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A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will be
- a)0.847
- b)0.628
- c)0.734
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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A load of 1 KΩ is connected to a diode detector which is shunted...
We know,
On putting these values in (1) we will get,
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A load of 1 KΩ is connected to a diode detector which is shunted...
Given:
- Load resistance (RL) = 1 KΩ
- Capacitor (C) = 10 k pF
- Forward resistance of diode (Rf) = 1 Ω
- Modulating signal frequency (fm) = 10 kHz
Calculating maximum permissible depth of modulation:
1. The maximum depth of modulation (m) can be calculated using the formula:
m = (1 - Vr/Vm) * 100%
2. The voltage at the output of the diode detector (Vr) can be given as:
Vr = Vm/(1 + RL/Rf)
3. The voltage at the output of the detector without modulation is:
Vm = Vc
where Vc is the carrier voltage
4. The carrier voltage (Vc) can be calculated using the formula:
Vc = (RL/(RL + Rf)) * Vm
5. The voltage across the capacitor (Vc) can be given as:
Vc = (RL/(RL + Rf)) * Vm = (RL/(RL + Rf)) * Vm * sin(2πfmt)
6. Substituting the values and simplifying, we get:
m = (1 - 1/(1 + RL/Rf)) * 100%
7. Plugging in the values of RL, Rf, and simplifying, we get:
m = (1 - 1/(1 + 1/1)) * 100% = 0.847
Therefore, the maximum permissible depth of modulation to avoid diagonal clipping is 0.847 or option 'A'.
- Load resistance (RL) = 1 KΩ
- Capacitor (C) = 10 k pF
- Forward resistance of diode (Rf) = 1 Ω
- Modulating signal frequency (fm) = 10 kHz
Calculating maximum permissible depth of modulation:
1. The maximum depth of modulation (m) can be calculated using the formula:
m = (1 - Vr/Vm) * 100%
2. The voltage at the output of the diode detector (Vr) can be given as:
Vr = Vm/(1 + RL/Rf)
3. The voltage at the output of the detector without modulation is:
Vm = Vc
where Vc is the carrier voltage
4. The carrier voltage (Vc) can be calculated using the formula:
Vc = (RL/(RL + Rf)) * Vm
5. The voltage across the capacitor (Vc) can be given as:
Vc = (RL/(RL + Rf)) * Vm = (RL/(RL + Rf)) * Vm * sin(2πfmt)
6. Substituting the values and simplifying, we get:
m = (1 - 1/(1 + RL/Rf)) * 100%
7. Plugging in the values of RL, Rf, and simplifying, we get:
m = (1 - 1/(1 + 1/1)) * 100% = 0.847
Therefore, the maximum permissible depth of modulation to avoid diagonal clipping is 0.847 or option 'A'.
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A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer?
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A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A load of 1 KΩ is connected to a diode detector which is shunted by a 10kpF capacitor. The diode has a forward resistance of 1 Ω. The maximum permissible depth of modulation, so as to avoid diagonal clipping and modulating signal frequency of 10kHz will bea)0.847b)0.628c)0.734d)None of theseCorrect answer is option 'A'. Can you explain this answer?.
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