Latest revision as of 09:11, 1 October 2010

Discussion related to midterm 1

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Possible formula sheet for exam 1 Add things or suggest items? Side note: the formula sheet on the practice exam seems to be suitable. Will we see something similar?

It will not be the same formula sheet. --Mboutin 09:07, 1 October 2010 (UTC)

Midterm 1 Spring 2009 Question 3

a) $ H(w) = \frac{1}{3}[1 + e^{-jw} + e^{-j2w}] $

b) $ G(w) = rect(w\frac{3}{\pi}) $

$ A(w) = \frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{w-2\pi k}{6}) $

$ B(w) = A(w)H(w) = \frac{1}{3}[1 + e^{-jw} + e^{-j2w}] \cdot \frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{w-2\pi k}{6}) $

$ C(w) = B(6w) = \frac{1}{3}[1 + e^{-j(6w)} + e^{-j2(6w)}] \cdot \frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{6w-2\pi k}{6}) $

$ F(w) = C(w)G(w) = \frac{1}{3}[1 + e^{-j(6w)} + e^{-j2(6w)}] \cdot\frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{6w-2\pi k}{6}) \cdot rect(w\frac{3}{\pi}) $

Is this correct?

• I think the limits of the summation during downsampling go from 0 to D-1. This is because in the frequency domain you are trying to insert D copies of the signal every 2π.

• Yes, I agree with the previous statement. What ends up having is that you repeat the rect 6 times (k goes from 0 to D-1 = 5). Also, notice that since you're downsampling by 6, your down-sampled rect goes from − p'i to p'i. Repeat that 6 times and you basically get a rect that goes from − p'i to 11 * p'i. When you then up-sample that, you basically compress everything in the frequency domain by a factor of 6 (hence why you have 6*w/6). That means your rect will now go from − p'i / 6 to 11 * p'i / 6. (Note: I'm ignoring the [1 + e ^{− j'w} + e ^{− j2w}] for now to simplify the concept, but I don't think it affects the reasoning here). And finally sending it through a low pass filter, the "extra" rects get filtered out so when you end up with non-zero frequencies only between − p'i / 6 and 11 * p'i / 6. I end up with a final answer of
• $ F(w) = C(w)G(w) = \frac{1}{3}[1 + e^{-j(6w)} + e^{-j2(6w)}] \cdot\frac{1}{6} rect(w\frac{3}{\pi}) $
• Also, I think this should make sense logically, downsampling and upsampling should cancel each other out (except for the 1/D factor), so you should have the intial rect function multiplied by 1/D and H(w).

• WARNING: The function you get MUST be periodic with period 2π. This one is not. --Mboutin 09:24, 1 October 2010 (UTC)

• The answer is actually F(ω) = G(ω)H(6ω), for − π < ω < π. --Mboutin 09:24, 1 October 2010 (UTC)
  • correction to Mboutin's comment above: The LPF was Unity gain, not a full interpolator( Upscaling + LPF) Therefore the equation is F(ω) = (1/6)G(ω)H(6ω), for − π < ω < π.

• I'm a bit confused: Is G(ω) periodic to begin with? Is this because frequency itself is periodic for discrete signals? Or do we indeed take k to go from $ -\infty $ to $ \infty $ instead of from 0 to D-1
  • The DTFT is always periodic with a period $ 2\pi $. --Mboutin 13:11, 1 October 2010 (UTC)
• Also, shouldn't the answer have a factor of 1/D since downsampling the function changes the amplitude by 1/D but upsampling doesn't multiply it by anything. So shouldn't we have $ F(\omega) = \frac{1}{6} G(\omega ) H(6 \omega ) $, for − π < ω < π
  • Yes, you are correct. --Mboutin 13:11, 1 October 2010 (UTC)

Does anyone know what the trick is for doing 1A and 1c? I know there is a trick because doing integration by parts is just too damn long.

• Yes, there is a function that breaks down the system. "sin(x)cos(y)=(sin(x+y)+sin(x-y))/2". You can then simply take the system as 2 separate sin functions.
• Actually, there is a simpler way to answer that question a) : Just use the multiplication property of the CTFT. Then look in the table for the CTFT of cosine, ant the CTFT of sinc. --Mboutin 08:48, 1 October 2010 (UTC)

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