GATE ECE 2023 | Question: 53

Question

In a given sequential circuit, initial states are $Q_1=1$ and $Q_2=0$. For a clock frequency of $1 \; \mathrm{MHz}$, the frequency of signal $\mathrm{Q}_2$ in $\mathrm{kHz}$, is __________ (rounded off to the nearest integer).

 

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Answer 1

We can interpret the question in two ways

Conider $Q_1,Q_2$ as current output

$D_1,D_2$ current input

$Q_1^{+},Q_2^{+}$ as next output

$1MHz=1000kHz$

Case 1:

considering $D_2=\overline{Q_1}$, $D_1=Q_1$

$Q_1$	$Q_2$	$Q_1^+$	$Q_2^+$	$D_1$	$D_2$
1	0	0	0	0	0
0	0	0	1	0	1
0	1	1	1	1	1
1	1	1	0	1	0
1	0	0	0	0	0

From the table we can see that the $Q_2$ complete 1 clock on 4 input clock or the frequency is $\frac{1000}{4}=250kHz$

(Repeat itself in 4 clock)

Case 2:

considering $D_2=\overline{\overline{Q_1}}=Q_1$, $D_1=Q1$

$Q_1$	$Q_2$	$Q_1^+$	$Q_2^+$	$D_1$	$D_2$
1	0	0	1	0	1
0	1	1	0	1	0
1	0	0	1	0	1

In this case the output $Q_2$ changes every 2 clock hence frequency is $\frac{1000}{2}=500kHz$