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.
Detailed Video Solution, with Two Methods to Solve: https://youtu.be/x7nDZzEtdjo
$$\begin{array}{|ccc|ccc|} \hline \textbf{$Q_0$} & \textbf {$Q_1$} &\textbf {$Q_2$} & \textbf {$Q_{0N}$} & \textbf{$Q_{1N }$}&\textbf{$Q_{2N}$} \\\hline \text{0}& \text{0} & \text{0} & \text{0} & \text{0} &\text{1}\\\hline \text{0}& \text{0} & \text{1} & \text{1} & \text{1} &\text{0}\\\hline \text{0}& \text{1} & \text{0} & \text{1} & \text{0} &\text{0} \\\hline\text{0}& \text{1} & \text{1} & \text{1} & \text{0} &\text{0} \\\hline\text{1}& \text{0} & \text{0} & \text{0} & \text{0} &\text{0} \\\hline \text{1}& \text{0} & \text{1} & \text{0} & \text{1} &\text{0} \\\hline \text{1}& \text{1} & \text{0} & \text{0} & \text{1} &\text{0} \\\hline \text{1}& \text{1} & \text{1} & \text{0} & \text{1} &\text{0} \\\hline \end{array}$$ $Q_{0N} = Q_0 \implies J_0 = Q_1 + Q_2, K_0 = 1$
$Q_{1N} = Q_1 \implies J_1 = Q_2, K_1 = \bar{Q_0}$
$Q_{2N} = Q_2 \implies J_2 = \bar{Q_1}.\bar{Q_0}, K_2 = 1$
$$0 - 1 - 6 - 2- 4-0$$
So, MOD $5$ counter.
@akash.dinkar12 but sir I have learned that if MSB is not given highest number output is given as MSB. Is it not correct?
@rajatmyname YES this is correct convention. If MSB is not given by default highest subscript is the MSB.
@Rishav Kumar Singh ,yeah I am getting same
Mod "5"
Ans. is 0-4-3-2-1- repeat
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