Question

A signal m(t) is applied to a FM modulator with frequency modulation constant $K_f=24kHz/V.$ If the Fourier Transform of m(t) is $M\left(\omega \right)=Sa\left(2\omega \right)$ then which of the following statement(s) is/are correct?

• A. The total frequency swing of FM signal is 12 kHz.
• B. The maximum phase deviation of FM signal is $12\pi \times {10}^{3}rad$.
• C. If the carrier frequency, $f_C=1MHz$ then rnaximum instantaneous frequency will be 1006 kHz.
• D. The maximum frequency deviation of FM signal is 6 kHz.

Answer 1

Answer: (C), (D)

The maximum frequency deviation of FM signal is 6 kHz.

$FMsignal,S_{FM}\left(t\right)=A_x\cos (2\pi f_{c}t+2\pi K_f\int m(t\left)dt\right)$
Given: $m\left(t\right)\leftrightarrow M\left(\omega \right)=Sa\left(2\omega \right)$


Maximum frequency deviation,
$|\Delta f{|}_{max}=K_{f}m\left(t\right){|}_{max}=\frac{K_f}{4}=6kHz$
$\text{Maximum phase deviation,}[\Delta \varphi {]}_{max}=max|(2\pi K_f\int m\left(t\right)dt)|$

$[\varphi {]}_{max}=2\pi K_f=10\pi \times {10}^{3}rad$
$\text{Total frequency swing}=f_{imax}{f}_{1min}$
$=[f_c+K_{f}m(t\left){|}_{max}\right]-[f_c+K_{f}m(t\left){|}_{min}\right]=[K_f\times \frac{1}{4}]-[K_f\times 0]$
$=K_f\times \frac{1}{4}=\frac{24}{4}kHz=6kHz$
Maximum instantaneous frequency,
$f_{1,max}=f_c+K_{f}m\left(t\right){|}_{max}$
$f_{i,max}=1000kHz+6kHz=1006kHz$