# Is $\mathcal{Z}\{4\delta[n-8]\delta[n-8]\} = 4z^{-16}$?

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?

• Hi! What is this d[n-8]^2 ? something like $\delta[n-8]^2 = \delta[n-8] \delta[n-8]$? Jan 7 '18 at 14:49
• Yes, it is like that.
– Bay
Jan 7 '18 at 15:00

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 $z \to \infty$