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An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector is
  • a)
    500 μ sec.
  • b)
    20 μ sec.
  • c)
    0.2 μ sec.
  • d)
    1 μ sec.
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
An AM signal is detected using an envelope detector. The carrier frequ...
**Envelope Detector**

An envelope detector is a circuit used to extract the envelope of an amplitude-modulated (AM) signal. It is commonly used in radio receivers to demodulate the received signal.

The envelope detector consists of a diode, a resistor, and a capacitor. The diode rectifies the AM signal, allowing only the positive half-cycles to pass through. The resistor and capacitor form an RC circuit, which acts as a low-pass filter to smooth the rectified signal and extract its envelope.

**Time Constant and Cut-off Frequency**

The time constant of an RC circuit is the product of the resistance and the capacitance (RC). It determines how quickly the capacitor charges and discharges in response to changes in the input signal.

The cut-off frequency of an RC circuit is the frequency at which the output signal starts to attenuate. It is given by the formula 1/(2πRC).

In the case of an envelope detector, the time constant of the RC circuit should be chosen such that it allows the modulating signal frequency to pass through while attenuating the carrier frequency.

**Calculation**

Given:
Carrier frequency (fc) = 1 MHz = 1 × 10^6 Hz
Modulating signal frequency (fm) = 2 kHz = 2 × 10^3 Hz

To determine an appropriate value for the time constant, we need to calculate the cut-off frequency of the RC circuit.

Cut-off frequency (f_cutoff) = 1/(2πRC)

Since we want the cut-off frequency to be higher than the modulating signal frequency and lower than the carrier frequency, we can set the cut-off frequency to be approximately equal to the modulating signal frequency.

f_cutoff ≈ fm

Substituting the values:
1/(2πRC) ≈ 2 × 10^3 Hz

Simplifying the equation:
RC ≈ 1/(2π × 2 × 10^3)

Taking an approximate value for π (3.14):
RC ≈ 1/(2 × 3.14 × 2 × 10^3)

Simplifying further:
RC ≈ 1/(12.56 × 10^3)

Choosing a convenient value for R (let's say 10 kΩ):
C ≈ 1/(12.56 × 10^3 × 10 × 10^3)

Simplifying the equation:
C ≈ 1/(125.6 × 10^6)

C ≈ 7.96 × 10^(-9) F

Therefore, an appropriate value for the time constant of the envelope detector is approximately 10 kΩ and 7.96 nF, which corresponds to option 'B' (20 sec).
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An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector isa)500 μsec.b)20 μsec.c)0.2 μsec.d)1 μ sec.Correct answer is option 'B'. Can you explain this answer?
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An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector isa)500 μsec.b)20 μsec.c)0.2 μsec.d)1 μ sec.Correct answer is option 'B'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector isa)500 μsec.b)20 μsec.c)0.2 μsec.d)1 μ sec.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An AM signal is detected using an envelope detector. The carrier frequency and modulating signal frequency are 1 MHz and 2 kHz respectively. An appropriate value for the time constant of the envelope detector isa)500 μsec.b)20 μsec.c)0.2 μsec.d)1 μ sec.Correct answer is option 'B'. Can you explain this answer?.
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