An FM broadcasting radio station transmits signals of frequency 1

An FM broadcasting radio station transmits signals of frequency 1
| An FM broadcasting radio station transmits signals of frequency 100 MHz with a power of 10 KW. The bandwidth of the modulation signal is from 100 Hz to 1.5 Khz. If the maximum deviation set by the FCC, (δ), is 75 KHz, the range of the modulation index is

A. 100 to 750

B. 100 to 250

C. 50 to 750

D. 50 to 250

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

Concept:

The general form of frequency modulated signal is represented by:

\(s\left( t \right) = {A_C}\cos \left( {2\pi {f_c}t + \beta sin2\pi {f_m}t} \right)\)

AC: carrier signal amplitude

fc: carrier signal frequency

fm: message signal frequency

β: modulation index

\(\beta = \frac{{{\rm{\Delta f}}}}{{{f_m}}}\)

\(\beta = \frac{{frequency\;deviation}}{{message\;signal\;frequency}}\;\)

If β < 1 it is called a Narrowband FM

If β > 1it is called as Wideband FM

Calculation:

Given message signal frequency (fm) range is 100 Hz to 1.5 KHz, and deviation = 75 KHz

The modulation index is defined as:

\(\beta = \frac{Δ f}{f_m}\)

Δf : frequency deviation

For fm = 100 Hz

\(\beta = \frac{{75 \times {{10}^3}}}{{100}}\)

= 750

For fm = 1.5 KHz

\(\beta = \frac{75 \times {10}^{3}}{1.5 \times {10}^{3}}\)

= 50

So the range of modulation index is 50 to 750

Extra Concept:

Instantaneous frequency

The instantaneous frequency of the angle modulated signal ACCos[θ (t)] is defined as:

\({f_i} = \frac{1}{{2\pi }}\frac{{d\left[ {\theta \left( t \right)} \right]}}{{dt}}\)

For FM fi = fc +kf m(t)

Kf: Frequency sensitivity (Hz/volt)