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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. | ||
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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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[[2010_Fall_ECE_438_Boutin|Back to ECE438 Fall 2010 Prof. Boutin]] | [[2010_Fall_ECE_438_Boutin|Back to ECE438 Fall 2010 Prof. Boutin]] | ||
Revision as of 15:57, 30 September 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?
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?
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.
