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===Answer 3=== | ===Answer 3=== | ||
| − | + | <math> e^{j \omega} = cos( \omega) + i*sin( \omega) </math> | |
| + | <math>\left| e^{j \omega} \right| = \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 </math> | ||
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[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
[[ECE301|Back to ECE438]] | [[ECE301|Back to ECE438]] | ||
Revision as of 10:34, 19 September 2011
Contents
What is the norm of a complex exponential?
After class today, a student asked me the following question:
$ \left| e^{j \omega} \right| = ? $
Please help answer this question.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
By Euler's formular
$ e^{j \omega} = cos( \omega) + i*sin( \omega) $
hence,
$ \left| e^{j \omega} \right| = \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 $
- TA's comments: Is this true for all $ \omega \in R $? The answer is yes.
Answer 2
becasue: $ e^{jx} =cos(x)+ jsin(x) $
$ | e^{j \omega}|=|cos(\omega) + i*sin(\omega)|=\sqrt{cos(\omega)^2 +sin(\omega)^2}=1 $
- TA's comments: The point here is to use Euler's formula to write a complex exponential as a complex number. Then the norm(magnitude) and angle(phase) of this complex number can be easily computed.
Answer 3
$ e^{j \omega} = cos( \omega) + i*sin( \omega) $ $ \left| e^{j \omega} \right| = \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 $
