Question

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
1. 100 to 750
2. 100 to 250
3. 50 to 750
4. 50 to 250

Answer 1

Correct Answer - Option 3 : 50 to 750

Concept:

The general form of frequency modulated signal is represented by:

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

A_C: carrier signal amplitude

f_c: carrier signal frequency

f_m: 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 (f_m) 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 f_m = 100 Hz

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

= 750

For f_m = 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)