View source for Midterm1 discussion ECE438F10 - Rhea

=Discussion related to midterm 1=

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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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Midterm 1 Spring 2009 Question 3

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>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>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?

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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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