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.

The task:

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: enter image description here

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.




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


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.


1 Answer 1


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


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


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.

  • $\begingroup$ 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! $\endgroup$
    – Aedrha
    Jun 15, 2021 at 17:16

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