# Finite moving average filter

I am trying to solve this problem but I need a lot of help. Below are my answers for the separate parts, please check and tell me where I am wrong because I am weak on the fundamental concepts of this.

• b. FIR, Causal, Unstable
• c. band-pass.

• Could you also elaborate on your answers for b and c and also add what you have tried (or at least your thoughts) about the other questions? Nov 13 '17 at 10:13

The best way to approach problems like this is to understand conceptually what's going on. Think about what you're doing when you apply an averaging filter, or simply take averages across a signal (like an image). Basically , it gets "blurry", and you lose lots of high frequency components like edges and details. So you know you should expect a lowpass response in the frequency domain. This response should have high values around 0 (low freq), and low values around pi (high freqs). Understanding this will help you check your work when you actually go through the problems, so you know if you're doing things correctly.

But let's just go through these problems briefly

a. Another name for a filter $h[n]$ is the impulse response; i.e. $h[n]$ is just what the output $y[n]$ of your system would be if the input was a delta impulse ($x[n] = \delta[n]$).

b. Just definitions.
- A signal is causal when your output only depends on values of the signal at the present ($x[n]$) or in the past ($x[n-n_0]$). Check $y[n]$ to see if this is the case. If you're given only an $h[n]$, check if it is nonzero only when $n\geq0$.
Perform the Z-Transform on both sides of your system; you should get something like $Y(z) = X(z) \cdot H(z)$ (recall, your system in the time domain is just a convolution of $x[n]$ and $h[n]$, which becomes a multiplication of $X(z)$ and $H(z)$ in the Z domain. Find $H(z)$ simply by dividing $X(z)$ on both sides.
Now you can find the zeros (zeros of the numerator) and poles (zeros of the denominator) of this $H(z)$, determine if it's FIR/IIR (FIR has no poles), its magnitude response in Fourier ($|H(e^{j\omega})|$), and how it changes with respect to $q$.
g. Again, think about what's actually happening. If your q is 1, then your response is just $y[n] = \frac{1}{1}(x[n]) = x[n]$. Nothing happens to your image; you're literally letting in all frequencies. When q is 2, you actually start taking some averages and your signal gets more blurred. As you increase q, more and more values are averaged, leading to a more blurry image (try it on MATLAB and see what happens). This is equivalent to your filter having a smaller and smaller range in the frequency domain (it's attenuating more and more high frequencies).
h. The cutoff frequency of $H(z)$ in this case is its first (positive) zero in the frequency domain.