Calculation of Fourier coefficients

3 Calculation of Fourier coefficients

Consider the Fourier series for a function $f (t)$ of period $2 π$ :

$f (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} cos n t + b_{n} sin n t)$ (7)

To obtain the coefficients $a_{n}$ ( $n = 1, 2, 3, \dots$ ), we multiply both sides by $cos m t$ where $m$ is some positive integer and integrate both sides from $- π$ to $π$ .

For the left-hand side we obtain

$\int_{- π}^{π} f (t) cos m t d t$

For the right-hand side we obtain

$\frac{a_{0}}{2} \int_{- π}^{π} cos m t d t + \sum_{n = 1}^{\infty} \{a_{n} \int_{- π}^{π} cos n t cos m t d t + b_{n} \int_{- π}^{π} sin n t cos m t d t\}$

The first integral is zero using (5).

Using the orthogonality relations all the integrals in the summation give zero except for the case $n = m$ when, from Key Point 3

$\int_{- π}^{π} {cos}^{2} m t d t = π$

Hence

$\int_{- π}^{π} f (t) cos m t d t = a_{m} π$

from which the coefficient $a_{m}$ can be obtained.

Rewriting $m$ as $n$ we get

$a_{n} = \frac{1}{π} \int_{- π}^{π} f (t) cos n t d t for n = 1, 2, 3, \dots$ (8)

Using (6), we see the formula also works for $n = 0$ (but we must remember that the constant term is $\frac{a_{0}}{2}$ .)

From (8)

$a_{n} = 2 \times$ average value of $f (t) cos n t$ over one period.

Task!

By multiplying (7) by $sin m t$ obtain an expression for the Fourier Sine coefficients $b_{n}$ , $n = 1, 2, 3, \dots$

A similar calculation to that performed to find the $a_{n}$ gives

\begin{array}{rcl} \int_{- π}^{π} f (t) sin m t d t = \frac{a_{0}}{2} \int_{- π}^{π} sin m t d t + \sum_{n = 1}^{\infty} \{\int_{- π}^{π} a_{n} cos n t sin m t d t + \int_{- π}^{π} b_{n} sin n t sin m t d t\} \end{array}

All terms on the right-hand side integrate to zero except for the case $n = m$ where

$\int_{- π}^{π} b_{m} {sin}^{2} m t d t = b_{m} π$

Relabelling $m$ as $n$ gives

$b_{n} = \frac{1}{π} \int_{- π}^{π} f (t) sin n t d t n = 1, 2, 3, \dots$ (9)

(There is no Fourier coefficient $b_{0}$ .)

Clearly $b_{n} = 2 \times$ average value of $f (t) sin n t$ over one period.

Key Point 4

A function $f (t)$ with period $2 π$ has a Fourier series

f (t) = \frac{a_{0}}{2} + \sum_{n = 1}^{\infty} (a_{n} cos n t + b_{n} sin n t)

The Fourier coefficients are

\begin{array}{rcl} a_{n} & = & \frac{1}{π} \int_{- π}^{π} f (t) cos n t d t n = 0, 1, 2, \dots \\ b_{n} & = & \frac{1}{π} \int_{- π}^{π} f (t) sin n t d t n = 1, 2, \dots \end{array}

In the integrals any convenient integration range extending over an interval of $2 π$ may be used.

3 Calculation of Fourier coefficients

Task!

Answer

Key Point 4