# An FM detector produces a peak - to - peak output voltage of 1.2 V from an FM signal that is modulated to 10 kHz deviation by a sine wave. (a) The pea

48 views
in General
closed

An FM detector produces a peak - to - peak output voltage of 1.2 V from an FM signal that is modulated to 10 kHz deviation by a sine wave.

(a) The peak voltage is 0.6 V

(b) The detector sensitivity is 6 μV/Hz

(c) The peak voltage is 2.4 V

(d) The detector sensitivity is 60 μV/Hz

Which of the above are correct?

1. (a) and (b)
2. (b) and (c)
3. (c) and (d)
4. (a) and (d)

by (35.4k points)
selected

Correct Answer - Option 4 : (a) and (d)

Concept:

General equation of FM modulated signal is expressed as:

(S)FM = AC cos [ωct + kf ∫m(t) dt]

m(t) = Modulating signal

kf = Frequency sensitivity (Hz/volt)

Ac = Carrier Amplitude

Total peak to peak voltage = 2Ac

Also, the frequency deviation in FM is defined as:

Δf = |kf.m(t)|max

Calculation:

Given 2AC = 1.2

So the peak voltage is:

Ac = 0.6 volts

Option (a) is therefore correct.

Given frequency deviation Δf = 10 kHz

Since m(t) is a sin wave with 0.6 V as peak amplitude, frequency deviation can be written as:

⇒ (kf)(0.6) = 10 kHz

$\Rightarrow {k_f} = \frac{{10}}{{0.6}}\;\frac{{kHz}}{{Volts}}$

The detector, however, senses voltage or ampere per Hz or rad/sec. It is therefore expressed as Volts/Hz

Taking the reciprocal of frequency sensitivity we get the detector sensitivity (D) as:

$D = \frac{{0.6}}{{10}}\frac{{Volts}}{{kHz}}$

D = 60 μV/Hz

Hence both Option (a) and (d) are therefore correct.