DTFT and Inverse DTFT Homework Problem I'm trying to solve this signals homework problem:



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.



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
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).