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[[Category:ECE301Spring2011Boutin]] | [[Category:ECE301Spring2011Boutin]] | ||
[[Category:problem solving]] | [[Category:problem solving]] | ||
| − | = Compute the Magnitude of the following discrete-time signals | + | [[Category:ECE301]] |
| + | [[Category:complex numbers]] | ||
| + | [[Category:Euler's formula]] | ||
| + | |||
| + | <center><font size= 4> | ||
| + | '''[[Signals_and_systems_practice_problems_list|Practice Question on "Signals and Systems"]]''' | ||
| + | </font size> | ||
| + | |||
| + | |||
| + | [[Signals_and_systems_practice_problems_list|More Practice Problems]] | ||
| + | |||
| + | |||
| + | Topic: Review of Complex Numbers | ||
| + | </center> | ||
| + | ---- | ||
| + | ==Question== | ||
| + | |||
| + | Compute the Magnitude of the following discrete-time signals | ||
| + | |||
a) <math>x[n]=e^{2n}</math> | a) <math>x[n]=e^{2n}</math> | ||
| Line 15: | Line 33: | ||
===Answer 1=== | ===Answer 1=== | ||
a) <math>|e^{2n}|=\sqrt{(e^{2n})^2}=e^{2n}</math> | a) <math>|e^{2n}|=\sqrt{(e^{2n})^2}=e^{2n}</math> | ||
| + | :<span style="color:green"> Instructor's comments: Yes, this is correct, but I would like to propose an alternative justification: </span> | ||
| + | :<math>{\color{green}|e^{2n}|=\sqrt{(e^{2n})(e^{2n})^*}=\sqrt{e^{2n}e^{2n}}=e^{2n}}</math> | ||
| + | :<span style="color:green">where <math>{\color{green}~^*}</math> denotes the complex conjgate.-pm </span> | ||
b) <math>|e^{2jn}|=|cos(2n)+jsin(2n)|=\sqrt{cos(2n)^2+sin(2n)^2}=\sqrt{1}=1</math> | b) <math>|e^{2jn}|=|cos(2n)+jsin(2n)|=\sqrt{cos(2n)^2+sin(2n)^2}=\sqrt{1}=1</math> | ||
| + | :<span style="color:green"> 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 </span> | ||
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> | |
| + | :<span style="color:green"> 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 </span> | ||
===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]] | ||
Latest revision as of 15:18, 26 November 2013
Practice Question on "Signals and Systems"
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?
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.
