Revision as of 05:56, 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 $ 2\pi $.

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 $ -pi $ to $ pi $. Repeat that 6 times and you basically get a rect that goes from $ -pi $ to $ 11*pi $. 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 $ -pi/6 $ to $ 11*pi/6 $. (Note: I'm ignoring the $ [1 + e^{-jw} + 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 $ -pi/6 $ and $ 11*pi/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 $ 2\pi $. This one is not. --Mboutin 09:24, 1 October 2010 (UTC)

The answer is actually $ F(\omega) = G(\omega ) H(6 \omega ) $, for $ -\pi < \omega < \pi $. --Mboutin 09:24, 1 October 2010 (UTC)

I'm a bit confused: Is $ G(\omega ) $ 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

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 $ -\pi < \omega < \pi $

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 convolution 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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