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Official PYQ · 2025

GATE CS 2025 Set 2 Q24 — Sequential Circuits

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In a 44-bit ripple counter, if the period of the waveform at the last flip-flop is 6464 microseconds, then the frequency of the ripple counter in kHz is _________. (Answer in integer)
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Answer: 250:250

A 44-bit ripple counter consists of 44 cascaded flip-flops. Each stage divides the frequency of its input signal by 22.

For a 44-bit ripple counter, the frequency of the waveform at the last (fourth) flip-flop, foutf_{\text{out}}, is related to the input clock frequency, finf_{\text{in}}, by:

fout=fin24=fin16f_{\text{out}} = \frac{f_{\text{in}}}{2^4} = \frac{f_{\text{in}}}{16}

In terms of time periods, since T=1fT = \frac{1}{f}:

Tout=16×TinT_{\text{out}} = 16 \times T_{\text{in}}

Given that the period of the waveform at the last flip-flop is Tout=64 μsT_{\text{out}} = 64\ \mu\text{s}:

64 μs=16×Tin64\ \mu\text{s} = 16 \times T_{\text{in}}Tin=64 μs16=4 μsT_{\text{in}} = \frac{64\ \mu\text{s}}{16} = 4\ \mu\text{s}

The frequency of the ripple counter (which is the input clock frequency finf_{\text{in}}) is:

fin=1Tin=14 μs=0.25 MHz=250 kHzf_{\text{in}} = \frac{1}{T_{\text{in}}} = \frac{1}{4\ \mu\text{s}} = 0.25\ \text{MHz} = 250\ \text{kHz}

Thus, the frequency of the ripple counter is 250 kHz.

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