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===Answer 1===
 
===Answer 1===
  
By Euler's formular
+
By [[More_on_Eulers_formula|Euler's formula]]
  
 
<math> e^{j \omega}  = cos( \omega) + i*sin( \omega) </math>
 
<math> e^{j \omega}  = cos( \omega) + i*sin( \omega) </math>
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:<span style="color:green">TA's comments: Is this true for all <math>\omega \in R</math>? The answer is yes.</span>
 
:<span style="color:green">TA's comments: Is this true for all <math>\omega \in R</math>? The answer is yes.</span>
 +
 +
:<span style="color:purple">Instructor's comment: I would like to propose a more straightforward way to compute this norm using the fact that <math>|z|^2=z \bar{z}</math>. Can you try it out? -pm </span>
 +
  
 
===Answer 2===
 
===Answer 2===
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<math>| e^{j \omega}|=|cos(\omega) + i*sin(\omega)|=\sqrt{cos(\omega)^2 +sin(\omega)^2}=1</math>
 
<math>| e^{j \omega}|=|cos(\omega) + i*sin(\omega)|=\sqrt{cos(\omega)^2 +sin(\omega)^2}=1</math>
  
:<span style="color:green">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.</span>
+
:<span style="color:green">TA's comments: The point here is to use [[More_on_Eulers_formula|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.</span>
 +
 
 +
:<span style="color:purple">Instructor's comment: Again, I would argue that using the fact that <math>|z|^2=z \bar{z}</math> is more straightforward. Can you try it out? -pm </span>
  
 
===Answer 3===
 
===Answer 3===
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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>
 
<math>\left| e^{j \omega} \right| =  \left|cos( \omega) + i*sin( \omega) \right| = \sqrt{cos^2( \omega) + sin^2( \omega)} = 1 </math>
  
 +
:<span style="color:purple">Instructor's comment: Can you think of a way to compute this norm without using [[More_on_Eulers_formula|Euler's formula]]? -pm </span>
 
----
 
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[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]
 
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]]
  
 
[[ECE301|Back to ECE438]]
 
[[ECE301|Back to ECE438]]

Revision as of 04:15, 23 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 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

Back to ECE438

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