adv math[unit 3]

Advanced Engineering Mathematics

Fourier Series, Integrals and Transforms

FOURIER SERIES, INTEGRALS AND TRANSFORMS

Fourier series are infinite series designed to represent general periodic functions in terms of cosines and

sines. They constitute a very important tool, in particular in solving problems that involve ODEs and PDEs.

The theory of Fourier series is complicated, but we shall see that the application of these series is rather

simple. Fourier series are in a certain sense more universal than Taylor series in calculus because many

discontinuous periodic functions of practical interest can be developed in Fourier series but do not have

Taylor series representations.

In the latter part, we extend the ideas and techniques of Fourier series to nonperiodic functions which give

rise to Fourier integrals and Fourier transforms. These have applications to solving PDEs.

3.1 Fourier Series

Fourier series are the basic tool for representing periodic functions, which play an important role in

applications. A function f�t� is called a periodic function if f�t� is defined for all real t (except perhaps for some points) and if there is some positive number p, called a period of f�t� such that

f�t + p� = f�t� (3.1)

The graph of such a function is obtained by periodic repetition of its graph in any interval of length p. If f�t� has period p, it also has the period 2p. Thus for any integer n = 1, 2, 3…

f�t + np� = f�t� (3.2)

Furthermore, if f�t� and g�t� have period p , then af�t� + bg�t� with any constants a and b also has period p. We will represent the periodic function f�t� of period 2π in terms of simple functions whose period is also 2π. The series to be obtained will be a trigonometric series of the form

a� + a� cos t + b� sin t + a� cos2t + b� sin 2t +⋯

Thus, the Fourier series expansion of a periodic function f�t� with p = 2π is f�t� = a� +��a� cosnt + b� sin nt��

�� (3.3)

provided the right term converges to f�t�. The constants a�, a�, b�, a�, b�, etc., are called the coefficients of the series or the Fourier coefficients given by the following Euler formulas



a� = 12π� f�t�dt!

(3.4a)

a� = 1π� f�t� cos nt dt!

(3.4b)

b� = 1π� f�t� sin nt dt!

(3.4c)

where n = 1, 2, 3,⋯ and the limits of the integral defined for the whole period of f�t�. Example 3.1 Find the Fourier coefficients of the periodic function f�t� = "−k − π < t < 0k0 < t < π and determine its Fourier series expansion. Answer: Fourier coefficients: a� = 0 a� = 0 b� = 2knπ '1 − �−1��(

b� = 4kπ b� = 0 b* = 4k3π b+ = 0 b, = 4k5π b. = 0 b/ = 4k7π b1 = 0 b2 = 4k9π b�� = 0 ⋮ ⋮

Fourier series expansion: f�t� = 4kπ 5sin t + 13 sin 3t + 15 sin 5t + 17 sin 7t + ⋯6



Figure 3.1. The given periodic function f(t). Periodic rectangular wave.

Figure 3.2. Partial sums of the Fourier series expansion for f(t)



Drill Problems 3.1

Find the Fourier series expansion of the given f�t� which is assumed to have the period 2π . In the expansion, terms involving cos5t and sin 5t must be included. 1.

2.

3.

4.

5.



6.

7.

8.

9. f�t� = t��−π < t < π� 10. f�t� = t��0 < t < 2π�

3.2. Functions of any Period p = 2L The functions considered so far had period 2π, for the simplicity of formulas. However, periodic functions of

practical interest will generally have other periods. By change of scale, we find that the Fourier series

expansion of function f�t� whose period is p = 2L is f�t� = a� +�8a� cosπnL t + b� sin πnL t9�

�� (3.5)

and the Fourier coefficients are found by the formulas



a� = 12L� f�t�dt!

(3.6a)

a� = 1L� f�t� cosπnL t dt!

(3.6b)

b� = 1L� f�t� sin πnL t dt!

(3.6c)

Example 3.2

Find the Fourier series expansion of the function whose period is p = 4 f�t� = "−k − 2 < t < 0k0 < t < 2 Answer f�t� = 4kπ 5sin πt2 + 13 sin 3πt2 + 15 sin 5πt2 + 17 sin 7πt2 + ⋯6

Drill Problems 3.2 Find the Fourier series expansion of the following periodic functions. Terms involving cos 5t and sin 5t must be included.

1. f�t� = "−1 − 2 < t < 010 < t < 2 , p = 4

2. f�t� = "0 − 2 < t < 040 < t < 2, p = 4

3. f�t� = t� − 1 < t < 1, p = 2

4. f�t� = :;<� − 1 < t < 1, p = 2

5. f�t� = sin πt 0 < t < 1, p = 1

6. f�t� = cos πt − �� < t < ��, p = 1 7. f�t� = |t| − 1 < t < 1, p = 2



8. f�t� = "1 + t − 1 < t < 01 − t0 < t < 1, p = 2

9. f�t� = 1 − t� − 1 < t < 1, p = 2

10. f�t� = "0 − 2 < t < 0t0 < t < 2 , p = 4

3.3 Even and Odd Functions.

A function f�t� is even if f�−t� = f�t� and is symmetric with respect to the vertical axis.

Figure 3.3. An even function.

A function f�t� is odd if f�−t� = −f�t� and is symmetric with respect to the origin.

Figure 3.4. An odd function



The Fourier series of an even function of period p = 2L is a Fourier cosine series f�t� = a� +�a� cosπnL t�

�� (3.7)

with coefficients (integration from 0 to L only!) a� = 1L� f�t�dt>

� (3.8a)

a� = 2L� f�t� cos πnL t dt>�

(3.8b)

The Fourier series of an odd function of period p = 2L is a Fourier sine series f�t� = �b� sin πnL t�

�� (3.9)

with coefficients

b� = 2L� f�t� sin πnL t dt>�

(3.10)

The function in Examples 3.1 and 3.2 is an odd function. Therefore, it has a Fourier sine series expansion

given by 3.9 and coefficients given by 3.10.



Example 3.3 Find the Fourier series expansion of the function

f�t� = ?0 − 2 < t < −1k − 1 < t < 101 < t < 2

whose period is p = 2L = 4 Answer f�t� = k2 + 2kπ 5cosπt2 − 13 cos3πt2 + 15 cos 5πt2 − ⋯+⋯6

Figure 3.5. The function f(t) in Example 3.3

Drill Problems 3.3 Determine whether the following functions are even or odd then find its Fourier series expansion. Terms

involving cos5t and sin 5t must be included.

1. f�t� = π − |t| − π < t < π

2. f�t� = 2t|t| − 1 < t < 1

3. f�t� = ? t − :� < t < :�π − t :� < t < *:�

4. f�t� = "πeA; −π < t < 0πe; 0 < t < π

5. f�t� = " 2 −2 < t < 00 0 < t < 2

3.4 Fourier Integral

Fourier series are powerful tools for problems involving functions that are periodic or are of interest on a

finite interval only. Many problems involve functions that are nonperiodic and are of interest on the whole x-

axis, thus, we extend the method of Fourier series to such functions. This idea will lead to Fourier integrals.



We will start from a function whose period is p = 2L, derive its Fourier series expansion, and then let L → ∞. For a rectangular pulse, Figure 3.6 illustrates the idea

Figure 3.6. Waveform and amplitude spectrum of a rectangular pulse

For a periodic function f>�t� whose period is p = 2L, its Fourier series expansion is given by f>�t� = a� +��a� cosw�t + b� sinw�t��

�� (3.11)

where w� = �:> . If we let L → ∞, then we will haveA, for nonperiodic f�t�, f�t� = � 'A�w� coswt + B�w� sinwt(dw�

� (3.12)



with

A�w� = 1π� f�v� coswvdv�A� (3.13a)

B�w� = 1π� f�v� sinwvdv�A� (3.13b)

Equation 3.12 is the Fourier integral representation of f�t�. It can be seen that this representation is continuous, unlike the Fourier series for periodic function which is defined for integer values of n, and thus the quantity w� = �:> becomes continuous also.

Fourier Cosine Integral and Fourier Sine Integral. For an even or odd function, the Fourier integral

becomes simpler. If f�t� is an even function, then B�w� = 0 in 3.13b and A�w� = 1π� f�v� coswvdv�

A� (3.13a)

and the Fourier integral then reduces to the Fourier cosine integral. Similarly if f�t� is an odd function, A�w� = 0 in 3.13a and B�w� = 1π� f�v� sinwvdv�

A� (3.13b)

The Fourier integral becomes the Fourier sine integral.

3.5 Fourier Transform

The Fourier cosine and sine transforms are real. We now consider a third one, called the Fourier transform,

which is complex. From the real form of the Fourier integral

f�t� = � 'A�w� coswt + B�w� sinwt(dw�� (3.12)

with

A�w� = 1π� f�v� coswvdv�A� (3.13a)



B�w� = 1π� f�v� sinwvdv�A� (3.13b)

Substituting these into the integral, we have

f�t� = 1π� � f�v�'coswv coswt + sinwv sinwt(dvdw�A�

�� (3.14)

The terms inside the bracket is the addition formula for cosine, or cos�wx −wv�. Thus, f�t� = 1π� I� f�v� cos�wx − wv� dv�

A� J�� dw (3.15)

We shall call the term inside the bracket F�w� since cos�wx −wv� is an even function of w, and we

integrate with respect to v. We extend the limit of the integral of F�w� from w = 0 to ∞ to w = −∞ to +∞; thus a factor of �� is multiplied to the integral to account for this extension. Thus,

f�t� = 12π� I� f�v� cos�wx − wv� dv�A� J�

A� dw (3.16)

We claim that the integral of 3.16 with sine instead of a cosine is zero,

12π� I� f�v� sin�wx − wv� dv�A� J�

A� dw = 0 (3.17)

This is true since sin�wx − wv� is an odd function of w, which makes the integral inside the bracket,

calling it G�w�, equal to zero. Taking the integrand of 3.16 plus j times the integrand of 3.17 and use the Euler’s formula

eN; = cos t + j sin t (3.18)

Thus, the complex Fourier integral can now be written as

f�t� = 12π� � f�v�eNO�PAQ��A� dv�

A� dw (3.19)



Writing 3.19 as a product of exponentials

f�t� = 1√2π� I 1√2π� f�v�eANOQdv�A� J�

A� eNO;dw (3.20)

The expression in brackets is a function of w - for application purposes, we let w = ω – is denoted by F�ω� and is called the Fourier transform of f�t�; writing v = t. Thus, F�ω� = 1√2π� f�t�eANO; dt�

A� (3.21)

With this, 3.20 becomes

f�t� = 1√2π� F�w��A� eNO;dw (3.22)

which is the inverse Fourier transform of F�w�. We will use the notation

ℱ'f�t�( = F�ω� (3.23a)

to indicate the Fourier transform of f�t� and ℱA�'F�ω�( = f�t� (3.23b)

as the inverse Fourier transform of F�ω�. Example 3.4 Find the Fourier transform of f�t� = 1 if |x| < 1 and f�t� = 0 otherwise. Answer:

F�ω� = Uπ2 sinωω

Example 3.5

Find the Fourier transform of f�t� = eAV; for t > 0 and f�t� = 0 for t < 0. Answer: F�ω� = 1√2π�a + jω�



Drill Problems 3.4 Find the Fourier transform of the following functions using the Fourier transform integral.

1. f�t� = X1 − |;|V |t| ≤ a0otherwise

2. f�t� = eAV|;|

3. f�t� = "teA; − 1 < t < 00otherwise

4. f�t� = "|t| − 1 ≤ t ≤ 10otherwise

5. f�t� = ? 10 ≤ t ≤ a/2−1 − a/2 ≤ t ≤ 00otherwise

3.6 Properties of Fourier Transform

Linearity. Let f�t� and g�t� be two functions with Fourier transforms F�ω� and G�ω� respectively. Then, for constants a and b,

ℱ'af�t� + bg�t�( = aF�ω� + bG�ω� (3.24)

Shift in the frequency domain. Let f�t� be a function with Fourier transform F�ω�. Then ℱ]eNV;f�t�^ = F�ω − a� (3.25)

Example 3.6 If the function f�t� has the Fourier transform F�ω�, use the linearity and shifting in the frequency domain

properties of the Fourier transform to find ℱ'f�t� cos at(. Answer:

F�ω� = F�ω − a�2 + F�ω + a�2

Time scaling. Let f�t� be a function whose Fourier transform is F�ω�. For a constant c, ℱ'f�ct�( = F 8ωc 9c (3.26)



A special case of the scaling is called time-reversal, that is c = −1. Equation 3.26 becomes then,

ℱ'f�−t�( = −F�−ω� (3.27)

Example 3.7

If f�t� = "1 − 1 ≤ t ≤ 10otherwise, plot f�2t� and f 8;�9and find their Fourier transform. Use the results of

Example 3.4. Answer:

ℱ'f�2t�( = _2π sinω2ω

ℱ `f 5t26a = _2π sin 2ωω

Differentiation in time domain. Let f�t� be a continuously differentiable function with Fourier transform F�ω� and assume that lim;→±� f�t� = 0. Then, ℱ'f e�t�( = jωF�ω� (3.28)

and in general, if f�t� can be differentiated m times,

ℱ]f �f��t�^ = �jω�fF�ω� (3.29)

This property of Fourier transform can be used in the same manner as the Laplace transform in solving

ordinary differential equations.

Differentiation in frequency domain. If the function f�t� has the Fourier transform F�ω� and if tf�t� is absolutely integrable, then F�ω� is differentiable and

ℱ'tf�t�( = −jFe�ω� (3.30)

and in general

ℱ'�jt�ff�t�( = F�f��ω� (3.31)



Example 3.8

Find the Fourier transform of f�t� = "t − 1 ≤ t ≤ 10otherwise , using the results of Example 3.4 and the

differentiation property of the Fourier transform. Answer:

F�ω� = _2πωcosω − sinωω�

Integration in time-domain. If f�t� is a continuous and absolutely integrable function with Fourier transform F�ω�, and lim;→� g f�τ�dτ;A� = 0. Then

ℱ I� f�τ�dτ;A� J = F�ω�jω (3.32)

for ω ≠ 0. The following table summarizes the Fourier transform of various functions.

f�t� ℱ'f�t�( = F�ω� 3.1.1 "1 − b < t < b0otherwise _2πsin bωω

3.1.2 "1b < t < c0otherwise eANjk −eANlkjω√2π

3.1.3 1x� + a� �a > 0� Uπ2eAV|k|a

3.1.4 ? t0 < t < b2t − bb < t < 2b0otherwise −1 + 2eNjk − eAN�jkω�√2π

3.1.5 "eAV;t > 00otherwise �a > 0� 1√2π�a + jω� 3.1.6 "eV; b < t < c0otherwise e�VANk�l − e�VANk�j√2π�a − jω�

3.1.7 meNV; − b < t < b0otherwise _2πsin b�ω − a��ω − a�

3.1.8 m eNV; b < t < c0otherwise j√2π eNj�VAk� − eNl�VAk�a − ω

3.1.9 eAV;n �a > 0� 1√2πeAkn/+V



f�t� ℱ'f�t�( = F�ω� 3.1.10

sin att �a > 0� oUπ2|ω| < a0|ω| > a Table 3.1. Table of Fourier Transforms

The following table summarizes the Fourier transform of distributed functions, i.e. functions that are not

square-integrable but have Fourier transforms.

f�t� ℱ'f�t�( = F�ω� Remarks

3.2.1 1 √2πδ�ω� The transform of a constant function is the Dirac delta function.

3.2.2 δ�t� 1√2π The dual of rule 3.2.1. The transform of the Dirac delta function is a constant.

3.2.3 eNV; √2πδ�ω − a� Applying frequency shift to 3.2.1.

3.2.4 cos at √2πδ�ω − a� + δ�ω + a�2 Using rules 3.2.1, 3.2.3 and the Euler’s formula for cosine

3.2.5 sin at j√2πδ�ω + a� + δ�ω − a�2 Using rules 3.2.1, 3.2.3 and the Euler’s formula for sine

Table 3.2. Table of Fourier Transform for distributed functions

Drill Problems 3.5 Prove rules 3.1.1 through 3.1.10 using the integral for the Fourier transform and its properties.

adv math[unit 3]

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