How do I find the Fourier Series for F(t) = sin(wt)?

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SUMMARY

The discussion focuses on finding the Fourier Series for the function F(t) = sin(wt) defined on the interval 0 < t < pi/w and 0 when pi/w < t < 2pi/w. The Fourier coefficients are calculated using the formula ck = (w/2pi) * (integral from 0 to pi/w of sin(wt)e^(-ikt) dt) + (w/2pi) * (integral from pi/w to 2pi/w of sin(wt)e^(-ikt) dt). The user expresses difficulty in performing the integration due to the differing arguments of the trigonometric functions. The solution involves rewriting sin(wt) using exponential functions and then performing the necessary integrals.

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Homework Statement



Consider F(t) = sin(wt) when 0 < t < pi/w and 0 when pi/w < t < 2pi/w. Where w is the frequency and t is the time. Find the Fourier Series

Homework Equations




F(t) = sum of (ck e^ikt)

See attached doc with math type; its a lot more readable.

The Attempt at a Solution




ck = (w/2pi integral from 0 to pi/w of sin(wt)*e^-ikx) + (w/2pi integral from pi/w to 2pi/w of sin(wt)*e^-ikx)

I’m not sure how to do this integration. If all the trig functions had the same argument, it would be do-able, but they don’t. w is a constant, right? And k isn’t, right? :confused: We never dealt with stuff like this back in calc. I’m very grateful for any help or suggestions. I tried to use trig addition formulas, but that didn’t work. Thanks! :smile:
 

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Your integral is over t, not x, so it should have exp(-ikt).
Write sin(wt) as [exp(iwt)-exp(-iwt)]/2i, and do the exponential integrals.
 

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