# Identify out signal from in signal and impulse response

Hi there I am studying a course in signal processing and systems. I was given an excecise which I am having a great deal of trouble solving. I am allowed to solved using matlab, but for top marks I must also be able to give an account of a analytical solution.

Given the in signal $$x(n)=(0,4)^nu(n)$$

And the impulse response $$h(n)=(0,8)^ncos(\dfrac{\pi n}{3})$$

compare the plots blow and select the correct alternative:

Since i needed to give an account of a analytical solution for top marks i decided to try and use z-transformations to solve this. My thinking was to use the fact that the out signal $$Y(z)=X(z)H(z)$$, and thus transform the insignal and the impulse response do the multiplication and then transform the results back.

$$X(z)=\dfrac{1}{1-0,4z^{-1}}$$

and

$$H(z)=\dfrac{1-0,8cos(\pi/3)z^{-1}}{1-2z^{-1}0,8cos(\pi/3)+(0,8)^2z^{-2}}=\dfrac{1-0,4z^{-1}}{1-0,8z^{-1}(0.8)^2z^{-2}}$$

Which gives $$Y(z)=\dfrac{1}{1-0,8z^{-1}(0.8)^2z^{-2}}$$

This I have no clue how to transform back, so i decided to let Matlab do the heavy lifting

f=cos(pi*n/3)*(0.8)^n
a=ztrans(f)
F=(0.4)^n
b=ztrans(F)
c=a*b
g=iztrans(c)


Which gave me the output:

a =(5z((5z)/4 - 1/2))/(4((25z^2)/16 - (5z)/4 + 1))

b =z/(z - 2/5)

c =(5z^2((5z)/4 - 1/2))/(4(z - 2/5)((25z^2)/16 - (5*z)/4 + 1))

g =((-1)^n3^(1/2)25^(1 - n)(- 10 - 3^(1/2)10i)^(n - 1)4i)/15 - ((-1)^n3^(1/2)25^(1 - n)(- 10 + 3^(1/2)10i)^(n - 1)4i)/15 + (2(-1)^n16^ncos((2pi*n)/3))/20^n

I have worked on this problem for about a week and can't figure it out. If i am supposed to be able to compare a function with th graph and select the correct one the function have to be fairly simple. The answer sheet sasy the correct grach is A. Feels like I overcomplicate this somewhere. Help me, please and thank you.

I'd recommend to work in the time domain. The output $$y[n]$$ is given by the convolution of $$x[n]$$ and $$h[n]$$:

\begin{align}y[n]&=\sum_{k=-\infty}^{\infty}x[n-k]h[k]\\&=u[n]\sum_{k=0}^n(0.4)^{n-k}(0.8)^k\cos\left(\frac{k\pi}{3}\right)\\&=u[n](0.4)^n\sum_{k=0}^n2^k\cos\left(\frac{k\pi}{3}\right)\tag{1}\end{align}

It's clear that the first sample of the output signal is just the multiplication of $$x[0]$$ and $$h[0]$$, i.e., $$y[0]=x[0]h[0]=1\cdot 1=1$$. These leaves us with options $$A$$ and $$C$$. Looking at the second output sample

$$y[1]=x[0]h[1]+x[1]h[0]=h[1]+x[1]=0.8\cos\left(\frac{\pi}{3}\right)+0.4=0.8\tag{2}$$

reveals that it must be option $$A$$.

A closed-form expression for $$y[n]$$ can be derived from $$(1)$$ by splitting the cosine into two complex exponentials, and using the formula for the geometric series.

• Thanks a bunch! I tried to use convolution but used h(n-k) and got stuck again. You really cleared this up for me, thanks again! Jun 15, 2021 at 17:16