Difference between revisions of "HW 3 Question 4 mlo" - Rhea

Revision as of 21:02, 10 February 2009

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)