Revision as of 10:34, 19 September 2011 by Zhang372 (Talk | contribs)

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

Back to ECE438

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang