Question

Consider the cascade of two LTI systems as shown below
where



$h_1\left[n\right]=sin4n$ and $h_2\left[n\right]=\left(\frac{1}{2}{)}^{n}u\right[n]$ and the input is $x\left[n\right]=\delta \left[n\right]-\frac{1}{2}\delta [n-1].$
Then the output $y\left[n\right]$ is

• A. $sin2n$
• B. $2sin4n$
• C. $sin4n$
• D. $2sin2n$

Answer 1

The correct option is C $sin4n$

Given LTI systems



$y\left[n\right]=x\left[n\right]\ast h_2\left[n\right]\ast h_1\left[n\right]$

$=x\left[n\right]\ast h_2\left[n\right]\ast h_1\left[n\right]$

$x\left[n\right]\ast h_2\left[n\right]=\left[\right(\frac{1}{2}{)}^{n}u\left[n\right]\ast \left(\delta \right[n]-\frac{1}{2}\delta [n-1\left]\right)]$

$=\left(\frac{1}{2}{)}^{n}u\right[n]-\frac{1}{2}(\frac{1}{2}{)}^{n-1}u[n-1]$

$=\left(\frac{1}{2}{)}^{n}u\right[n]-(\frac{1}{2}{)}^{n-1}u[n-1]$

$=\left(\frac{1}{2}{)}^n\right[u\left[n\right]-u[n-1]]$

$x\left[n\right]\ast h_2\left[n\right]=\delta \left[n\right]$

$y\left[n\right]=\delta \left[n\right]\ast sin4n$

$y\left[n\right]=sin4n$