Question

Consider the discrete time signal,

$x\left(n\right)={\left(\frac{1}{3}\right)}^{n}u\left[n\right]$ where $u\left[n\right]=\{\begin{array}{cc}1& n\ge 0\\ 0& n<0\end{array}$

$y\left[n\right]=x[-n],-\infty <x[-n]<\infty$

Then, ${\sum }_{n=-\infty }^{\infty }y\left(n\right)=$ ___.

• A. 1.5

Answer 1

The correct option is A 1.5
$x\left[n\right]={\left(\frac{1}{3}\right)}^{n}u\left[n\right]$

$y\left[n\right]=x[-n]={\left(\frac{1}{3}\right)}^{-n}u[-n]$

${\sum }_{n=-\infty }^{\infty }y\left(n\right)={\sum }_{n=-\infty }^{\infty }{\left(\frac{1}{3}\right)}^{-n}u[-n]$

$u[-n]=\{\begin{array}{cc}1,& n\le 0\\ 0,& n>0\end{array}$

So, ${\sum }_{n=-\infty }^{\infty }y\left[n\right]={\sum }_{n=-\infty }^0{\left(\frac{1}{3}\right)}^{-n}$

Let n = -m, then we get,

${\sum }_{n=-\infty }^{\infty }y\left[n\right]={\sum }_{m=0}^{\infty }{\left(\frac{1}{3}\right)}^m$

$=1+\frac{1}{3}+{\left(\frac{1}{3}\right)}^2+...=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}$