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For the synchronous counter shown in Fig$.3,$ write the truth table of $Q_{0}, Q_{1}$, and $Q_{2}$ after each pulse, starting from $Q_{0}=Q_{1}=Q_{2}=0$ and determine the counting sequence and also the modulus of the counter.

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MSB as $Q_{0} - (Q_{0}Q_{1}Q_{2})$ answer is 0-1-6-2-4-repeat (all the state is in decimal representation form)

MSB as $Q_{2} - (Q_{2}Q_{1}Q_{0})$ answer is 0-4-3-2-1-repeat
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although we could take any sequence but in general see the order in which it is given like here the order is Q0,Q1,Q2 not Q2,Q1,Q0. values will be count in that order.
q0 q1 q2
0   0    1  = 1
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Detailed Video Solution, with Two Methods to Solvehttps://youtu.be/x7nDZzEtdjo 

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5 Answers

28 votes
28 votes
Best answer

$$\begin{array}{|ccc|ccc|} \hline \textbf{$Q_0$} & \textbf {$Q_1$} &\textbf {$Q_2$} & \textbf {$Q_{0N}$} & \textbf{$Q_{1N }$}&\textbf{$Q_{2N}$} \\\hline \text{0}& \text{0} & \text{0}  & \text{0} & \text{0} &\text{1}\\\hline \text{0}& \text{0} & \text{1}  & \text{1} & \text{1} &\text{0}\\\hline \text{0}& \text{1} & \text{0}  & \text{1} & \text{0} &\text{0} \\\hline\text{0}& \text{1} & \text{1}  & \text{1} & \text{0} &\text{0} \\\hline\text{1}& \text{0} & \text{0}  & \text{0} & \text{0} &\text{0} \\\hline \text{1}& \text{0} & \text{1}  & \text{0} & \text{1} &\text{0} \\\hline \text{1}& \text{1} & \text{0}  & \text{0} & \text{1} &\text{0} \\\hline \text{1}& \text{1} & \text{1}  & \text{0} & \text{1} &\text{0} \\\hline \end{array}$$
$Q_{0N} = Q_0 \implies J_0 =  Q_1 + Q_2, K_0 = 1$

$Q_{1N} = Q_1 \implies J_1 =   Q_2, K_1 = \bar{Q_0}$

$Q_{2N} = Q_2 \implies J_2 =  \bar{Q_1}.\bar{Q_0}, K_2 = 1$

$$0 - 1 - 6 - 2- 4-0$$

So, MOD $5$ counter.

edited by

3 Comments

now got it
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@akash.dinkar12 but sir I have learned that if MSB is not given highest number output is given as MSB. Is it not correct?

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@rajatmyname YES this is correct convention. If MSB is not given by default highest subscript is the MSB.

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7 votes
7 votes
${\begin{array}{|c|c|c|}\hline
\bf{Q_0}&    \bf{Q_1}&  \bf{Q_2} \\\hline
0&0&0\\ 0&0&1 \\    1&1&0  \\   0&1&0\\   1&0&0  \\   0&0&0 \\ \hline
\end{array}}$

Counting sequence: $ 0-1-6-2-4$

As there are $5$ different states, so $5$ modulus
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4 Comments

@Rishav Kumar Singh ,yeah I am getting same

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is it not $0\rightarrow 4\rightarrow 3\rightarrow 2\rightarrow 1$

if we interchange MSB and LSB?
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yes, srestha u are correct...
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5 votes
5 votes

Mod "5" 

0 votes
0 votes

Ans. is 0-4-3-2-1- repeat 

1 comment

Same I getting

@prashant Dubey

Your table is correct but transition diagram wrong
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