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My lecturer has decided that in order to pass our signal analysis class, we have to give a presentation on the blackboard in front of the entire class, so its kinda critical that i get this right. (Issue is i'm pretty bad at this class). I managed to convolute the subsystems, but i would appreciate some confirmation that its correct.

Then, i have to check that it's BIBO stable, and i'm not quite sure how to proceed.

Here is my work so far enter image description here

EDIT: It was pretty easy showing that each cascade in the system was BIBO, as Hilmar hinted, but this shows me that my response on the convolution is wrong, as I can't seem to be able to get that to converge. I still don't get what i am doing wrong. I see the geometric sum similarity, but i dont get how the unit step affects it

enter image description here

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HINTS:

  1. Concerning the result of the convolution, think about the range of $n$ for which your result is valid.

  2. This will have consequences on the sum $\sum|h[n]|$. First of all, the sum index should be $n$, not $k$. Second, from the previous point you should now know over which values of $n$ you need to sum.

  3. I assume you know the formula for $$\sum_{n=0}^{\infty}|a|^n,\qquad |a|<1$$ If you differentiate both sides of the corresponding equation you obtain the formula you need. This will give you an exact value for the sum, which is actually more than you need because you only want to show that the sum converges.

  4. You might as well consider showing BIBO-stability by looking at the poles of the total system.

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    $\begingroup$ Another hint: The cascade is BIBO stable if each of the cascaded systems is BIBO stable. $\endgroup$
    – Hilmar
    Apr 16, 2023 at 12:53
  • $\begingroup$ I don't quite get what you mean with point 1 and 2, about what n's are valid/ should be summed over. Does this have something to do with the unit step function ? Does this also affect the absolute summation ? $\endgroup$
    – Iscariot
    Apr 16, 2023 at 19:32
  • $\begingroup$ @Iscariot: You wrote that $h[n]=(n+1)a^n$. If that were valid for all possible values of $n$ then $h[n]$ would be unstable. But the fact is that that formula is only valid for $n$ in a certain range. Another hint: is $h[-10]$ really given by $-9a^{-10}$? $\endgroup$
    – Matt L.
    Apr 17, 2023 at 6:44

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