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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.