एक समानांतर अनुनादक बैंडपास फ़िल्टर की अनुनादक आवृत्ति 20 kHz है और इसका बैंडविड्थ 2 kHz है। तो इसकी अधिकतम विच्छेद आवृत्ति ___________है। 

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LPSC (ISRO) Technician B (Electronic Mechanic): Previous Year Paper (Held on 4 March 2018)
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  1. 19 kHz
  2. 22 kHz
  3. 18 kHz
  4. 21 kHz

Answer (Detailed Solution Below)

Option 4 : 21 kHz
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Detailed Solution

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संकल्पना:

एक समानांतर RLC परिपथ के प्रतिबद्ध Z और आवृत्ति के बीच का आलेख:

F1 Jai 21.11.20 Pallavi D2

यहाँ, 

f1 न्यूनतम विच्छेद आवृत्ति है। 

fअधिकतम विच्छेद आवृत्ति है। 

fr अनुनादक आवृत्ति है। 

BW बैंडविड्थ है। 

सूत्र:

BW = f2 – f1

\({f_1} = {f_r} - \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

गणना:

दिया गया है

अनुनादक आवृत्ति fr = 20 kHz

बैंडविड्थ = 2 kHz

अधिकतम विच्छेद आवृत्ति को निम्न रूप में ज्ञात किया गया है:

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {20kHz} + \left( {\frac{{2kHz}}{2}} \right)\)

f2 = 21 kHz

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