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When I try to calculate the $\mathcal{Z}$-transform of $4\delta[n-8]\delta[n-8]$, I put the statement into the formula of $\mathcal{Z}$-transform from $-\infty$ to $+\infty$, and I get the result $4z^{-16}$, but I can't be sure about it.

I think that the ROC changes when I square the $\delta[n-8]$ location on the number chart. What is the ROC?

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  • $\begingroup$ Hi! What is this d[n-8]^2 ? something like $\delta[n-8]^2 = \delta[n-8] \delta[n-8] $? $\endgroup$ – Fat32 Jan 7 '18 at 14:49
  • $\begingroup$ Yes, it is like that. $\endgroup$ – Bay Jan 7 '18 at 15:00
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The notation is a little obscure but the answer is simple.

Since $\delta[n-8]^2$ can be represented as : $$\delta[n-8]^2 = \delta[n-8] \delta[n-8]$$ and since the multiplication of the impulses can be simplified as : $$\delta[n-8] \delta[n-8] = \delta[n-8]$$ , then you can conlcude that the $\mathcal{z}$ transform of the signal $4 \delta[n-8]^2 $ will be:

$$ \mathcal{Z}\{ 4 \delta[n-8]^2 \} = 4 z^{-8} $$ whose region of convergence is all $z$ except $z = 0$.

There is a pole at $z = 0$ and a zero at $ z \to \infty $

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