 0& 0& 1 \\\hline 0& 0& 1 & 0& 1& 1  \\\hline 0& 1& 1 & 0& 1& 0 \\\hline 0& 1& 0 & 1& 1& 0 \\\hline 1& 1& 0 & 1& 1& 1 \\\hline 1& 1& 1 & 1& 0& 1 \\\hline 1& 0& 1 & 1& 0& 0  \\\hline 1& 0& 0 & 0& 0& 0 \\\hline \end{array}$$

Comments

i believe the K-MAP's in the answer are wrong

this is a same question https://gateoverflow.in/17408/gate1992-04-c

with same table but with different K-MAP's drawn in answer

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I guess, the answer is different from the answer of question https://gateoverflow.in/17408/gate1992-04-c because we are also trying to avoid hazards which may cause more than one state change during transition from one state to other, seems more appropriate.