A Fourier series is an expansion of a periodic function
\(\begin{array}{l}f(x)\end{array} \)
in terms of an infinite sum of sines and cosines. It decomposes any periodic function or periodic signal into the sum of a set of simple oscillating functions, namely sines and cosines.
\[\large f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx\]
Where
ao =
\(\begin{array}{l}\frac{1}{\pi} \int_{- \pi}^{\pi} f(x) dx\end{array} \)
an =Â
\(\begin{array}{l}\frac{1}{\pi} \int_{-\pi}^{\pi} f(x)cos\;nx\;dx\end{array} \)
bn =Â
\(\begin{array}{l}\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)sin\;nx\;dx\end{array} \)
n = 1, 2, 3…..
Solved Example
Question: Expand the function f(x) = ekx in the interval [ –
\(\begin{array}{l}\pi\end{array} \)
,
\(\begin{array}{l}\pi\end{array} \)
]Â using fourier series.
Solution:
Let the Fourier series for f(x) be
f(x) = \(\begin{array}{l}\frac{1}{2}\end{array} \)
ao +
\(\begin{array}{l}\sum_{n=1}^{\infty}\end{array} \)
(a
ncos(nx) + b
nsin(nx))
Here
a0 = \(\begin{array}{l}\frac{1}{\pi}\end{array} \)
\(\begin{array}{l}\int_{-\pi}^{\pi}\end{array} \)
f(x) dx
= \(\begin{array}{l}\frac{1}{\pi}\end{array} \)
\(\begin{array}{l}\int_{-\pi}^{\pi}\end{array} \)
e
kxdx
= \(\begin{array}{l}\frac{1}{k \pi}\end{array} \)
[ekx
\(\begin{array}{l}]_{-\pi}^{\pi}\end{array} \)
= \(\begin{array}{l}\frac{1}{k \pi}\end{array} \)
(ekπ – e-kπ )
= \(\begin{array}{l}\frac{2}{k \pi}\end{array} \)
sinh( k
\(\begin{array}{l}\pi\end{array} \)
)
Now,
an = \(\begin{array}{l}\frac{1}{\pi}\end{array} \)
\(\begin{array}{l}\int_{- \pi}^{\pi}\end{array} \)
e
kx cos (nx) dx
= \(\begin{array}{l}\frac{e^{(kx)}}{\pi(k^2+n^2))}\end{array} \)
\(\begin{array}{l}[(k cos(nx) + n sin(nx))]_{-\pi}^{\pi}\end{array} \)
= \(\begin{array}{l}\frac{1}{\pi(k^2 + n^2)}\end{array} \)
\(\begin{array}{l}[e^{k \pi} (k cos(n \pi) + n sin(n \pi)) – e^{-k \pi} [k cos(n \pi) – n sin (n \pi) ]\end{array} \)
= \(\begin{array}{l}\frac{k cos(n \pi)}{\pi (k^2 + n^2)}\end{array} \)
Â
\(\begin{array}{l}[e^{k \pi} – e^{-k \pi}]\end{array} \)
= 2k (-1)n\(\begin{array}{l}\frac{sinh(k \pi)}{\pi (k^2 + n^2)}\end{array} \)
and
bn = \(\begin{array}{l}\frac{1}{\pi}\end{array} \)
\(\begin{array}{l}\int_{-\pi}^{\pi}\end{array} \)
e
kx sin(nx) dx
= \(\begin{array}{l}\frac{e^{(kx)}}{\pi(k^2 + n^2)}\end{array} \)
\(\begin{array}{l}[k sin(nx) – n cos(nx) ]_{-\pi}^{\pi}\end{array} \)
= \(\begin{array}{l}\frac{1}{\pi(a^2 + n^2)}\end{array} \)
\(\begin{array}{l}[e^{k \pi}(k sin(n \pi) – n cos(n \pi)) – e^{-k \pi} (-k sin (n \pi) – n cos(n \pi)) ]\end{array} \)
= \(\begin{array}{l}\frac{-n cos(n \pi)}{\pi(k^2 + n^2)}\end{array} \)
\(\begin{array}{l}[e^{k \pi} – e^{-k \pi}]\end{array} \)
= \(\begin{array}{l}\frac{-2n(-1)^n sinh(k \pi)}{\pi(k^2 + n^2)}\end{array} \)
Substituting these values of ao ,an ,bn we get
f(x) = ekx = \(\begin{array}{l}\frac{1}{k \pi}\end{array} \)
sinh(k
\(\begin{array}{l}\pi\end{array} \)
) +
\(\begin{array}{l}\sum_{n=1}^{\infty}\end{array} \)
\(\begin{array}{l}\frac{2 (-1)^{n} sinh(k \pi)}{\pi(k^{2} + n^{2})}\end{array} \)
 (a
\(\begin{array}{l}\;\end{array} \)
cos(nx) – n
\(\begin{array}{l}\;\end{array} \)
sin(nx))
\(\begin{array}{l}\frac{2 sinh(k \pi)}{\pi}\end{array} \)
\(\begin{array}{l}\frac{1}{2k}\end{array} \)
+
\(\begin{array}{l}\sum_{1}^{\infty}\end{array} \)
\(\begin{array}{l}\frac{(-1)^{n}}{k^{2} + n^{2}}\end{array} \)
(k
\(\begin{array}{l}\;\end{array} \)
cos(nx) – n
\(\begin{array}{l}\;\end{array} \)
sin(nx))]
or
ekx = \(\begin{array}{l}\frac{2sinh(k \pi)}{\pi}\end{array} \)
\(\begin{array}{l}\frac{1}{2k}\end{array} \)
– k [\(\begin{array}{l}\frac{cos x}{k^{2} + 1^{2}}\end{array} \)
– \(\begin{array}{l}\frac{cos 2x}{k^{2} + 2^{2}}\end{array} \)
+ \(\begin{array}{l}\frac{cos 3x}{k^{2} + 3^{2}}\end{array} \)
– ….] + [\(\begin{array}{l}\frac{sin x}{k^{2} + 1^{2}}\end{array} \)
– \(\begin{array}{l}\frac{2\;sin\;2x}{k^{2} + 2^{2}}\end{array} \)
+ \(\begin{array}{l}\frac{3\; sin\; 3x}{k^{2} + 3^{2}}\end{array} \)
-…..) ]
Comments