Fourier Series Solved Problems

Fourier Series Solved Problems-56
\[\begin = \frac\int_^ & = \frac\left[ \right]\ & = \frac\left[ \right]\end\] As with the coefficients for the cosines will probably be easier to do each of these individually. \right|_^0 = \frac\left( \right) = \frac\left( \right)\] \[\begin\int_^ & = \left.\right|_0^L\ & = \left( \right)\left( \right)\ & = \left( \right)\left( \right) = - \frac\end\] So, if we put all of this together we have, \[\begin = \frac\int_^ & = \frac\left[ \right]\ & = \frac\left[ \right] = - \frac\left( \right)\hspace\hspacen = 1,2,3, \ldots \end\] So, after all that work the Fourier series is, \[\beginf\left( x \right) & = \sum\limits_^\infty \sum\limits_^\infty \ & = \sum\limits_^\infty \sum\limits_^\infty \ & = L \sum\limits_^\infty - \sum\limits_^\infty \end\] As we saw in the previous example there is often quite a bit of work involved in computing the integrals involved here.

\[\begin = \frac\int_^ & = \frac\left[ \right]\ & = \frac\left[ \right]\end\] As with the coefficients for the cosines will probably be easier to do each of these individually. \right|_^0 = \frac\left( \right) = \frac\left( \right)\] \[\begin\int_^ & = \left.\right|_0^L\ & = \left( \right)\left( \right)\ & = \left( \right)\left( \right) = - \frac\end\] So, if we put all of this together we have, \[\begin = \frac\int_^ & = \frac\left[ \right]\ & = \frac\left[ \right] = - \frac\left( \right)\hspace\hspacen = 1,2,3, \ldots \end\] So, after all that work the Fourier series is, \[\beginf\left( x \right) & = \sum\limits_^\infty \sum\limits_^\infty \ & = \sum\limits_^\infty \sum\limits_^\infty \ & = L \sum\limits_^\infty - \sum\limits_^\infty \end\] As we saw in the previous example there is often quite a bit of work involved in computing the integrals involved here.

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Therefore, this is the only form of the coefficients for the Fourier series.\[ = \frac\int_^ = \frac\int_^ = L\] \[\begin &= \frac\int_^ = \frac\int_^\ & = \frac\left.\right|_^L\ & = \frac\left( \right)\left( \right) = 0\hspace\hspace\hspace\hspacen = 1,2,3, \ldots \end\] \[\begin &= \frac\int_^ = \frac\int_^\ & = \frac\left.Doing this gives, \[\int_^ = \sum\limits_^\infty \sum\limits_^\infty \] We can now take advantage of the fact that the sines and cosines are mutually orthogonal.The integral in the second series will always be zero and in the first series the integral will be zero if \(n \ne m\) and so this reduces to, \[\int_^ = \left\{ \right.\] Solving for \(\) gives, \[\begin& = \frac\int_^\ & = \frac\int_^\hspace\hspacem = 1,2,3, \ldots \end\] Now, do it all over again only this time multiply both sides by \(\sin \left( \right)\), integrate both sides from –\(L\) to \(L\) and interchange the integral and summation to get, \[\int_^ = \sum\limits_^\infty \sum\limits_^\infty \] In this case the integral in the first series will always be zero and the second will be zero if \(n \ne m\) and so we get, \[\int_^ = \left( L \right)\] Finally, solving for \(\) gives, \[ = \frac\int_^\hspace\hspacem = 1,2,3, \ldots \] In the previous two sections we also took advantage of the fact that the integrand was even to give a second form of the coefficients in terms of an integral from 0 to \(L\).Next here is the integral for \(\) \[ = \frac\int_^ = \frac\int_^ = \frac\int_^\] In this case we’re integrating an even function (\(x\) and sine are both odd so the product is even) on the interval \(\left[ \right]\) and so we can “simplify” the integral as shown above.The reason for doing this here is not actually to simplify the integral however.(Then t-x = 0 which means the integral can be further simplified) According to my understanding, t is the time domain and any domain x can be ...I have been tooling around with the discrete Fourier transform, and have noticed that when I try to oversample the inverse transform, I get aliasing effects. The important thing to note here is that the answer that we got in that example is identical to the answer we got here.If you think about it however, this should not be too surprising.

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