મોડ્યુલસ-5, મોડ્યુલસ-8 અને મોડ્યુલસ-10 કાઉન્ટર ધરાવતા કેસ્કેડેડ કાઉન્ટર પર 12 MHz ક્લોક આવૃત્તિ લગાડવામાં આવે છે. લઘુતમ શક્ય આઉટપુટ આવૃત્તિ ________ છે.
જવાબ અને પગલાવાર ઉકેલ
કેસ્કેડેડ કાઉન્ટર ક્લોકને તેમના મોડ્યુલસના ગુણાકારથી ભાગે છે. કુલ ભાગાકાર અવયવ 5×8×10 = 400, તેથી 12 MHz / 400 = 30 kHz.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept & Reasoning
For cascaded divide-by-N stages, the output of one stage clocks the next, so frequency division factors multiply. The lowest available frequency is at the final stage. This is a direct modulus-product calculation.
How to Solve It in the Exam
Multiply all moduli first, then divide the input frequency.
Important Exam Point
Use the final-stage output for the lowest frequency.
Quick Trick
Multiply all moduli first, then divide the input frequency.
Common Mistake
Adding the moduli instead of multiplying their division ratios.
A 12 MHz clock frequency is applied to a cascaded counter containing a modulus-5 counter, a modulus-8 counter and a modulus-10 counter. The lowest output frequency possible is ________
Answer & step-by-step solution
Cascaded counters divide the clock by the product of their moduli. Here the total division factor is 5×8×10 = 400, so 12 MHz / 400 = 30 kHz.
Concept & Reasoning
For cascaded divide-by-N stages, the output of one stage clocks the next, so frequency division factors multiply. The lowest available frequency is at the final stage. This is a direct modulus-product calculation.
How to Solve It in the Exam
Multiply all moduli first, then divide the input frequency.
Important Exam Point
Use the final-stage output for the lowest frequency.
Quick Trick
Multiply all moduli first, then divide the input frequency.
Common Mistake
Adding the moduli instead of multiplying their division ratios.
मॉड्यूलस-5, मॉड्यूलस-8 और मॉड्यूलस-10 काउंटर वाले कैस्केडेड काउंटर पर 12 MHz क्लॉक आवृत्ति लगाई जाती है। न्यूनतम संभव आउटपुट आवृत्ति ________ है।
उत्तर और चरणबद्ध हल
कैस्केडेड काउंटर क्लॉक को उनके मॉड्यूलस के गुणनफल से विभाजित करते हैं। कुल विभाजन गुणांक 5×8×10 = 400, अतः 12 MHz / 400 = 30 kHz।
विस्तृत हल (मूल भाषा में)
Concept & Reasoning
For cascaded divide-by-N stages, the output of one stage clocks the next, so frequency division factors multiply. The lowest available frequency is at the final stage. This is a direct modulus-product calculation.
How to Solve It in the Exam
Multiply all moduli first, then divide the input frequency.
Important Exam Point
Use the final-stage output for the lowest frequency.
Quick Trick
Multiply all moduli first, then divide the input frequency.
Common Mistake
Adding the moduli instead of multiplying their division ratios.
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