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| + | <math>Y(e^{jw}) = \frac{1}{N}(X(e^{jw})+e^{-jw}X(e^{jw})+.....+X(e^{jw})e^{(N-1)jw})</math> | ||
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| + | <math> = \frac{1}{N} \sum_{k=0}^{N-1} e^{-jwk} X(e^{jw})</math> | ||
| + | <math> = \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})</math> | ||
| + | <math> = \frac{1}{N} e^{-jw(\frac{N-1}{2})} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}}</math> | ||
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| + | <math> \left | H(e^{jw}) \right \vert (Magnitude) = \frac{1}{N} \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}}</math> | ||
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| + | <math> \angle H(e^{jw}) (Phase) = -w(\frac{N-1}{2}) (when \frac{\sin \frac{wN}{2}} {\sin \frac{w}{2}} > 0)</math> | ||
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| + | *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. | ||
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| + | --[[User:Kim415|Kim415]] 06:03, 4 February 2009 (UTC) | ||
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| + | [[Category:ECE438Spring2009mboutin]] | ||
Latest revision as of 05:54, 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)
