1. Periodic Functions
A function f is periodic if and only if there exists a minimum positive number 2p
such that for
every t in the domain of f,
. The number 2p is called a period of f.
2. Fourier Series
It can be proved that for a periodic function f with period 2p , it can be expanded in
periodic
series with sine and cosine terms
(1.1)
The coefficients
are known as the Fourier coefficients.
The Fourier coefficients can be obtained by considering the following integrals:
(1.2)
(1.3)
(1.4)
(1.5)
(1.6)
(1.7)
1
(1.8)
To find ao, we assume that the series (1.1) can legitimately be integrated term by
term from
to
. Then,
(1.9)
The first term on the right is simply
or
(1.10)
To find
integrated
term by term from
justified.
This gives
, we multiply each side of (19.1) by
to
and then
, assuming again that term-by-term integration is
(1.11)
2
All integrals on the right vanishes except the one involving
. Hence
or
(1.12)
To find
integrated
term by term from
justified.
This gives
, we multiply each side of (1.1) by
to
and then
, assuming again that term-by-term integration is
(1.13)
As before, every integral on the right vanishes but one, leaving
or
(1.14)
Formulae (1.10), (1.12) and (1.14) for the Fourier coefficients are known as the
Euler or Euler-Fourier formulas, and the series (1.1) is known as the Fourier series of
.
3
We must be careful at this stage not to conclude that we have proved that every
periodic
function
has a Fourier expansion that converges to it. The convergence problem can
be solved
by the famous theorem of Dirichlet
Theorem of Dirichlet
If
is a bounded periodic function which in any one period has at most a finite
number of
local maxima and minima and a finite number of points of discontinuity, then the Fourier
series of
converges to
average of the
at all points where
left- and right-hand limits of
is continuous and converges to the
at each point where
is discontinuous.
The conditions of Theorem 1.1 under which a periodic function possesses a valid
Fourier
expansion are referred to collectively as the Dirichlet conditions.
Definition:
A function
is said to satisfy the Dirichlet conditions on an interval I if and
only if
is bounded and has at most a finite number of local maxima and minima and a finite
number of
discontinuities on I.
Example
Find the Fourier series of the following periodic function whose definitions on one period
is
Solution 1.1
The period of this function is
(i.e.
and
4
). Using Eq.(19.12), we have
The constant term of the Fourier series is given by
Continuing by using Eq.(1.14), we have
Thus the function can be represented as
3. Complex Exponential Fourier Series
By using the Euler formula
complex exponential terms
,
and
and
can be expressed as
(1.15)
Therefore the Fourier coefficients of a Fourier series can be rewritten by using the
complex
exponential terms as follows
5
(1.16)
(1.17)
The standard Fourier series (1.1) can be converted into a complex exponential form by
substituting
their exponential equivalents for the cosine and sine terms:
(1.18)
Collecting terms on the various exponentials and noting that 1/i = -i, we obtain
If we now define
(1.19)
the last series can be written in the more symmetric form
(1.20)
The coefficients
can easily be found from their definitions:
(1.21)
6
(1.22)
(1.23)
Clearly, whether the index n is positive, negative, or zero, cn is correctly given by the
single formula
(1.24)
A series of the form (19.20), with its coefficients determined by (19.24) is called a
complex
exponential Fourier series.
Example 1.2
Find the complex exponential Fourier series of the periodic function whose definitions in
one period is
.
Solution 1.2
The period of this function is 2 (i.e.
and
7
). Using Eq.(1.24), we have
Now
The Fourier series of
, and thus
. Hence
is therefore
4. Application of Fourier Series in Solving the Differential Equations
Let us consider a uniform beam of length l, supported at each end, bears an
arbitrary load per
unit length given by the function
. Neglecting the weight of the beam, the transverse
deflection
of the beam satisfies the following differential equation
(1)
and then imposing the end conditions of a simple supported beam, namely
(2a)
(2b)
(No deflection at either end)
(No moment at either end)
Our objective is to obtain a solution satisfying the differential equation (1) and the
boundary
conditions (2). Since a Fourier series will vanish at
and
only if each of its
terms does
also, it is clear that we must assume for the deflection y a half-range sine expansion
8
in which the bn's have yet to be determined. Guided by this, our next step is to expand the
given load
function
in a half-range sine series, getting, by now familiar steps,
The differential equation becomes
If we now substitute the series we assumed for y, we obtain
The coefficients of like terms on each side must be equal. Hence for every
,
and
With Bn determined by the given load function
and our
problem is solved.
, bn is now completely determined
Problem 1.1
Find the Fourier series of the periodic functions whose definitions on one period are
(a)
9
(b)
(c)
Problem 1.2
Find the complex exponential Fourier series of the periodic functions whose definitions
in one period are
(a)
(b)
(c)
Problem 1.3
(a) Find the Fourier series of the periodic function
is
whose definition on one period
(b) By substituting a suitable value in the Fourier series in (a), show that
Problem 1.4
A triangular wave is represented by
Represent
by a Fourier series.
Problem 1.5
10
The static deflection y of a uniform horizontal string of length l stretched under tension
T and bearing a vertical load per unit length
satisfies the equation
provided the deflections are small. Find the static deflection of a string stretched
between
and
due to the given load.
Problem 1.6
Fig. P19.6 shows a LCR series circuit which is driven by a variable electromotive force
.
Fig. P19.6
The response of the system, namely the current
differential equation
, is known to satisfy the
If the electromotive force is periodic (of period T) with Fourier expansion
where the Fourier coefficients
are constant. Show that in steady state, the
response of the system is given by
11
where
and
.
Problem 19.7
A damped harmonic oscillator under the influence of an external periodic force
obeys the differential equation
Assuming a steady-state solution, solve the problem by the Fourier series method if
is the triangular wave in the form
Problem 19.8
Assuming the Fourier coefficients of a periodic function
with period 2p are
, show that
The above identity is known as Parseval's identity.
Fourier Integral
1. Fourier Integral
To begin, we let f be a nonperiodic function with domain
and such that
(a) On every finite interval, f satisfies the Dirichlet conditions.
(b) The improper integral
exists.
We then define a periodic function fp of period 2p in terms of f as follows:
12
(1)
Clearly, f(t) is the limit fp(t) as
. Because f satisfies the Dirichlet conditions on
every finite
interval, it follows that for every value of p the function fp satisfies the same conditions in
each period
and hence possesses a valid Fourier series. Then we can expand fp in the complex
exponential form
and we can therefore write
(2)
where
Substituting cn into the expression for fp(t) gives
(3)
Now, let us denote the frequency of the general term by
frequency
between successive terms by
.
Then
(4)
If we now set
(5)
and for each p define a function Fp by
(6) Fp() = Cp()exp(it)
13
and the difference in
Eq.(6) becomes simply
(7)
In the limiting case, as
and
, we have
and
(8)
and
(9)
the nonperiodic limit f(t) of fp(t) is correctly given by the formula
(10)
Therefore we do have the following theorem:Theorem 2.1
If on every finite interval, f satisfies the Dirichlet conditions and if the improper
integral
exists, the Fourier integral
(2.1)
gives the value f at every point where f is continuous.
The Fourier integral (20.1) can be written as the integral
(2.2)
in which C is the coefficient function
14
(2.3)
2. Fourier Integral Representation
Just as in the case in Fourier series, the above Fourier integral can be written in
alternative
form. To do this, we change the dummy variable of integration from t to  in the inner
integral in (20.1)
and then move eit across the integral sign, which we can do because it does not involve 
. This
gives
(2.4)
In this, we can replace the exponential by its trigonometric equivalent, getting
(2.5)
If the function f is purely real, the second term of the R.H.S. will vanish. This gives
(2.6)
Since the integral of Eq.(20.6) is an even function of  , we need perform the 
integration only
between 0 and , provided we multiply the result by 2. This gives us the modified form
(2.7)
which, when cos ( - t) is expanded, becomes
15
(2.8)
The two integrals in R.H.S. of the above equality are called the coefficient functions.
(2.9)
Hence we arrive at the standard Fourier integral representation of f
(2.10)
If f is an even function, B() = 0
f is even
(2.11)
This is so-called Fourier cosine integral of f.
If f is an odd function, A() = 0
f is odd
(2.12)
This is so-called Fourier sine integral of f.
Theorem 20.2
If for t < 0, f(t) = 0, then the Fourier cosine integral Eq.(2.11) and the Fourier sine
integral
Eq.(20.12) are, respectively, just twice the first and second terms in the trigonometric
Fourier integral
Eq.(20.10) of f.
Example 2.1
16
Find the complex exponential Fourier integral representation of the following function
Solution 20.1
Using Eq.(20.3), we have
Substituting the coefficient function into Eq.(20.2) gives
as the required Fourier integral representation.
Example 2.2
Find the Fourier cosine and sine integral representations of the following function
Solution 2.2
Using Eq.(20.9), we have
17
Applying Theorem 2.2 we see at once that the required cosine integral is
and the sine integral is
Fourier Transform
1. Fourier Transform
If on every finite interval, f satisfies the Dirichlet conditions and if the improper
integral
exists, the following integral
(2.13)
is known as the Fourier transform, whose value at f(t) is the function F(). Denoting
this
transformation by F, we write
(2.14)
The function f can be recovered by using the Fourier integral
(2.15)
The function f(t) is in turn referred to as the inverse Fourier transform of F() and is
denoted by
(2.16)
18
Several important transforms are listed in the following table:
F()
f(t)
a.
b.
c.
2. Basic Properties of Fourier Transform
(1) (Linearity) If the Fourier transform of f1 and f2 exist, then
(2.17)
Proof
(2) (Symmetry) If F() is the Fourier transform of f(t), then 2f() is the Fourier
transform of F(t).
Proof
By hypothesis
and inversely
If in the latter integral we replace t by -t, it becomes
Finally, by interchanging the dummy variables  and t, we obtain
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(3) (Change of Time Scale) If a is a positive real constant,
(2.18)
Proof
By definition,
Under the substitution at = z we have
becomes
, and the integral
(4) (Time Shifting)
(2.19)
Proof
By definition,
If we put t - t0 = z we have dt = dz, we have
(5) (Frequency Shifting) If  o is a real constant, then
(2.20)
Proof
20
(6) (Time Differentiation) If f(t) is continuous and f '(t) is at least piecewise continuous
on
, and if
and
exist, then
(2.21)
Proof
By definition,
. Integrating by
parts, with u = exp(-i t) and dv = f '(t)dt, we have
Since f(t) is assumed to be absolutely integrable, f(t) must vanish at
both  and . Hence the integrated portion of the last expression is
equal to zero. What remains is just iF().
N.B.In general
for positive integer n.
Definition 20.1
The function h defined by the integral
(2.22)
is called the convolution of the functions f and g and it is denoted by h = f * g.
(7) (Frequency Convolution) If f and g satisfy the Dirichlet conditions on every finite
interval and are absolutely integrable on
, then
(2.23)
21
Proof
By definition
and
, the product of f and g has the
representation
The change of variable  gives dd , and
Therefore
Directly from the definition of inverse Fourier transform and property (7), we have
(8) (Time Convolution) The inverse of the product of two transforms is the convolution
of their inverses; that is,
(2.24)
Example 2.3
Find the Fourier transform of the following function
22
Solution 2.3
The Fourier transform is given by Eq.(20.13)
Example 2.4
Find the inverse Fourier transform of the function:
F() = exp(-k22)
Solution 20.4
The Fourier transform is given by Eq.(2.13)
By using the integral
, we have
Example 2.5
23
Find the inverse Fourier transform of the following Fourier transform
Solution 20.5
The inverse Fourier transform of the above function is
3. Applications of Fourier Integrals and Transforms
Fourier transform can be used to find particular solutions of the following linear
differential
equation
(2.25)
If
and assuming f(t) has the transform F() we obtain
(2.26)
Solving this equation for Y() and introducing the so-called transfer function
(2.27)
We have for the Fourier transform of y(t),
Y() = W(i )F()
(2.28)
24
Hence by property (10), assuming
y(t) = w(t)* f(t)
,
(2.29)
is a particular solution of the differential equation Eq.(2.25).
Example 2.5
Find a particular solution for
by means of Fourier transform.
Solution 2.5
Writing the Fourier transforms of both members of the equation, we have
[(i)2+ 3i+ 2]F() = F()
Solving for Y(), then using the partial fractions, we get
Bu Eq.(2.24) and from the definition of f(t),
where the inverse transform g(t) of G() is given by Example 2.4
and
Depending on how t relates to the limits of integration on  , we have
25
Finally, performing the indicated integrations and simplifying, we obtain
Problem 2.1
Find the complex exponential Fourier integral of the following functions
(a)
(b)
Problem 2.2
Find the Fourier cosine and sine integral representations of the following functions
(a)
(b)
Problem 2.3
Find the Fourier transform of the following functions
(a)
(b)
26
Problem 2.4
Let F() be the Fourier transform of f(t) and G() the Fourier transform of
g(t) = f(t + a). Show that
G() = e-iaF()
Problem 2.5
This problem is about an improper integral
.
Consider a function
(a) Show that the Fourier cosine representation of f(x) is given by
.
(b) By substituting a suitable value to t, show that
.
Problem 2.6
In a resonant cavity an electromagnetic oscillation of frequency 0 dies out as
(Take A(t) = 0 for t < 0)
The parameter Q is a measure of the ratio of stored energy to energy loss per cycle.
Show that the frequency distribution of the oscillation,
27
where a() is the Fourier transform of A(t).
Problem 2.7
Find the inverse Fourier transform of the following Fourier transforms
(a)
(b)
Problem 2.8
Find a particular solution for the following differential equations by means of Fourier
transform.
(a) y'' + 3y' + 2y = 6et
(b) 2y'' + 7y' + 3y = 100e2t
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