(New page: <math>x(t)=2t^2</math> ---- <math>E_{\infty}</math> <math>E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt</math> <math>E_{\infty}=\int_{-\infty}^\infty |2t^2|^2\,dt</math> <math>E_{\inft...) |
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| + | [[Category:energy]] | ||
| + | [[Category:signal]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:practice problem]] | ||
| + | |||
| + | |||
| + | ==Problem== | ||
| + | Calculate the energy <math>E_\infty</math> and the average power <math>P_\infty</math> for the CT signal | ||
<math>x(t)=2t^2</math> | <math>x(t)=2t^2</math> | ||
---- | ---- | ||
| − | + | ==Solution 1== | |
<math>E_{\infty}</math> | <math>E_{\infty}</math> | ||
| Line 32: | Line 41: | ||
<math>P_{\infty}=\infty</math> | <math>P_{\infty}=\infty</math> | ||
| + | |||
| + | <span style="color:green"> Both of your final answer are correct, but there are small mistakes (constant multipliers) in your computation. </span> | ||
| + | ---- | ||
| + | ==Solution 2== | ||
| + | <math>E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt =\infty</math> | ||
| + | |||
| + | <math>P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\lim_{T \to \infty} \ \frac{2}{5T}t^5|_{-T}^{T}=\lim_{T \to \infty} \ \frac{2}{5T}[T^5 - (-T^5)] =\lim_{T \to \infty} \ \frac{4T^4}{5}=\infty</math> | ||
| + | |||
| + | |||
| + | <span style="color:green"> Looks pretty good!</span> | ||
| + | ---- | ||
| + | ==Solution 3== | ||
| + | <math>E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt = 4 \frac{t^5}{5}=\infty.</math> | ||
| + | |||
| + | <math>P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\frac{\infty}{\infty}=1.</math> | ||
| + | |||
| + | |||
| + | <span style="color:red"> The energy computation looks good. But in the power computation you distributed the limit too early and so your final answer is wrong. </span> | ||
| + | ---- | ||
| + | [[Signal_energy_CT|Back to CT signal energy page]] | ||
Latest revision as of 17:05, 25 February 2015
Contents
Problem
Calculate the energy $ E_\infty $ and the average power $ P_\infty $ for the CT signal $ x(t)=2t^2 $
Solution 1
$ E_{\infty} $
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt $
$ E_{\infty}=\int_{-\infty}^\infty |2t^2|^2\,dt $
$ E_{\infty}=2\int_{-\infty}^\infty t^4\,dt $
$ E_{\infty}=\frac{2}{5}t^5|_{-\infty}^\infty $
$ E_{\infty}=\infty $
$ P_{\infty} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{T}\int_{-T}^{T}t^4\,dt $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}t^5|_{-T}^{T} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{5T}[T^5 - (-T^5)] $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^5}{5T} $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{2T^4}{5} $
$ P_{\infty}=\infty $
Both of your final answer are correct, but there are small mistakes (constant multipliers) in your computation.
Solution 2
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt =\infty $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\lim_{T \to \infty} \ \frac{2}{5T}t^5|_{-T}^{T}=\lim_{T \to \infty} \ \frac{2}{5T}[T^5 - (-T^5)] =\lim_{T \to \infty} \ \frac{4T^4}{5}=\infty $
Looks pretty good!
Solution 3
$ E_{\infty}=\int_{-\infty}^\infty |x(t)|^2\,dt =\int_{-\infty}^\infty |2t^2|^2\,dt= 4\int_{-\infty}^\infty t^4\,dt = 4 \frac{t^5}{5}=\infty. $
$ P_{\infty}=\lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|x(t)|^2\,dt= \lim_{T \to \infty} \ \frac{1}{2T}\int_{-T}^{T}|2t^2|^2\,dt =\lim_{T \to \infty} \ \frac{2}{T}\int_{-T}^{T}t^4\,dt =\frac{\infty}{\infty}=1. $
The energy computation looks good. But in the power computation you distributed the limit too early and so your final answer is wrong.
