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Topic: Review of Complex Numbers
 
  
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[[Signals_and_systems_practice_problems_list|More Practice Problems]]
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Topic: Review of Complex Numbers
 
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==Question==
 
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Compute the Magnitude of the following discrete-time signals
 
Compute the Magnitude of the following discrete-time signals
  

Latest revision as of 15:18, 26 November 2013


Practice Question on "Signals and Systems"


More Practice Problems


Topic: Review of Complex Numbers


Question

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} $

Instructor's comments: Yes, this is correct, but I would like to propose an alternative justification:
$ {\color{green}|e^{2n}|=\sqrt{(e^{2n})(e^{2n})^*}=\sqrt{e^{2n}e^{2n}}=e^{2n}} $
where $ {\color{green}~^*} $ denotes the complex conjgate.-pm

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

Instructor's comments: Again, the answer and justification are correct. But can somebody propose a different justification? One that is similar to the one I proposed above? -pm

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

b) $ |e^{2jn}| = \sqrt{(e^{2jn})(e^{2jn})^*} =\sqrt{(e^{2jn})(e^{-2jn})} = 1 $

Instructor's comments: Good, this is what I was talking about above. I would like to argue that this approach 1. is more general, and 2. is oftentimes more straightforward (i.e. easier computations). -pm

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