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Tweedle_Dee Group Title

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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  1. niki Group Title
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    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]/(4|dw:1331380539415:dw|80*pi)

    • 2 years ago
  2. Microarray Group Title
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    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. \[\int\limits_{0}^{2\pi} {\cos}^2(x)\ dx = A = \int\limits_{0}^{2\pi} 1-{\sin}^2(x)\ dx = 2\pi-A\] \[2A = 2\pi\] \[\frac{A}{2\pi} = \frac{1}{2}\] 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(1-sin^(x)) = 2*pi-A because cos^2(x)+sin^2(x)=1.

    • 2 years ago
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