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If f(k)=4^k U(k)and g(k)=5^k U(k), then find the z-transform of f(k).g(k) I know that \( Z[a^k U(k)] = 1/(1-az^{-1})\) |z|>|a|
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if it was f(k) * g(k) * = convolution then it would have been easy
Just so we're on the same page... Is it \[U[k-c]=\begin{cases}1&\text{for }k\ge c\\0&\text{for }k<c\end{cases}~~?\]
I suppose we can say this: \[\left(4^kU[k]\right)\left(5^kU[k]\right)=20^k\left(U[k]\right)^2=20^kU[k]\]
yes, U is that only.... oh, and is that multiplication directly possible ?
I would think so. Wolfram would agree: \[Z[a^kU[k]]=\frac{1}{1-\dfrac{a}{z}}=\frac{z}{z-a}\] http://www.wolframalpha.com/input/?i=ZTransform%5B%284%5Ek*Heaviside%5Bk%5D*5%5Ek*Heaviside%5Bk%5D%29%2Ck%2Cz%5D
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