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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_{-\...


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: $$ ...


I propose the following answer for $$ X(z)=\mathcal{Z} \Big\{-a^nu[-(n+1)] \Big\} $$ given $$ U(z) = \mathcal{Z} \Big\{u[n] \Big\} = \frac{1}{1-z^{-1}} $$ Let $$ \{X_1, X_2, X_3\}(z) $$ denote the transformation stages under which each property applies Using the time reversal property $$ X_1(z) = U(z^{-1}) $$ Applying the time shifting property: $$ X_2(z) =...


Just because $f(z)$ converges for $|z|\le|z_0|$ for your choice of $\{ z_0 : f(z_0) = 0\}$ doesn't imply that $f(z)$ can't converge for $|z| \gt |z_0|$ also... up until the pole of next greatest magnitude.

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