Difference between revisions of "Question 5 (ECE438Spring2009mboutin)" - Rhea

Latest revision as of 04:54, 4 February 2009

Homework 2 Question 5.a

$ Y(e^{jw}) = \frac{1}{N}(X(e^{jw})+e^{-jw}X(e^{jw})+.....+X(e^{jw})e^{(N-1)jw}) $

      $  = \frac{1}{N} \sum_{k=0}^{N-1} e^{-jwk} X(e^{jw}) $
      $  = \frac{1}{N} e^{-jw(\frac{N-1}{2})} \frac{e^{jw(\frac{N}{2})} - e^{-jw(\frac{N}{2})}} {e^{j(\frac{w}{2})} -e^{-j(\frac{w}{2})}} X(e^{jw}) $
      $  = \frac{1}{N} e^{-jw(\frac{N-1}{2})} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} $

$ \left | H(e^{jw}) \right \vert (Magnitude) = \frac{1}{N} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} $

$ \angle H(e^{jw}) (Phase) = -w(\frac{N-1}{2}) (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0) $

• Warning:The answer is completed by one of the ECE 438 Students, not by Professor. The answer is not guaranteed whether or not it is right.

--Kim415 06:03, 4 February 2009 (UTC)