Question

$\text{Let x[n] be discrete time signal whose discrete time Fourier transform is represented as}X\left(e^{j\omega }\right)=e^{\frac{-j\omega M}{2}}\frac{\sin \left(\frac{\omega N}{2}\right)}{\sin \left(\frac{\omega }{2}\right)}\text{and x[n] is represented as shown below.}$

$\text{Then the value of}(M\times N)\text{is\_\_\_\_\_\_\_\_.}$

• A. 30

Answer 1

The correct option is A 30
Given,

$\text{We can write,}x\left[n\right]=u\left[n\right]-u[n-6]$
$\therefore x\left[n\right]\stackrel{DTFT}{⟷}X\left(e^{j\omega }\right)$

$u\left[n\right]\stackrel{DTFT}{⟷}\frac{1}{1-e^{-j\omega }}$


$u[n-6]\stackrel{DTFT}{⟷}\frac{e^{-j6\omega }}{1-e^{-j\omega }}$


$\therefore X\left(e^{j\omega }\right)=\frac{1-e^{-j6\omega }}{1-e^{-j\omega }}=\frac{e^{-j3\omega }}{e^{\frac{-j\omega }{2}}}\frac{[e^{j3\omega }-e^{-j3\omega }]}{[e^{\frac{j\omega }{2}}-e^{\frac{-j\omega }{2}}]}$
$X\left(e^{j\omega }\right)=e^{\frac{-j5\omega }{2}}\frac{\sin \left(3\omega \right)}{\sin \left(\frac{\omega }{2}\right)}$
by comparing with the given DTFT

$X\left(e^{j\omega }\right)=e^{\frac{-j\omega M}{2}}\frac{\sin \left(\omega N\right)}{\sin \left(\frac{\omega }{2}\right)}$
$\therefore M=5;$

$\frac{N}{2}=3$

$\Rightarrow N=6$

$\therefore M\times N=30$