Question

When the message signal m(t) is applied to PM modulator , the maximum frequency deviation of the resulting PM signal is 10kHz. If m(2t) is applied as the message signal to the same PM modulator, then the maximum frequency deviation of the resulting PM signal will be

• A. can't be determined
• B. 20 kHz
• C. 10 kHz
• D. 5 kHz

Answer 1

The correct option is B 20 kHz

(c)
When m(t) is applied as message signal:

$\Delta f_{max}=\frac{k_p}{2\pi }{\left|\frac{dm\left(t\right)}{2\pi }\right|}_{max}$

$\frac{k_p}{2\pi }{\left|\frac{dm\left(t\right)}{2\pi }\right|}_{max}=10kHz$

When x(t) = m (2t) is applied as message signal;

$\frac{dx\left(t\right)}{dt}=\frac{dm\left(2t\right)}{dt}$

$Let,\tau =2t\Rightarrow d\tau =2dt$

So, $\frac{dx\left(t\right)}{dt}=\frac{dm\left(\tau \right)}{d\tau }\times \frac{d\tau }{dt}=2\frac{dm\left(\tau \right)}{d\tau }$

${\left|\frac{dx\left(t\right)}{dt}\right|}_{max}=2{\left|\frac{dm\left(\tau \right)}{d\tau }\right|}_{max}=2{\left|\frac{dm\left(t\right)}{dt}\right|}_{max}$

So, $\Delta f_{max}=\frac{k_p}{2\pi }{\left|\frac{dx\left(t\right)}{dt}\right|}_{max}=2\left[\frac{k_p}{2\pi }{\left|\frac{dm\left(t\right)}{dt}\right|}_{max}\right]=20kHz$