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I derived a an impulse response of $h[n] = (3/4) (-j3/4)^{n} u[n]$, where $u[n]$ is the unit step function. I have an input $x[n] = u[n-5]$. I can find a vector representation of the convolution of these two functions. But I this wouldn't be a closed form solution since the vector goes on infinitely. How can I find the closed form convolution of this function? I'm not sure how to graph it because it contains real and complex parts mixed in. thanks

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Please ask the moderators to migrate this to dsp.SE where it is a much better fit. You can contact them by clicking on the flag link below your question. – Dilip Sarwate Mar 2 '13 at 3:08
I already did Dilip – user1945925 Mar 2 '13 at 19:36
up vote 1 down vote accepted

A closed form solution is straightforward to obtain. You just have to actually do the computation.

Let $\theta = -i \frac{3}{4}$. (To avoid confusing myself, I will use $H$, as in Heaviside, to denote the unit step function.)

Then \begin{eqnarray} y_n = (h * x)_n &=& \sum_{k \in \mathbb{Z}} x_k h_{n-k} \\ &=& \sum_{k \in \mathbb{Z}} H_{k-5} \frac{3}{4} \theta^{n-k} H_{n-k} \\ &=& H_{n-5} \frac{3}{4} \sum_{k=5}^n \theta^{n-k} \\ &=& H_{n-5} \frac{3}{4} \sum_{k=0}^{n-5} \theta^k \\ &=& H_{n-5} \frac{3}{4} \frac{1-\theta^{n-4}}{1-\theta} \end{eqnarray}

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so the imaginary part $i$ has no effect in taking the geometric series? Just treat it like any other number? – user1945925 Mar 2 '13 at 6:14
It is a finite sum. As long as $\theta \neq 1$ (which is true here), this works. So, to answer your question, yes, treat it like any other number. Since $|\theta| <1$, you can see that $\lim_{n \to \infty} y_n = \frac{3}{4} \frac{1}{1-\theta}$. – copper.hat Mar 2 '13 at 6:16

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