3. Frequency-Domain Analysis of Continuous-Time Signals and Systems.pdf

3. Frequency-Domain Analysis of Continuous-
Time Signals and Systems
3.1. Definition of Continuous-Time Fourier Series (3.3-3.4)
3.2. Properties of Continuous-Time Fourier Series (3.5)
3.3. Definition of Continuous-Time Fourier Transform (4.0-4.2)
3.4. Properties of Continuous-Time Fourier Transform (4.3-4.6)
3.5. Frequency Response (3.2, 3.8, 4.4)
3.6. Linear Constant-Coefficient Differential Equations (4.7)

3.1. Definition of Continuous-Time Fourier Series
A continuous-time signal x(t) with period T can be represented by
a continuous-time Fourier series, i.e.,
X(k) is called the spectrum of x(t).
(3.1) and (3.2) show that a continuous-time periodic signal can be
decomposed into a set of continuous-time elementary signals. Any
continuous-time elementary signal, X(k)exp(j2kt/T), is periodic and
has the frequency 2k/T and the coefficient X(k).
,
kt
T
2
j
exp
)
k
(
X
)
t
(
x
k









 

.
dt
kt
T
2
j
exp
)
t
(
x
T
1
)
k
(
X
T
 




 


where X(k) is given by
(3.1)
(3.2)

3.1.1. Derivation of Continuous-Time Fourier Series
Assume that x(t) can be represented by (3.1). We show that X(k) is
given by (3.2). Substituting k for k in (3.1), we obtain
.
t
k
T
2
j
exp
)
k
(
X
)
t
(
x
k










 



Next, (3.3) is multiplied by exp(j2kt/T), integrated over one period
and divided by T. That is,
(3.3)
 






 












 








 





T
k
T
.
dt
kt
T
2
j
exp
t
k
T
2
j
exp
)
k
(
X
T
1
dt
kt
T
2
j
exp
)
t
(
x
T
1
(3.4)
Changing the order of the integration and the summation on the right
side of (3.4), we obtain

.
dt
t
)
k
k
(
T
2
j
exp
T
1
)
k
(
X
dt
kt
T
2
j
exp
)
t
(
x
T
1
T
k
T



















 





(3.5)
Since
,
k
k
0,
k
k
,
1
dt
t
)
k
k
(
T
2
j
exp
T
1
T

















 (3.6)
the right side of (3.5) equals X(k) and (3.2) is derived.
3.1.2. Convergence of Continuous-Time Fourier Series
The integral in (3.2) converges when the following conditions are
satisfied.
(1) In any period, x(t) is absolutely integrable. That is, there exists
a finite constant B such that

.
B
dt
|
)
t
(
x
|
T

 (3.7)
(2) In any period, x(t) has a finite number of maxima and minima.
(3) In any period, x(t) has a finite number of discontinuities, and
has both the left-sided limit and the right-sided limit at each of these
discontinuities.
The above conditions are called the Dirichlet conditions. It should
be noted that they are sufficient for the convergence of the integral in
(3.2) but unnecessary.
Suppose that the Dirichlet conditions are satisfied and the integral
in (3.2) converges. Then, the series in (3.1) converges but may not
converge to x(t) everywhere. At a continuity, it converges to x(t). At
a discontinuity, it converges to the average of the left-sided limit and
the right-sided limit of x(t).
Example. Determine the Fourier series coefficients for each of the

following signals:
(1) x(t)=sin(0t).
(2) x(t)=1+sin(0t)+2cos(0t)+cos(20t+/4).
(3) Over period [T/2, T/2), x(t) is defined as
.
t
|
t
|
0,
t
|
t
|
,
1
x(t)
0
0






3.2. Properties of Continuous-Time Fourier Series
3.2.1. Linearity
Suppose that x1(t) and x2(t) have the same period, and a1 and a2 are
two arbitrary constants. If x1(t)X1(k) and x2(t)X2(k), then
.
)
nT
t
(
)
t
(
x
(4)
n








a1x1(t)+a2x2(t)a1X1(k)+a2X2(k). (3.8)
3.2.2. Differentiation
If x(t)X(k), then
).
k
(
kX
T
2
j
dt
)
t
(
dx 
 (3.9)
3.2.3. Shifting
,
kt
T
2
j
exp
)
k
(
X
)
t
t
(
x 0
0 




 


 (3.10)
where t0 is an arbitrary real number.
),
k
k
(
X
t
k
T
2
j
exp
)
t
(
x 0
0 






 
(3.11)

where k0 is an arbitrary integer.
3.2.4. Scaling
x(at)X[sgn(a)k], (3.12)
where a is a nonzero real number.
When a>0, (3.12) becomes
x(at)X(k). (3.13)
Thus, only X(k) cannot specify x(t) uniquely. T is also required.
Letting a=1 in (3.12), we obtain
x(t)X(k), (3.14)
i.e., the reversal property of the continuous-time Fourier series. From
(3.14), the following conclusions can be drawn:

(1) x(t) even  X(k) even.
(2) x(t) odd  X(k) odd.
3.2.5. Conjugation
x*(t)X*(k). (3.15)
From (3.15), the following conclusions can be drawn:
(1) Im[x(t)]=0  X(k)=X*(k).
(2) Re[x(t)]=0  X(k)=X*(k).
(3) Im[X(k)]=0  x(t)=x*(t).
(4) Re[X(k)]=0  x(t)=x*(t).
3.2.6. Convolution
Assume that x1(t) and x2(t) have the same period T. If x1(t)X1(k)

),
k
(
X
)
k
(
TX
d
)
t
(
x
)
(
x 2
1
T
2
1 




 (3.16)
,
)
m
k
(
X
)
m
(
X
)
t
(
x
)
t
(
x
m
2
1
2
1 




 (3.17)
where the integral is called the periodic convolution integral of x1(t)
and x2(t).
Assume that x1(t) and x2(t) have the same period T. If x1(t)X1(k)
and x2(t)X2(k), then
where the sum is called the convolution sum of X1(k) and X2(k).
3.2.7. Parseval’s Equation
.
)
k
(
X
dt
|
)
t
(
x
|
T
1
k
2
T
2





 (3.18)
and x2(t)X2(k), then

3.3. Definition of Continuous-Time Fourier Transform
A continuous-time signal x(t) can be represented by a continuous-
time Fourier integral, i.e.,
  ,
d
t
j
exp
)
(
X
2
1
)
t
(
x 







 (3.19)
(3.20) is called the continuous-time Fourier transform, and (3.19) is
called the inverse continuous-time Fourier transform. X() is called
the spectrum of x(t).
It can be seen from (3.19)-(3.20) that a continuous-time signal can
be decomposed into a set of continuous-time elementary signals. Any
continuous-time elementary signal, X()exp(jt)d/(2), is periodic
and has the frequency  and the coefficient X()d/(2).
  .
dt
t
j
exp
)
t
(
x
)
(
X 







where X() is given by
(3.20)

3.3.1. Derivation of Continuous-Time Fourier Transform
Assume that x(t) is x(t) extended with period T. Then,
.
kt
T
2
j
exp
d
k
T
2
j
exp
)
(
x
T
1
)
t
(
x
k
T
 








 



















 (3.21)
Letting T=2/, we obtain
Letting 0, we obtain
    .
t
jk
exp
d
jk
exp
)
(
x
2
1
)
t
(
x
k
2
 


 

 

















 (3.22)
    .
d
t
j
exp
d
j
exp
)
(
x
2
1
)
t
(
x  













 




 (3.23)
(3.23) shows that x(t) can be expressed as
  ,
d
t
j
exp
)
(
X
2
1
)
t
(
x 







 (3.24)

  .
dt
t
j
exp
)
t
(
x
)
(
X 







where
(3.25)
3.3.2. Convergence of Continuous-Time Fourier Transform
When the following conditions are satisfied, the integral in (3.20)
converges.
(1) Over (, ), x(t) is absolutely integrable. That is, there exists
a finite constant B such that
.
B
dt
|
)
t
(
x
| 




(3.26)
(2) Over (, ), x(t) has a finite number of maxima and minima.
(3) Over (, ), x(t) has a finite number of discontinuities, and
has both the left-sided limit and the right-sided limit at each of these
discontinuities.

The above conditions are called the Dirichlet conditions. It should
be noted that they are sufficient for the convergence of the integral in
(3.20) but unnecessary.
Assume that the Dirichlet conditions are satisfied and the integral
in (3.20) converges. Then, the integral in (3.19) converges but may
not converge to x(t) everywhere. At the continuities, it converges to
x(t), but at the discontinuities, it converges to the average of the left-
sided limit and the right-sided limit of x(t).
3.3.3. Examples of Continuous-Time Fourier Transform
The continuous-time Fourier transform can be used to represent
both aperiodic continuous-time signals and periodic continuous-time
signals. First we consider aperiodic continuous-time signals.
Example. Find the Fourier transforms of the following signals:
(1) x(t)=(t).

.
t
|
t
|
0,
t
|
t
|
,
1
x(t)
(2)
0
0






(3) x(t)=eatu(t), Re(a)<0.
(4) x(t)=eatu(t), Re(a)>0.
(5) x(t)=ea|t|, Re(a)<0.
Next we consider periodic continuous-time signals.
Example. Prove
exp(j0t)2(0). (3.27)
Example. Find the Fourier transform of x(t)=sin(0t).
Example. Assume that x(t) is a continuous-time signal with period
T, and X(k) is the Fourier series coefficient of x(t). Determine X(),
the Fourier transform of x(t).

3.4. Properties of Continuous-Time Fourier Transform
3.4.1. Linearity
If x1(t)X1() and x2(t)X2(), then
a1x1(t)+a2x2(t)a1X1()+a2X2(), (3.28)
where a1 and a2 are two arbitrary constants.
3.4.2. Differentiation
If x(t)X(), then
.
d
)
(
dX
)
t
(
jtx



 (3.30)
).
(
X
j
dt
)
t
(
dx


 (3.29)

Example. Evaluate the derivative of x(t) at t=0. Assume that the
Fourier transform of x(t) is given as follows:
.
0
),
0
(
)
0
(
X
0
,
j
)
(
X
)
(
Y
d
)
(
x
)
t
(
y
t


















  

(3.31)
Example. Let x(t)X(). Prove
.
1
|
|
,
0
1
|
|
5
.
0
,
5
.
0
|
|
,
2
/
)
(
X
)
1
(


















.
otherwise
,
0
1
0
,
j
0
1
,
j
)
(
X
)
2
(




















.
d
)
(
X
)
(
Y
0
t
),
0
(
)
0
(
x
0
t
,
jt
)
t
(
x
)
t
(
y 


















 (3.32)
Example. Let x(t)X(). Prove
3.4.3. Shifting
x(tt0)X()exp(jt0), (3.33)
where t0 is an arbitrary real number.
x(t)exp(j0t)X(0), (3.34)
where 0 is an arbitrary real number.
Example. A modulator is a basic unit in a communication system.

It converts a signal into another one which can be transmitted more
effectively and more efficiently. A sinusoidal amplitude modulator is
described as
r(t)=s(t)p(t). (3.35)
s(t) is called the modulating signal and bears the wanted information.
Since it is a low-frequency signal, s(t) cannot be transmitted
effectively and efficiently. p(t) is called the carrier signal. It is a high-
frequency sinusoidal signal. Here, we assume
p(t)=cos(0t). (3.36)
r(t) is called the modulated signal. Since it is a high-frequency signal,
r(t) can be transmitted effectively and efficiently. Let the spectrum of
s(t) be S(). Find R(), the spectrum of r(t).
Example. A demodulator recovers the modulating signal from the
modulated signal. In the last example, s(t) can be recovered from r(t)

by a sinusoidal amplitude demodulator, which is characterized by
g(t)=r(t)p(t) (3.37)
followed by a low-pass filter. Find G(), the spectrum of g(t), and its
expression after the low-pass filtering.
3.4.4. Scaling
,
a
X
|
a
|
1
)
at
(
x 




 
 (3.38)
where a is a nonzero real number.
Letting a=1 in (3.38), we obtain
x(t)X(), (3.39)
the reversal property of the continuous-time Fourier transform.

(1) x(t) even  X() even.
(2) x(t) odd  X() odd.
3.4.5. Conjugation
x*(t)X*(). (3.40)
(1) Im[x(t)]=0  X()=X*().
(2) Re[x(t)]=0  X()=X*().
(3) Im[X()]=0  x(t)=x*(t).
(4) Re[X()]=0  x(t)=x*(t).

3.4.6. Symmetry
X(t)2x(). (3.41)
Proof. Substituting  for t in
  .
d
j
exp
)
(
x
)
(
X 








 (3.43)
  .
d
jt
exp
)
(
x
)
t
(
X 







 (3.44)
  ,
dt
t
j
exp
)
t
(
x
)
(
X 






 (3.42)
we obtain
Substituting t for  in (3.43), we obtain
Substituting  for  in (3.44), we obtain (3.41).

(3.45)
Example. Find the Fourier Transform of x(t)=2/(1+t2).
3.4.7. Convolution
x1(t)x2(t)X1()X2(). (3.46)
This property is proved as follows:
 
 
 
 


















 












 



d
dt
t
j
exp
)
t
(
x
)
(
x
dt
t
j
exp
d
)
t
(
x
)
(
x
2
1
2
1
.
|
|
,
0
|
|
,
1
t
)
t
sin(
0
0
0












Example. Prove

 
).
(
X
)
(
X
d
j
exp
)
(
x
)
(
X
2
1
1
2








 



(3.47)
Example. Find x1(t)x2(t), where
(1) x1(t)=eatu(t), Re(a)<0, x2(t)=ebtu(t), Re(b)<0, and ab.
(3.48)
.
t
)
t
sin(
)
t
(
x
and
t
)
t
sin(
)
t
(
x
)
2
( 2
2
1
1






)].
(
X
)
(
X
[
2
1
)
t
(
x
)
t
(
x 2
1
2
1 




Example. Let
(3.49)
.
t
)
2
/
t
sin(
)
t
sin(
)
t
(
x 2



Find the Fourier transform of x(t).
3.4.8. Parseval’s Equation
.
d
|
)
(
X
|
2
1
dt
|
)
t
(
x
| 2
2











 (3.50)
This property is proved as follows:
 
 
.
d
|
)
(
X
|
2
1
d
)
(
X
)
(
X
2
1
d
dt
t
j
exp
)
t
(
x
)
(
X
2
1
dt
d
t
j
exp
)
(
X
2
1
)
t
(
x
dt
)
t
(
x
)
t
(
x
dt
|
)
t
(
x
|
2
*
*
*
*
2


 
 









































 
















(3.51)

Example. Evaluate the integral of |x(t)|2 over (,). Assume that
the Fourier transform of x(t) is given as follows:
3.5. Frequency Response
A linear time-invariant continuous-time system can be described
by the frequency response, which is defined as the Fourier transform
of the impulse response.
.
1
|
|
,
0
1
|
|
5
.
0
,
5
.
0
|
|
,
2
/
)
(
X
)
1
(


















.
otherwise
,
0
1
0
,
j
0
1
,
j
)
(
X
)
2
(




















3.5.1. Response to exp(j0t)
Let the input of a linear time-invariant continuous-time system be
x(t)=exp(j0t). Then, the output of the system will be
y(t)=exp(j0t)H(0), (3.52)
where H() is the frequency response of the system.
Proof. Let h(t) be the impulse response of the system. Then,
 
).
(
H
)
t
j
exp(
d
)
j
exp(
)
(
h
)
t
j
exp(
d
)
t
(
j
exp
)
(
h
)
t
(
y
0
0
0
0
0
























(3.53)
When the input is a weighted sum of signals with form exp(j0t),
the output can be determined according to (3.52) and the linearity of
the system.

3.5.2. Response to a Periodic Signal
Let the input of a linear time-invariant continuous-time system be
periodic. Then the output has the same period. The relation between
the input and the output can be expressed as
.
k
T
2
H
)
k
(
X
)
k
(
Y 




 
 (3.54)
X(k) and Y(k) are the Fourier series coefficients of the input and the
output, respectively. H() is the frequency response of the system. T
is the period.
Proof. x(t) can be expressed as
.
kt
T
2
j
exp
)
k
(
X
)
t
(
x
k









 
 (3.55)
According to (3.52) and the linearity of the system, we obtain

,
kt
T
2
j
exp
k
T
2
H
)
k
(
X
)
t
(
y
k









 





 
 (3.56)
and thus (3.54) is derived.
3.5.3. Response to a General Signal
The I/O relation of a linear time-invariant continuous-time system
can be expressed by the frequency response, i.e.,
Y()=X()H(), (3.57)
where X() and Y() are the Fourier transforms of the input and the
output, respectively, and H() is the frequency response.
Proof. x(t) can be expressed as
  .
d
t
j
exp
)
(
X
2
1
)
t
(
x 







 (3.58)
According to (3.52) and the linearity of the system, we obtain

  ,
d
t
j
exp
)
(
H
)
(
X
2
1
)
t
(
y 








 (3.59)
and thus (3.57) is derived. Also, (3.57) can be directly derived from
the convolution property of the Fourier transform.
Example. Let a linear time-invariant continuous-time system have
the input x(t)=cos(0t) and the frequency response H()=exp(2).
Find the output.
Example. A distortionless transmission system is described as
y(t)=Ax(tt0). (3.60)
Find the frequency response.
Example. The phase of the frequency response is referred to as the
phase response. The minus derivative of the phase response is called
the group delay. Show that a distortionless transmission system has a
constant group delay.

Example. An ideal low-pass filter has the frequency response
.
|
|
0,
|
|
,
1
)
H(
0
0










 (3.61)
Find the impulse response.
Example. Find the responses of the above system to the following
excitations:
Example. An ideal filter is best in frequency selectivity. However,
it is noncausal, and its impulse response is oscillatory. These defects
.
2
,
2
period
over
/
|
t
|
,
0
/
|
t
|
,
1
x(t)
)
1
(
0
0
0
0






















.
t
)
t
2
sin(
)
t
(
x
)
2
( 0




can be overcome by a non-ideal filter. The frequency response of a
typical non-ideal low-pass filter is
,
j
)
(
H
0
0





 (3.62)
Example. A band-pass filter can be implemented using a low-pass
filter, as shown in figure 3.1. Explain how it works.
Figure 3.1. Implementation of a Band-Pass Filter.
x(t)
e(t) r(t)
y(t)
exp(j0t) exp(j0t)
Low-Pass
Filter

3.6. Linear Constant-Coefficient Differential Equations
The zero-state response of a linear constant-coefficient differential
equation can be found using the Fourier transform.
Example. A causal, stable continuous-time system is given by
),
t
(
x
)
t
(
y
2
dt
)
t
(
dy

 (3.63)
where x(t)=etu(t) and y(0)=2. Find the zero-input response, the
zero-state response and the complete response.
First, using the method in section 2.5, we can obtain the zero-input
response
yzi(t)=2e2t. (3.64)
Then, let us consider the zero-state response. The zero-state response
satisfies (3.63), i.e.,

).
t
(
x
)
t
(
y
2
dt
)
t
(
dy
zs
zs

 (3.65)
Taking the Fourier transform of (3.65), we obtain
.
1
j
1
)
(
Y
2
)
(
Y
j zs
zs






 (3.66)
From (3.66), we obtain
.
1
j
1
2
j
1
)
(
Yzs







 (3.67)
The inverse Fourier transform of Yzs() is
yzs(t)=(e2t+et)u(t). (3.68)
The complete response is the sum of the zero-input response and the
zero-state response, i.e.,
y(t)= 2e2t+(e2t+et)u(t). (3.69)

Example. A stable continuous-time system is given by
).
t
(
x
)
t
(
y
2
dt
)
t
(
dy

 (3.70)
The zero-state response satisfies (3.70), i.e.,
).
t
(
x
)
t
(
y
2
dt
)
t
(
dy
zs
zs

 (3.71)
Taking the Fourier transform of (3.71), we obtain
.
2
j
1
)
(
X
)
(
Yzs





(3.72)
Thus,
.
2
j
1
)
(
H



 (3.73)

The inverse Fourier transform of H() is
h(t)=e2tu(t). (3.74)

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