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The text of my exercise ask :

Determine the closed form of the $\mathcal{Z}$-transform for this $x(n)$ $$ x(n) = \begin{cases} |n-N| & \text{if 0<$n$<2N} \\ 0 & \text{elsewhere} \end{cases} $$

I see $x(n)$ like a composition of two signals

$$ x(n)= x'(n)+x''(n) $$ I use for greater semplicity $x_1(n)$ $$x_1(n)= (u(n)-(u-(N+1))$$ whence $$x'(n)= (n-N) x_1(n) $$ $$x''(n)= x'(n-N)$$

Determination $X(Z)$ it easy, but if the above equations are correct, for this reason i need post this question here.

$$X_1(Z)= \frac{z^{-(N+1)}-1}{Z^{-1} -1}$$ $$X'(Z)= z^{-1} \frac{{d}}{dz^{-1}} X_1(Z)- N X_1(Z)$$ $$X''(Z)= X_1 (\frac{1}{Z})z^{-N}$$

This is my solution.

$$X(Z)= X'(Z)+X''(Z)$$ enter image description here

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Or you can proceed this way: \begin{align}\mathcal Z\left\{x(n)\right\}= X(z) &= \sum_{n=-\infty}^\infty x(n)z^{-n}\\ & = 0 + \sum_{n=0}^{2N} \left|n - N\right|z^{-n} + 0\\ & = \sum_{n=0}^{N} \left(-n + N\right)z^{-n} + \sum_{n=N}^{2N} \left(n - N\right)z^{-n}\\ & = N\sum_{n=0}^{N}z^{-n} - N\sum_{n=N}^{2N}z^{-n} - \sum_{n=0}^{N} nz^{-n} + \sum_{n=N}^{2N} nz^{-n} \end{align}

And changing the boundaries for the second and fourth term, putting $k = n - N$, you get :

$$X(z) = N\sum_{n=0}^{N}z^{-n} - Nz^{-N}\sum_{k=0}^{N}z^{-k} - \sum_{n=0}^{N} nz^{-n} + z^{-N}\sum_{k=0}^{N} kz^{-k} $$

You can continue from there.

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  • $\begingroup$ Mmmh.. i don't verify if the result is same because my teacher want a resolution with the method that I wrote, without explicit calculation. if i use this method, your method instead there are many calculations. $\endgroup$ – P_B Feb 10 '16 at 14:58
  • $\begingroup$ Your exercise text does NOT specify the method to use. And your now-edited question has "your" $X(z)$ solution included. What is your question right now ? $\endgroup$ – Gilles Feb 10 '16 at 15:07
  • $\begingroup$ My question is about my solution..is correct? indeed i wrote "Determination X(Z) it easy, but if the above equations are correct, for this reason i need post this question here." Perhaps I have not explained well , but my doubt was about the equations that I had written. That is why now I have added my solution for completeness: $$x''(n)= x'(n-2N)$$ $\endgroup$ – P_B Feb 10 '16 at 15:22
  • $\begingroup$ Or is : $$x''(n)= x'(n-N)$$ . $\endgroup$ – P_B Feb 10 '16 at 15:29
  • $\begingroup$ Btw, "your method instead there are many calculations", it's just using the properties of the $\mathcal{Z}$-transform. You have to edit your question and specify clearly the teacher's suggested method (or yours if this is the way you want to go). And include clearly the doubts you're having. What do you want to confirm. And I will edit/delete my answer since you're having a different question. $\endgroup$ – Gilles Feb 10 '16 at 17:34

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