Question

For identical total transmitted power, the ratio of S / N for wideband FM to S / N of DSB-AM (with m_a = 1) is:

(Given f_m = 15 kHz & frequency deviation = ± 75 kHz)

1. 112.5
2. 300
3. 75
4. 450

Answer 1

Correct Answer - Option 1 : 112.5

Concept:

Signal-to-Noise Ratio (SNR):

It is the ratio of the signal power to noise power. The higher the value of SNR, the greater will be the quality of the received output.

\(\left ( SNR \right )= \frac{Average \:\: power \:\:of \:\:modulated \:\:signal}{Average\:\: power \:\:of \:\:noise \:\:in \:\:message \:\:bandwidth}\)

The figure of Merit:

The ratio of (SNR) at the output to the (SNR) at the input is known as the figure of merit.

It is denoted by F.

\(FOM = \frac{{S/N}_{output}}{{S/N}_{input}}\)

For wideband FM

\((F)_{FM}=\frac{3β ^2}{2}\)  ----(1)

Where,

β = modulation index.

For DSB - AM,

\((F)_{DSB-AM}=\frac{μ ^2}{2+μ^2}\)   ----(2)

Where,

μ = modulation index

Calculation:

Given:

f_m = 15 kHz

Δ_f = ± 75 kHz

\(\beta =\frac{\Delta_f}{f_m}\)

\(\beta =\frac{75}{15}=5\)

As transmitted power is the same for both wideband FM and DSB-AM

So (S/N)_input is the same for both.

From equation (1):

\((FOM)_{FM}=\frac{3 \ \times \ 25}{2}=\frac{75}{2}\)    ---(3)

Putting μ = 1 in equation (2):

\((FOM)_{DSB-AM}=\frac{1}{3}\) ----(4)

Dividing equation (3) and (4) we get:

\(\frac{S/N_{FM}}{S/N_{DSB-AM}}=\frac{225}{2}=112.5\)