Difference between revisions of "Template for Text practice problem ECE201" - Rhea

Revision as of 20:49, 11 January 2015

Question

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 formula

$ 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.

Instructor's comment: I would like to propose a more straightforward way to compute this norm using the fact that $ |z|^2=z \bar{z} $. Can you try it out? -pm

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.

Instructor's comment: Again, I would argue that using the fact that $ |z|^2=z \bar{z} $ is more straightforward. Can you try it out? -pm

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 $

Instructor's comment: Can you think of a way to compute this norm without using Euler's formula? -pm

Back to ECE438 Fall 2011 Prof. Boutin

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