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===Answer 2=== | ===Answer 2=== | ||
| − | + | becasue: <math> e^{jx} =cos(x)+ jsin(x) </math> | |
| + | |||
| + | <math>| e^{j \omega}|=|cos(\omega) + i*sin(\omega)|=\sqrt{cos(\omega)^2 +sin(\omega)^2}=1</math> | ||
| + | |||
===Answer 3=== | ===Answer 3=== | ||
Write it here | Write it here | ||
Revision as of 10:56, 10 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(j \omega) + i*sin(j \omega) $
hence,
$ \left| e^{j \omega} \right| = \left|cos(j \omega) + i*sin(j \omega) \right| = \sqrt{cos^2(j \omega) + sin^2(j \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
