We have $g(k)=\{ [(1/5)^k]u(k)\text{ for $1 \le k\le3$ and $0$ for other }k\}$
The input is $x(k)=\delta(k) +3\delta(k-1)+ \delta(k-2) $
Using Z transform we have to find the output $y(k)$ and the region of convergence of $Y(z)$.
The problem is that although I have read about the input and output and impulse in the Alan V. Oppenheim book about signals,I dont know how to solve this one,especially the part about finding the output.
$\endgroup$ 21 Answer
$\begingroup$In this case using the $\mathcal{Z}$-transform seems almost like an overkill, but let's see. I use $a=1/5$:
$$G(z)=\sum_{k=0}^3a^kz^{-k}=\frac{1-(az^{-1})^4}{1-az^{-1}}\\ X(z)=1+3z^{-1}+z^{-2}$$
So we have
$$Y(z)=X(z)G(z)=G(z)+3G(z)z^{-1}+G(z)z^{-2}\tag{1}$$
Since $Y(z)$ has the same pole as $G(z)$ (which is $z_{\infty}=a$), its region of convergence is $|z|>|a|=1/5$. (Since $Y(z)$ corresponds to a causal sequence, the region of convergence is the region outside a circle).
The inverse transform of (1) is simply
$$y(k)=g(k)+3g(k-1)+g(k-2)$$
I'm sure you can fill in the numerical values yourself if necessary.
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