FS9

From Example Problems
Jump to: navigation, search

Find the Fourier series for f(t)={\begin{cases}{\frac  {4}{\pi }}t&0\leq t<{\frac  {\pi }{2}},\\{\frac  {-4}{\pi }}t&{\frac  {-\pi }{2}}\leq t\leq 0\end{cases}}

This is the general Fourier Series:

f(x)={a_{0} \over 2}+\sum _{{n=1}}^{\infty }\left[a_{n}\cos({2n\pi x \over T})+b_{n}\sin({2n\pi x \over T})\right]\,

a_{n}={2 \over T}\int _{c}^{{c+T}}f(x)\cos({2n\pi x \over T})\,dx\,

b_{n}={2 \over T}\int _{c}^{{c+T}}f(x)\sin({2n\pi x \over T})\,dx\,

So the given function can be replaced by its Fourier expansion:

f(t)={a_{0} \over 2}+\sum _{{n=1}}^{\infty }\left[a_{n}\cos \left({\frac  {2n\pi x}{\pi }}\right)+b_{n}\sin \left({\frac  {2n\pi x}{\pi }}\right)\right]\,

a_{0}={2 \over \pi }\int _{{{\frac  {-\pi }{2}}}}^{0}{\frac  {-4}{\pi }}t\,dt+{2 \over \pi }\int _{0}^{{\frac  {\pi }{2}}}{\frac  {4}{\pi }}t\,dt=2\,

a_{n}={\frac  {2}{\pi }}\int _{{-\pi  \over 2}}^{0}{\frac  {-4}{\pi }}t\cos(2nt)\,dt+{\frac  {2}{\pi }}\int _{0}^{{\frac  {\pi }{2}}}{\frac  {4}{\pi }}t\cos(2nt)\,dt={\frac  {4((-1)^{n}-1)}{\pi ^{2}n^{2}}}\,

b_{n}={\frac  {2}{\pi }}\int _{{{\frac  {-\pi }{2}}}}^{0}{\frac  {-4}{\pi }}t\sin(2nt)\,dt+{\frac  {2}{\pi }}\int _{0}^{{{\frac  {\pi }{2}}}}{\frac  {4}{\pi }}t\sin(2nt)\,dt=0\,

So the solution is

f(t)=1+\sum _{{n=1}}^{\infty }\left[{\frac  {4}{n^{2}\pi ^{2}}}((-1)^{n}-1)\cos(2nt)\right]\,

=1+\sum _{{n=1}}^{\infty }\left[{\frac  {-4\cdot 2}{(2n-1)^{2}\pi ^{2}}}\cos(2(2n-1)t)\right]\,

=1-{\frac  {8}{\pi ^{2}}}\sum _{{n=1}}^{\infty }{\frac  {\cos \left[2(2n-1)t\right]}{(2n-1)^{2}}}\,

Fourier Series

Main Page