f x a an nx bn ( ) 0 cos (1) - t U · 1 1 Fourier Series Periodic Function A function is called...
Transcript of f x a an nx bn ( ) 0 cos (1) - t U · 1 1 Fourier Series Periodic Function A function is called...
1
1 Fourier Series
Periodic Function A function is called “periodic” function of period p if f(x+np) = f(x), x, n >0, n: integer.
If both f and g are functions of period p, then af + bg is also a function of period p.
Fourier Series Period functions of period 2 can be represented in terms of trigonometric series as
1
0 sincos)(n
nn nxbnxaaxf (1)
If the series converges, it is called “Fourier series” of f(x), where a0, an, bn are called “Fourier coefficients”.
a0, an, bn are given by the “Euler formulas”:
K,2,1,sin
cos)(
1;)(
2
10
ndx
nx
nxxf
b
adxxfa
n
n
(2)
Derivation of the Euler Formulas
Preliminary Orthogonality of trigonometric system
(1) nmmn
mnmxdxnx
0
coscos
Proof
mn
mndxxmnxmnmxdxnx
0)cos()cos(
2
1coscos
(2) nmmn
mnmxdxnx
0
sinsin
Proof
mn
mndxxmnxmnmxdxnx
0)cos()cos(
2
1sinsin
(3) 0cossin
mxdxnx
Proof 0)sin()sin(2
1cossin
dxxmnxmnmxdxnx
Now, integrating both sides of (1) from – to yields
0
1
0 2sincos)( adxnxbnxaadxxfn
nn
dxxfa )(
2
10
Likewise, multiplying both sides of (*) by cos nx and integrating both sides from – to yields
n
m
mm adxmxbmxanxnxanxdxxf
1
0 sincoscoscoscos)(
Likewise, n
m
mm bdxmxbmxanxnxanxdxxf
1
0 sincossinsinsin)( .
Clearly, a0 = 0 and
EX1
Odd function
“Kronecker Delta”
2
evenn
oddnnk
:0
:/4
Lxxxk
xf 5sin5
13sin
3
1sin
4)(
.
NOTE The integral range does not need to be from – to , it can be “arbitrary” as long as it covers an entire
period, e.g., 0 to 2 and vice versa.
2 Functions of any period p = 2L
Assume g(v) is a function of period 2, then
1
0 sincos)(n
nn nvbnvaavg .
Now, let x = Lv/ (or v=x/L) (called “change of scale”), then let
nLxL
gnxL
gnvgvg 22)2()(
Let g(v) = f(x), then f(x+2nL) = f(x), i.e., periodic function of 2L.
1
0 sincos)(n
nnL
xnb
L
xnaaxf
with K,2,1,
sin
cos
)(1
;)(2
10
ndx
L
xn
L
xn
xfLb
adxxf
La
L
Ln
nL
L
(3)
a0=2k/4=k/2, bn=0
evenn
noddnnk
:0
12,:/2)1( ll
2/xv
Odd functions include only sine terms
Period 2L with respect to x
EX2
EX3 EX1 with period changed to 4
Even function
Even functions include only cosine terms
EX4 half-wave rectifier
Clearly, n=1 a1=0
3
21
2
1
2
1
2
2,:;0,:
n
E
nn
Eaevennaoddn nn
;
else
nEtdtntEbn
0
12/sinsin
/
0
3 Even, Odd Functions & Cosine, Sine Series
EVEN: fe(-x)=fe(x), bn=0,
1
0 cos)(n
neL
xnaaxf
with K,2,1,cos)(2
;)(1
000 ndx
L
xnxf
Ladxxf
La
L
en
L
e
ODD: fo(-x)=-fo(x), an=0,
1
sin)(n
noL
xnbxf
with K,2,1,sin)(
2
0 ndx
L
xnxf
Lb
L
on
.
{Even, Odd} functions become Fourier {Cosine, Sine} series, respectively. Thus, for any function f,
4434421444 3444 21odd
n
n
even
n
noeL
xnb
L
xnaaxfxfxf
11
0 sincos)()()(
EX4b Consider w(t) = u(t)-(E/2)sin t, one can see that w(t) = (E/2)sin t, p = / becomes an even function.
It follows that K,2,1,2cossin2
2;sin
2
/
0
/
00 ntdtnt
Eatdt
Ea n
Consider f(x) = f1(x) + f2(x) where f1(x) = x, p=2 (f1 is an odd function), f2(x) = , then
Condition for Existence and Convergence
Existence L
Ldxxf )( (Absolutely integrable)
Convergence
EX5 Sawtooth wave
4
PARSEVAL’s IDENTITY states that
1
222
0
22)(
1
n
nn
L
Lbaadxxf
L if an and bn are the Fourier
coefficients corresponding to f(x) and if f(x) satisfies the Dirichlet conditions. For the proof, consider
dxxfL
xnb
L
xnaadxxf
L
Ln
nn
L
L)(sincos)(
1
0
2
, then use Euler’s formula.
Gibbs Phenomenon
The sinusoidal components of the signal that occur at
multiples of the fundamental frequency are called
harmonics. In general, for well-behaved (continuous)
periodic signals, a sufficiently large number of harmonics can
be used to approximate the signal reasonably well. For
periodic signals with discontinuities, however, such as a
periodic square wave, even a large number of harmonics will
not be sufficient to reproduce the square wave exactly,
resulting in the appearance of so-called “overshoot”. This
effect is known as Gibbs phenomenon and it manifests itself
in the form of ripples of increasing frequency and closer to
the transitions of the square signal. An illustration of Gibbs
phenomenon is shown in the figure on the right. The figure shows the result of adding 5, 9 and 13 harmonics.
The overshoots tend to shift toward the discontinuities, but their magnitude do not change much.
EX4c (Half-wave rectifier)
LLp
tL
Ltttu ,2,
00
0sin)(
LtttE
tE
tu
6cos35
24cos
15
22cos
3
21sin
2)(
…(*)
LLp
tL
Ltttu ,2,
00
0cos)('
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-0.2
0
0.2
0.4
0.6
0.8
1
1.2
Original
N=5
N=9
N=13
5
else
nEtdtntEatdtEa n
0
12/coscos;0cos
2
/
0
/
00
/
0
/
0
2
/
0
2/
0
)1(
)1cos(
)1(
)1cos(
)1sin()1sin(2
sincos
n
tn
n
tnE
dttntnE
tdtntEbn
1
2
1
2
1
2
2,:;0,:
2
n
nE
nn
Ebevennboddn nn
LtttE
tE
tu
6sin35
64sin
15
42sin
3
22cos
2)('
EX Consider f(x) = x on the interval [-,] Fourier series:
1
1
sin)1(2
)(n
n
nxn
xf …(*)
From Parseval’s formula: 6
1
3
2
3
)(
3
114
1)1(2 2
12
233
12
2
1
21
nnn
n
nndxx
n
EX Consider f(x)=x,22
)()(22
xududuufxg
xx
Fourier series:
12
2
12
22
cos)1(2
3cos
)1(2
26)(
n
n
n
n
nxn
nxn
xg
Integrating term by term of (*) in the previous example yields
3cos
)1(2
2cos
)1(2
cos)1(2sin
)1(2)(
2
12
12
12
1
1
1
1
n
n
nn
n
n
xn
n
xn
x
nxnn
nxnn
nu
nnudu
nduuf
4 Application of Fourier Series to ODE for steady-state solution
Consider ODE when the input (external force) becomes periodic, e.g., my''+cy'+ky = r(t), where r(t) is periodic.
The idea is to represent r(t) in terms of Fourier series, and then solve the ODE for each term. For example,
Alternative Approach Using the transfer function Q(s) = (s2+0.05s+25)-1. For sinusoidal steady state response,
222
212
05.025
05.0252505.0)(
nn
njnnjnjnQ
;
nt
D
nnt
D
n
nejnQ
ny
nn
jnt
n sin05.0
cos254
)(Re4
2
22
Example
First Fourier series
Next solve
with
nth term becomes
6
The total solution is given by y = y1 + y3 + y5 + …
5 Application to PDE
(a) Heat Equation
Consider the heat conduction along a heat-conducting homogeneous rod with temperature distribution u(x, t).
Assumptions for one dimensional (1-D) case:
1. homogeneous rod with uniform cross section A and constant density ρ [kg/m3], specific heat [J/kgK],
thermal conductivity κ [W/mK]
2. insulated laterally so heat flows only in x-direction only
3. temperature is constant at all points of a cross section
The heat equation in this case is given by
2
2
22 );1()0,0(, cLxt
x
uc
t
uK [m2/s] (for three dimensional (3-D) case: )0(,22
tuct
u)
with initial condition: u(x, 0) = f(x) (0 ≤ x ≤ L)…(2),
and boundary conditions : u(0, t) = 0, u(L, t) = 0, (t ≥ 0)… (3)
By method of separation of variables: u(x, t) = F(x)G(t)
{ {
tpc
tpc
xt CetGGpcGp
CtGG
CetGGpcGp
F
F
Gc
GGFcGF
22
22
)(
)(00
)(''
'')1(222
222
dependence-dependence-
2
2
&
&
&&
&
)5(0),4(0''222
K&K GpcGFpF
0sin,00)()0(0)()(,0)()0(sincos)()4()3( pLALFFtGLFtGFpxBpxAxF
KK ,2,1,sin)(,2,1, nL
xnxFn
L
npnpL n
K& ,2,1,)(,0)3(2
2 neBtG
L
cnGG
t
nnnnn
Eigenfunction K,2,1,sin)()(),(2
ne
L
xnBtGxFtxu
t
nnnnn
with corresponding eigenvalue n
dxL
xnxf
LB
L
xnBxfxue
L
xnBtxutxu
L
n
n
n
n
t
n
n
nn
0
1
)2(
11
sin)(2
sin)()0,(sin),(),(2
.
(b) Wave Equation
Consider a 1-D wave equation, which represents an up-down movement of a “thin” string.
);1()0,0(,2
22
2
2
KLxtx
uc
t
u
(for three dimensional (3-D) case: )0(,22
2
2
tuct
u)
with initial conditions (initial deflection, velocity): u(x, 0) = f(x) …(2a), ∂u(x,0)/∂t = g(x)…(2b), (0 ≤ x ≤ L)
and boundary conditions : u(0, t) = 0, u(L, t) = 0, (t ≥ 0)… (3)
By method of separation of variables: u(x, t) = F(x)G(t)
{ {
jpxjpx
pxpx
xt BeAexFFpFp
babaxxFF
BABeAexFFpFp
F
F
Gc
GGFcGF
)(''
0)(0''0
0)(''''
'')1(22
)3(
)3(22
dependence-dependence-
2
2&&
&&
)5(0),4(0''222
K&&K GpcGFpF
0sin,00)()0(0)()(,0)()0(sincos)()4()3( pLALFFtGLFtGFpxBpxAxF
KK ,2,1,sin)(,2,1, nL
xnxFn
L
npnpL n
K&& ,2,1,sincos)(,0)3( 2 ntBtAtGL
cnGG nnnnnnn
A constant is required for
the RHS in order to satisfy
for all t and x.
Fourier sine series
7
Eigenfunction K,2,1,sinsincos),( nL
xntBtAtxu nnnnn
with corresponding eigenvalue n
11
sinsincos),(),(n
nnnn
n
nL
xntBtAtxutxu
dxL
xnxf
LA
L
xnAxfxu
L
n
n
n
a
0
1
)2(sin)(
2sin)()0,(
dxL
xnxg
LB
L
xnBxg
x
u L
nn
n
nn
t
b
0
10
)2(sin)(
2sin)(
6 Complex Fourier Series
Using Euler’s Formula, given by 2
sin;2
cosj
eex
eex
jxjxjxjx
, then
n
xL
nj
n
n
xL
nj
nn
n
xL
nj
nn
n
xL
njx
L
nj
n
xL
njx
L
nj
n
n
nn
ecejba
ejba
a
j
eeb
eeaax
L
nbx
L
naaxf
11
0
1
0
1
0
2222
22sincos)(
where c0 = a0, cn=(an-jbn)/2, c-n=cn*, n=1,2,…
L
L
xL
njL
L
L
L
nnn
L
Ldxexf
Ldx
L
xnxf
L
jdx
L
xnxf
L
jbacdxxf
Lac
)(
2
1sin)(
2cos)(
2
1
2;)(
2
100
EX4c Half-wave rectifier with E=1. Note that L = .
/
0
)1()1(/
0
/
0 422sin
2dtee
jdte
j
eedttec
tnjtnjtjntjtj
tjn
n
/
0
)1(/
0
)1(/
0
)1()1(
)1()1(44 nj
e
nj
e
jdtee
jc
tnjtnjtnjtnj
n n: odd, n≠1 cn=0,
n = 1; 44
14
/
0
2
1
j
jdte
jc tj
; a1 = 0, b1 = ½
n:even 1
12;
1;
1
11
1
1
1
1
2
1
)1(
2
)1(
2
42002
nccaca
nnnnjnjjc nnnn
7 Fourier Transform
Consider a periodic function fL(x),
n
xL
nj
nL ecxf
)( , p=2L. Now, set
w
LLwww
L
nw nnn
1,, 1 .
Let L∞, assuming )(lim)( xfxf LL
is “absolutely integrable”, i.e.,
dxxfdxxfdxxf
b
baa)(P.V.)(lim)(lim
0
0
exists.
then w0, wnw (becomes “continuous”), the infinite sum of the Fourier series of fL(x) becomes infinite
integral as wdw in integrand:
n
xjwL
L
vjw
LL
n
xL
njL
L
vL
nj
LL
LL
nn edvevfw
edvevfL
xfxf )(2
lim)(2
1lim)(lim)(
dwedvevfedvevfdw
xfjwx
wf
jwvjwxjwv
44 344 21)(ˆ
)(2
1)(
2)(
P.V. stands for Cauchy’s principal value
8
In engineering, Fourier Transform pair :
dwewfxffdxexfwffjwxjwx )(ˆ
2
1)(]ˆ[)()(ˆ][ 1
FF
In mathematics, generally define
dwewfxfdxexfwf jwxjwx )(ˆ2
1)()(
2
1)(ˆ
Note also that conventionally variable pairs (x, w), (t (time), (frequency)) are used.
EX1 unit gate function
5.00
5.05.0
5.01
)()(
x
x
x
xrectxf
)2/( sinc2/
)2/sin()2/sin(2)(ˆ
2/2/2/2/5.0
5.0
5.0
5.0w
w
w
w
w
jw
ee
jw
ee
jw
edxewf
jwjwjwjwjwxjwx
.
EX2 0,0
0)(
aelse
xexf
ax
=e-axu(x), u(x) : unit step function
jwajwa
edxedxeewf
xjwaxjwajwxax
1
)()(ˆ
0
)(
0
)(
0.
Some Properties
Linearity : F[af + bg] = aF[f] + bF[g]
Symmetry or Duality : )(2)(ˆ)(ˆ)( wfxfwfxf
Proof
dxexfwfdxexfwfdwewfxfjwxjwxxwwxjwx )(ˆ)(2)(ˆ
2
1)()(ˆ
2
1)(
,
.
Scaling :
a
wf
aaxfwfxf ˆ
||
1)()(ˆ)(
Proof a>0:
a
wf
advevf
aa
dvevfdxeaxfaxf ajwvajwv
axvjwx ˆ
||
1)(
1)()()( //
F
a<0, a=-b, b>0:
a
wf
advevf
bb
dvevfdxeaxfaxf ajwvbjwv
bxvjwx ˆ
||
1)(
1)()()( /)/(
F
EX1b gate function
2/0
2/5.0
2/1
)(
ax
ax
ax
a
xrectxf
)2/( sinc2/
)2/sin()2/sin(2)(ˆ
2/2/2/2/2/
2/
2/
2/waa
wa
waa
w
wa
jw
ee
jw
ee
jw
edxewf
jwajwajwajwaa
a
jwxa
a
jwx
Time shift : 321shift phase
00ˆ)()(ˆ)(
jwxewfxxfwfxf
Frequency shift : 0
modulation
ˆ)()(ˆ)( 0 wwfexfwfxfxjw
43421
Special Cases
Delta Function :
1)(ˆ);()( dxexwfxxf jwx 1)( x
Furthermore,
dwexjwx
2
1]1[)( 1
F ;
Likewise, from duality )(2)(211)( wwx
dxewjwx
2
1]1[
2
1)( F
9
Harmonics : 0
)(2)(ˆ)( 000 wwdxedxeewfexf
xwwjjwxxjwxjw
Unit step function : f(x) = u(x) [= 1 (x≥0) and = 0 (x < 0)]
Let )(lim)(0
xuexuax
a
,
2200000lim
1limlim)](lim[)(ˆ
wa
jwa
jwadweexuewu
aa
jwxax
a
ax
aF
Imaginary part : jwwa
jw
a
1lim
220
Real part : )(lim;tan220
1
22w
wa
a
a
wdw
wa
a
a
jwwwu /1)()(ˆ
Parseval’s Theorem :
dwwfdxxf
22)(ˆ
2
1)(
dwwfwfdwdvwvvfwfdwdvdxevfwf
dxdvevfdwewfdxxfxfdxxf
xwvj
xf
jvx
xf
jwx
)(ˆ)(ˆ2
1)()(ˆ
2
1)(ˆ
2
1)(ˆ
2
1)(ˆ
)(ˆ2
1)(ˆ
2
1)()()(
**)(*
)(
*
)(
*2
*
444 3444 21444 3444 21
Significance Total energy in x (or time) domain = total energy in transform (or frequency) domain
Fourier transform of derivative : )(ˆ)(')(ˆ)( wfjwxfwfxf
Proof )(ˆ)()()(')]('[
0)(
wfjwdxexfjwexfdxexfxf jwx
xfx
jwxjwx
43421F
It follows that )(ˆ)(ˆ)]('[)('')(ˆ)( 2wfwwfjwjwxfjwxfwfxf F
EX Consider 2
)(x
xexf ; ][
2])'[(
2
1)(ˆ)'(
2
1 2222 xxxx ejw
ewfexe FF
44)
2(
44)
2(
2
2
22
222
22
][
w
z
wjwx
wwjwx
jwxxxedzeedxeedxeedxeee
F
Thus, 4
2
2
2][
2)(ˆ
w
xe
jwe
jwwf
F
Convolution
dppxgpfxgf )()())(*( gfgfwgxgwfxf ˆˆ*)(ˆ)(),(ˆ)(
)(ˆ)(ˆ)()(ˆ)()()()(
)()()()(]*[
)(wfwgdppfewgdppfedqeqgdppfdqeqg
dppfdxepxgdxedppxgpfgf
jwpjwpjwqpqjwqpx
jwxjwx
F
Duality of convolution : gffgwgxgwfxf ˆ*ˆ2
1)(ˆ)(),(ˆ)(
Exercise Prove this property.
EX Let f(x) = g(x) = rect(x), then for |x| > 1, f*g=0, -1< x<0, 1)(*5.0
5.0
xdpxgf
x
, 0<x<1,
xdpxgfx
1)(*
5.0
5.0. 321
1
0
0
1)1()1(]*[ IIIdxexdxexgf jwxjwx
F
10
w
w
jw
ee
jw
edxeI
jwjwjwxjwx sin
1
1
1
11
;2
0
1
0
1
0
12
11
w
e
jw
edxe
jwjw
xedxxeI
jwjwjwx
jwxjwx
2
1
0
1
0
1
03
11
w
e
jw
edxe
jwjw
xedxxeI
jwjwjwx
jwxjwx
2sinc
2sin
)2/(
1
2sin
4
2sin2
1
12
111]*[
2
2
2
2
2
2
2
22/2/
2222
ww
w
w
w
wj
w
eew
eeww
e
jw
e
w
e
jw
e
jw
eegf
jwjwjwjwjwjwjwjwjwjw
F
Using the “even” function property yields
2
2
2
1
0
2
1
0
1
0
1
0
1
0
)2/(sin4cos22cos2sin
2sin2sin2cos)1(2]*[
w
w
w
w
w
wxwxdx
ww
wxx
w
wxwxdxxgf
F
Alternatively, since )2/(sinc]*[),2/(sinc][ 2wrectrectwrect FF .
8 Application of Fourier Transform to PDE
(a) Heat Equation Consider
)1(),0(,2
22
K
xtx
uc
t
uwith initial condition: u(x, 0) = f(x) …(2),
Recall ][)](''[],[)]('[,)()]([)(ˆ 2 uwxuujwxudxexuxuwu jwxFFFFF
Take Fourier transform of (1) yields
uwcx
uc
t
uFFF
22
2
22
; ][][ ut
dxuet
dxet
u
t
u jwxjwxFF
4444 34444 21444 3444 21
ODEPDE
twuwct
twuuwc
t
u),(ˆ
),(ˆ 2222
FF …(3)
Take Fourier transform of (2) yields )(ˆ)0,(ˆ wfwu …(4)
From (3) twc
Cetwu22
),(ˆ ; from (4) )(ˆ)0,(ˆ wfCwu twc
ewftwu22
)(ˆ),(ˆ
Take inverse Fourier transform yields
dweewfdwetwutwutxujwxtwcjwx 22
)(ˆ2
1),(ˆ
2
1)],(ˆ[),( 1
F
(b) Wave Equation Consider
);1(),0(,2
22
2
2
K
xtx
uc
t
usubject to
with initial conditions : u(x, 0) = f(x) …(2a), ∂u(x,0)/∂t = 0…(2b),
and boundary conditions : u 0, ux 0, as |x|∞ (t ≥ 0)… (3)
where f(x) assumed to have Fourier transform
uwcucuu xxttttˆˆ 222 FF uwcutt
ˆˆ 22 cwtwBcwtwAtwu sin)(cos)(),(ˆ
For t=0, )(ˆ)()0,(ˆ wfwAwu , 0)()0,(ˆ wcwBwut , Thus 2
)(ˆcos)(ˆ),(ˆjcwtjcwt ee
wfcwtwftwu
)()(2
1)],(ˆ[),(
shiftfrequency 1
ctxfctxftwutxu F “d’Alembert’s solution”
9 Poisson’s Sum Formula
11
Consider a “periodic” impulse train with period p=2L,
k
kLxxs )2()( (Called comb, Ш (Shah)
function). Fourier series of s(x) is given by
LdxekLx
Ldxexs
Lc
L
L
xL
nj
k
L
L
xL
nj
n2
1)2(
2
1)(
2
1
and
kn
xL
nj
kLxeL
xs )2(2
1)(
Now, consider Fourier transform of s(x)
nnn
xL
nwj
jwx
n
xL
nj
L
nw
LL
nw
Ldxe
Ldxee
Lws )()(
2
2
2
1
2
1)(ˆ
)(
Let
kk
kLxgkLxxgxsxgxh )2()2(*)()(*)()( ”periodic version” of g(x)
FT of h(x) :
nnn L
nw
L
ng
LL
nwwg
LL
nw
Lwgwswgwh )()(ˆ)()(ˆ)()(ˆ)(ˆ)(ˆ)(ˆ
)(ˆ wg sampled at n/L. Take inverse FT of )(ˆ wh yields
n
L
xnj
k
eL
ng
LkLxg
)(ˆ
2
1)2( Poisson’s Sum Formula
Fourier series vs Fourier transform
Since
n
L
nj
n
L
nj
n
k
eL
ng
LeckLxgxh
)(ˆ
2
1)2()( , )(ˆ
2
1
L
ng
Lcn
.
For an example, consider a half-wave rectifier. First, let g(t) = sin t[u(t)-u(t-/)],
/
0
)()(/
0
/
0 2
1
2sin)(ˆ dtee
jdte
j
eedttewg
twjtwjjwttjtj
jwt
)(
1
)(
1
2
1
)()(2
1 /)(/)(/
0
)(/
0
)(
wj
e
wj
e
jwj
e
wj
e
j
wjwjtwjtwj
)()(2
1 2/)(2/)(2/)(
2/)(2/)(2/)(
wj
eee
wj
eee
j
wjwjwj
wjwjwj
)(
2/)sin(
)(
2/)sin(1 2/)(2/)(
w
we
w
we
j
wjwj
2/)(
2/)sin(
2/)(
2/)sin(
2
1 2/)(2/)(
w
we
w
we
j
wjwj
2
)(sinc
2
)(sinc
2
1 2/)(2/)( we
we
j
wjwj
Alternatively, let
)2/(
2
)2/(sin)(
trect
j
eetrectttg
tjtj
)()(2
2
2
wwjj
eetjtj
F ;
2sinc
)2/( 2/ we
trect
jw
F
Using gffg ˆ*ˆ2
1
yields
2
)(sinc
2
)(sinc
2
1)(ˆ 2/)(2/)( w
ew
ej
wgwjwj
Now, let
LkLtgkLttgtstgthkk
;)2()2(*)()(*)()(
12
2
)(sinc
2
)(sinc
2
1
2)(ˆ
2)(ˆ
2
1 2/)(2/)( ne
ne
jng
L
ng
Lc
njnj
n
2
)1(sinc
2
)1(sinc
4
12
)1(
2
)1( n
en
ej
nj
nj
n=0,
1
2/2/4
1
2sinc
2sinc
4
122
0
jj
jee
jc
jj
4
1
2
2sinc0 sinc
4
12
2
1j
ej
cj
, n: odd, n≠1 cn=0,
n:even: 1
11
)1(
2
)1(
2
4
1
2
)1(sinc
2
)1(sinc
4
12
2
)1(
2
)1(
nn
j
n
j
j
ne
ne
jc
nj
nj
n
Discrete (Sampled) Signal
Let fC(x) be a “continuous” signal, and fS(x) be a “sampled” signal with sampling period p=2L.
k
C
k
CCS kLxxfkLxxfxsxfxf )2()()2()()()()(
Using gffg ˆ*ˆ2
1
,
n
C
n
CCSL
nwwf
LL
nw
Lwfwswfwf )(*)(ˆ
2
1)(*)(ˆ
2
1)(ˆ*)(ˆ
2
1)(ˆ
which means )(ˆ wfC is repeated every /L, in other words, )(ˆ wfS is a “periodic” version of )(ˆ wfC . This is
called Discrete Time Fourier Transform (DTFT).
Summary
x (or time)
domain
Example Waveform Transform
(frequency)
domain
Example Spectrum
Continuous,
aperiodic
Continuous,
aperiodic
Continuous,
periodic
2L
Discrete (Fourier
series)
π/L Discrete (sampled
signal)
2L
Continuous,
periodic (Periodic
version of Fourier
transform)
π/L
10 Laplace Transform and Fourier Transform
Laplace to Fourier: Recall that Laplace transforms exist only for functions that satisfy the “growth restriction”
condition, which requires region of convergence (ROC) consideration (i.e., region of s where the transform
exists). Now, the conventional Laplace transform is defined for f(t), t≥0, as
0)()( dtetfsF
st,
which is sometimes called “unilateral” Laplace transform. Since f(t) is assumed to be 0 for t<0,
dtetfsFst
)()( .(called “bilateral” Laplace transform)
Comparing to the Fourier transform for (t,) pair given by
dtetff tj )()(ˆ ,
13
it follows that by substituting s=j in F(s), provided that s=j (imaginary axis) is in the ROC, one can obtain
the Fourier transform, i.e., )()(ˆ jFf .
EX1 Consider f(t)=e-at, a>0, then F(s)=(s+a)-1. 1)()(ˆ ajjFf .
EX2 Consider f(t) = sin t[u(t)-u(t-/)]. Note that w is used as the variable for frequency domain here.
/
2222
/2/2/2/
2/2/
2/)(2/)(
12
2
111
2
2/)(
2/cos
2/)(
2/cos
2
1
2
)(sinc
2
)(sinc
2
1)(ˆ
jwjwjwjw
jw
jwjw
wjwj
eww
e
ww
eee
w
wje
w
wje
j
we
we
jwf
)()(sin)(sin)(sin)()(sin)(22
tuts
ttuttutututsF LLLL
s
ess
2222
wjwj
ew
eww
jwFwf 1)()(ˆ222222
NOTE u(t)1/s, but )/(1)(ˆ ju ! This is because s > 0 (Re s > 0) for u(t).
Fourier to Laplace: Fourier transforms exist for functions that are “absolutely integrable”. Now, consider a
the Fourier transform of f(t) which satisfies f(t)=0 for t<0. It follows that
0)()(ˆ dtetff tj .
This integral does not exist unless f(t) is absolutely integrable. Thus, an additional term is introduced to
guarantee the existence of integral, namely
)()()()(ˆ00
sFdtetfdteetff sttjt .
Therefore, s=+j becomes complex and the ROC for s needs to be considered.
Inverse Laplace Transform
Recall the growth restriction dictates that Laplace transform F(s) exists if |f(t)|<Mekt, for t>0, ∃k,M. Can
choose >k such that 0)( ttkMe
. Therefore, for f(t)=0, t<0,
ddefeetfe
tfe
jtjt
t
444 3444 21)]([
0)(
2
1)(
F
.
Moving the et term to the right hand side yields
j
j
sF
sstjs
jtjtdsdefe
jddeefeetf
4434421
)(
00)(
2
1)(
2
1)( .
Thus,
j
j
stdsesFj
tf
)(
2
1)( ”Bromwich integral”
where kk
sRemax , sk denotes the kth pole of F(s), i.e., > Maximum of Real part of poles.
Brief Introduction of Poles Let F(s)=A(s)/B(s), and sk denote a pole of F(s), then B(sk)=0.
(Recall eat(s-a)-1, and a becomes a pole)
EX 326
)(2
3/1
1
2/1
1
6/1
22
1)(
2
23
ttt eeetf
sssssssF
Thus, s>1 is the ROC, and it follows that > 1.
14
11 Fourier Cosine and Sine Transform
Recall that
0
0
00
)(
)sin)(cos)(
sinsincoscos)(1
)(cos2)(cos,0)(sin)(cos)(1
)(sin)(cos)(2
1
)(2
1)(
2
1)(
dwwxwBwxwA
dwdvwvwxwvwxvf
dwvxwdwvxwdwvxwdwdvvxwvf
dwdvvxwjvxwvf
dwdvevfdwdvevfexfvxjwjwvjwx
Q where
wvdvvfwBwvdvvfwA sin)(
1)(;cos)(
1)(
If f(x) is an “even” function, B(w)=0.
00
cos)()(;cos)(2
)( wxdwwAxfwvdvvfwA
If f(x) is an “odd” function, A(w)=0.
00
sin)()(;sin)(2
)( wxdwwBxfwvdvvfwB
Let ),(ˆ2)( wfwA c then
00
cos)(ˆ2)(cos)()(ˆ wxdwwfxfwxdxxfwf cc
c
FFFF
Let ),(ˆ2)( wfwB s then
00
sin)(ˆ2)(sin)()(ˆ wxdwwfxfwxdxxfwf ss
s
FFFF
Note that in mathamatics:
00
cos)(ˆ2)(cos)(
2)(ˆ wxdwwfxfwxdxxfwf cc
c
FFFF
,
00
sin)(ˆ2)(sin)(
2)(ˆ wxdwwfxfwxdxxfwf ss
s
FFFF
.
Some Examples
EX1
ax
axkxf
0
0)(
)( sincsinsin
cos)(ˆ
00
wakaw
wak
w
wxkwxdxkwf
aa
c .
waw
k
w
wxkwxdxkwf
aa
s cos1cos
sin)(ˆ
00
.
EX2 f(x) = e-kx, x > 0, k > 0
“Fourier Integral”
“Fourier Cosine Integral”
“Fourier Sine Integral”
“Fourier Cosine Transform (FCT)”
“Fourier Sine Transform (FST)”
15
0 2
2
0
2
00
0
coscos
sinsin
cos)(ˆ
wxdxew
k
w
wxke
wxdxew
k
w
wxewxdxewf
kxkx
kxkx
kx
c
2222
2
0202
2
/1
/cos)(ˆcos1
wk
k
wk
wkwxdxewf
w
kwxdxe
w
k kx
c
kx
.
0,0,cos2
cos)(ˆ2)(
0 220
kxdw
wk
wxkwxdwwfexf c
kx
.
It follows that 0,0,2
cos
0 22
kxek
dwwk
wx kx… (*)
Likewise,
0 2
2
0
2
00
0
sinsin1
coscos
sin)(ˆ
wxdxew
k
w
wxke
w
wxdxew
k
w
wxewxdxewf
kxkx
kxkx
kx
s
2222002
2
/1
/1sin)(ˆ1
sin1wk
w
wk
wwxdxewf
wwxdxe
w
k kx
s
kx
0,0,sin2
sin)(ˆ2)(
0 220
kxdw
wk
wxwwxdwwfexf s
kx
It follows that 0,0,2
sin
0 22
kxedwwk
wxw kx… (**)
Equations (*), (**) are called “Laplace Integrals”
Cosine transform of derivative : )0()(ˆ)(')(ˆ)( fwfwxfwfxf sc
Proof )0()(ˆsin)(cos)(cos)(')]('[
0)(
00fwfwwxdxxfwwxxfwxdxxfxf s
xfx
c
4434421F
Sine transform of derivative : )(ˆ)(')(ˆ)( wfwxfwfxf cs
Proof )(ˆcos)(sin)(sin)(')]('[
0)(
00wfwwxdxxfwwxxfwxdxxfxf c
xfx
s
4434421F
It follows that
)0(')]([)0(')]('[)](''[ 2 fxfwfxfwxf csc FFF
)0()]([)]('[)](''[ 2 wfxfwxfwxf scs FFF .
Application to PDE
Consider the heat equation given by:
)1()0,0(,2
22
K
xtx
uc
t
uwith conditions u(x, 0) = f(x) …(2), u(0,t) = 0…(3)
Recall ][)](''[],[)]('[,)()]([)(ˆ 2 uwxuujwxudxexuxuwu jwxFFFFF
Take Fourier sine transform of (1) yields
ss
ssss
s uwct
uuwctwuuwc
x
uc
t
u
t
uˆ
ˆˆ),0(
ˆ 222222
2
22
FFF …(4)
16
Take Fourier sine transform of (2) yields )(ˆ)0,(ˆ wfwu ss …(5)
From (4) twc
s Cetwu22
),(ˆ ; from (5) )(ˆ)0,(ˆ wfCwu ss twc
ss ewftwu22
)(ˆ),(ˆ
Take inverse Fourier sine transform yields
00
1sin)(ˆ2
sin),(ˆ2
)],(ˆ[),(22
wxdwewfwxdwtwutwutxutwc
ss F .
12 Practice Problems
f(x) or f(t) ODE
1
3
5
2
4
6
7
f(x) = x2 (-1<x<1), p = 2
f(x) = x|x| (-1<x<1), p = 2
a. y'+y = f(t)
b. y'+2y = f(t)
c. y'+3y = f(t)
d. y'+4y = f(t)
e. y''+3y'+2y = f(t)
f. y''+6y'+5y = f(t)
g. y''+5y'+4y = f(t)
h. y''+5y'+6y = f(t)
i. y''+4y'+4y = f(t)
Question
1. Find the Fourier series of f(x) (or f(t)), then find the “steady-state” solution for the ODE.
2. Find the Fourier transform of f(x) assuming that p∞.
3. Compare the Fourier transform found in prob. 2 with the Fourier series found in prob. 1.
4. Prove gffgwgxgwfxf ˆ*ˆ2
1)(ˆ)(),(ˆ)(
.
5. Find the Fourier cosine transforms of:
(a) –x + a, 0 < x < a (b) 2
axe(a > 0)
7. Show that
6. Find the Fourier sine transforms of:
(a) –x + a, 0 < x < a (b) x2
axe(a > 0)
a.
b.
c. d.
8. Solve the following integral equations; (i.e., find f(x))
(a) awewxdxxf 4/
0
2
cos)( (b)
2/
0
2
sin)( wwewxdxxf
(c) awjwx edxexf 4/2
)(
(d) wadxexf jwx sinc)(