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| − | =Discussion related to midterm 1= | + | = Discussion related to midterm 1 = |
| − | Ask your questions here! | + | 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) | ||
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| − | Midterm 1 Spring 2009 Question 3 | + | Midterm 1 Spring 2009 Question 3 |
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| + | a) <math>H(w) = \frac{1}{3}[1 + e^{-jw} + e^{-j2w}]</math> | ||
| + | |||
| + | b) <math>G(w) = rect(w\frac{3}{\pi})</math> | ||
| − | + | <math>A(w) = \frac{1}{6} \Sigma_{k=-\infty}^{\infty} rect(\frac{3}{\pi}\cdot\frac{w-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> | |
| − | <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> |
| − | <math> | + | <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> |
| − | + | Is this correct? | |
| − | < | + | <br> |
| − | + | *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>. | |
| + | *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 | ||
| + | *<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> | ||
| + | *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 <span class="texhtml">2π</span>. This one is not. --[[User:Mboutin|Mboutin]] 09:24, 1 October 2010 (UTC) |
| − | + | *The answer is actually <span class="texhtml">''F''(ω) = ''G''(ω)''H''(6ω)</span>, for <span class="texhtml"> − π < ω < π</span>. --[[User:Mboutin|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 <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 | ||
| + | **The DTFT is always periodic with a period <math>2\pi</math>. --[[User:Mboutin|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 <math>F(\omega) = \frac{1}{6} G(\omega ) H(6 \omega ) </math>, for <span class="texhtml"> − π < ω < π</span> | ||
| + | **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. | + | 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. | + | *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. --[[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]] |
Latest revision as of 08:11, 1 October 2010
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 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)
