Question

A Gaussian message signal $m\left(t\right)=e^{-t^2}V$ is applied to a phase modulator. If the phase sensitivity of the modulator is $k_p=8000\pi rad/V$, then the maximum frequency deviation of the resultant PM signal will be ______kHz

• A. 3.43

Answer 1

The correct option is A 3.43
For a PM signal,

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

$m\left(t\right)=e^{-t^2}V$

$\frac{dm\left(t\right)}{dt}=-2t{e}^{-t^2}V$

Let, $x\left(t\right)=\frac{dm\left(t\right)}{dt}=-2t{e}^{-t^2}$

Finding the maximum value of x(t)

$\frac{dx\left(t\right)}{dt}=\frac{d}{dt}(-2t{e}^{-t^2})=-(2-4t^2)e^{-t^2}=0$

$2-4t^2=0$

$t=\pm \frac{1}{\sqrt{2}}$

So, at $t=\frac{1}{\sqrt{2}}\left|\frac{dm\left(t\right)}{dt}\right|$will attain its maximum value

${\left|\frac{dm\left(t\right)}{dt}\right|}_{max}{=\left(2t{e}^{-t^2}\right)|}_{t=\frac{1}{\sqrt{2}}}=\sqrt{2}{e}^{-0.5}$

So,$\Delta f_{max}=\frac{8000\pi }{2\pi }\left(\sqrt{2}{e}^{-0.5}\right)Hz$

$=4\sqrt{2}\left(e^{-0.5}\right)kHz=3.43kHz$