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:: 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 <math> 2\pi </math>. | :: 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 <math> 2\pi </math>. | ||
| − | :: 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 = | + | :: 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 = 6). Also, notice that since you're downsampling by 6, your down-sampled rect goes from <math> -pi </math> to <math> pi </math>. Repeat that 6 times and you basically get a rect that goes from <math> -pi </math> to <math> 11*pi </math>. 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 <math> -pi/6 </math> to </math> 11*pi/6 </math>. (Note: I'm ignoring the <math>[1 + e^{-jw} + e^{-j2w}]</math> 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 <math> -pi/6 </math> and <math> 11*pi/6 </math>. I end up with a final answer of |
<math>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})</math> | <math>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})</math> | ||
Revision as of 22:17, 30 September 2010
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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?
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 = 6). 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 </math> 11*pi/6 </math>. (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}) $
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.
