| Line 17: | Line 17: | ||
By Euler's formular | By Euler's formular | ||
| − | <math> e^{j \omega} = cos( | + | <math> e^{j \omega} = cos( \omega) + i*sin( \omega) </math> |
hence, | hence, | ||
| − | <math>\left| e^{j \omega} \right| = \left|cos( | + | <math>\left| e^{j \omega} \right| = \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 </math> |
===Answer 2=== | ===Answer 2=== | ||
Revision as of 08:24, 11 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 $
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 $
Answer 3
Write it here
