GATE CSE 1991 | Question: 5-a

Question

Analyse the circuit in Fig below and complete the following table
$${\begin{array}{|c|c|c|}\hline
\textbf{a}&    \textbf{b}& \bf{ Q_n} \\\hline
0&0\\\ 0&1 \\     1&0 \\    1&1 \\ \hline  
 \end{array}}$$

Comments

where is c) question?

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yes.. Actually that was moved to a new question- answer is yet to be moved. Each linked question was made separate so as to be included in exam when created.

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Part (c) is here

https://gateoverflow.in/26442/gate1991_5-c

Answer 1

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

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

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

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

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

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

$${\begin{array}{cc|c}
\textbf{a}&    \textbf{b}& \bf{ Q_n} \\\hline
0&0&0\\ 0&1& Q_{n-1} \\    1&0& Q_{n-1} \\   1&1 &1\\
\end{array}}$$

Comments

11 =>  Q_n-1(1+1) + 1.1 = Q_n-1 (1) + 1 = Q_n-1+1 = 1 hwo it is equals to 1

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1+x=1 here x=Q n-1 .

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your part c answer is wrong