फ्रीक्वेंसी काउंटर में आउटपुट आवृत्ति सैंपलिंग अंतराल से _____________ संबंधित होती है
उत्तर और चरणबद्ध हल
C — फ्रीक्वेंसी काउंटर निश्चित गेट/सैंपलिंग अंतराल में इनपुट चक्र गिनकर आवृत्ति निकालता है (f = N/T_g)। लंबा अंतराल अधिक चक्र गिनता है, इसलिए आवृत्ति रेज़ोल्यूशन और परिशुद्धता बढ़ती है।
विस्तृत हल (मूल भाषा में)
Concept & Reasoning
If N complete cycles are counted during a gate interval T_g, the measured frequency is N/T_g. The ±1-count uncertainty corresponds to an approximate frequency resolution of 1/T_g. Thus increasing the sampling interval reduces the frequency quantization step and improves precision.
How to Solve It in the Exam
Longer gate time → finer frequency resolution.
Important Exam Point
Separate actual signal frequency from measurement resolution; the signal itself does not change when gate time changes.
Quick Trick
Longer gate time → finer frequency resolution.
Common Mistake
Interpreting a longer sampling interval as lowering the actual input frequency.
The output frequency related to the sampling interval of a frequency counter as _____________
Answer & step-by-step solution
A frequency counter estimates frequency by counting input cycles during a fixed gate/sampling interval. A longer interval counts more cycles and therefore improves frequency resolution and precision; the basic count-to-frequency relation is f = N/T_g.
Concept & Reasoning
If N complete cycles are counted during a gate interval T_g, the measured frequency is N/T_g. The ±1-count uncertainty corresponds to an approximate frequency resolution of 1/T_g. Thus increasing the sampling interval reduces the frequency quantization step and improves precision.
How to Solve It in the Exam
Longer gate time → finer frequency resolution.
Important Exam Point
Separate actual signal frequency from measurement resolution; the signal itself does not change when gate time changes.
Quick Trick
Longer gate time → finer frequency resolution.
Common Mistake
Interpreting a longer sampling interval as lowering the actual input frequency.
ફ્રિક્વન્સી કાઉન્ટરમાં આઉટપુટ આવૃત્તિ સેમ્પલિંગ અંતરાલ સાથે _____________ સંબંધિત છે
જવાબ અને પગલાવાર ઉકેલ
C — ફ્રિક્વન્સી કાઉન્ટર નિશ્ચિત ગેટ/સેમ્પલિંગ અંતરાલમાં ઇનપુટ ચક્રો ગણીને આવૃત્તિ મેળવે છે (f = N/T_g). લાંબો અંતરાલ વધુ ચક્રો ગણે છે, તેથી આવૃત્તિ રિઝોલ્યુશન અને ચોકસાઈ વધે છે.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
Concept & Reasoning
If N complete cycles are counted during a gate interval T_g, the measured frequency is N/T_g. The ±1-count uncertainty corresponds to an approximate frequency resolution of 1/T_g. Thus increasing the sampling interval reduces the frequency quantization step and improves precision.
How to Solve It in the Exam
Longer gate time → finer frequency resolution.
Important Exam Point
Separate actual signal frequency from measurement resolution; the signal itself does not change when gate time changes.
Quick Trick
Longer gate time → finer frequency resolution.
Common Mistake
Interpreting a longer sampling interval as lowering the actual input frequency.
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