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

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

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

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

Finding the maximum value of x(t) :

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

$2–4t^2=0\Rightarrow t=\pm \frac{1}{\sqrt{2}}$

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, $\Delta f_{max}=\frac{8000\pi }{2\pi }\left(\sqrt{2}{e}^{-0.5}\right)Hz=4\sqrt{2}\left(e^{-0.5}\right)kHz=3.43kHz$