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+ | [[Category:practice problem]] | ||
+ | =[[Homework_1_ECE301Fall2008mboutin|HW1]], [[ECE301]], Fall 2008, [[user:mboutin|Prof. Boutin]]= | ||
+ | ---- | ||
+ | ==Question== | ||
+ | |||
+ | Compute the power and energy of the signal | ||
+ | |||
+ | <math>x(t)=cos(t)</math>. | ||
+ | ---- | ||
==Energy== | ==Energy== | ||
We will find the energy in one cycle of the cosine waveform. | We will find the energy in one cycle of the cosine waveform. | ||
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<math>=\frac{1}{2}</math> | <math>=\frac{1}{2}</math> | ||
+ | ---- | ||
+ | [[Homework_1_ECE301Fall2008mboutin|Back to HW1, ECE301]] |
Latest revision as of 13:45, 24 February 2015
HW1, ECE301, Fall 2008, Prof. Boutin
Question
Compute the power and energy of the signal
$ x(t)=cos(t) $.
Energy
We will find the energy in one cycle of the cosine waveform.
$ E=\int_0^{2\pi}{|cos(t)|^2dt} $
$ =\frac{1}{2}\int_0^{2\pi}(1+cos(2t))dt $
$ =\frac{1}{2}(t+\frac{1}{2}sin(2t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{2}(2\pi+0-0-0) $
$ =\pi $
Energy
We will find the average power in one cycle of the cosine waveform.
$ E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(t)|^2dt} $
$ =\frac{1}{2\pi-0}\frac{1}{2}\int_0^{2\pi}(1+cos(2t))dt $
$ =\frac{1}{4\pi}(t+\frac{1}{2}sin(2t))|_{t=0}^{t=2\pi} $
$ =\frac{1}{4\pi}(2\pi+0-0-0) $
$ =\frac{1}{2} $