Derivation of Fourier Coefficients

Derivation of Fourier CoefficientsSwapnil Sunil JainDecember 28, 2006

Derivation of Fourier Coefficients

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

$f(t)$$=$$c_0+{\displaystyle \sum _{n=1}^{\mathrm{\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

$f(t)=$$\mathrm{\hspace{0.33em}}{c}_0+a_1\cos ({\omega }_{1}t)+a_2\cos ({\omega }_{2}t)+\mathrm{\dots }+a_k\cos ({\omega }_{k}t)+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+b_1\sin ({\omega }_{1}t)+b_2\sin ({\omega }_{2}t)+\mathrm{\dots }+b_k\sin ({\omega }_{k}t)+\mathrm{\dots }$(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=$$\mathrm{\hspace{0.33em}}{\displaystyle {\int }_{\tau }}{c}_0𝑑t+{\displaystyle {\int }_{\tau }}{a}_1\cos ({\omega }_{1}t)𝑑t+{\displaystyle {\int }_{\tau }}{a}_2\cos ({\omega }_{2}t)𝑑t+\mathrm{\dots }+{\displaystyle {\int }_{\tau }}{a}_k\cos ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+{\displaystyle {\int }_{\tau }}{b}_1\sin ({\omega }_{1}t)𝑑t+{\displaystyle {\int }_{\tau }}{b}_2\sin ({\omega }_{2}t)𝑑t+\mathrm{\dots }+{\displaystyle {\int }_{\tau }}{b}_k\sin ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

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=$$\mathrm{\hspace{0.33em}}{c}_0{\displaystyle {\int }_{\tau }}𝑑t$

$\Rightarrow {\displaystyle {\int }_{\tau }}f(t)𝑑t=$$\mathrm{\hspace{0.33em}}{c}_0(T)$

$\Rightarrow c_0=$$\mathrm{\hspace{0.33em}}{\displaystyle \frac{1}{T}}{\displaystyle {\int }_{\tau }}f(t)𝑑t={\displaystyle \frac{1}{T}}{\displaystyle {\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

$f(t)\cos ({\omega }_{k}t)=$$\mathrm{\hspace{0.33em}}{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)+\mathrm{\dots }+a_k{\cos }^2({\omega }_{k}t)+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+b_1\sin ({\omega }_{1}t)\cos ({\omega }_{k}t)+b_2\sin ({\omega }_{2}t)\cos ({\omega }_{k}t)+\mathrm{\dots }+b_k\sin ({\omega }_{k}t)\cos ({\omega }_{k}t)+\mathrm{\dots }$

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=$$\mathrm{\hspace{0.33em}}{\displaystyle {\int }_{\tau }}{c}_0\cos ({\omega }_{k}t)𝑑t+{\displaystyle {\int }_{\tau }}{a}_1\cos ({\omega }_{1}t)\cos ({\omega }_{k}t)𝑑t+{\displaystyle {\int }_{\tau }}{a}_2\cos ({\omega }_{2}t)\cos ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+{\displaystyle {\int }_{\tau }}{a}_k{\cos }^2({\omega }_{k}t)𝑑t+\mathrm{\dots }+{\displaystyle {\int }_{\tau }}{b}_1\sin ({\omega }_{1}t)\cos ({\omega }_{k}t)𝑑t+{\displaystyle {\int }_{\tau }}{b}_2\sin ({\omega }_{2}t)\cos ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+{\displaystyle {\int }_{\tau }}{b}_k\sin ({\omega }_{k}t)\cos ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

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=$$\mathrm{\hspace{0.33em}}{\displaystyle {\int }_{\tau }}{a}_k{\cos }^2({\omega }_{k}t)𝑑t$

$\Rightarrow {\displaystyle {\int }_{\tau }}f(t)\cos ({\omega }_{k}t)𝑑t=$$\mathrm{\hspace{0.33em}}{a}_k({\displaystyle \frac{T}{2}})$

$\Rightarrow a_k=$$\mathrm{\hspace{0.33em}}{\displaystyle \frac{2}{T}}{\displaystyle {\int }_{\tau }}f(t)\cos ({\omega }_{k}t)𝑑t={\displaystyle \frac{2}{T}}{\displaystyle {\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

$f(t)\sin ({\omega }_{k}t)=$$\mathrm{\hspace{0.33em}}{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)+\mathrm{\dots }+a_k\cos ({\omega }_{k}t)\sin ({\omega }_{k}t)+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+b_1\sin ({\omega }_{1}t)\sin ({\omega }_{k}t)+b_2\sin ({\omega }_{2}t)\sin ({\omega }_{k}t)+\mathrm{\dots }+b_k{\sin }^2({\omega }_{k}t)+\mathrm{\dots }$

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=$$\mathrm{\hspace{0.33em}}{\displaystyle {\int }_{\tau }}{c}_0\cos ({\omega }_{k}t)𝑑t+{\displaystyle {\int }_{\tau }}{a}_1\cos ({\omega }_{1}t)\cos ({\omega }_{k}t)𝑑t+{\displaystyle {\int }_{\tau }}{a}_2\cos ({\omega }_{2}t)\cos ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+{\displaystyle {\int }_{\tau }}{a}_k\cos ({\omega }_{k}t)\sin ({\omega }_{k}t)𝑑t+\mathrm{\dots }+{\displaystyle {\int }_{\tau }}{b}_1\sin ({\omega }_{1}t)\cos ({\omega }_{k}t)𝑑t+{\displaystyle {\int }_{\tau }}{b}_2\sin ({\omega }_{2}t)\cos ({\omega }_{k}t)𝑑t+\mathrm{\dots }$

$\mathrm{\hspace{0.33em}}+{\displaystyle {\int }_{\tau }}{b}_k{\sin }^2({\omega }_{k}t)𝑑t+\mathrm{\dots }$

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=$$\mathrm{\hspace{0.33em}}{\displaystyle {\int }_{\tau }}{b}_k{\sin }^2({\omega }_{k}t)𝑑t$

$\Rightarrow {\displaystyle {\int }_{\tau }}f(t)\sin ({\omega }_{k}t)𝑑t=$$\mathrm{\hspace{0.33em}}{b}_k({\displaystyle \frac{T}{2}})$

$\Rightarrow b_k=$$\mathrm{\hspace{0.33em}}{\displaystyle \frac{2}{T}}{\displaystyle {\int }_{\tau }}f(t)\sin ({\omega }_{k}t)𝑑t={\displaystyle \frac{2}{T}}{\displaystyle {\int }_{t_0}^{t_0+T}}f(t)\sin ({\omega }_{k}t)𝑑t$