Topic: Communication Systems
For identical total transmitted power, the ratio of $S/N$ for wide band FM to $S/N$ of DSB-AM (with $m_{a}=1$) is (Given FM = 15 kHz and frequency deviation $=\pm75~kHz$)
- A. 112.5
- B. 300
- C. 75
- D. 450
Answer: A. 112.5
Explanation: $(FOM)_{wide~bandFM}=\frac{3}{2}\beta^{2}=\frac{(S/N)_{o}}{(S/N)_{i}}$. $(FOM)_{DSBAM}=\frac{1}{3}=\frac{(S/N)_{o}}{(S/N)_{i}}$. Since the total transmitted power is same for both wide band FM and DSB-AM, then $(\frac{S}{N})_{input}$ for both will also be same. $\beta=\frac{\Delta f}{f_{m}}=\frac{75}{15}=5$. $\frac{(S/N)_{wide~band~FM}}{(S/N)_{DSB-AM}}=\frac{3/2\beta^{2}}{1/3} = \frac{9}{2}\times25=112.5$.