Part 4a
Given
$ y(n) = \frac{x(n) + x(n-1) + x(n-2)}{3} \quad (1) $
to find H(w), we let
$ x(n)\,\! = \exp(jwn) \quad (2) $
such that
$ y(n)\,\! = H(w)\exp(jwn) $
substituting equation (1) into (2)
$ y(n)\,\! = \frac{\exp(jwn) + \exp(jw(n-1)) + \exp(jw(n-2))}{3} \quad (3) $
factoring out the $ \,\! \exp(jwn) $ from equation (3), we obtain H(w)
$ \,\! H(w) = \frac{1 + \exp(-jw) + \exp(-jw2)}{3} $
Part 4b
$ \left| H(w)\right| $
Part 4c
To find the the overall frquency response F(w) for this system, I assumed the up/down samplers canceled each other out and so we were left with the LPF and original H(w).
Combining terms
$ \,\! F(w) = G(w)H(w)G(w) $
- It is important to realize that upsampling is not the inverse of downsampling: when you upsample after having downsampled, you introduce zeros in the signal that were not previously there. To undo a downsampling, you have to use an interpolator (as defined in class a couple of lectures ago). --Mboutin 11:46, 10 February 2009 (UTC)
- Ah, gotcha, thanks! --Mlo 12:03, 10 February 2009 (UTC)
