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===Answer 1===
 
===Answer 1===
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a) <math>|e^{2n}|=\sqrt{(e^{2n})^2}=e^{2n}</math>
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b) <math>|e^{2jn}|=|cos(2n)+jsin(2n)|=\sqrt{cos(2n)^2+sin(2n)^2}=\sqrt{1}=1</math>
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c) <math>|j^n|=|e^{nj\pi/2}|=|cos(n\pi/2)+jsin(n\pi/2)|=\sqrt{cos(n\pi/2)^2+sin(n\pi/2)^2}=\sqrt{1}=1</math>
 
===Answer 2===
 
===Answer 2===
 
write it here.
 
write it here.

Revision as of 13:04, 11 January 2011

Compute the Magnitude of the following discrete-time signals

a) $ x[n]=e^{2n} $

b) $ x[n]=e^{2jn} $

c) $ x[n]=j^n $

What properties of the complex magnitude can you use to check your answer?


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

a) $ |e^{2n}|=\sqrt{(e^{2n})^2}=e^{2n} $

b) $ |e^{2jn}|=|cos(2n)+jsin(2n)|=\sqrt{cos(2n)^2+sin(2n)^2}=\sqrt{1}=1 $

c) $ |j^n|=|e^{nj\pi/2}|=|cos(n\pi/2)+jsin(n\pi/2)|=\sqrt{cos(n\pi/2)^2+sin(n\pi/2)^2}=\sqrt{1}=1 $

Answer 2

write it here.

Answer 3

write it here.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva