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 frequency 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)
Trying part 4c again
To find F(w), I stayed in the time domain and wrote out the output for an input x(n).
After the first G(w) block, my output is
$ \,\! a(n) = x(n) * g(n) $
after the downsampler, we have
$ \,\!b(n) = a(Dn) $
after the H(w) block we have
$ \,\!c(n) = b(n)*h(n) $
after the upsampler we have
$ \,\!d(n) = c(n/D) $
after the second G(w) we have
$ \,\!y(n) = d(n)*g(n) $
combining terms and using linearity, we say
$ \,\!f(n) = g(n)*h(n/D)*g(n) $
and taking the DTFT of that we get
$ \,\!F(w) = G(w)G(w)H(Dw) $
I am not sure if i follow your logic here. Where did the a(Dn) go?
If i use linearity as you suggest I do not get the same answer. --Jwromine 01:02, 11 February 2009 (UTC)
