# Derivation of Fourier Coefficients

Derivation of Fourier Coefficients Swapnil Sunil Jain December 28, 2006

Derivation of Fourier Coefficients

As you know, any periodic function $f(t)$ can be written as a Fourier series like the following

 $\displaystyle f(t)$ $\displaystyle=$ $\displaystyle c_{0}+\sum_{n=1}^{\infty}a_{n}\cos(\omega_{n}t)+b_{n}\sin(\omega% _{n}t)$ (1)

where $\omega_{n}=n\omega_{0}$ and $\omega_{0}=\frac{2\pi}{T}$

In the process to find an explicit expression for the coefficients $c_{0},a_{n},b_{n}$ in terms of $f(t)$, we write (1) in a slightly different way as the following

 $\displaystyle f(t)=$ $\displaystyle\;c_{0}+a_{1}\cos(\omega_{1}t)+a_{2}\cos(\omega_{2}t)+...+a_{k}% \cos(\omega_{k}t)+...$ $\displaystyle\;+b_{1}\sin(\omega_{1}t)+b_{2}\sin(\omega_{2}t)+...+b_{k}\sin(% \omega_{k}t)+...$ (2)

where $k$ is a positive integer.

In order to derive the coefficient $c_{0}$, we take the integral of both sides of (2) over one period.

 $\displaystyle\int_{\tau}f(t)dt=$ $\displaystyle\;\int_{\tau}c_{0}dt+\int_{\tau}a_{1}\cos(\omega_{1}t)dt+\int_{% \tau}a_{2}\cos(\omega_{2}t)dt+...+\int_{\tau}a_{k}\cos(\omega_{k}t)dt+...$ $\displaystyle\;+\int_{\tau}b_{1}\sin(\omega_{1}t)dt+\int_{\tau}b_{2}\sin(% \omega_{2}t)dt+...+\int_{\tau}b_{k}\sin(\omega_{k}t)dt+...$

where $\tau=[t_{0},t_{0}+T]$. After evaluating the above equation, all the integrals on the right side with a sine or a cosine term drop out (since the integral of a sine or cosine over one period is zero) and we get

 $\displaystyle\int_{\tau}f(t)dt=$ $\displaystyle\;c_{0}\int_{\tau}dt$ $\displaystyle\Rightarrow\int_{\tau}f(t)dt=$ $\displaystyle\;c_{0}(T)$ $\displaystyle\Rightarrow c_{0}=$ $\displaystyle\;\frac{1}{T}\int_{\tau}f(t)dt=\frac{1}{T}\int_{t_{0}}^{t_{0}+T}f% (t)dt$

Now, in order to find $a_{k}$, we multiply both sides of (2) by $\cos(\omega_{k}t)$ and we arrive at

 $\displaystyle f(t)\cos(\omega_{k}t)=$ $\displaystyle\;c_{0}\cos(\omega_{k}t)+a_{1}\cos(\omega_{1}t)\cos(\omega_{k}t)+% a_{2}\cos(\omega_{2}t)\cos(\omega_{k}t)+...+a_{k}{\cos}^{2}(\omega_{k}t)+...$ $\displaystyle\;+b_{1}\sin(\omega_{1}t)\cos(\omega_{k}t)+b_{2}\sin(\omega_{2}t)% \cos(\omega_{k}t)+...+b_{k}\sin(\omega_{k}t)\cos(\omega_{k}t)+...$

Then we take the integral of both sides of the above equation over one period and we get

 $\displaystyle\int_{\tau}f(t)\cos(\omega_{k}t)dt=$ $\displaystyle\;\int_{\tau}c_{0}\cos(\omega_{k}t)dt+\int_{\tau}a_{1}\cos(\omega% _{1}t)\cos(\omega_{k}t)dt+\int_{\tau}a_{2}\cos(\omega_{2}t)\cos(\omega_{k}t)dt% +...$ $\displaystyle\;+\int_{\tau}a_{k}{\cos}^{2}(\omega_{k}t)dt+...+\int_{\tau}b_{1}% \sin(\omega_{1}t)\cos(\omega_{k}t)dt+\int_{\tau}b_{2}\sin(\omega_{2}t)\cos(% \omega_{k}t)dt+...$ $\displaystyle\;+\int_{\tau}b_{k}\sin(\omega_{k}t)\cos(\omega_{k}t)dt+...$

By using orthogonality relationships or by literally evaluating the above integrals, we get the following

 $\displaystyle\int_{\tau}f(t)\cos(\omega_{k}t)dt=$ $\displaystyle\;\int_{\tau}a_{k}{\cos}^{2}(\omega_{k}t)dt$ $\displaystyle\Rightarrow\int_{\tau}f(t)\cos(\omega_{k}t)dt=$ $\displaystyle\;a_{k}(\frac{T}{2})$ $\displaystyle\Rightarrow a_{k}=$ $\displaystyle\;\frac{2}{T}\int_{\tau}f(t)\cos(\omega_{k}t)dt=\frac{2}{T}\int_{% t_{0}}^{t_{0}+T}f(t)\cos(\omega_{k}t)dt$

Now, the process of finding $b_{k}$ is similar  . We multiply both sides of (2) by $\sin(\omega_{k}t)$ and we get

 $\displaystyle f(t)\sin(\omega_{k}t)=$ $\displaystyle\;c_{0}\cos(\omega_{k}t)+a_{1}\cos(\omega_{1}t)\sin(\omega_{k}t)+% a_{2}\cos(\omega_{2}t)\sin(\omega_{k}t)+...+a_{k}\cos(\omega_{k}t)\sin(\omega_% {k}t)+...$ $\displaystyle\;+b_{1}\sin(\omega_{1}t)\sin(\omega_{k}t)+b_{2}\sin(\omega_{2}t)% \sin(\omega_{k}t)+...+b_{k}{\sin}^{2}(\omega_{k}t)+...$

Then we take the integral of both sides of the above equation over one period and we arrive at

 $\displaystyle\int_{\tau}f(t)\sin(\omega_{k}t)dt=$ $\displaystyle\;\int_{\tau}c_{0}\cos(\omega_{k}t)dt+\int_{\tau}a_{1}\cos(\omega% _{1}t)\cos(\omega_{k}t)dt+\int_{\tau}a_{2}\cos(\omega_{2}t)\cos(\omega_{k}t)dt% +...$ $\displaystyle\;+\int_{\tau}a_{k}\cos(\omega_{k}t)\sin(\omega_{k}t)dt+...+\int_% {\tau}b_{1}\sin(\omega_{1}t)\cos(\omega_{k}t)dt+\int_{\tau}b_{2}\sin(\omega_{2% }t)\cos(\omega_{k}t)dt+...$ $\displaystyle\;+\int_{\tau}b_{k}{\sin}^{2}(\omega_{k}t)dt+...$

By using orthogonality relationships or by literally evaluating the above integrals, we get the following

 $\displaystyle\int_{\tau}f(t)\sin(\omega_{k}t)dt=$ $\displaystyle\;\int_{\tau}b_{k}{\sin}^{2}(\omega_{k}t)dt$ $\displaystyle\Rightarrow\int_{\tau}f(t)\sin(\omega_{k}t)dt=$ $\displaystyle\;b_{k}(\frac{T}{2})$ $\displaystyle\Rightarrow b_{k}=$ $\displaystyle\;\frac{2}{T}\int_{\tau}f(t)\sin(\omega_{k}t)dt=\frac{2}{T}\int_{% t_{0}}^{t_{0}+T}f(t)\sin(\omega_{k}t)dt$
Title Derivation of Fourier Coefficients DerivationOfFourierCoefficients1 2013-03-11 19:30:41 2013-03-11 19:30:41 swapnizzle (13346) (0) 1 swapnizzle (0) Definition