@Arjun Sir here question is incomplete

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Analyse the circuit in Fig below and complete the following table

$${\begin{array}{|c|c|c|}\hline

\textbf{a}& \textbf{b}& \bf{ Q_n} \\\hline

0&0\\\ 0&1 \\ 1&0 \\ 1&1 \\ \hline

\end{array}}$$

### 5 Comments

## 1 Answer

Best answer

The output of the circuit given as $: Q=aQ_{n-1}+ab+bQ_{n-1}$

Hence, $Q_{n}=Q_{n-1}(a+b)+ab$

$00 \implies Q_{n-1}(0+0) + 0.0 = Q_{n-1}(0) + 0 = 0+0 = 0$

$01 \implies Q_{n-1}(0+1) + 0.1= Q_{n-1} (1)+ 0 = Q_{n-1}+0 = Q_{n-1}$

$10 \implies Q_{n-1}(1+0) + 1.0 = Q_{n-1} (1) + 0 = Q_{n-1}+0= Q_{n-1}$

$11 \implies Q_{n-1}(1+1)+ 1.1 =Q_{n-1}(1) + 1 =Q_{n-1}+1 = 1$

$${\begin{array}{cc|c}

\textbf{a}& \textbf{b}& \bf{ Q_n} \\\hline

0&0&0\\ 0&1& Q_{n-1} \\ 1&0& Q_{n-1} \\ 1&1 &1\\

\end{array}}$$

Hence, $Q_{n}=Q_{n-1}(a+b)+ab$

$00 \implies Q_{n-1}(0+0) + 0.0 = Q_{n-1}(0) + 0 = 0+0 = 0$

$01 \implies Q_{n-1}(0+1) + 0.1= Q_{n-1} (1)+ 0 = Q_{n-1}+0 = Q_{n-1}$

$10 \implies Q_{n-1}(1+0) + 1.0 = Q_{n-1} (1) + 0 = Q_{n-1}+0= Q_{n-1}$

$11 \implies Q_{n-1}(1+1)+ 1.1 =Q_{n-1}(1) + 1 =Q_{n-1}+1 = 1$

$${\begin{array}{cc|c}

\textbf{a}& \textbf{b}& \bf{ Q_n} \\\hline

0&0&0\\ 0&1& Q_{n-1} \\ 1&0& Q_{n-1} \\ 1&1 &1\\

\end{array}}$$