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=Discussion related to midterm 1=
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= Discussion related to midterm 1 =
  
Ask your questions here!
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Ask your questions here!  
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Possible [[2010 Fall ECE 438 Boutin/ECE438Mid1FormulaSheet Work|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?
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:It will not be the same formula sheet. --[[User:Mboutin|Mboutin]] 09:07, 1 October 2010 (UTC)
  
Possible [[2010_Fall_ECE_438_Boutin/ECE438Mid1FormulaSheet_Work|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?
 
 
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Midterm 1 Spring 2009 Question 3
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Midterm 1 Spring 2009 Question 3  
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a) <math>H(w) = \frac{1}{3}[1 + e^{-jw} + e^{-j2w}]</math>
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b) <math>G(w) = rect(w\frac{3}{\pi})</math>
  
a) <math>H(w) = \frac{1}{3}[1 + e^{-jw} + e^{-j2w}]</math>
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<math>A(w) = \frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{w-2\pi k}{6})</math>  
  
b) <math>G(w) = rect(w\frac{3}{\pi})</math>
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<math>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})</math>  
  
<math>A(w) = \frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{w-2\pi k}{6})</math>
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<math>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})</math>  
  
<math>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})</math>
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<math>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})</math>  
  
<math>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})</math>
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Is this correct?
  
<math>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})</math>
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<br>  
  
Is this correct?
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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 <span class="texhtml">2π</span>.
  
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*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 <span class="texhtml"> − ''p''''i''</span> to <span class="texhtml">''p''''i''</span>. Repeat that 6 times and you basically get a rect that goes from <span class="texhtml"> − ''p''''i''</span> to <span class="texhtml">11 * ''p''''i''</span>. 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 <span class="texhtml"> − ''p''''i'' / 6</span> to <span class="texhtml">11 * ''p''''i'' / 6</span>. (Note: I'm ignoring the <span class="texhtml">[1 + ''e''<sup> − ''j''''w''</sup> + ''e''<sup> − ''j''2''w''</sup>]</span> 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 <span class="texhtml"> − ''p''''i'' / 6</span> and <span class="texhtml">11 * ''p''''i'' / 6</span>. I end up with a final answer of
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*<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>
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*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).
  
:: 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>.
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*WARNING: The function you get MUST be periodic with period <span class="texhtml"></span>. This one is not. --[[User:Mboutin|Mboutin]] 09:24, 1 October 2010 (UTC)
  
:: 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 [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 <math> -pi/6 <math> and <math> 11*pi/6 </math>. I end up with a final answer of
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*The answer is actually <span class="texhtml">''F''(ω) = ''G''(ω)''H''()</span>, for <span class="texhtml"> − π &lt; ω &lt; π</span>. --[[User:Mboutin|Mboutin]] 09:24, 1 October 2010 (UTC)
<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>
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**correction to Mboutin's comment above: The LPF&nbsp;was Unity gain, not a full interpolator(&nbsp;Upscaling&nbsp;+ LPF) Therefore the equation is&nbsp;F(ω) = (1/6)G(ω)H(), for − π &lt; ω &lt; π.
  
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*I'm a bit confused: Is <span class="texhtml">''G''(ω)</span> periodic to begin with? Is this because frequency itself is periodic for discrete signals? Or do we indeed take k to go from <math>-\infty</math> to <math>\infty</math> instead of from 0 to D-1
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**The DTFT is always periodic with a period <math>2\pi</math>. --[[User:Mboutin|Mboutin]] 13:11, 1 October 2010 (UTC)
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*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 <math>F(\omega) = \frac{1}{6}  G(\omega ) H(6 \omega ) </math>, for <span class="texhtml"> − π &lt; ω &lt; π</span>
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**Yes, you are correct. --[[User:Mboutin|Mboutin]] 13:11, 1 October 2010 (UTC)
 
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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.
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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.
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*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.  
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*Actually, there is a simpler way to answer that question a)&nbsp;: Just use the multiplication property of the CTFT. Then look in the table for the CTFT of cosine, ant the CTFT of sinc. --[[User:Mboutin|Mboutin]] 08:48, 1 October 2010 (UTC)
  
 
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[[2010_Fall_ECE_438_Boutin|Back to ECE438 Fall 2010 Prof. Boutin]]
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[[2010 Fall ECE 438 Boutin|Back to ECE438 Fall 2010 Prof. Boutin]]

Latest revision as of 09:11, 1 October 2010

Discussion related to midterm 1

Ask your questions here!

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 .
  • 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 + ej'w + ej2w] 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 . 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)

Back to ECE438 Fall 2010 Prof. Boutin

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