Question

Consider Two LTI system in cascade as shown below, y(n) is output .

if $h_1\left(n\right)=3^{n}u\left(n\right)$ and $h_2\left(n\right)=3\delta \left(n\right)-9\delta (n-1)$

if x(n) = sin(n) then

• A. $y\left(n\right)\ast \left[\delta \right(n)-\delta (n-1\left)\right]=3sin\left(n\right)-3sin(n-1)$
• B. $y\left(n\right)=5sin\left(\frac{n}{2}\right)+3sin\left(n\right)$
  
• C. $y\left(n\right)\ast \left[\delta \right(n)-\delta (n-1\left)\right]=5sin\left(n/2\right)-3sin(n-1)$
• D. y(n) = 3sin(n)

Answer 1

The correct option is A $y\left(n\right)\ast \left[\delta \right(n)-\delta (n-1\left)\right]=3sin\left(n\right)-3sin(n-1)$

(a,c)
In the given question two LTI system are in cascade the overall impulse response will be

$h\left(n\right)=h_2\left(n\right)\ast h_1\left(n\right)$

= $\left[3\delta \right(n)-9\delta (n-1\left)\right]\ast 3^{n}u\left(n\right)$

=$3^{n+1}u\left(n\right)-3^{n+1}u(n-1)$

=$3^{n+1}u\left(n\right)-u(n-1)$

=$=3^{n+1}\delta \left(n\right)$

h(n) = 3$\delta \left(n\right)$

Since input is sin(n) so output will be
sin(n)*h(n) = 3$\delta \left(n\right)$ * sin(n)
= 3 sin (n)