Question

Consider a discrete time signal x[n] whose discrete time Fourier transform is $X\left(e^{j\omega }\right)$ where $X\left(e^{j\omega }\right)=\frac{1}{1-e^{-j\omega }}\left(\frac{sin\frac{3}{2}\omega }{sin\frac{\omega }{2}}\right)+5\pi \delta \left(\omega \right);-\pi \le \omega \le \pi$

Then the value of x[n] at n=3 is

• A. 4

Answer 1

The correct option is A 4
Given, $X\left(e^{j\omega }\right)=\frac{1}{1-e^{-j\omega }}\left(\frac{sin\frac{3}{2}\omega }{sin\frac{\omega }{2}}\right)+5\pi \delta \left(\omega \right);-\pi \le \omega \le \pi$

$X_1\left(e^{j\omega }\right)=\frac{sin\frac{3}{2}\omega }{sin\frac{\omega }{2}}$

$X_1\left(e^{j\omega }\right)=\stackrel{IDTFT}{⟷}{x}_1\left[n\right]$

$\frac{sin\frac{3}{2}\omega }{sin\frac{\omega }{2}}=\stackrel{IDTFT}{⟷}{x}_1\left[n\right]$

$\frac{sin\frac{3}{2}\omega }{sin\frac{\omega }{2}}=\stackrel{IDTFT}{⟷}\{\begin{array}{cc}1,& |n|\le 1\\ 0,& |n|>1\end{array}$

Using the accumulation property,
${\sum }_{k=-\infty }^n{x}_1\left(k\right)\stackrel{IDTFT}{\leftrightarrow }\frac{1}{1-e^{-j\omega }}{X}_1\left(e^{j\omega }\right)+\pi X_1\left(e^{j_0}\right){\sum }_{k=-\infty }^{\infty }\delta (\omega -2\pi )$

Therefore in the range $-\pi <\omega \le \pi$
${\sum }_{k=-\infty }^n{x}_1\left(k\right)\stackrel{IDTFT}{⟷}\frac{1}{1-e^{-j\omega }}{X}_1\left(e^{j\omega }\right)+3\pi \delta \left(\omega \right)$

Also in the range $-\pi <\omega \le \pi$
$1\stackrel{FT}{⟷}2\pi \delta \left(\omega \right)$
$\therefore$ In the range $-\pi <\omega \le \pi$

$x\left[n\right]=1+{\sum }_{k=-\infty }^n{x}_1\left[k\right]\stackrel{DTFT}{⟷}\frac{1}{1-e^{-j\omega }}{X}_1\left(e^{j\omega }\right)+5\pi \delta \left(\omega \right)$

$\therefore$ The discrete signal x[n] may be expressed as
$x\left[n\right]=1+{\sum }_{k=-\infty }^n{x}_1\left[k\right]=\{\begin{array}{cc}1,& n\le -2\\ n+3,& -1\le n\le 1\\ 4,& n\ge 2\end{array}$

$x\left[n\right]atn=3isx\left[3\right]=4$