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My integral calculus is a little rusty. There was an exercise in the first lecture series where we calculated the average power by integrating the COS waveform over one cycle. I'm sorry I don't have the problem in front of me. Can anyone show me how this is done? I'm a little confused.
For example: how do you integrate (cos(2*pi*60t))^2 over one cycle?
I fill in more detail when I get home tonight if needed.
 2 years ago
 2 years ago
My integral calculus is a little rusty. There was an exercise in the first lecture series where we calculated the average power by integrating the COS waveform over one cycle. I'm sorry I don't have the problem in front of me. Can anyone show me how this is done? I'm a little confused. For example: how do you integrate (cos(2*pi*60t))^2 over one cycle? I fill in more detail when I get home tonight if needed.
 2 years ago
 2 years ago

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nikiBest ResponseYou've already chosen the best response.1
let y=(cos(2*pi*60*t))^2 cos^2t=(1+cos2t)/2 hence y=(1+cos2(2*pi*60t))/2 refer my solving thus we obtain y=t/2+sin[240*pi*t]/(4dw:1331380539415:dw80*pi)
 2 years ago

MicroarrayBest ResponseYou've already chosen the best response.0
This can be done only knowing that the integral of 1 over 0 to 2*pi is 2*pi. We want to compute the value of cos^2(x) over one cycle. Note that cos^2(x) = 1  sin^2(x), and as cos(x) and sin(x) are equal by a translation of pi radians, the integrals A of cos^2(x) and sin^2(x) over one cycle are equal. Thus 2*pi  A = A, and so A/(2*pi) = 1/2. \[\int\limits_{0}^{2\pi} {\cos}^2(x)\ dx = A = \int\limits_{0}^{2\pi} 1{\sin}^2(x)\ dx = 2\piA\] \[2A = 2\pi\] \[\frac{A}{2\pi} = \frac{1}{2}\] More detail: Let the integral of cos^2(x) over one cycle be A; this is equal to sin^2(x) over one cycle as the cos(x) and sin(x) are the same curve (just shifted). Then A = integral(1sin^(x)) = 2*piA because cos^2(x)+sin^2(x)=1.
 2 years ago
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