GATE CSE 2025 | Set 2 | Question: 24

Question

In a $4$-bit ripple counter, if the period of the waveform at the last flip-flop is $64$ microseconds, then the frequency of the ripple counter in kHz is _________. (Answer in integer)

Comments

 

Might help!
Giga (G)		$10^9$	$1 \text{ GHz} = 10^9 \text{ Hz}$
Mega (M)		$10^6$	$1 \text{ MHz} = 10^6 \text{ Hz}$
Kilo (k)		$10^3$	$1 \text{ kHz} = 10^3 \text{ Hz}$
Milli (m)		$10^{-3}$	$1 \text{ ms} = 10^{-3} \text{ s}$
Micro ($\mu$)		$10^{-6}$	$1 \text{ }\mu\text{s} = 10^{-6} \text{ s}$
Nano (n)		$10^{-9}$	$1 \text{ ns} = 10^{-9} \text{ s}$

Answer 1

ANSWER :  250KHZ

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Comments

beautiful (answer + handwriting)

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But how, we are calculating clock frequency here not the frequency of the ripple counter

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Counter Type	FF Clock Source	Output Frequency	Time Period per FF	Total Cycle Time
Asynchronous	Previous FF’s output	Halves each FF	Doubles each FF	2n⋅T2^n \cdot T
Synchronous	Same clock for all FFs	Same for all	Same for all	n⋅Tn \cdot T

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corrected it.

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Nice explanation.

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nice explaination..

Answer 2

\[f_{\text{last}} = \frac{1}{T_{\text{last}}}\]\[f_{\text{last}} = \frac{1}{64 \times 10^{-6}}\]\[f_{\text{last}} = 15625 \text{ Hz} = 15.625 \text{ kHz}\]\[f_{\text{last}} = \frac{f_{\text{input}}}{2^4} = \frac{f_{\text{input}}}{16}\]\[f_{\text{input}} = f_{\text{last}} \times 16\]\[f_{\text{input}} = 15625 \times 16 = 250000 \text{ Hz} = 250 \text{ kHz}\]
 

Comments

Thank you so much!

Answer 3

Answer = 250KHz

Comments

thnks this is best than any above

Answer 4

Hence -> 250Khz