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So I think I have the answer I just want to make sure I'm doing things correctly.

Let a be in the open unit disk, $F_s > 0$, and $0<f_0 < \frac{F_s}{2}$. Let $h[n]=0$ for $n < 0$ and $h[n] = a^n\cos(2\pi\frac{n}{F_s}f_0)$ for $n \geq 0$ be the impulse response of an LTI system. What is the output for the input sequence $x[n]=0$ for $n < 0$ and $x[n]=\cos(\frac{2\pi f_1 n}{F_s})$ for $n \geq 0$. Assume that $0<f_1 < \frac{F_s}{2}$.

So basically I form the convolution sum:

$y[k] = \sum_{n=0}^{\infty}a^n \cos(2\pi\frac{n}{F_s}f_0) \cos(\frac{2\pi f_1 (n-k)}{F_s})$

It would take far too long to type it all up and latex but basically I split the cosines out into their exponential forms, FOIL them, pull the exponential functions which only depend on k outside of the infinite series, and then split up the series and perform the z-transform for each one to obtain:

$$y[k] = \frac{1}{4}e^{\tfrac{2\pi ikf_1}{F_s}}\left[ H \left(\frac{1}{a}e^{\tfrac{-2\pi ik(f_1-f_0)}{F_s}}\right) + H\left(\frac{1}{a}e^{\tfrac{-2\pi ik(f_0+f_1)}{F_s}}\right)\right] + \\ \frac{1}{4}e^{\tfrac{-2\pi ikf_1}{F_s}}\left[ H\left(\frac{1}{a}e^{\tfrac{-2\pi ik(f_0+f_1)}{F_s}}\right) + H\left(\frac{1}{a}e^{\tfrac{-2\pi ik(f_0-f_1)}{F_s}}\right)\right]$$

Is this correct? Thanks.

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