GATE CS 2003 Q47 — Combinational Circuits

Let $A(t)$ be the input signal. The circuit consists of two cascaded stages of XOR gates with delayed feedback paths:

• The output of the first XOR gate is:
  $X(t) = A(t) \oplus A(t - 2)$
• The output of the second XOR gate (the final output $B$) is:
  $B(t) = X(t) \oplus X(t - 4)$

The input waveform $A(t)$ has a single transition from $0$ to $1$ at $t = 1\text{ ns}$:

We can now determine the intermediate signal $X(t)$:

• For $t < 1\text{ ns}$: $A(t) = 0$ and $A(t-2) = 0 \implies X(t) = 0 \oplus 0 = 0$
• For $1\text{ ns} \le t < 3\text{ ns}$: $A(t) = 1$ and $A(t-2) = 0 \implies X(t) = 1 \oplus 0 = 1$
• For $t \ge 3\text{ ns}$: $A(t) = 1$ and $A(t-2) = 1 \implies X(t) = 1 \oplus 1 = 0$

Thus, $X(t)$ is a pulse of duration $2\text{ ns}$ active during the interval $[1, 3]\text{ ns}$.

Next, we find $B(t) = X(t) \oplus X(t-4)$:

• $X(t)$ is active ($1$) during $[1, 3]\text{ ns}$.
• $X(t-4)$ is active ($1$) during $[5, 7]\text{ ns}$.

Since these two intervals do not overlap, the XOR output $B(t)$ is active during both intervals:

The transitions at $B$ occur at $t = 1\text{ ns}$, $t = 3\text{ ns}$, $t = 5\text{ ns}$, and $t = 7\text{ ns}$. This results in exactly 4 transitions during the interval from $0$ to $10\text{ ns}$.