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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> | 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=== | ||
| − | + | b) <math>|e^{2jn}| = \sqrt{(e^{2jn})(e^{2jn})^*} =\sqrt{(e^{2jn})(e^{-2jn})} = 1</math> | |
===Answer 3=== | ===Answer 3=== | ||
write it here. | write it here. | ||
---- | ---- | ||
[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Revision as of 16:26, 11 January 2011
Contents
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?
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 $
Answer 3
write it here.
