Chapter3 - Fourier Series Representation of Periodic Signals
1. 241-306 Fourier Series Representation of Periodic Signals
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Chapter 3
Fourier Series
Representation of
Periodic Signals
2. 241-306 Fourier Series Representation of Periodic Signals
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Outline
1 The Response of LTI Systems to Complex
Exponentials
2 Fourier Series Representation of Continuous-Time
Periodic Signals
3 Convergence of The Fourier Series
4 Properties of Continuous-Time Fourier Series
5 Fourier Series Representation of Discrete-Time
Periodic Signals
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6 Properties of Discrete-Time Fourier Series
7 Fourier Series and LTI Systems
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In Chapter 2, We discuss about a representation of
signals as linear combination of a set of basic
signals. Two properties of the basic signals that we
desire :
3.1 The Response of LTI Systems to Complex
Exponentials
1 The set of basic signals can be used to
construct a broad and useful class of signals.
2 The response of an LTI system to each signal
should be simple enough in structure to provide
us with a convenient representation for the
response of the system to any signal constructed
as a linear combination of the basic signals.
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Eigenfunction and Eigenvalue of the system
LTI CT
e
st
H se
st
Eigenvalue Eigenfunction
yt=∫−∞
∞
hxt−d
=∫−∞
∞
he
st−
d
=e
st
∫−∞
∞
he
s
d
H(s)
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LTI DT
z
n
H z z
n
Eigenvalue Eigenfunction
y[n]= ∑
k=−∞
∞
h[k ] x[n−k ]
= ∑
k=−∞
∞
h[k ] z
n−k
=z
n
∑
k=−∞
∞
h[k ]z
−k
H(z)
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If the input to a continuous-time LTI system is
represented as a linear combination of complex
exponentials
xt=∑
k
ak e
sk t
then the output will be
yt=∑
k
ak H sk e
sk t
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For discrete-time LTI system,it the input is
represented as a linear combination of complex
exponentials
x[n]=∑
k
ak zk
n
then the output will be
y[n]=∑
k
ak H zk zk
n
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Example 3.1
yt=xt−3
yt=e
j2t−3
=e
− j6
e
j2t
H s=∫−∞
∞
−3e−s
d =e−3s
Consider the LTI system
If the input to this system is x(t) = e j2t
then
We would expect, since e j2t
is eigenfunction. The
associated eigenvalue is H(j2) = e -j6
. The impulse
response of the system is h(t) = δ(t-3)
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yt=cos4t−3cos7t−3
xt=
1
2
e
j4t
1
2
e
− j4t
1
2
e
j7t
1
2
e
− j7t
xt=cos4tcos7t
Consider the input signal
The output is
By using Euler's relation
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yt=
1
2
e
− j12
e
j4t
1
2
e
j12
e
− j4t
1
2
e
− j21
e
j7t
1
2
e
j21
e
− j7t
yt=
1
2
e
j4t−3
1
2
e
− j4t−3
1
2
e
j7 t−3
1
2
e
j7t−3
=cos4t−3cos7t−3
then
or
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Problem section 3.1
1 Consider three continuous-time systems S1
, S2
and
S3
whose response to a complex exponential input ej5t
are specified as
S1 :e
j5t
te
j5t
S2 :e
j5t
te
j5t−1
S3 :e
j5t
cos5t
For each system, determine determine whether the
given information is sufficient to conclude that the
system is definitely not LTI.
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2 Consider three discrete-time systems S1
, S2
and S3
whose response to a complex exponential input ejπn/2
are specified as
S1 :e jn/2
e j n/2
u[n]
S2 :e
j n/2
e
j3n/2
S3 :e
j n/2
2e
j5 n/2
For each system, determine determine whether the
given information is sufficient to conclude that the
system is definitely not LTI.
Problem section 3.1Problem section 3.1 (Cont.)
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3.2 Fourier Series Representation of Continuous-
Time Periodic Signals
Linear Combination of Harmonically Related
Complex Exponentials
xt=xtT
xt=cos0 t
xt=e
j
0
t
Definition of periodic signal
two basic periodic signals
ω0
– fundamental frequency
T = 2π/ω0
– fundamental period
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Set of harmonically related complex exponentials
k=e
jk 0 t
=e
jk 2/T t
k=0,±1,±2,...
xt= ∑
k=−∞
∞
ak e
jk 0 t
= ∑
k=−∞
∞
ak e
jk 2/T t
A linear combination of harmonically related
exponentials of the from
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Example 3.2
xt= ∑
k=−3
3
ak e
jk2t
Consider the signal x(t), with fundamental
frequency 2π.
where
a0=1 a1=a−1=
1
4
a2=a−2=
1
2 a3=a−3=
1
3
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xt=1
1
4
e
j2t
e
− j2t
1
2
e
j4t
e
− j4t
1
3
e
j6 t
e
− j6t
xt=1
1
2
cos2tcos4t
2
3
cos6t
Rewrite the x(t)
Using Euler's relation, we can write x(t) in the from
We illustrate graphically how the signal x(t) is built
up from its harmonic component
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xt= ∑
k=−∞
∞
ak
∗
e
− jk 0 t
Suppose that x(t) is real and can be represented in the
from
Fourier series of real periodic signals
xt= ∑
k=−∞
∞
ak e
jk 0 t
since x*(t) = x(t), we obtain
(*)
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xt= ∑
k=−∞
∞
a−k
∗
e
jk 0 t
Replacing k by -k :
by comparision with (*) in slide 19
ak=a−k
∗
or
ak
∗
=a−k
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xt=a0∑
k=1
∞
[ak e
jk 0 t
a−k e
− jk 0 t
]
xt=a0∑
k=1
∞
[ak e
jk 0 t
ak
∗
e
− jk 0 t
]
Alternative forms of the Fourier series
Suppose that x(t) is real and can be represented in the
from
xt= ∑
k=−∞
∞
ak e
jk 0 t
(*)
rearrange the summation
substituting ak
* for a-k
:
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xt=a0∑
k=1
∞
2ℜe {ak e
jk 0 t
}
ak=Ak e
jk
xt=a0∑
k=1
∞
2ℜe {Ak e
jk 0 t−k
}
since the two term inside the summation are
complex conjugates of each other.
If ak
is express in polar form
That is
xt=a02∑
k=1
∞
Ak cosk 0 t−k
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ak=Bk jCk
xt=a02∑
k=1
∞
[Bk cosk 0 t−Ck sink 0 t]
But if ak
is express in rectangular form
where Bk
and Ck
are both real. We can write :
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Determination of the Fourier Series Representation
of a Continuous-Time Periodic Signal
Assuming that a given periodic signal can be
represented with the series
xt= ∑
k=−∞
∞
ak e
jk 0 t
We need the procedure for determining the
coefficients ak
.
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Multiplying both sides by e
− jn
0
t
xte
− jn0 t
= ∑
k=−∞
∞
ak e
jk 0 t
e
− jn0 t
Integrating both sides from 0 to T=2π/ω0
:
∫
0
T
xte
− jn0 t
dt=∫
0
T
∑
k=−∞
∞
ak e
jk 0 t
e
− jn0 t
dt
or
∫
0
T
xte
− jn0 t
dt= ∑
k=−∞
∞
ak
[∫
0
T
e
jk−n0 t
dt
] (*)
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By using Euler's relation
∫
0
T
e
jk−n0 t
dt =∫
0
T
cosk−n0 tdt
j∫
0
T
sink−n0 tdt
For k ≠ n, the integral over one period of
sinusoidal signal is zero. For k=n, the integral
equal to T
∫
0
T
e
jk−n0 t
dt =
{T , k=n
0, k≠n
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Consequently, the equation
∫
0
T
xte
− jn0 t
dt= ∑
k=−∞
∞
ak
[∫
0
T
e
jk−n0 t
dt
]
reduce to
∫
0
T
xte
− jn0 t
dt=an T
or
an=
1
T
∫
0
T
xte
− jn0 t
dt=
1
T
∫T
xte
− jn0 t
dt
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xt= ∑
k=−∞
∞
ak e
jk 0 t
= ∑
k=−∞
∞
ak e jk 2/T t
ak=
1
T
∫T
xte
− jk 0 t
dt=
1
T
∫T
xte− jk 2/T t
dt
a0=
1
T
∫T
xtdt
Equations of Fourier series of a periodic
continuous-time signal
For k=0
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Example 3.3
Consider the signal
xt=sin0 t
Determine the Fourier series coefficients for this
signal
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Solution
We can express sin ω0
t as
sin0 t=
1
2j
e
j
0
t
−
1
2 j
e
− j
0
t
Comparing to the equation of Fourier series
a1=
1
2j
a−1=−
1
2j
ak=0 k≠1 o r −1
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∣ak∣
∢ak
10-1
●● ●
10
-1
●● ●
k
k
Graph of the magnitude and phase of ak
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Example 3.4
xt=1sin 0 t2cos0 tcos20 t
4
Determine the Fourier series coefficients for the
signal
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Solution
Express this signal in the term of complex
exponentials
xt=1
1
2j
[e
j
0
t
−e
− j
0
t
][e
j
0
t
e
− j
0
t
]
1
2
[e
j2
0
t/4
e
− j2
0
t/4
]
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Collecting terms :
xt=11
1
2je
j
0
t
1−
1
2je
− j
0
t
1
2
e
j/4
e
j 2
0
t
1
2
e
− j/4
e
− j 2
0
t
The Fourier series coefficients are :
a0=1 a1=1
1
2j=1−
1
2
j
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a2=
1
2
e
j/4
=2
4
1 j
a−2=
1
2
e
− j/4
=
2
4
1− j
ak=0, ∣k∣2
a−1=1−
1
2j=1
1
2
j
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Example 3.5
Determine the Fourier series coefficients for the
signal
xt=
{1, ∣t∣T1
0, T1∣t∣T /2
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Solution
For k = 0 :
a0=
1
T
∫
−T 1
T 1
dt=
2T1
T
For k ≠ 0 :
ak=
1
T
∫
−T 1
T 1
e
− jk 0 t
dt=−
1
jk 0 T
e
− jk 0 t
∣−T1
T 1
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ak=
2
k 0 T [e
jk
0
T
1
−e
− jk
0
T
1
2j ]
ak=
2sink T1
k T
=
sink T1
k
, k≠0
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Plot of the scaled Fourier series coefficients Tak
with T = 4T1
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Plot of the scaled Fourier series coefficients Tak
with T = 8T1
Plot of the scaled Fourier series coefficients Tak
with T = 16T1
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3.4 Convergence of the Fourier Series
The question of the validity of Fourier series
representation.
xt= ∑
k=−∞
∞
ak e
jk 0 t
Is the FS can approximate the periodic signal x(t)?
We can prove the validity by using the difference
in energy of x(t) and its FS
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et=xt−∑−∞
∞
ak e
− jk 0 t
If x(t) and its FS are the same, there is no energy in
the difference.
∫T
∣et∣
2
dt=0
One class of periodic signals is the signal that have
finite energy over a signal period.
∫T
∣xt∣
2
dt∞
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Condition 1. x(t) must be absolutely integrable over
one period.
The set of conditions (The DiriChlet condition)
guarantees that x(t) equal its Fourier series
representation
∫T
∣xt∣dt∞
xt=
1
T
violates the first
condition
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Condition 2. In a finite time interval, x(t) has a finite
number of maxima and minima.
xt=sin2
t , 0t≤1 meets condition 1
but not condition 2
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Condition 3. In a finite time interval, x(t) has only a
finite number of discontinuities.
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3.5 Properties of Continuous-Time Fourier Series
The notation
FS
xt ↔ ak
mean ak
are the Fourier series coefficients of the
periodic signal signal x(t) which have a period T
and fundamental frequency ω0
=2π/T
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Linearity
FS
xt ↔ ak
FS
xt ↔ bk
FS
zt=AxtByt ↔ ck=AakBbk
Let x(t) and y(t) are periodic signals with period T
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Time Shifting
FS
xt ↔ ak
FS
xt−t0 ↔ e
− jk 0 t0
ak=e
− jk2/T t0
ak
Let x(t) is a periodic signal with period T
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Time Reversal
FS
xt ↔ ak
FS
x−t ↔ a−k
Let x(t) is a periodic signal with period T
Time Scaling
The Fourier series coefficient dose not changed
but the Fourier series has changed because of the
change in the fundamental frequency.
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Multiplication
FS
xt ↔ ak
FS
yt ↔ bk
xt yt ↔ hk= ∑
l=−∞
∞
al bk−l
FS
Let x(t) is a periodic signal with period T
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Conjugation and Conjugate Symmetry
FS
xt ↔ ak
FS
x∗
t ↔ a−k
∗
Let x(t) is a periodic signal with period T
If x(t) is real, x(t) = x*(t), the Fourier series
coefficients will be conjugate symmetric.
a−k = ak
∗
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Parseval's Relation for Continuous-Time Periodic
Signals
1
T
∫T
∣xt∣
2
dt= ∑
k=−∞
∞
∣ak∣
2
1
T
∫T
∣ak e
jk 0t
∣
2
dt=
1
T
∫T
∣ak∣
2
dt=∣ak∣
2
Parseval 's relation for continuous-time periodic
signals is
In one period of the periodic signal x(t)
The total average power in a periodic signal equal
the sum of the average powers in all of its harmonic
component.
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Example 3.6
Determine the Fourier Series representation of g(t)
with a fundamental period of 4.
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Solution
From the example 3.5 with T = 4 and T1
= 1 we
can write the function g(t) in the form of x(t).(add
dc offset -1/2 and shift the signal to right)
g t=xt−1−1/2
From shift property, the Fourier coefficients of
x(t-1) may be express as :
bk=ak e− jk /2
ak
is FS coefficient of x(t)
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The Fourier coefficient of the dc offset is given by
ck=
{
0, for k≠0
−
1
2
, for k=0
By using linearity property, the coefficients for g(t)
may be expressed as
d k=
{
ak e
− jk /2
, for k≠0
a0−
1
2
, for k=0
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by replacing ak
from example 3.5
d k=
{
sin k /2
k
e
− jk /2
, for k≠0
0, for k=0
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Example 3.7
Consider the triangular signal x(t) with period T = 4
and fundamental frequency ω0
= π/2. Determine
the Fourier coefficient from g(t) of example 3.6
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Solution
The signal g(t) is the derivative of x(t) or x(t) is
the integration of signal g(t).
dk= jk /2ek
dk
is the Fourier coefficient of g(t) and ek
is the
Fourier coefficient of x(t)
Solution
ek=
2dk
jk
=
2sin k /2
jk 2 e− jk /2
, k≠0
for k=0, e0
is determine by finding the area under
one period of x(t) and dividing by the period. e0
=1/2
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3.6 Fourier Series Representation of Discrete-Time
Periodic Signals
Linear Combination of Harmonically Related
Complex Exponentials
The definition of discrete-time periodic signals
with period N
x[n]=x[nN ]
The fundamental period is the smallest positive
integer of N for which the equation above holds and
ω0
=2π/N
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Set of harmonically related complex exponentials
k=e
jk 0 n
=e
jk 2/N n
k=0,±1,±2,...
There are N distinct signals in the set. The
consequence is the discrete-time complex
exponential which differ in frequency by multiple
of 2π are identical.
k [n]=krN [n]
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x[n]=∑
k
ak e
jk 0 n
=∑
k
ak e
jk 2/N n
A linear combination of harmonically related
exponentials of the from
Since the sequence Φk
[n] are distinct only over a
range of N successive values of k. we can
express the limits of the summation as k = <N>.
x[n]= ∑
k=〈N 〉
ak e
jk 0 n
= ∑
k=〈N 〉
ak e jk2/ N n
k could take on values, for example, k = 1,2,...,N-1
or k = 3,4,...,N+2
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Determination of the Fourier Series
Representation of a Periodic Signal
x[n]= ∑
k=〈N 〉
ak e
jk 0 n
= ∑
k=〈N 〉
ak e jk 2/ N n
The equation of the discrete-time Fourier series
ak
is the Fourier series coefficients
If we evaluate the equation above for N
successive values of n corresponding to one
period of x[n] :
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x[0]= ∑
k=N
ak
x[1]= ∑
k= N
ak e
j2k/ N
x[N −1]= ∑
k=N
ak e
j2 k N −1/ N
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∑
n=〈 N 〉
e jk 2/ N n
=
{N , k=0,±N ,±2N ,...
0, otherwise
∑
n=〈 N 〉
x[n]e
− jr2/N n
= ∑
n=〈N 〉
∑
k=〈 N 〉
ak e
jk−r2/ N n
The basic result is
Multiplying both side of x[n] by e-jr(2π/N)n
and
summing over N terms
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∑
n=〈 N 〉
x[n]e
− jr2/N n
= ∑
k=〈 N 〉
ak ∑
n=〈 N 〉
e
jk−r2/ N n
ar=
1
N
∑
n=〈 N 〉
x[n]e
− jr2/N n
Interchanging the order of summation on right-
hand side
N , k−r=0,±N ,±2N ,...
∑
n=〈 N 〉
x[n]e
− jr2/N n
=ar N
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x[n]= ∑
k=〈N 〉
ak e
jk 0 n
= ∑
k=〈N 〉
ak e
jk2/ N n
ak=
1
N
∑
n=〈N 〉
x[n]e
− jk 0 n
=
1
N
∑
n=〈N 〉
x[n]e
− jr2/N n
Discrete-time Fourier Series pair
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Example 3.10
x[n]=sin 0 n
0=
2
N
Fine the Fourier series coefficient for the signal
when
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x[n]=
1
2j
e
j2/N n
−
1
2j
e
− j2/N n
a1=
1
2j
a−1=−
1
2j
Expanding the signal as a sum of two complex
exponentials
We see by inspection that
Solution
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These coefficient repeat with period N thus aN+1
also equal to 1/2j and aN-1
equals (-1/2j)
For example with N = 5
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Consider 0=
2M
N
Assuming that M and N do not have any common
factors
x[n]=
1
2j
e jM 2/N n
−
1
2j
e− jM 2/N n
By inspection aM
= (1/2j), a-M
= (-1/2j) and the
remaining coefficients over one period of length
N are zero
75. 241-306 Fourier Series Representation of Periodic Signals
75
For example with M = 3 and N = 5
76. 241-306 Fourier Series Representation of Periodic Signals
76
Example 3.11
x[n]=1sin2
N n3cos2
N n
cos4
N
n
2
Fine the Fourier series coefficient for the signal
77. 241-306 Fourier Series Representation of Periodic Signals
77
x[n]=1
1
2j
[e
j2/N n
−e
− j2/N n
]
3
2
[e
j2/N n
e
− j2/N n
]
1
2
[e
j4n/ N /2
e
− j4n/N /2
]
Expanding the signal as a sum of two complex
exponentials
Solution
78. 241-306 Fourier Series Representation of Periodic Signals
78
x[n]=13
2
1
2je
j2/ N n
3
2
−
1
2je
− j2/N n
1
2
e
j/2
e
j22/ N n
1
2
e
− j/2
e
− j22/N n
Collecting terms
The Fourier series coefficient are
80. 241-306 Fourier Series Representation of Periodic Signals
80
Fourier series coefficient for N = 10
81. 241-306 Fourier Series Representation of Periodic Signals
81
Fourier series coefficient for N = 10
82. 241-306 Fourier Series Representation of Periodic Signals
82
Example 3.12
Evaluate the Fourier series for the signal
83. 241-306 Fourier Series Representation of Periodic Signals
83
ak=
1
N
∑
m=0
2N1
e
− jk 2/N m−N1
=
1
N
e
− jk 2/N N1
∑
m=0
2N1
e
− jk 2/N m
ak=
1
N
∑
n=−N 1
N 1
e− jk 2/ N n
For this signal, the Fourier coefficient become
Let m = n+N1
84. 241-306 Fourier Series Representation of Periodic Signals
84
ak=
1
N
e
jk 2/ N N 1
1−e
− jk22N11/ N
1−e
− jk 2/N
=
1
N
e− jk 2/2N
[e
jk2N 11/2/N
−e
− jk2N11/2/ N
]
e
− jk 2/2N
[e
jk 2/2N
−e
− jk 2/2N
]
=
1
N
sin[2 k N11/2/ N ]
sin k / N
, k≠0,±N ,±2N ,...
ak=
2N11
N
, k=0,±N ,±2N ,...
and
85. 241-306 Fourier Series Representation of Periodic Signals
85
Plot of Nak
for 2N1
+1 = 5 (a) N = 10; (b) N = 20;
(c) N = 40
86. 241-306 Fourier Series Representation of Periodic Signals
86
Multiplication
FS
x[n] ↔ ak
FS
y[n] ↔ bk
x[n] y[n] ↔ dk= ∑
l=〈N 〉
al bk−l
FS
Let x(t) is a periodic signal with period T
3.6 Properties of Discrete-Time Fourier Series
87. 241-306 Fourier Series Representation of Periodic Signals
87
First Difference
FS
x[n] ↔ ak
FS
x[n]−x[n−1] ↔ 1−e− jk 2/N
ak
The first-difference operation define as x[n] -x[n-1]
88. 241-306 Fourier Series Representation of Periodic Signals
88
Parseval's Relation for Discrete-Time Periodic
Signals
1
N
∑
n=〈 N 〉
∣x[n]∣2
= ∑
k=〈 N 〉
∣ak∣2
Where the ak
are the Fourier series coefficients of
x[n] and N is the period
90. 241-306 Fourier Series Representation of Periodic Signals
90
Example 3.13
Find the Fourier series coefficient of the x[n] below
solution
By linearity property, x[n] = x1
[n]+x2
[n], ak
= bk
+ck
when ak
FS coefficients of x[n], bk
FS coefficients
of x1
[n], ck
FS coefficients of x2
[n]
92. 241-306 Fourier Series Representation of Periodic Signals
92
bk=
{
1
5
sin3k /5
sink /5
, for k≠0,±5,±10,...
3
5
, for k=0,±5,±10,...
c0=
1
5
∑
n=0
4
x2 [n]=1
From example 3.12, with N1
= 1 and N = 5
The sequence x2[n] has only DC value, which is
captured by its zeroth Fourier series coefficient.
93. 241-306 Fourier Series Representation of Periodic Signals
93
ak=
{
1
5
sin3k /5
sink /5
, for k≠0,±5,±10,...
8
5
, for k=0,±5,±10,...
The Fourier series coefficient of x[n] are :
94. 241-306 Fourier Series Representation of Periodic Signals
94
Example 3.14
Suppose we given the following facts about a
sequence x[n].
1 x[n] is periodic with period N = 6
2
3
4 x[n] has the minimum power per period
among the set of signals satisfying the
preceding three conditions.
∑n=0
5
x[n]=2
∑n=2
7
−1
n
x[n]=1
Find the sequence x[n]
95. 241-306 Fourier Series Representation of Periodic Signals
95
p=∑
k=0
5
∣ak∣
2
From fact 2, we conclude that a0
= 1/3
From fact 3, (-1)n
= e-jπn
= e-j(2π/6)3n
, we conclude
that a3
= 1/6
From fact 4, the average power in x[n] using
Parseval's relation is
96. 241-306 Fourier Series Representation of Periodic Signals
96
x[n]=a0a3 e
jn
=
1
3
1
6
−1n
The value of P is minimized by choosing
a1
=a2
=a4
=a5
=0
We can sketch x[n] :
97. 241-306 Fourier Series Representation of Periodic Signals
97
3.7 Fourier Series and LTI Systems
H s=∫−∞
∞
he
−s
d
H z= ∑
k=−∞
∞
h[k ]z
−k
From section 3.2, if x(t) = est
is the input to a
continuous-time LTI system, then the output is
given by y(t) = H(s)est
, where
If x[n] = zn
is the input to a continuous-time LTI
system, then the output is given by y[n]= H(z)zn
,
where
98. 241-306 Fourier Series Representation of Periodic Signals
98
Where s and z are complex number, H(s) and
H(z) are referred as the system functions of the
corresponding systems.
For continuous-time signals and systems, we
focus of the specific case in which Re{s} = 0, so
that s = jω. The system function is :
H j =∫−∞
∞
hte
− j t
dt
This system function is referred to as the
frequency response of the system
99. 241-306 Fourier Series Representation of Periodic Signals
99
H e
j
= ∑
n=−∞
∞
h[n]e
− jn
For discrete-time signals and systems, we focus of
the specific case in which |z| = 1, so that z = ejω
.
The system function is :
This system function is referred to as the
frequency response of the system
100. 241-306 Fourier Series Representation of Periodic Signals
100
xt= ∑
k=−∞
∞
ak e
jk 0 t
yt= ∑
k=−∞
∞
ak H jk 0e
jk 0 t
For continuous-time system, Let x(t) be a
periodic signal apply as the input to the
system:
The output of the system is
101. 241-306 Fourier Series Representation of Periodic Signals
101
x[n]= ∑
k=〈N 〉
ak e jk2/ N n
For discrete-time system, Let x[n] be a periodic
signal apply as the input to the system:
The output of the system is
y[n]= ∑
k=〈 N 〉
ak H e
jk2/ N
e
jk2/ N n
102. 241-306 Fourier Series Representation of Periodic Signals
102
Example 3.16
ht=e
−t
ut
Suppose that the periodic signal x(t) in example
3.2 is the input to LTI system with impulse
response
Find the Fourier series coefficient of output y(t)
103. 241-306 Fourier Series Representation of Periodic Signals
103
H j =∫
0
∞
e
−
e
− j
d
=−
1
1 j
e
−
e
− j
∣0
∞
=
1
1 j
solution
To calculate the Fourier series coefficients of
the output y(t), we first compute the frequency
response
104. 241-306 Fourier Series Representation of Periodic Signals
104
= ∑
k=−3
3
bk e jk2t
yt= ∑
k=−∞
∞
ak H jk 0e
jk 0 t
The output of the system is
xt= ∑
k=−3
3
ak e
jk2t
The input of the system is
105. 241-306 Fourier Series Representation of Periodic Signals
105
b0=1
b1=
1
4 1
1 j2, b−1=
1
4 1
1− j2
b2=
1
2 1
1 j4 , b−2=
1
2 1
1− j4
b3=
1
3 1
1 j6 , b−3=
1
3 1
1− j6
With bk
= ak
H(jk2π), so that
106. 241-306 Fourier Series Representation of Periodic Signals
106
yt=12∑
k=1
3
Dk cos jk2tk
yt=12∑
k=1
3
[Ek cos2kt−Fk sin2kt]
bk=Dk e
j
k
=Ek jFk , k=1,2,3,...
y(t) must be a real-valued signal, since it is the
convolution of x(t) and h(t) which are both real.
Therefore, y(t) can be expressed in the form
or
where
107. 241-306 Fourier Series Representation of Periodic Signals
107
D1=∣b1∣=
1
414
2
, 1=∡b1=−tan
−1
2
E1=ℜe {b1}=
1
414
2
,
F1=ℑm{b1}=−
214
2
This coefficients can be evaluated from bk
, for
example
108. 241-306 Fourier Series Representation of Periodic Signals
108
Example 3.17
x[n]=cos2n
N
Find the Fourier series coefficient of output y(t)
Consider an LTI system with impulse response
h[n] = αn
u[n], -1< α <1, and with the input
109. 241-306 Fourier Series Representation of Periodic Signals
109
x[n]=
1
2
e
j2/ N n
1
2
e
− j2/ N n
solution
x[n] can be written in Fourier series form as
Frequency response of system
H e
j
=∑
n=0
∞
n
e
− j n
=∑
n=0
∞
e
− j
n
110. 241-306 Fourier Series Representation of Periodic Signals
110
H e
j
=
1
1−e
− j
y[n]=
1
2
H e j2/N
e j2/N n
1
2
H e− j2/N
e− j2/N n
The summation of this series is
We obtain the Fourier series for the output :
111. 241-306 Fourier Series Representation of Periodic Signals
111
y[n]=
1
2 1
1−e
− j2/ N e
j2/N n
1
2 1
1− e
j2/ N e
− j2/N n
1
1−e− j2/ N
=r e
j
If we write
The output become
y[n]=r cos2
N
n
112. 241-306 Fourier Series Representation of Periodic Signals
112
1
1−e
− j2/4
=
1
1 j
=
1
1
2
e
j−tan
−1
y[n]=
1
1
2
cosn
2
−tan
−1
For example, if N = 4,
and thus