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===Answer 3===
 
===Answer 3===
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<math> e^{j \omega}  = cos( \omega) + i*sin( \omega) </math>
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<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]]
 
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Revision as of 11:34, 19 September 2011

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.


Share your answers below

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 $


Back to ECE438 Fall 2011 Prof. Boutin

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