Worked examples · Sheet 1

Fourier series of square waves and step functions

Theory used: Fourier coefficients Cosine-only and sine-only series Fourier series

Exercise 1

Find the Fourier series of the $2 π$ -periodic extension of

f (x) = {- 1 1 if - π \leq x < 0 if 0 \leq x < π

f odd ⟹ a_{n} = 0, b_{n} = \frac{2}{π} \int_{0}^{π} sin (n x) d x

b_{n} = \frac{2}{π} [- \frac{cos ( n x )}{n}]_{0}^{π} = \frac{2}{nπ} (1 - cos (nπ)) = \frac{2 ( 1 - ( - 1 ) ^{n} )}{nπ} = ⎩ ⎨ ⎧ \frac{4}{nπ} 0 if n is odd if n is even

f (x) \sim k = 0 \sum \infty \frac{4}{( 2 k + 1 ) π} sin ((2 k + 1) x)

The function equals $- 1$ to the left of the origin and $1$ to the right, therefore it is odd (the value at $x = 0$ does not affect the coefficients, which are integrals). Since the function is odd, all the $a_{n}$ are zero and the series contains only sines, with $b_{n}$ given by twice the integral over $[0, π]$ . On $[0, π]$ the function equals $1$ , hence what remains is the integral of $sin (n x)$ , which equals $\frac{1 - cos ( nπ )}{n}$ . Since $cos (nπ) = (- 1)^{n}$ , the coefficient is zero for even $n$ and equals $\frac{4}{nπ}$ for odd $n$ ; in the final series only the odd harmonics survive, written with the index $n = 2 k + 1$ .

Exercise 2

Find the Fourier series of the $2 π$ -periodic extension of

f (x) = {01 if - π \leq x < 0 if 0 \leq x < π

a_{0} = \frac{1}{π} \int_{0}^{π} 1 d x = 1

a_{n} = \frac{1}{π} \int_{0}^{π} cos (n x) d x = \frac{1}{π} [\frac{sin ( n x )}{n}]_{0}^{π} = 0 (n \geq 1)

b_{n} = \frac{1}{π} \int_{0}^{π} sin (n x) d x = \frac{1 - ( - 1 ) ^{n}}{nπ}

f (x) \sim \frac{1}{2} + k = 0 \sum \infty \frac{2}{( 2 k + 1 ) π} sin ((2 k + 1) x)

y = \dfrac{1}{2}

y = \dfrac{1}{2} + \dfrac{2}{\pi} \left(\sin x + \dfrac{\sin \left(3 x\right)}{3} + \dfrac{\sin \left(5 x\right)}{5}\right)

Periodic extension (

0

[- π, 0)

1

[0, π)

), the mean

\frac{1}{2}

(dashed) and the partial sum up to

sin (5 x)

(red), which oscillates around the function

The function is neither even nor odd, therefore all the coefficients are computed with the integrals over $[- π, π]$ ; since the function is zero on $[- π, 0)$ , every integral reduces to the interval $[0, π]$ . The coefficient $a_{0}$ equals $1$ , therefore the constant term of the series is $\frac{a _{0}}{2} = \frac{1}{2}$ , which is the mean of the function over one period. The $a_{n}$ with $n \geq 1$ are zero because $sin (nπ) = sin 0 = 0$ . The $b_{n}$ are those of the previous exercise divided by $2$ , hence the series is the constant $\frac{1}{2}$ plus half the series of the square wave: the result is consistent with the identity $f (x) = \frac{1}{2} + \frac{1}{2} g (x)$ , where $g$ is the square wave of exercise 1.

Exercise 3

Find the Fourier series of the $2 π$ -periodic extension of the odd function defined on $(0, π)$ by

f (x) = ⎩ ⎨ ⎧ 10 if 0 < x < \frac{π}{2} if \frac{π}{2} \leq x < π

f odd ⟹ a_{n} = 0, b_{n} = \frac{2}{π} \int_{0}^{π /2} sin (n x) d x

b_{n} = \frac{2}{π} [- \frac{cos ( n x )}{n}]_{0}^{π /2} = \frac{2}{nπ} (1 - cos \frac{nπ}{2})

cos \frac{nπ}{2} = ⎩ ⎨ ⎧ 01 - 1 if n is odd if n = 4, 8, 12, \dots if n = 2, 6, 10, \dots

f (x) \sim n = 1 \sum \infty \frac{2}{nπ} (1 - cos \frac{nπ}{2}) sin (n x) = \frac{2}{π} sin x + \frac{2}{π} sin (2 x) + \frac{2}{3 π} sin (3 x) + \frac{2}{5 π} sin (5 x) + \dots

y = \dfrac{2}{\pi} \left(\sin x + \sin \left(2 x\right) + \dfrac{\sin \left(3 x\right)}{3}\right)

Odd

2 π

-periodic extension:

1

(0, \frac{π}{2})

0

(\frac{π}{2}, π)

- 1

(- \frac{π}{2}, 0)

0

(- π, - \frac{π}{2})

; dashed, the partial sum up to

sin (3 x)

The function is given on $(0, π)$ and extended by oddness, therefore the $a_{n}$ are zero and the $b_{n}$ are computed with twice the integral over $[0, π]$ . The function is zero on $[\frac{π}{2}, π)$ , hence the integral reduces to $(0, \frac{π}{2})$ . The value $cos \frac{nπ}{2}$ follows a cycle of length four, therefore the coefficients are no longer split into even and odd but into four classes according to the remainder of $n$ modulo $4$ : for odd $n$ , $b_{n} = \frac{2}{nπ}$ ; for $n$ a multiple of $4$ the coefficient is zero; for even $n$ not a multiple of $4$ , $b_{n} = \frac{4}{nπ}$ . In the series written out in full the term in $sin (2 x)$ has coefficient $\frac{4}{2 π} = \frac{2}{π}$ and the term in $sin (4 x)$ does not appear.