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I'm trying to solve this signals homework problem:

enter image description here

So for part a, since multiplication in the time domain is convolution in the frequency domain, I just used a DTFT table, found the DTFT for $\left(\frac12\right)^n$ and $\cos(\pi n/2)$, convolved them, and solved for $H(\Omega)$. I got the same answer they have in part a.

enter image description here

I don't quite understand their solution though, which leads me to believe there is a more intuitive way to think about this.

Anyways...for part b, you would think you'd use the DTFT of $\cos(n \pi/2)$, multiply that by $H(\Omega)$ and take that whole result and Inverse DTFT it back to the time domain, then solve for $y[n]$.

However, I can't get the math to work, and I can't seem to follow their solution.

Can anyone show me mathematically or intuitively how they get that final $\frac43\cos(n\pi/2)$?

Thanks, Clint

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  • $\begingroup$ Hi, any chance of getti LaTeX instead of screenshoots and plain-text equations? $\endgroup$ – jojek Jun 25 '16 at 18:09
  • $\begingroup$ Sorry, I wish I could, but I'm taking this from an old PDF. $\endgroup$ – robertneville777 Jun 25 '16 at 18:11
  • $\begingroup$ then learn to do $\LaTeX$ here at dsp.se . also you should show us that you tried to compute the Z transform of $h[n]$. $\endgroup$ – robert bristow-johnson Jun 25 '16 at 18:18
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    $\begingroup$ @robertneville777: It is extremely easy. Here is some tutorial. $\endgroup$ – jojek Jun 25 '16 at 18:53
  • $\begingroup$ @robertneville777: Check what I did to your text (click the edit button), so you can see how you can easily Latexify your questions. Doesn't take much time but makes it much more readable. $\endgroup$ – Matt L. Jun 25 '16 at 18:58
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The only thing you need to know is that an LTI system's response to a sinusoidal input signal is a sinusoidal output signal with the same frequency as the input signal, with its amplitude and phase modified by the magnitude and phase of the system's transfer function evaluated at the input frequency, or, in math:

$$x[n]=A\cos(\omega_0n)\Longrightarrow y[n]=A|H(\omega_0)|\cos(\omega_0n+\phi(\omega_0))\tag{1}$$

where $x[n]$ and $y[n]$ are the input and output signals, respectively, and $H(\omega)=|H(\omega)|e^{j\phi(\omega)}$ is the system's frequency response.

So if you got $H(\omega)$, you just need to evaluate it at the given input frequency $\omega_0=\pi/2$ and apply $(1)$. Splitting up the cosine into two complex exponentials, as suggested by the given solution, is not necessary (and not even easier I'd say).

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  • $\begingroup$ Wow, yeah that made it way easier to compute the output. I guess the solution decided to split it up into complex exponentials because in their notes, they use complex exponentials instead of sines and cosines. Either way, this makes more sense than how I was going about it. Thanks! $\endgroup$ – robertneville777 Jul 2 '16 at 18:01

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