Adv math[unit 3]

Advanced Engineering Mathematics
Fourier Series, Integrals and Transforms Page 1
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 fሺtሻ is called a periodic function if fሺtሻ is defined for all real t (except perhaps for
some points) and if there is some positive number p, called a period of fሺtሻ such that
fሺt + pሻ = fሺtሻ (3.1)
The graph of such a function is obtained by periodic repetition of its graph in any interval of length p.
If fሺtሻ has period p, it also has the period 2p. Thus for any integer n = 1, 2, 3 …
fሺt + npሻ = fሺtሻ (3.2)
Furthermore, if fሺtሻ and gሺtሻ have period p, then afሺtሻ + bgሺtሻ with any constants a and b also has
period p.
We will represent the periodic function fሺtሻ 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଴ + aଵ cos t + bଵ sin t + aଶ cos 2t + bଶ sin 2t + ⋯
Thus, the Fourier series expansion of a periodic function fሺtሻ with p = 2π is
fሺtሻ = a଴ + ෍ሺa୬ cosnt + b୬ sin ntሻ
ஶ
୬ୀଵ
(3.3)
provided the right term converges to fሺtሻ. The constants a଴, aଵ, bଵ, aଶ, bଶ, etc., are called the coefficients
of the series or the Fourier coefficients given by the following Euler formulas

a଴ =
1
2π
න fሺtሻ dt
୮
(3.4a)
a୬ =
1
π
න fሺtሻcos nt dt
୮
(3.4b)
b୬ =
1
π
න fሺtሻ sin nt dt
୮
(3.4c)
where n = 1, 2, 3, ⋯ and the limits of the integral defined for the whole period of fሺtሻ.
Example 3.1
Find the Fourier coefficients of the periodic function
fሺtሻ = ቄ
−k − π < t < 0
k 0 < t < π
and determine its Fourier series expansion.
Answer:
Fourier coefficients:
a଴ = 0
a୬ = 0
b୬ =
2k
nπ
ሾ1 − ሺ−1ሻ୬ሿ
bଵ =
4k
π
bଶ = 0
bଷ =
4k
3π
bସ = 0
bହ =
4k
5π
b଺ = 0
b଻ =
4k
7π
b଼ = 0
bଽ =
4k
9π
bଵ଴ = 0
⋮ ⋮
Fourier series expansion:
fሺtሻ =
4k
π
൬sin t +
1
3
sin 3t +
1
5
sin 5t +
1
7
sin 7t + ⋯ ൰

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 fሺtሻ which is assumed to have the period 2π. In the
expansion, terms involving cos 5t and sin 5t must be included.
1.
2.
3.
4.
5.

6.
7.
8.
9. fሺtሻ = tଶ
ሺ−π < t < πሻ
10. fሺtሻ = tଶ
ሺ0 < t < 2πሻ
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 fሺtሻ whose period is p = 2L is
fሺtሻ = a଴ + ෍ ቀa୬ cos
πn
L
t + b୬ sin
πn
L
tቁ
ஶ
୬ୀଵ
(3.5)
and the Fourier coefficients are found by the formulas

a଴ =
1
2L
න fሺtሻ dt
୮
(3.6a)
a୬ =
1
L
න fሺtሻ cos
πn
L
t dt
୮
(3.6b)
b୬ =
1
L
න fሺtሻ sin
πn
L
t dt
୮
(3.6c)
Example 3.2
Find the Fourier series expansion of the function whose period is p = 4
fሺtሻ = ቄ
−k − 2 < t < 0
k 0 < t < 2
Answer
fሺtሻ =
4k
π
൬sin
πt
2
+
1
3
sin
3πt
2
+
1
5
sin
5πt
2
+
1
7
sin
7πt
2
+ ⋯ ൰
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. fሺtሻ = ቄ
−1 − 2 < t < 0
1 0 < t < 2
, p = 4
2. fሺtሻ = ቄ
0 − 2 < t < 0
4 0 < t < 2
, p = 4
3. fሺtሻ = tଶ
− 1 < t < 1, p = 2
4. fሺtሻ =
஠୲య
ଶ
− 1 < t < 1, p = 2
5. fሺtሻ = sin πt 0 < t < 1, p = 1
6. fሺtሻ = cos πt −
ଵ
ଶ
< t <
ଵ
ଶ
, p = 1
7. fሺtሻ = |t| − 1 < t < 1, p = 2

8. fሺtሻ = ቄ
1 + t − 1 < t < 0
1 − t 0 < t < 1
, p = 2
9. fሺtሻ = 1 − tଶ
− 1 < t < 1, p = 2
10. fሺtሻ = ቄ
0 − 2 < t < 0
t 0 < t < 2
, p = 4
3.3 Even and Odd Functions.
A function fሺtሻ is even if fሺ−tሻ = fሺtሻ and is symmetric with respect to the vertical axis.
Figure 3.3. An even function.
A function fሺtሻ is odd if fሺ−tሻ = −fሺtሻ 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
fሺtሻ = a଴ + ෍ a୬ cos
πn
L
t
ஶ
୬ୀଵ
(3.7)
with coefficients (integration from 0 to L only!)
a଴ =
1
L
න fሺtሻ dt
୐
଴
(3.8a)
a୬ =
2
L
න fሺtሻ cos
πn
L
t dt
୐
଴
(3.8b)
The Fourier series of an odd function of period p = 2L is a Fourier sine series
fሺtሻ = ෍ b୬ sin
πn
L
t
ஶ
୬ୀଵ
(3.9)
with coefficients
b୬ =
2
L
න fሺtሻ sin
πn
L
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
fሺtሻ = ൝
0 − 2 < t < −1
k − 1 < t < 1
0 1 < t < 2
whose period is p = 2L = 4
Answer
fሺtሻ =
k
2
+
2k
π
൬cos
πt
2
−
1
3
cos
3πt
2
+
1
5
cos
5πt
2
− ⋯ + ⋯ ൰
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 cos 5t and sin 5t must be included.
1. fሺtሻ = π − |t| − π < t < π
2. fሺtሻ = 2t|t| − 1 < t < 1
3. fሺtሻ = ൝
t −
஠
ଶ
< t <
஠
ଶ
π − t
஠
ଶ
< t <
ଷ஠
ଶ
4. fሺtሻ = ቄπeି୲
−π < t < 0
πe୲
0 < t < π
5. fሺtሻ = ቄ
2 −2 < t < 0
0 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୐ሺtሻ whose period is p = 2L, its Fourier series expansion is given by
f୐ሺtሻ = a଴ + ෍ሺa୬ cos w୬t + b୬ sin w୬tሻ
ஶ
୬ୀଵ
(3.11)
where w୬ =
୬஠
୐
. If we let L → ∞, then we will haveA, for nonperiodic fሺtሻ,
fሺtሻ = න ሾAሺwሻcos wt + Bሺwሻsin wtሿdw
ஶ
଴
(3.12)

with
Aሺwሻ =
1
π
න fሺvሻ cos wv dv
ஶ
ିஶ
(3.13a)
Bሺwሻ =
1
π
න fሺvሻ sin wv dv
ஶ
ିஶ
(3.13b)
Equation 3.12 is the Fourier integral representation of fሺtሻ. 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 w୬ =
୬஠
୐
becomes continuous also.
Fourier Cosine Integral and Fourier Sine Integral. For an even or odd function, the Fourier integral
becomes simpler. If fሺtሻ is an even function, then Bሺwሻ = 0 in 3.13b and
Aሺwሻ =
1
π
ஶ
ିஶ
(3.13a)
and the Fourier integral then reduces to the Fourier cosine integral. Similarly if fሺtሻ is an odd function,
Aሺwሻ = 0 in 3.13a and
Bሺwሻ =
1
π
ஶ
ିஶ
(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
fሺtሻ = න ሾAሺwሻcos wt + Bሺwሻsin wtሿdw
ஶ
଴
(3.12)
with
Aሺwሻ =
1
π
ஶ
ିஶ
(3.13a)

Bሺwሻ =
1
π
ஶ
ିஶ
(3.13b)
Substituting these into the integral, we have
fሺtሻ =
1
π
න න fሺvሻሾcoswv coswt + sin wv sin wtሿdv dw
ஶ
ିஶ
ஶ
଴
(3.14)
The terms inside the bracket is the addition formula for cosine, or cosሺwx − wvሻ. Thus,
fሺtሻ =
1
π
න ቈන fሺvሻ cosሺwx − wvሻ dv
ஶ
ିஶ
቉
ஶ
଴
dw (3.15)
We shall call the term inside the bracket Fሺwሻ since cosሺwx − wvሻ is an even function of w, and we
integrate with respect to v. We extend the limit of the integral of Fሺwሻ from w = 0 to ∞ to w = −∞ to
+∞; thus a factor of
ଵ
ଶ
is multiplied to the integral to account for this extension. Thus,
fሺtሻ =
1
2π
න ቈන fሺvሻ cosሺwx − wvሻ dv
ஶ
ିஶ
቉
ஶ
ିஶ
dw (3.16)
We claim that the integral of 3.16 with sine instead of a cosine is zero,
1
2π
න ቈන fሺvሻ sinሺwx − wvሻ dv
ஶ
ିஶ
቉
ஶ
ିஶ
dw = 0 (3.17)
This is true since sinሺwx − wvሻ is an odd function of w, which makes the integral inside the bracket,
calling it Gሺwሻ, equal to zero.
Taking the integrand of 3.16 plus j times the integrand of 3.17 and use the Euler’s formula
e୨୲
= cost + j sin t (3.18)
Thus, the complex Fourier integral can now be written as
fሺtሻ =
1
2π
න න fሺvሻe୨୵ሺ୶ି୴ሻ
ஶ
ିஶ
dv
ஶ
ିஶ
dw (3.19)

Writing 3.19 as a product of exponentials
fሺtሻ =
1
√2π
න ቈ
1
√2π
න fሺvሻeି୨୵୴
dv
ஶ
ିஶ
቉
ஶ
ିஶ
e୨୵୲
dw (3.20)
The expression in brackets is a function of w - for application purposes, we let w = ω – is denoted by
Fሺωሻ and is called the Fourier transform of fሺtሻ; writing v = t. Thus,
Fሺωሻ =
1
√2π
න fሺtሻeି୨୵୲
dt
ஶ
ିஶ
(3.21)
With this, 3.20 becomes
fሺtሻ =
1
√2π
න Fሺwሻ
ஶ
ିஶ
e୨୵୲
dw (3.22)
which is the inverse Fourier transform of Fሺwሻ. We will use the notation
ℱሾfሺtሻሿ = Fሺωሻ (3.23a)
to indicate the Fourier transform of fሺtሻ and
ℱିଵሾFሺωሻሿ = fሺtሻ (3.23b)
as the inverse Fourier transform of Fሺωሻ.
Example 3.4
Find the Fourier transform of fሺtሻ = 1 if |x| < 1 and fሺtሻ = 0 otherwise.
Answer:
Fሺωሻ = ට
π
2
sin ω
ω
Example 3.5
Find the Fourier transform of fሺtሻ = eିୟ୲
for t > 0 and fሺtሻ = 0 for t < 0.
Answer:
Fሺωሻ =
1
√2πሺa + jωሻ

Drill Problems 3.4
Find the Fourier transform of the following functions using the Fourier transform integral.
1. fሺtሻ = ቊ1 −
|୲|
ୟ
|t| ≤ a
0 otherwise
2. fሺtሻ = eିୟ|୲|
3. fሺtሻ = ቄteି୲
− 1 < t < 0
0 otherwise
4. fሺtሻ = ቄ
|t| − 1 ≤ t ≤ 1
0 otherwise
5. fሺtሻ = ൝
1 0 ≤ t ≤ a/2
−1 − a/2 ≤ t ≤ 0
0 otherwise
3.6 Properties of Fourier Transform
Linearity. Let fሺtሻ and gሺtሻ be two functions with Fourier transforms Fሺωሻ and Gሺωሻ respectively. Then,
for constants a and b,
ℱሾafሺtሻ + bgሺtሻሿ = aFሺωሻ + bGሺωሻ (3.24)
Shift in the frequency domain. Let fሺtሻ be a function with Fourier transform Fሺωሻ. Then
ℱൣe୨ୟ୲
fሺtሻ൧ = Fሺω − aሻ (3.25)
Example 3.6
If the function fሺtሻ has the Fourier transform Fሺωሻ, use the linearity and shifting in the frequency domain
properties of the Fourier transform to find ℱሾfሺtሻcos atሿ.
Answer:
Fሺωሻ =
Fሺω − aሻ
2
+
Fሺω + aሻ
2
Time scaling. Let fሺtሻ be a function whose Fourier transform is Fሺωሻ. For a constant c,
ℱሾfሺctሻሿ =
F ቀ
ω
c
ቁ
c
(3.26)

A special case of the scaling is called time-reversal, that is c = −1. Equation 3.26 becomes then,
ℱሾfሺ−tሻሿ = −Fሺ−ωሻ (3.27)
Example 3.7
If fሺtሻ = ቄ
1 − 1 ≤ t ≤ 1
0 otherwise
, plot fሺ2tሻ and f ቀ
୲
ଶ
ቁand find their Fourier transform. Use the results of
Example 3.4.
Answer:
ℱሾfሺ2tሻሿ = ඨ
2
π
sin
ω
2
ω
ℱ ൤f ൬
t
2
൰൨ = ඨ
2
π
sin 2ω
ω
Differentiation in time domain. Let fሺtሻ be a continuously differentiable function with Fourier transform
Fሺωሻ and assume that lim୲→±ஶ fሺtሻ = 0. Then,
ℱሾfᇱሺtሻሿ = jωFሺωሻ (3.28)
and in general, if fሺtሻ can be differentiated m times,
ℱൣfሺ୫ሻሺtሻ൧ = ሺjωሻ୫
Fሺωሻ (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 fሺtሻ has the Fourier transform Fሺωሻ and if t fሺtሻ is
absolutely integrable, then Fሺωሻ is differentiable and
ℱሾt fሺtሻሿ = −j Fᇱሺωሻ (3.30)
and in general
ℱሾሺjtሻ୫
fሺtሻሿ = Fሺ୫ሻሺωሻ (3.31)

Example 3.8
Find the Fourier transform of fሺtሻ = ቄ
t − 1 ≤ t ≤ 1
0 otherwise
, using the results of Example 3.4 and the
differentiation property of the Fourier transform.
Answer:
Fሺωሻ = ඨ
2
π
ωcos ω − sin ω
ωଶ
Integration in time-domain. If fሺtሻ is a continuous and absolutely integrable function with Fourier
transform Fሺωሻ, and lim୲→ஶ ‫׬‬ fሺτሻ dτ
୲
ିஶ
= 0. Then
ℱ ቈන fሺτሻ dτ
୲
ିஶ
቉ =
Fሺωሻ
jω
(3.32)
for ω ≠ 0.
The following table summarizes the Fourier transform of various functions.
fሺtሻ ℱሾfሺtሻሿ = Fሺωሻ
3.1.1 ቄ1 − b < t < b
0 otherwise
ඨ
2
π

sin bω
ω
3.1.2 ቄ1 b < t < c
0 otherwise
eି୨ୠன
− eି୨ୡன
jω√2π
3.1.3
1
xଶ + aଶ
ሺa > 0ሻ ට
π
2

eିୟ|ன|
a
3.1.4 ൝
t 0 < t < b
2t − b b < t < 2b
0 otherwise
−1 + 2e୨ୠன
− eି୨ଶୠன
ωଶ√2π
3.1.5 ቄeିୟ୲
t > 0
0 otherwise
ሺa > 0ሻ
1
√2π ሺa + jωሻ
3.1.6 ቄeୟ୲
b < t < c
0 otherwise
eሺୟି୨னሻୡ
− eሺୟି୨னሻୠ
√2π ሺa − jωሻ
3.1.7 ൜e୨ୟ୲
− b < t < b
0 otherwise
ඨ
2
π

sin bሺω − aሻ
ሺω − aሻ
3.1.8 ൜ e୨ୟ୲
b < t < c
0 otherwise
j
√2π
e୨ୠሺୟିனሻ
− e୨ୡሺୟିனሻ
a − ω
3.1.9 eିୟ୲మ
ሺa > 0ሻ
1
√2π
eିனమ/ସୟ

fሺtሻ ℱሾfሺtሻሿ = Fሺωሻ
3.1.10
sin at
t
ሺa > 0ሻ ቐ
ට
π
2
|ω| < a
0 |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.
fሺtሻ ℱሾfሺtሻሿ = Fሺωሻ Remarks
3.2.1 1 √2πδሺωሻ
The transform of a constant
function is the Dirac delta
function.
3.2.2 δሺtሻ
1
√2π
The dual of rule 3.2.1. The
transform of the Dirac delta
function is a constant.
3.2.3 e୨ୟ୲
√2π δሺω − aሻ
Applying frequency shift to
3.2.1.
3.2.4 cos at √2π
δሺω − aሻ + δሺω + aሻ
2
Using rules 3.2.1, 3.2.3 and
the Euler’s formula for cosine
3.2.5 sin at j√2π
δሺω + aሻ + δሺω − aሻ
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.

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