(New page: == The following signals are shown to be either an energy signal or a power signal ==)
 
 
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== The following signals are shown to be either an energy signal or a power signal ==
 
== The following signals are shown to be either an energy signal or a power signal ==
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<math>\,\!x(t)=e^{-at}u(t)</math>  for a > 0
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solution:
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since <math>Energy(\infty) = \int_{-\infty}^{\infty} \! |x(t)|^2\ dt</math> ,
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<math> = \int_{0}^{\infty}\!e^{-2at}dt</math> <math>=\frac{1}{2a} < {\infty}</math>
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therefore x(t) is an energy function because the energy is finite, and not a power function.
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A consequence of this is that P=0.  If the energy of the signal was infinite, then the power would be
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found to have a finite value.

Latest revision as of 07:24, 5 September 2008

The following signals are shown to be either an energy signal or a power signal

$ \,\!x(t)=e^{-at}u(t) $ for a > 0

solution:

since $ Energy(\infty) = \int_{-\infty}^{\infty} \! |x(t)|^2\ dt $ ,

$ = \int_{0}^{\infty}\!e^{-2at}dt $ $ =\frac{1}{2a} < {\infty} $

therefore x(t) is an energy function because the energy is finite, and not a power function.

A consequence of this is that P=0. If the energy of the signal was infinite, then the power would be found to have a finite value.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva