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| + | [[Category:energy]] | ||
| + | [[Category:signal]] | ||
| + | [[Category:ECE]] | ||
| + | [[Category:ECE301]] | ||
| + | [[Category:practice problem]] | ||
| + | |||
| + | =Question= | ||
| + | Compute the energy and the average power of the following signal: | ||
| + | |||
<math>x(t)=\sqrt{t}</math> | <math>x(t)=\sqrt{t}</math> | ||
---- | ---- | ||
| + | ==Answer 1== | ||
<math>E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt</math> | <math>E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt</math> | ||
| Line 15: | Line 25: | ||
lim T<math>\to</math><math>\infty</math> = <math>\infty</math> = P<math>\infty</math> | lim T<math>\to</math><math>\infty</math> = <math>\infty</math> = P<math>\infty</math> | ||
| + | *<span style="color:green"> Be careful! The stuff inside the integral should always be positive. You are integrating "t", which is sometimes positive and sometimes negative. So there must be a mistake somewhere. </span> | ||
| + | *<span style="color:blue"> The key is to take the '''norm''' of the signal squared. Here the signal is <math>\sqrt{t}</math>, so taking the norm of the signal squared gives <math>|t|</math>. </span> | ||
| + | ---- | ||
| + | ==Answer 2== | ||
| + | * | ||
| + | ---- | ||
| + | [[Signal_energy_CT|Back to CT signal energy page]] | ||
Revision as of 13:52, 24 February 2015
Question
Compute the energy and the average power of the following signal:
$ x(t)=\sqrt{t} $
Answer 1
$ E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt $
E$ \infty $ = $ \int_{-\infty}^{\infty} tdt $
E$ \infty $ = $ \frac{1}{2} t^2 $ evaluated from -$ \infty $ to +$ \infty $ = $ \infty $
P$ \infty $ = lim T$ \to $$ \infty $ $ \frac{1}{2T} $ $ \int_{-T}^{T}\ tdt $
$ \frac{1}{2T} (.5t^2)|_{-T}^{T} = \frac{T}{4} $
lim T$ \to $$ \infty $ = $ \infty $ = P$ \infty $
- Be careful! The stuff inside the integral should always be positive. You are integrating "t", which is sometimes positive and sometimes negative. So there must be a mistake somewhere.
- The key is to take the norm of the signal squared. Here the signal is $ \sqrt{t} $, so taking the norm of the signal squared gives $ |t| $.
