| Line 22: | Line 22: | ||
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! | 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! | ||
---- | ---- | ||
| − | ==Answer 1 | + | ==Answer 1== |
<math> | <math> | ||
P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T | x(t) |^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} 4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} 8T = 4 | P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T | x(t) |^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} 4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} 8T = 4 | ||
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*<span style="color:blue">You made a limit manipulation error. </span> | *<span style="color:blue">You made a limit manipulation error. </span> | ||
---- | ---- | ||
| − | ==Answer 3 | + | ==Answer 3== |
<math> | <math> | ||
P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T (2j)^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T -4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} -4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} -8T = -4 | P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T (2j)^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T -4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} -4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} -8T = -4 | ||
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*<span style="color:blue">Power cannot be negative, so your answer cannot be correct. </span> | *<span style="color:blue">Power cannot be negative, so your answer cannot be correct. </span> | ||
---- | ---- | ||
| − | ==Answer 4 | + | ==Answer 4== |
<math>|x|^2=4 \quad 4 \frac{\infty}{\infty}=4</math> | <math>|x|^2=4 \quad 4 \frac{\infty}{\infty}=4</math> | ||
*<span style="color:blue">Sorry but I don't understand your explanation. </span> | *<span style="color:blue">Sorry but I don't understand your explanation. </span> | ||
---- | ---- | ||
[[Signal_power_CT|Back to Signal Power]] | [[Signal_power_CT|Back to Signal Power]] | ||
Latest revision as of 15:29, 21 April 2015
Practice Question on "Signals and Systems"
Topic: Signal Power
Contents
Question
Compute the power of the signal $ x(t)= 2j $
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!
Answer 1
$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T | x(t) |^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T 4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} 4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} 8T = 4 $
- looks good!
Answer 2
$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-\infty}^\infty| x(t) |^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-\infty}^\infty 4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \infty= \frac{\infty}{\infty}=1 $
- You made a limit manipulation error.
Answer 3
$ P_{\infty}=\lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T (2j)^2 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} \int_{-T}^T -4 dt = \lim_{T\rightarrow \infty} {1 \over {2T}} -4t \Big| ^T _{-T} = \lim_{T\rightarrow \infty} {1 \over {2T}} -8T = -4 $
- Power cannot be negative, so your answer cannot be correct.
Answer 4
$ |x|^2=4 \quad 4 \frac{\infty}{\infty}=4 $
- Sorry but I don't understand your explanation.
