Line 6: Line 6:
 
----
 
----
  
Question: When the ideal low pass filter has a cutoff of <math>\frac{\pi}{6}</math>, what is the variable for rect(?)? Is it <math>rect(\frac{6w}{\pi})</math>?
+
Midterm 1 Spring 2009 Question 3
  
Yes we know it's 202 material but we are getting old =D
+
a) <math>H(w) = \frac{1}{3}[1 + e^{-jw} + e^{-j2w}]</math>
  
:: I think the LPF is going to be <math>rect(w)= rect(\frac{3w}{\pi})</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?
  
 
----
 
----

Revision as of 15:18, 30 September 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?


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.


Back to ECE438 Fall 2010 Prof. Boutin

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva