# Fourier series

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.1a

Show that the Fourier series coefficients, ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$ in equation (9.1a), are given by equations (9.1d,e).

### Background

If ${\displaystyle g(t)}$ is a periodic function, that is, one that repeats exactly at intervals ${\displaystyle T}$ (called the period) and has a finite number of maxima, minima, and discontinuities in each interval ${\displaystyle T}$, then it can be expressed in a Fourier series in any one of the three following equivalent forms, ${\displaystyle n}$ being an integer:

 {\displaystyle {\begin{aligned}g(t)={\frac {1}{2}}a_{0}+\sum \limits _{n=1}^{+\infty }(a_{n}\cos \omega _{n}t+b_{n}\sin \omega _{n}t)\end{aligned}}} (9.1a)
 {\displaystyle {\begin{aligned}={\frac {1}{2}}c_{0}+\sum \limits _{n=1}^{+\infty }c_{n}\cos(\omega _{n}t-\gamma _{n})\end{aligned}}} (9.1b)
 {\displaystyle {\begin{aligned}=\sum \limits _{n=-\infty }^{+\infty }\alpha _{n}e^{{\hbox{j}}\omega _{n}t},\end{aligned}}} (9.1c)

where ${\displaystyle \omega _{0}=2\pi f_{0}=2\pi /T=}$ fundamental angular frequency, ${\displaystyle \omega _{n}=n\omega _{0}=}$ harmonic frequency, and the amplitudes and phases are given by the following equations where ${\displaystyle 0, except for equation (9.1h) where ${\displaystyle -\infty \leq n\leq +\infty }$:

 {\displaystyle {\begin{aligned}a_{n}={\frac {2}{T}}\int _{-T/2}^{+T/2}g(t)\cos \,\omega _{n}t\,\mathrm {d} t,\end{aligned}}} (9.1d)
 {\displaystyle {\begin{aligned}b_{n}={\frac {2}{T}}\int _{-T/2}^{+T/2}g(t)\sin \,\omega _{n}t\,\mathrm {d} t,\end{aligned}}} (9.1e)
 {\displaystyle {\begin{aligned}c_{n}={\frac {2}{T}}\int _{-T/2}^{+T/2}g(t)\cos \,(\omega _{n}t-\gamma _{n})\,\mathrm {d} t,\end{aligned}}} (9.1f)
 {\displaystyle {\begin{aligned}\gamma _{n}=\tan ^{-1}(b_{n}/a_{n})=\mathrm {phase} ,\gamma _{0}=0,\end{aligned}}} (9.1g)
 {\displaystyle {\begin{aligned}\alpha _{n}={\frac {1}{T}}\int _{-T/2}^{+T/2}g(t)e^{-{\hbox{j}}\omega _{n}t}\mathrm {d} t.\end{aligned}}} (9.1h)

### Solution

We multiply both sides of equation (9.1a) by ${\displaystyle {\rm {\;cos\;}}\omega _{r}t}$, ${\displaystyle r}$ being any nonzero positive integer, and integrate between the limits ${\displaystyle -T/2}$ and ${\displaystyle +T/2}$. (We omit the limits on the integrals in the following proofs since they are all the same.) Equation (9.1a) now becomes

{\displaystyle {\begin{aligned}\int g(t)\cos \omega _{r}t\mathrm {d} t&=\sum \limits _{n=1}^{+\infty }a_{n}\int \cos \,\omega _{r}t\cos \omega _{n}t\,\mathrm {d} t\\&+\sum \limits _{n=1}^{+\infty }b_{n}\int \cos \omega _{r}t\sin \omega _{n}t\,\mathrm {d} t.\end{aligned}}}

By using the identities

{\displaystyle {\begin{aligned}\cos x\cos y={\frac {1}{2}}[\cos(x+y)+\cos(x-y)]\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{and}}\quad \quad \quad \cos x\sin y={\frac {1}{2}}[\sin(x+y)-\sin(x-y)],\end{aligned}}}

the first integrand reduces to the sum of two cosines of the form ${\displaystyle \cos(\omega _{r}\pm \omega _{n})t}$ while the second integrand becomes the difference between two sines of the form ${\displaystyle \sin(\omega _{r}\pm \omega _{n})t.}$ If ${\displaystyle r\neq n}$, the integrals are zero, because at the limits the arguments of the cosines and sines are equal to ${\displaystyle 2p\omega _{0}T/2=2p\pi }$, ${\displaystyle p}$ integral. When ${\displaystyle r=n}$, the first integrand becomes ${\displaystyle \cos ^{2}\omega _{n}t={\frac {1}{2}}(1+\cos \,2\omega _{n}t)}$ and the right-hand side equals ${\displaystyle a_{n}T/2}$. When ${\displaystyle r=n}$, the second integrand becomes ${\displaystyle \cos \omega _{n}t\sin \omega _{n}t\,\mathrm {d} t={\frac {1}{2}}\sin \,2\omega _{n}t\,\mathrm {d} t}$, which again gives zero upon integration.

Thus we are left with

{\displaystyle {\begin{aligned}\int g(t)\cos \omega _{n}t\,\mathrm {d} t=a_{n}T/2,\quad n=0,1,2,\ldots .\end{aligned}}}

Solving for ${\displaystyle a_{n}}$ gives equation (9.1d).

To verify equation (9.1e), we multiply both sides of equation (9.1a) by ${\displaystyle \sin \omega _{r}t}$ and integrate, all integrals on the right side of the equation giving zero except for the one with the integrand ${\displaystyle b_{n}\sin ^{2}\omega _{n}t=(b_{n}/2)(1-\cos \,2\omega _{n}t)}$. Proceeding as before, we arrive at equation (9.1e).

When we set ${\displaystyle n=0}$ in equations (9.1d,e) we get ${\displaystyle b_{0}=0}$ and

{\displaystyle {\begin{aligned}a_{0}/2={\frac {1}{T}}\int g(t)\,\mathrm {d} t=\mathrm {average\ value\ of} \,g(t).\end{aligned}}}

## Problem 9.1b

Verify equations (9.1f,g) for the Fourier series coefficients ${\displaystyle c_{n}}$ and ${\displaystyle \gamma _{n}}$.

### Solution

To verify equation (9.1f) for ${\displaystyle c_{n}}$, we multiply both sides of equation (9.1b) by ${\displaystyle \cos(\omega _{r}t-\gamma _{r})}$, ${\displaystyle r\neq 0}$, and proceed as in part (a). All integrals vanish except when ${\displaystyle r=n}$, and the result is equation (9.1f).

When ${\displaystyle r=0}$, we have

{\displaystyle {\begin{aligned}\int g(t)\,\mathrm {d} t={\frac {c_{0}}{2}}\int \mathrm {d} t={\frac {c_{0}T}{2}},\\{\hbox{so}}\quad \quad \quad {\frac {c_{0}}{2}}={\frac {1}{T}}\int g(t)\,\mathrm {d} t=\mathrm {average\ value\ of} \ g(t).\end{aligned}}}

To verify equation (9.1g) for ${\displaystyle \gamma _{n}}$, we compare terms in equations (9.1a) and (9.1b) that involve ${\displaystyle \omega _{n}t}$, ${\displaystyle n\neq 0}$. The result is

{\displaystyle {\begin{aligned}a_{n}\cos \omega _{n}t+b_{n}\sin \omega _{n}t&=c_{n}\cos(\omega _{n}t-\gamma _{n})\\&=c_{n}(\cos \omega _{n}t\,\cos \,\gamma _{n}+\sin \omega _{n}t\,\sin \,\gamma _{n}).\end{aligned}}}

Equating coefficients of ${\displaystyle \cos \omega _{n}t}$ and ${\displaystyle \sin \omega _{n}t}$, we get

{\displaystyle {\begin{aligned}a_{n}=c_{n}\cos \gamma _{n},\quad b_{n}=c_{n}\sin \gamma _{n},\quad c_{n}^{2}=a_{n}^{2}+b_{n}^{2},\quad \tan \gamma _{n}=b_{n}/a_{n}.\end{aligned}}}

When ${\displaystyle n=0}$, ${\displaystyle a_{0}=c_{0}}$, ${\displaystyle b_{0}=0}$, ${\displaystyle \gamma _{0}=0}$.

## Problem 9.1c

Verify the coefficients for the exponential form, equation (9.1h).

### Solution

To derive equation (9.1h), we use Euler’s equations (Sheriff and Geldart, 1995, problem 15.12a), namely,

{\displaystyle {\begin{aligned}\cos x={\frac {1}{2}}(e^{{\hbox{j}}x}+e^{-{\hbox{j}}x}),\quad \sin \,x={\frac {1}{2{\hbox{j}}}}(e^{{\hbox{j}}x}-e^{-{\hbox{j}}x}),\\{\hbox{or}}\quad \quad \quad e^{{\hbox{j}}x}=(\cos x+{\hbox{j}}\sin x),\quad e^{-{\hbox{j}}x}=(\cos x-{\hbox{j}}\sin x).\end{aligned}}}

Multiplying both sides of equation (9.1c) by ${\displaystyle e^{-{\hbox{j}}\omega _{r}t}}$, ${\displaystyle r\neq 0}$, and integrating, we get a series of integrals with integrands such as ${\displaystyle e^{{\hbox{j}}(\omega _{n}-\omega _{r})t}}$. Using Euler’s formula, ${\displaystyle e^{{\hbox{j}}(\omega _{n}-\omega _{r})t}={\hbox{cos}}(\omega _{n}-\omega _{r})t+{\hbox{j}}\ {\hbox{sin}}(\omega _{n}-\omega _{r})t}$, so the integrals vanish for all values of ${\displaystyle n}$ except when ${\displaystyle n=r}$; for this value we get

{\displaystyle {\begin{aligned}\int g(t)e^{-{\hbox{j}}\omega _{n}t}\,\mathrm {\rm {d}} t=\alpha _{n}\int \mathrm {d} t=\alpha _{n}T,\end{aligned}}}

which is equation (9.1h), ${\displaystyle n\neq 0}$. For ${\displaystyle n=0}$,

{\displaystyle {\begin{aligned}\alpha _{0}={\frac {1}{T}}\int g(t)\,\mathrm {d} t=\mathrm {average\ value\ of} \,g(t).\end{aligned}}}