Chapter 11 Fourier Analysis New
Chapter 11 Fourier Analysis New
Chapter 11 Fourier Analysis New
1
11.1 Fourier Series
2
Fourier Series
Fourier series are infinite series that represent periodic
functions in terms of cosines and sines. As such,
Fourier series are of greatest importance to the
engineer and applied mathematician.
A function f(x) is called a periodic function if f(x) is
defined for all real x, except possibly at some points,
and if there is some positive number p, called a period
of f(x), such that
(1)
3
Fourier Series
The graph of a periodic function has the characteristic
that it can be obtained by periodic repetition of its
graph in any interval of length p. (Fig. 258)
The smallest positive period is often called the
fundamental period.
4
Fourier Series
If f(x) has period p, it also has the period 2p because (1)
implies ; thus
for any integer n=1, 2, 3, …
(2)
Furthermore if f(x) and g(x) have period p, then af(x) +
bg(x) with any constants a and b also has the period p.
5
Fourier Series
Our problem in the first few sections of this chapter
will be the representation of various functions f(x) of
period in terms of the simple functions
(3)
All these functions have the period . They form the
so-called trigonometric system. Figure 259 shows the
first few of them
6
Fourier Series
The series to be obtained will be a trigonometric series,
that is, a series of the form
(4)
7
Fourier Series
Now suppose that f(x) is a given function of period
and is such that it can be represented by a series (4),
that is, (4) converges and, moreover, has the sum f(x).
Then, using the equality sign, we write
(5)
and call (5) the Fourier series of f(x). We shall prove
that in this case the coefficients of (5) are the so-called
Fourier coefficients of f(x), given by the Euler
formulas
(6.0)
(6.a)
8
(6.b)
Example 1
Find the Fourier coefficients of the period function f(x) in
Fig. 260. The formula is
(7)
9
(6.0)
(6.a)
Example 1 (6.b)
10
Example 1
We see that all these cosine coefficients are zero. That
is, the Fourier series of (7) has no cosine terms, just
sine terms, it is a Fourier sine series with coefficients
b1, b2, … obtained from (6b);
11
Example 1
Now, ; in general,
12
Example 1
Since the an are zero, the Fourier series of f(x) is
(8)
Thus
14
Example 1
15
Theorem 1 Orthogonality of the
Trigonometric System
The trigonometric system (3) is orthogonal on the
interval (hence also on or any
other interval of length because of periodicity); that
is, the integral of the product of any two functions in (3)
over that interval is 0, so that for any integers n and m,
(a)
(9) (b)
(c)
(3)
16
Proof of Theorem 1
Simply by transforming the integrands
trigonometrically from products into sums
17
Application of Theorem 1 to the Fourier
Series (5)
We prove (6.0). Integrating on both sides of (5) from
to , we get
18 (6.0)
Application of Theorem 1 to the Fourier
Series (5) (6.a)
21
Convergence and Sum of a Fourier
Series
The left-hand limit of f(x) at x0 is defined as the limit
of f(x) as x approaches x0 from the left and is
commonly denoted by f(x0-0). Thus
23
Summary
A Fourier series of a given function f(x) of period
is a series of the form (5) with coefficients given by
the Euler formulas (6). Theorem 2 gives conditions
that are sufficient for this series to converge and at
each x to have the value f(x), except at discontinuities
of f(x), where the series equals the arithmetic mean of
the left-hand and right-hand limits of f(x) at that point.
(5)
(6.0)
(6.a)
24
(6.b)
11.2 Arbitrary Period. Even and Odd
Functions. Half-Range Expansions
25
Introduction
This section concerns three topics:
Transition from period to any period 2L, for the
function f, simply by a transformation of scale on the x-
axis.
Simplifications. Only cosine terms if f is even (“Fourier
cosine series”). Only sine terms if f is odd (“Fourier sine
series”).
Expansion of f given for in two Fourier series,
one having only cosine terms and the other only sine
terms (“half-range expansions”).
26
Transition
Periodic functions in applications may have any period.
The notation p=2L for the period is practical because L
will be a length of a violin string in Sec. 12.2, of a rod
in heat conduction in Sec. 12.5, and so on.
The transition from period to be period p=2L is
effected by a suitable change of scale, as follows.
Let f(x) have period p = 2L. Then we can introduce a
new variable v such that f(x), as a function of v, has
period .
27
Transition
If we set
(1) so that
(3)
28
Transition
We could use these formulas directly, but the change to
x simplifies calculations. Since
(4) we have
29
Transition
Just as in Sec. 11.1, we continue to call (5) with any
coefficients a trigonometric series. And we can
integrate from 0 to 2L or over any other interval of
length p=2L.
(0)
(6) (a)
(b)
30
Example 1
Find the Fourier series of the function (Fig. 263)
p=2L=4, L=2
31 (6a)
(6b)
Example 1
From (6b) we find that bn=0 for n=1, 2, …. Hence the
Fourier series is a Fourier cosine series (that is, it has
no sine terms)
32
Example 2
Find the Fourier series of the function (Fig. 264)
p=2L=4, L=2
34
(6.0)
Example 3
From (6a), by using formula (11) in App. A3.1 with
and ,
35
Example 3
If n is odd, this is equal to zero, and for even n we have
36
Simplifications
If f(x) is an even function, that is, f(-x)=f(x) (see Fig.
266), its Fourier series (5) reduces to a Fourier cosine
series
(5*)
with coefficients (note: integration from 0 to L only!)
(6*)
37
Simplifications
If f(x) is an odd function, that is, f(-x)=-f(x) (see Fig.
267), its Fourier series (5) reduces to a Fourier sine
series
(5**)
with coefficients
(6**)
38
Simplifications
These formulas follow from (5) and (6) by
remembering from calculus that the definite integral
gives the net area (=area above the axis minus area
below the axis) under the curve of a function between
the limits of integration. This implies
(a) for even g
(7)
(b) for odd h
(0)
(6) (a)
39
(b)
Simplifications
Formula (7b) implies the reduction to the cosine series
(even f makes odd since sin is odd) and
to the sine series (odd f makes odd
since cos is even).
Similar, (7a) reduces the integrals in (6*) and (6**) to
integrals from 0 to L. These reductions are obvious
from the graphs of an even and an odd function.
(0)
(a) for even g
(7) (6) (a)
(b) for odd h
40
(b)
Summary
41
Example 4
The rectangular wave in Example 1 is even. Hence it
follows without calculation that its Fourier series is a
Fourier cosine series, the bn are all zero.
Similarly, it follows that the Fourier series of the odd
function in Example 2 is a Fourier sine series.
In Example 3 you can see that the Fourier cosine series
represents . This is an even
function.
42
Theorem 1
Sum and Scalar Multiple
The Fourier coefficients of a sum f1+f2 are the sums of
the corresponding Fourier coefficients of f1 and f2.
The Fourier coefficients of cf are c times the
corresponding Fourier coefficients of f.
43
Example 5
Find the Fourier series of the function (Fig. 268)
44
Example 5
Integrating by parts, we obtain
45
Half-Range Expansions
Half-range expansions are Fourier series. We want to
represent f(x) in Fig. 270.0 by a Fourier series, where
f(x) may be the shape of a distorted violin string or the
temperature in a metal bar of length L, for example.
46
Half-Range Expansions
We would extend f(x) as a function of period L and
develop the extended function into a Fourier series.
But this series would, in general, contain both cosine
and sine terms. We can do better and get simpler series.
Indeed, for our given f we can calculate Fourier
coefficients from (6*) or form (6**). And we have a
choice and can make what seems more practical.
(6*)
(6**)
47
Half-Range Expansions
If we use (6*), we get (5*). This is the even periodic
extension f1 of f in Fig. 270a. If we use (6**), we get
(5**). This is the odd periodic extension f2 of f in Fig.
270b.
Both extensions have period 2L. This motivates the
name half-range expansions: f is given only on half
the range, that is, on half the interval of periodicity of
length 2L.
48
Example 6
Find the two half-range expansion of the function (Fig.
271)
49
Example 6
Similarly, for the second integral we obtain
We insert these two results into the formula for an. The
sine terms cancel and so does a factor L2. This gives
Thus,
and an=0 if . Hence the first half-range
expansion of f(x) is Fig. (272a)
51
11.3 Forced Oscillations
52
Forced Oscillations
Fourier series have important application for both
ODEs and PDEs.
From Sec. 2.8 we know that forced oscillations of a
body of mass m on a spring of modulus k are governed
by the ODE
(1)
where is the displacement from rest, c the
damping constant, k the spring constant, and r(t) the
external force depending on time t.
53
Forced Oscillations
If r(t) is a sine or cosine function and if there is
damping (c>0), then the steady-state solution is a
harmonic oscillation with frequency equal to that of
r(t).
However, if r(t) is not a pure sine or cosine function
but is any other periodic function, then the steady-state
solution will be a superposition of harmonic
oscillations with frequencies equal to that of r(t) and
integer multiplies of these frequencies.
54
Forced Oscillations
And if one of these frequencies is close to the
(practical) resonant frequency of the vibrating system
(see Sec. 2.8), then the corresponding oscillation may
be the dominant part of the response of the system to
the external force.
This is what the use of Fourier series will show us. Of
course, this is quite surprising to an observer unfamilar
with Fourier series, which are highly important in the
study of vibrating systems and resonance.
55
Example 1
In (1), let m = 1 (g), c = 0.05 (g/sec), and k = 25
(g/sec2), so that (1) becomes
(2)
where r(t) is measured in . Let (Fig. 276)
56
Example 1
Solution. We represent r(t) by a Fourier series, finding
(3)
Then we consider the ODE
(4)
whose right side is a single term of the series (3). From
Sec. 2.8 we know that the steady-state solution yn(t) of
(4) is of the form
(5)
By substituting this into (4) we find that
(6)
57
Example 1
Since the ODE (2) is linear, we may expect the steady-
state solution to be
(7)
where yn is given by (5) and (6). In fact, this follows
readily by substituting (7) into (2) and using the
Fourier series of r(t), provided the termwise
differentiation of (7) is permissible.
(5)
58
Example 1
From (6) we find that the amplitude of (5) is (a factor
cancels out)
(5)
(6)
59
Example 1
Figure 277 shows the input (multiplied by 0.1) and the
output. For n=5 the quantity Dn is very small, the
denominator of C5 is small, and C5 is so large that y5 is the
dominating term in (7).
Hence the output is almost a harmonic oscillation of five
times the frequency of the driving force, a little distorted
due to the term y1, whose amplitude is about 25% of that of
y5. You could make the situation
still more extreme by decreasing
the damping constant c.
60
11.4 Approximation by
Trigonometric Polynomials
61
Approximation
Approximation theory: An area that is concerned
with approximating functions by other functions –
≈ + ∑ ( cos + sin )
Then the Nth partial sum of the Fourier series
(1)
is an approximation of the given f(x).
62
Approximation
In (1) we choose an arbitrary N and keep it fixed. Then
we ask whether (1) is the “best” approximation of f by
a trigonometric polynomial of the same degree N,
= + ∑ ( cos + sin )
that is, by a function of the form
(2)
Here, “best” means that the “error” of the
approximation is as small as possible.
We have to define error of such an approximation.
63
Approximation
on the whole interval − ≤ ≤ . This is preferable
Measure the goodness of agreement between f and F
"
= #" − !
choose
(3)
64
Approximation
function f on the interval − ≤ ≤ . Clearly ≥ 0
This is called the square error of F relative to the
2 + , we have
" " "
(4) = #" ! − 2 #" ! + #" !
We square (2), insert it into the last integral in (4), and
= (2 + + ⋯+ + + ⋯+ )
66
Approximation
With the expressions, (4) becomes
"
(5) =* ! −2 2 + + +
#"
+ 2 + +( + )
"
(6) ∗ = * ! − 2 + +( + )
67 #"
Approximation
−2 + =
We finally subtract (6) from (5). Then the integrals
− −
drop out and we get terms
and similar terms :
− ∗
= 2 − + + A − + −
− ≥0 ≥
cannot be negative,
∗ ∗
68
Theorem 1
Theorem 1: Minimum Square Error
"
(6) ∗ = * ! − 2 + +( + )
#"
70
Theorem 1
It can be shown that for such a function f, Parseval’s
theorem holds; that is, formula (7) holds with the
equality sign, so that it becomes Parseval’s identity
/
1 "
(8) 2 + +( + )= * !
#"
71
Example 1
Compute the minimum square error ∗ of F(x) with
= + (− < < )
N=1, 2, …, 10, 20, …, 100, and 1000 relative to
on the interval − ≤ ≤
Solution.
1 1 −1 2
= + 2 (sin − sin 2 + sin 3 − + ⋯ + sin 3 )
2 3 3
By Example 3 in Sec. 11.3. From this and (6),
"
1
∗ = * + ! − 2 +4+
#"
72
Example 1
Numeric values are
73
11.5 Sturm-Liouville Problems.
Orthogonal Functions
74
Sturm-Liouville Problem
Can we replace the trigonometric system by other
orthogonal systems (sets of other orthogonal functions)?
The answer is yes.
8 9: : + ; + 5< 9=0
Consider a second-order ODE of the form
≤ ≤ , satisfying equations of
(1)
on some interval
the form
6 9 + 6 9: = 0 =
7 9 + 7 9: = 0 =
(2a)
(2b)
Here 5 is a parameter, and 6 , 6 , 7 , 7 are given real
constants.
75
Sturm-Liouville Problem
At least one of each constant in each condition (2)
must be different from zero.
Equation (1) is known as a Sturm-Liouville equation.
Together with conditions 2(a), 2(b) it is known as the
Sturm-Liouville problem.
A boundary value problem consists of an ODE and
given boundary conditions referring to the two
interval ≤ ≤
boundary points (endpoints) x=a and x=b of a given
76
Eigenvalues, Eigenfunctions
Clearly, 9 ≡ 0 is a solution – the trivial solution – of
the problem (1), (2) for any 5 because (1) is
homogeneous and (2) has zeros on the right. This is of
9 :: + 59 = 0, 9 0 = 0, 9 =0
Liouville problem
(3)
This problem arises, for instance, if an elastic string (a
vibrate. Then 9
deflection >( , ?) of the string, assumed in the form
is the “space function” of the
78
(3) 9 :: + 59 = 0, 9 0 = 0, 9 =0
Example 1
Solution. From (1) and (2) we see that 8 = 1, ; =
0, < = 1 in (1), and = 0, = , 6 = 7 = 1, 6 =
7 = 0 in (2). For negative 5 = −A a general solution
of the ODE in (3) is 9 = B C DE + B C #DE .
From the boundary conditions we obtain B = B = 0,
so that 9 ≡ 0, which is not an eigenfunction. For
5 = 0 the situation is similar. For positive 5 = A a
general solution is 9 = cos A + sin A
Example 1
9 0 = = 0. The second boundary condition then
From the first boundary condition we obtain
80
Eigenvalues, Eigenfunctions
Note that the solution to this problem is precisely the
trigonometric system of the Fourier series considered
with = )
Note that this is the square root of the integral in (4)
82
Orthogonal Functions
The functions 9 , 9 , … are called orthonormal on
≤ ≤ if they are orthogonal on this interval and
all have norm 1.
O
0 G ) ≠
9N , 9 = * < 9N 9 ! = QN = R
P
1 G ) =
O O
9N , 9 = * 9N 9 ! =0 9N = (9N , 9N ) = * 9N !
() ≠ )
83 P P
Example 2
The functions 9N = sin ) , ) = 1, 2, … form an
orthogonal set on the interval − ≤ ≤ , because
for ) ≠ we obtain by integration () ≠ )
"
1 " 1 "
9N , 9 = * sin ) sin ! = * cos ) − ! − * cos ) + ! =0
#" 2 #" 2 #"
O
(6) 9N , 9 = * < 9N 9 ! =0 () ≠ )
P
85
Theorem 1
If 8 = 0, then (2a) can be dropped from the
problem. If 8 = 0, then (2b) can be dropped. [It is
then required that 9 and 9′ remain bounded at such a
point, and the problem is called singular, as opposed
87
Example 4
Legendre’s equation 1 − 9 :: − 2 9 : + +1 9 =0
1− 9 : : + 59 = 0 5 = ( + 1)
may be written
8 = 1 − , ; = 0, < = 1. Since 8 −1 = 8 1 = 0, we
Hence, this is a Sturm-Liouville equation (1) with
89
Orthogonal Series
Let 9 , 9 , 9 , ⋯ be orthogonal with respect to a weight
function < on an interval ≤ ≤ , and let ( )
be a function that can be represented by a convergent
series /
(1) = + N 9N ( )= 9 + 9 +⋯
N
90
/
(1) = + N 9N ( )= 9 + 9 +⋯
N
Orthogonal Series
Given ( ), we have to determine the coefficients in
to 9 , 9 , ….
(1), called the Fourier constants of with respect
91
Orthogonal Series
right are zero, except when ) = . Hence the whole
Because of the orthogonality all the integrals on the
( , 9N ) 1
the Fourier constants
O
= = * < 9N !
N
9N 9N
(2)
P
92
( = 0, 1, … )
Example 1
A Fourier-Legendre series is an eigenfunction
expansion
/
= + N ZN = Z + Z + Z +⋯= + + W
− +⋯
N
93
Example 1
We have < = 1 for Legendre’s equation, and (2)
2) + 1
gives
= * ZN ! ) = 0, 1, …
(3) N
2 #
94
Example 1
For instance, let = sin . Then we obtain the
2) + 1
coefficients
= * (sin )ZN !
N
2 #
3 3
= * (sin ) ! = = 0.95493
2 #
(14) F` = + N 9N ( )
N
96
Mean Square Convergence.
Completeness
An orthonormal set 9 , 9 , ⋯ on an interval
≤ ≤ is complete in a set of functions S defined
on ≤ ≤ if we approximate every belonging to
97
Mean Square Convergence.
Completeness
Performing the square in (13) and using (14), we first
have
O O O O
lim * < F` − ! = * <F` ! − 2 * < F` ! + * < !
`→/ P P P P
O ` ` O O
=* < + N 9N ! −2 + N* < 9N ! + * < !
P N N P P
98
Mean Square Convergence.
Completeness
Hence the first term on the right cancels half of the
second term, so that the right side reduces to
` O
−+ N +* < !
N P
99
Mean Square Convergence.
Completeness
Here we can let 6 → ∞, because the left sides form a
monotone increasing sequence that is bounded by the
right side, so that we have convergence by the familiar
Theorem 1 in App. A.3.3 Hence
/
(17) + N ≤
N
100
Mean Square Convergence.
Completeness
Hence in the case of completeness every in S
satisfies the so-called Parseval equality
/ O
(18) + N = = * <( ) ( ) !
N P
101
Theorem 2 Completeness
Let 9 , 9 , … be a complete orthonormal set on
≤ ≤ in a set of functions S. Then if a function
belongs to S and is orthogonal to every 9N , it must
have norm zero. In particular, if is continuous, then
must be identically zero.
102
11.7 Fourier Integral
103
Fourier Integral
Manny problems involve functions that are
nonperiodic and are of interest on the whole x-axis,
we ask what can be done to extend the method of
Fourier series to such functions. This idea will lead to
104
Example 1
g( ) of period
2h > 2 given by
Consider the periodic rectangular wave
0 G − h < < −1
g = i 1 G − 1 < < 1
0 G 1 < < h
105
Example 1
106
Example 1
coefficients of g as h increases. Since g is even,
We now explore what happens to the Fourier
107
Example 1
This sequence of Fourier coefficients is called the
amplitude spectrum of g because
maximum amplitude of the wave cos( /h).
is the
= + +( cos @ + sin @ ) @ =
g
h
+1
We now set
Δ@ = @ −@ = − =
2
h h h
110
From Fourier Series to Fourier Integral
Then = Δ@/ , and we may write the Fourier series
g
in the form
1 g
(1)
= * A !A
g
2h #g g
/
1 g g
+ + (cos @ )Δ@ * g A cos @ A !A + (sin @ )Δ@ * g A sin @ A !A
#g #g
111
From Fourier Series to Fourier Integral
We now let h → ∞ and assume that the resulting
nonperiodic function
= lim g( )
g→/
112
From Fourier Series to Fourier Integral
Then → 0, and the value of the first term on the right
g
"
side of (1) approaches zero. Also Δ@ = → 0 and it
g
1 g
= * A !A
(1) g
2h #g g
/
1 g g
+ + (cos @ )Δ@ * g A cos @ A !A + (sin @ )Δ@ * g A sin @ A !A
113 #g #g
From Fourier Series to Fourier Integral
1 1
If we introduce the notations
/ /
(4) @ = * A cos @A !A @ = * A sin @A !A
#/ #/
114
Theorem 1 Fourier Integral
If ( ) is piecewise continuous (see Sec. 6.1) in every
finite interval and has a right-hand derivative and a
115
Example 2
The main application of Fourier integrals is in solving
ODEs and PDEs.
Find the Fourier integral representation of the function
1 G 01
=R
0 G M1
116
1 /
(5) = * (@)cos @ + @ sin @ !@
Example 2
Solution. From (4) we obtain
1 /
1 sin @A 2 sin @
@ = * A cos @A !A = * cos @A !A = o =
#/ # @ # @
1
@ = * sin @A !A = 0
#
G 0 ≤ <1
/
cos @ sin @ 2
* !@ = G = 1
(7) @ 4
0 G > 1
118
Example 2
The case = 0 is of particular interest. If 0, then
(7) gives
sin @
/
* !@
@ 2
(8*)
2 P
cos @ sin @
integral
* !@
(9) @
approximates the right side in (6) and therefore ( ).
2 /
cos @ sin @
= * !@
(6) @
120
Example 2
discontinuity of ( ). We might expect that these
Figure 283 shows oscillations near the points of
121
Example 2
Using (11) in App. A3.1, we have
2 P
cos @ sin @ 1 P sin(@ + @ ) 1 P sin(@ − @ )
* !@ = * !@ + * !@
@ @ @
122
sin @ q
pG > = * !@
@
(8)
Example 2
Since sin −? = − sin ?, we thus obtain
2 P
cos @ sin @ 1 (E2 )P
sin ? 1 (E# )P
sin ?
* !@ = * !? − * !?
@ ? ?
From this and (8) we see that our integral (9) equals
1 1
pG +1 − pG( [ − 1])
1 /
1 /
(4) @ = * A cos @A !A @ = * A sin @A !A
#/ #/
1 /
(5) = * (@)cos @ + @ sin @ !@
125
11.8 Fourier Cosine and Sine
Transforms
126
Integral Transform
An integral transform is a transformation in the form
of an integral that produces from given function new
functions depending on a different variable.
One is mainly interested in these transforms because
they can be used as tools in solving ODEs, PDEs, and
integral equations and can often be of help in handling
and applying special functions.
Laplace transform is an example.
/
F =ℒ = * C #xt ? !?
127
Fourier Cosine Transform
( ). We obtain it from the Fourier cosine integral
The Fourier cosine transform concerns even functions
/
2 /
=* @ cos @ !@ @ = * A cos @A !A
129
/
2 /
(11) =* @ sin @ !@ @ = * A sin @A !A
130
Fourier Sine Transform
The process of obtaining x (@) from is also
called the Fourier sine transform or the Fourier sine
transform method.
ℱz = yz ℱx = yx
Other notations are
131
Example 1
Find the Fourier cosine and Fourier sine transforms of
6 G 0 < <
the function
=R
0 G >
Solution. From the definitions (1a) and (2a) we obtain
by integration
P
sin @
yz @ = 2/ 6 * cos @ ! = 2/ 6
@
P
1 − BkF @
yx @ = 2/ 6 * sin @ ! = 2/ 6
@
132
Example 2
Find ℱz C #E
Solution. By integration by parts and recursion
/
2 /
2 C #E 2/
ℱz C #E
= * C #E
cos @ ! = − cos @ + @ sin @ | =
1+@ 1+@
= 1.
This agrees with formula 3 in Table I, Sec. 11.10, with
133
Linearity, Transforms of Derivatives
These transforms have operational properties that
permit them to convert differentiations into algebraic
135
Theorem 1 Cosine and Sine Transforms
of Derivatives
→ 0 as → ∞. Then
be piecewise continuous on every
finite interval, and let
2
(4a) ℱz :( ) = @ℱx ( ) − (0)
(4b) ℱx :
( ) = −@ℱz ( )
136
Theorem 1 Cosine and Sine Transforms
of Derivatives
This follows from the definitions and by using
integration by parts, namely
/
ℱz :(
) = 2/ * :
cos @ !
/ /
= 2/ cos @ ~ + @ * sin @ !
2
=− 0 + @ℱx ( ( ))
/
ℱx :
( ) = 2/ * :
sin @ !
/ /
= 2/ sin @ ~ − @ * cos @ !
= 0 − @ℱz ( ( ))
137
Linearity, Transforms of Derivatives
Formula (4a) with ′ instead of gives (when ′, ′′
satisfy the respective assumptions for , ′ in Theorem
2
ℱz ( ) = @ℱx ( ) − ′(0)
1) :: :
hence by (4b)
2
(5a) ℱz ::
( ) = −@ ℱz ( ) − ′(0)
Similarly,
2
(5b) ℱx ::
( ) = −@ ℱx + @ (0)
138
Example 3
Find the Fourier cosine transform ℱz (C #PE ) of
= C #PE , where > 0.
Solution. By differentiation, C #PE :: = C #PE ; thus
= ′′( ) . From this, (5a)., and the linearity (3a)
2
ℱz = ℱz :: = −@ ℱz − : 0 = −@ ℱz + 2/
139
11.9 Fourier Transform. Discrete and
Fast Fourier Transforms
140
Complex Form of the Fourier Integral
The (real) Fourier integral is [see (4), (5), Sec. 11.7]
/
=* (@)cos @ + @ sin @ !@
1 1
where
/ /
@ = * A cos @A !A @ = * A sin @A !A
#/ #/
141
Complex Form of the Fourier Integral
By the addition formula for the cosine [(6) in App. 3.1]
142
Complex Form of the Fourier Integral
Hence the integral of (@) from @ = 0 to ∞ is times
the integral of (@) from −∞ to ∞. Thus (note the
change of the integration limit!)
1 / /
= * * (A) cos(@ − @A) !A !@
2 #/ #/
(1)
We claim that the integral of the form (1) with FG
instead of BkF is zero:
1 / /
(2) * * (A) sin(@ − @A) !A !@ = 0
2 #/ #/
143
Complex Form of the Fourier Integral
1 / /
(2) * * (A) sin(@ − @A) !A !@ = 0
2 #/ #/
This is true since sin(@ − @A) is an odd function of
@, which makes the integral in brackets an odd
function of @, call it € @ . Hence the integral of €(@)
from −∞ to ∞ is zero, as claimed.
144
Complex Form of the Fourier Integral
We now take the integrand of (1) plus G(= −1) times
the integrand of (2) and use the Euler formula [(11) in
C •E = cos + G sin
Sec. 2.2]
Taking @ − @A instead of
(3)
(A) gives
in (3) and multiplying by
146
Fourier Transform and Its Inverse
With this, (5) becomes
1 /
(7) = * y @ C •sE !@
2 #/
y = ℱ( ) so that = ℱ # ( y)
Another notation for the Fourier transform is
147
Theorem 1
148
Example 1
= 1 if < 1 and
= 0 otherwise.
Find the Fourier transform of
1
integration
/
ℱ C #PE = * C #PE C #•sE !
2
/
ƒ „ …†‡ˆ ‰
o
" # P2•s E "(P2•s)
150
Physical Interpretation: Spectrum
The nature of the representation (7) of ( ) becomes
clear if we think of it as a superposition of sinusoidal
oscillations of all possible frequencies, called a
spectral representation.
This name is suggested by optics, where light is such a
151
Physical Interpretation: Spectrum
152
Physical Interpretation: Spectrum
To make this plausible, we begin with a mechanical
system giving a single frequency, namely, the
harmonic oscillator (mass on a spring, Sec. 2.4)
)9 :: + 69 = 0
1 1 1 1
the equation for gives
= )A + 69 = ) G@ − + 6 +
2 2 2 2
154
Physical Interpretation: Spectrum
`
Here @ = as just stated; hence )@ = 6. Also
N
,
G = −1, so that
1
= 6− − + + = 26
2
= 26B C •s‹ E B# C #•s‹ E = 26B B# = 26 B
amplitude B
Hence the energy is proportional to the square of the
155
Physical Interpretation: Spectrum
a periodic solution 9 = ( ) that can be represented
As the next step, if a more complicated system leads to
156
Linearity. Fourier Transform of
Derivatives
Theorem 2. Linearity of the Fourier Transform. The
+ } exists, and
exist and any constants and , the Fourier transform
of
(8) ℱ + } = ℱ + ℱ(})
Proof.
1 /
ℱ ( ) + }( ) = * + } C #•sE !
2
1 1
#/
/ /
* C #•sE ! + * } C #•sE !
2 #/ 2 #/
ℱ + ℱ(} )
157
Linearity. Fourier Transform of
Derivatives
Œ(•)
Theorem 3. Fourier Transform of the Derivative of
158
Linearity. Fourier Transform of
Derivatives
Proof. From the definition of the Fourier transform we
1 /
:
ℱ =
have :
*
#•sE
C !
2 #/
Since
namely, ℱ : 0 + G@ℱ( ( )).
159
Linearity. Fourier Transform of
Derivatives
Two successive applications of (9) give ℱ :: =
G@ℱ : = G@ ℱ( ).
Since G@ = −@ , we have for the transform of the
second derivative of
(10) ℱ :: = −@ ℱ( ( ))
160
Example 3
C #E •
Find the Fourier transform of from Table III,
Sec. 11.10
Solution. We use (9). By formula 9 in Table III
1 #E• :
ℱ C #E =ℱ − C
•
2
1 :
=− ℱ C #E •
2
1
= − G@ ℱ C #E •
2
1 1 #s •
= − G@ C Ž
2 2
G@ #s •/Ž
=− C
2 2
161
Convolution
The convolution ∗ } of functions and } is defined
by / /
(11) ℎ ∗} * 8 } − 8 !8 * − 8 } 8 !8
#/ #/
162
Convolution
and }( ) are piecewise continuous, bounded, and
Theorem 4. Convolution Theorem. Suppose that
163
Discrete Fourier Transform (DFT)
Dealing with sampled values rather than with functions,
we can replace the Fourier transform by the so-called
164
Discrete Fourier Transform (DFT)
We now want to determine a complex trigonomentric
#
; = + B C• E•
polynomial
(15)
that interpolates ( ) at the nodes (14), that is,
; ` = ( ` ), written out, with ` denoting ` ,
#
(16) ` = ` =; ` = + B C• E• 6 = 0,1, … , 3 − 1
165