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1 3. Fourier series continuous-time
ResponseofLTISystemstoComplexExponentials
Outline
Consider an LTI system with the unit impulse response h t or h n .
Suppose the input signal is a complex exponential
is a complex number,
is a complex number.
st
n
x t e s
x n z z
Then we will see that the output signal is a complex exponential same as the input, multiplied by a constant factor that depends on s or z:
, where
, where
st st
n n
n
y t H s e H s h t e dt
y n H z z H z h n z
When the output signal is a constant times the input signal, the input signal is called an eigenfunction of the system. The constant, amplitude factor, is called the eigenvalue associated with the eigenfunction.
is the associated with the .
Similarly,
is the associated with the .
st
n
H s eigenvalue eigenfunction e
H z eigenvalue eigenfunction z
If the input signal is represented as a weighted sum of complex exponentials, then the output is the superposition of individual responses. For example,
1 1
2 2
1 2 1 2
1
2
1 2 1 1 2 2
for continuous-time signals,s t s t
s t s t
s t s t s t s t
e H s e
e H s e
a e a e a H s e a H s e
1 1 1
2 2 2
1 1 2 2 1 1 1 2 2 1
for discrete-time signals,n n
n n
n n n n
z H z z
z H z z
a z a z a H z z a H z z
2 3. Fourier series continuous-time
ResponsetoContinuous‐timeComplexExponential
Consider a continuous-time LTI system with the unit impulse response h t .
Suppose the input signal is
stx t e
Then the output is
s t
st s
st
y t x t h t h t x t
h x t d
e h d
e h e d
H s e
where
( ) ( )s stH s h e d h t e dt
is an eigenfunction of the LTI system.
is the eigenvalue associated with the eigenfunction.
ste
H s
ste stH s e h t
stH s h t e dt
3 3. Fourier series continuous-time
Example @3.1
Consider an LTI system which delays the input signal in time by 3: 3h t t .
Suppose the input signal is cos4 cos7 .
Then we know the output signal should be
y 3 cos4 3 cos7 3 .
x t t t
t x t t t
We could have shown this via convolution
y 3 3t x t h t x h t d x t d x t
In this example, we will find the output signal using eigenfunctions and eigenvalues.
3
The eigenvalue associated with the eigenfunction is
3
st
st st s
e
H s h t e dt t e dt e
4 4 7 7
4 4 7 7
Next, represent the input signal as a weighted sum of complex exponentials
cos4 cos7 =2 2
, , , are of the form . They are of LTI systems.
j t j t j t j t
j t j t j t j t st
e e e ex t t t
e e e e e eigenfunctons
4 4 7 7
12 4 12 4 21 7 21 7
4 3 4 3 7 3 7 3
Find the output signal by multiplying associated eigen values
4 4 7 7=
2
2
2
cos 4 3 cos 7 3
j t j t j t j t
j j t j j t j j t j j t
j t j t j t j t
H j e H j e H j e H j ey t
e e e e e e e e
e e e e
t t
equals 3 indeed.y t x t
4 3. Fourier series continuous-time
ResponsetoDiscrete‐timeComplexExponential
Consider a discrete-time LTI system with the unit impulse response h n .
Suppose the input is
nx n z
Then the output of an LTI system is
where
k
n k
k
n k
k
n
k
k
y n x n h n h n x n
h k x n k
h k z
z h k z
H z z
H z h k z
is an eigenfunction of the LTI system.
is the associated with the .
n
n
n
z
H z h n z eigenvalue eigenfunction
nz nH z z h n
n
n
H z h n z
5 3. Fourier series continuous-time
FourierSeriesRepresentationofContinuous‐TimePeriodicSignals
Outline
Consider a periodic signal with period .
The Fourier series represents as a weighted sum of periodic eigenfunctions
with the same period
x t T
x t
T
0
0
0
The Fourier series representation of is
2Synthesis Equation
are referred to as the Fourier coefficients or the spectral coefficients
Analysis Equation
jk tk
k
k
jk tk
T
x t
x t a eT
a
a x t e dt
02
indicates integration over any interval of length .
is the average power of the spectral component .
T
jk tthk k
T
a k a e
6 3. Fourier series continuous-time
DeterminationofFourierCoefficients
00
Assume we can represent a periodic signal as
2, 1
We want to find what should be.
jk tk
k
k
x t
x t a e eT
a
0
0 0
0
Multiplying to 1 and integrating over a period ,
2
jn t
jn t j k n tk
kT T
j k n tk
k T
e e T
x t e dt a e dt
a e dt e
0
0
0
0 0
0 0
Since cos sin ,
and both cos and sin are periodic with period for ,
0 for
and 2 is reduced to
j k n t
j n k t
T
jn tn
T
e k n t j k n t
Tk n t k n t k n
k n
e dt k n
e
x t e dt T a
00
Finally we have
1 2jn tn
T
a x t e dtT T
7
Examp
Consid
x
is x t
Set the
where
For k
For k
ple @3.5
1
der
1
0
tt
T
periodic w
fundament
kx t
1
1
1
1
kT
T
T
a xT
eT
0,1
1
2
1=
sin
k
j
ajk
jk
e
k
k
k
1
1
0
1
0,1
1
2.
T
T
T
aT
T
T
T
1
2
TT
t
ith period T
0
al frequency
jk tka e
0
0
cho
jk t
jk t
t e d
e dt
0
0 1
0 1
0
0 1
2
Tjk t
jk T
jk T j
eT
e e
e
j
T
0
1
jx t e
d
.T
02
y to T
oose the int
dt
1
1
0 1
0 1
T
T
jk T
jk T
e
0t dt
2 and repr
T
2terval T
noting
for k
resent the pe
2Tt
0 2
1, 2, .
T
3. Fourier
eriodic sign
r series continu
nal as
uous-time
8
When
Note th
T
1
0
,4
sin
1
2
hat is rea
k
k
TT
ka
k
a
a
0 1sin
al and even
kT
k
0 sin
4
. This will h
k T k
k
2 for
happen whe
k
k
0
en is rex t
3. Fourier
eal and even
r series continu
n.
uous-time
9
Spect
0
Consid
The Fo
For
jkka e
k
Examp
0
1
From e
sin
0
Let
As d
k
k
a
a
k
T
As the pfrequen
tralCom
0
0
der a continu
ourier series
is referre
0,
t
jk t
x t
e
0
1
2
3
k
ple
0 1
0
0
example 3.5
n.
for the first
indicate the
decreases,
k T
k
k
k
pulse widthncy domain.
mponen
uous-time s
s representat
d to as the
is t
kk
t a
k
he th hk
0
0
0
0
1
22
33
,
,
,
j t
j t
j t
terms
a
a e a
a e a
a e a
0 1
0
,
t time aroun
e first zero c
2
increases.
TT
h becomes n
ntsinth
0
ignal p
tion of
,
th spectral
jk tk
x t
x t
e
k
harmonic of
0
0
0
1
22
33
j t
j t
j t
e
a e
e
0 1
11
nd
crossing poi
.2
.
k T
T
TTT
narrower in t
heFouri
0
periodic wit
is
2
l componen
0f , andj te
1
2
3
compo
d - c
st har
nd ha
rd ha
.
int.
the time do
ierRepr
th period
2
nt.
T
T
d is periodic
onent
rmonic
armonic
armonic
main, the p
3. Fourier
resenta
.
c with perio
0
2
3
period
TT
T
ower spread
r series continu
ation
od .T
k
ds wider in
uous-time
the
10 3. Fourier series continuous-time
ConvergenceoftheFourierSeriesStudents are urged to read the section 3.4 of the textbook.
ExistenceoftheFourierCoefficients
is bounded if the signal is absolutely integrable over a single period.ka x t
0
0
0
.
1.
1
1
1
jk tk
T
jk tk
T
jk t
T
T
Proof
a x t e dtT
a x t e dtT
x t e dtT
x t dtT
However, existence of the Fourier coefficients does not necessarily mean the Fourier series representation is identical to the original periodic signal.
11 3. Fourier series continuous-time
ApproximateRepresentationoftheSignal
0
2
Approximate the periodic signal with harmonics
Define the error
( ) ,
and its energy in a period
( )
Njk t
N kk N
N N
N NT
x t N
x t a e
e t x t x t
E e t dt
We accept the following statements without proof.
2
If the signal has a finite energy over a single period, i.e.,
,
then
lim 0.
T
NN
x t
x t dt
E
As increases, decreases.NN E
0If is discontinuous at ,
then the Fourier series representation becomes the average of the values
on either side of the discontinuity.
x t t t
12
In the e
Howev
This ef
example bel
ver, as inc
ffect is know
N
low, the pea
creases, the
wn as the G
ak amplitud
energy in t
Gibbs pheno
de of the rip
the ripples d
.omenon
ples remain
diminishes.
3. Fourier
ns unchange
r series continu
ed.
uous-time
13 3. Fourier series continuous-time
PropertiesofContinuous‐TimeFourierSeries
02
Signals are periodic with period , and .TT
Linearity
kk k k
k
x t aAx t B y t c Aa Bb
y t b
FSFS
FS
TimeShift
0 00
jk tk k k
k k
x t a x t t b e a
b a
FS FS
0
0 0
0 0 0
0
0
1
shifting the integration interval by
1
1
jk tk
T
jk t t
T
jk t jk t
T
b x t t e dtT
t
x t e dtT
e x t e dtT
14 3. Fourier series continuous-time
Example: Consider example 3.5 again.
1
1 02
Obtain by substituting 2
2, , .
x t
T TT
2 2 sin1 22
k kj j
k
ke e
ak j k
1
0 210 2
jkk kx t x t b e a
FS
2 22
0 0
1
2
1
2
k kj j k
j
k
jk
e eb e
k j
ej
kb a
0
1 0 0
11
1.5 1.5 1 .
1.5 1.5 jkk k
x t x t x t
x t b b e
FS
23 1
, 0.4
01
2
Fourier coefficients are purely imaginary and odd, for the signal is real and odd.
jk
k
jk
ec j k
k
k evene
k odd
1 0
1
An alternative way is to consider
3 0.5 .
3 , 0k
x t x t
x t b k
FS
23 1
, 0. 2
1Note 1 for any integer .
2
jk
k
jkjk
ec j k
k
ee k
x t
t1
1
1 12 20
0x t
t1
1
1 0 2
1x t
t1
1.5
1 0 2
1.5
15 3. Fourier series continuous-time
TimeReversal
If is even; is even, .
If is odd; is odd,
k k k
k k k
k k k
x t a x t b a
x t a a a
x t a a a
FS FS
0
0
0
0
0
0
0
0
0
1
pick the period 0,1
let 1
1
1
1
jk tk
T
Tjk t
Tjk
jk
T
jk
T
j k
T
k
b x t e dtT
T
x t e dtT
t
x e dT
x e dT
x e dT
x e dTa
TimeScaling
remains unchanged.ka
0
0
2
2
0
Observe
,
is periodic with period .
The fundamental frequency becomes .
However remains unchanges.
T
jk tjk t Tk k
k k
jk tjk t
k k
k k
k
x t a e a e
x t a e a e
Tx t
a
16 3. Fourier series continuous-time
Multiplication
kk k k
k
x t ax t y t c a b
y t b
FSFS
FS
0
0
0
00 0Note .
Therefore must equal .
jk tk
k
jk tk
k
jk tk
k
j k tj t jk tk k
k k k k
x t a e
y t b e
x t y t c e
a e b e a b e
c a b a b
Conjugate
If is real, and thus .
If is real and even, . The Fourier coefficients are real and even;
If is real and odd, . The Fourier coeffic
k k k
k k k k
k k k
k k k
x t a x t b a
x t a a a a
x t a a a
x t a a a
FS FS
ients are purely imaginary
and odd.
0
0
0
1
1
1
jk tk
T
jk t
T
jk t
T
k
b x t e dtT
x t e dtT
x t e dtT
a
17 3. Fourier series continuous-time
Differentiation
0k k kd
x t a x t b jk adt
FS FS
0
0
00
jk tk
k
jk tk
k
jk tk
k
x t a e
d dx t a e
dt dt
jk a e
Integration
00
1 for 0
provided 0
tk
k kx t a
x d b a kjka
FS FS
Lemma
000 is periodic if and only if 0.
tj t
T Tx d a T x t e dt x t dt
0
0
Define .
Assume 0 so that is periodic.
.
Let and .
Then .
1For 0, .
t
T
k k
k k
k k
g t x d
x d g t
dx t g t
dt
x t a g t b
a jk b
k b ajk
FS FS
18 3. Fourier series continuous-time
Parseval’sRelation
0
2 2
2 2
1
The average power is the sum of average powers in all harmonics.
1 is the average power of the th harmonic component.
kT
k
jk tk k
T
x t dt aT
a e dt a kT
Problem 3.46 for proof
0
2 2
1
Set to zero to have
1
k k k
jk tk k k k k
T
T
x t a x t b a
x t x t e dt a b a a a aT
k
x t dt aT
FS FS
19
Examp
x t
1
1
1
kaT
T
T
Since x
q t
x
2
jk
jk
q e
e
j
T
Since q
q t
kq
gjk
1
is re
k
k
gjk
k
g
01
gT
ple @3.8 Im
k
t
2
2
1
T
T
j
Tx t e
t e
is real x t
1( )x t T x
t aFS
0 1
0 1
0 12
sin
jk Tk
k T jk
a e
e
T
k T
is real q t
dg t
dt
0 for kq
kk
0
0 1
2sin
sin
eal and even
jk
k T
k T
T
g t e
mpulse Tra
kT
0
0
jk t
jk t
dt
dt
and even, a
1( )x t T
ka x
0 1
0 1
jk Tk
k T
e a
and odd, kq
0.
0 1 ,
n, because
T
g
00j t dt
ain
is real anka
0t t FS
is purely ik
0 2
is real
T
g t
12T
T
nd even.
kb e S
imaginary a
and even.
0 0jk tka
and odd.
3. Fourierr series continuuous-time
