Line 1: Line 1:
 +
Homework 2 Question 5.a
 +
 
<math>Y(e^{jw}) = \frac{1}{N}(X(e^{jw})+e^{-jw}X(e^{jw})+.....+X(e^{jw})e^{(N-1)jw})</math>
 
<math>Y(e^{jw}) = \frac{1}{N}(X(e^{jw})+e^{-jw}X(e^{jw})+.....+X(e^{jw})e^{(N-1)jw})</math>
  
Line 9: Line 11:
  
 
<math> \angle H(e^{jw}) (Phase) =  e^{-jw(\frac{N-1}{2})}  (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0)</math>
 
<math> \angle H(e^{jw}) (Phase) =  e^{-jw(\frac{N-1}{2})}  (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0)</math>
 +
 +
 +
 +
*Warning:The answer is completed by one of the ECE 438 Students, not by Professor. The answer is not guaranteed whther or not it is right.
 +
 +
 +
 +
--[[User:Kim415|Kim415]] 06:03, 4 February 2009 (UTC)

Revision as of 02:03, 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) = e^{-jw(\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 whther or not it is right.


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

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett