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If $x[n]=(0.5)^nu[n]$ and $y[n]=(x*x)[n]$ then what is the value of $\sum\limits_{n=-\infty}^{\infty}y[n]$ ?

I calculated the $\mathcal{Z}$-transform of $x[n]$ and then applied the accumulation property of $\mathcal{Z}$-transform. $$X(z) = \frac{1}{1-0.5z^{-1}}\\ Y(z) = \bigg(\frac{1}{1-0.5z^{-1}}\bigg)^2\\ \sum\limits_{n=-\infty}^{\infty}y[n] \longleftrightarrow \bigg(\frac{1}{1-0.5z^{-1}}\bigg)^2.\bigg(\frac{1}{1-z^{-1}}\bigg)$$ Please point out the mistake I am doing. Thanks in advance for the help.

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  • $\begingroup$ What does your $*$ symbol mean? Convolution or multiplication ? $\endgroup$
    – Hilmar
    Oct 9 at 15:54
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    $\begingroup$ Interestingly enough the answer is the same for both multiplication or convolution, but it's a different path to get there. Your third equation is nonsense. Keep it mind that the sum is simply the z-transform at $ z = 1$ $\endgroup$
    – Hilmar
    Oct 9 at 16:03
  • $\begingroup$ @Hilmar the $\textbf{*}$ symbol is convolution. Can you please explain a little bit? Isn't the third equation valid for sums? $\endgroup$
    – edison
    Oct 9 at 16:15
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I was following you but I'm not sure where your last equation comes from. You want to use $Y(z)$ as I describe below.

This is the kind of problem that makes sense only if you "see it" in my opinion. The trick is that $\sum_{n=-\infty}^{\infty}y[n]$ can be computed by evaluating the z-transform at $z=1$. Remember that the z-transform equation is:

$$ Y(z)=\sum_{n=-\infty}^{\infty}y[n]z^{-n} $$

and so if we set $z=1$, we arrive at:

$$ Y(0)=\sum_{n=-\infty}^{\infty}y[n]$$

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Let's say we have a signal and it's z transform

$$x[n] \longleftrightarrow X(z)$$

and the convolution with itself

$$y[n] = (x*x)[n] \longleftrightarrow Y(z)$$

Convolution in the time domain is multiplication in the frequency domain, so we have

$$Y(z) = X^2(z)$$

The sum over each sequence is

$$\sum_{-\infty}^{\infty}x[n] = X(z)_{z=1}$$

Hence we have $$\sum_{-\infty}^{\infty}y[n] = Y(z)_{z=1} = X(z)^2_{z=1}$$

In the time domain this looks like.

$$\sum_{-\infty}^{\infty}y[n] = (\sum_{-\infty}^{\infty}x[n])^2$$

The simple answer: figure out the sum over $x[n]$ and square it. Hint: $2\cdot 2 = 4$.

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  • $\begingroup$ +1 Oh, shoot! I was always taught that it is $2+2$ that equals $4$, so presumably in this case, addition is the same as multiplication which is the same (cf. your comment on the OP's question) as convolution :-) $\endgroup$ Oct 10 at 20:33

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