Integration of a trig function > 0

  • Thread starter MechEMike
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Main Question or Discussion Point

I need to take the integral of the following function over one cycle.

[itex]y(t) = sin(wt)cos(wt-p)[/itex]


p is a phase angle that exist from 0 to pi. w is a frequency.

This would be pretty straightforward, but I need to take the integral only of the portions of the function that are greater than zero.
I need to do this for an efficiency analysis, where the negative portion of the cycle is waste energy and don't contribute to average power. I need an symbolic expression.

The zero crossings are at t1 = pi/w and t2 = pi/(2w) + p/w which I had been using as limits of integration.

[itex]\int ^{t_2}_{0} y(t) dt - \int ^{t_2}_{t_1} y(t) dt[/itex]

The resulting expression is correct for values of for 0<=p <= pi/2. Above p = pi/2 though, the expression does not give me the result I need, because for p> pi/2, t2<t1 and the result of

[itex] \int ^{t_2}_{t_1} y(t) dt[/itex]

changes sign.

I am not sure how to get around this. Any help would be...helpful. Thanks!
 
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Answers and Replies

  • #2
mathman
Science Advisor
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I would suggest an alternate approach. cos(wt-p) = coswtcosp + sinwtsinp, so you can then have two integrals from 0 to 2π/w. cosp∫sinwtcoswtdt + sinp∫sin2wtdt.
 

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