# SIGNALS AND SYSTEMS

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```					             Chapter 7 Fourier analysis for Continuous-Time systems

In preceding chapter, we have discussed the linear time-invariant Systems and have seen
that linear time-invariant systems can be described by constant coefficient differential
equation and may be characterized by its impulse response. So that, the linear time invariant
system may be analysis in time domain with both of the convolution operation and classical
Solving method.
In this chapter，we will discuss the Analysis for continuous Time invariant system in
frequency domain.
7.1 System Analysis by Fourier Transform
1. Principle of the Analysis in Frequency domain
The impulse response of the system is the solution to the differential equation when

forcing signals a Dirac delta function. Tat is, r t   ht  , when et    t  . This impulse

response may be used to obtain the system output when input is not an impulse, In this case,

the r t  can obtained by convolution operation,

r t   et  * ht                                                    7.1-1

that is , the impulse response can be used to characterize the response of the system in
time domain. The response of the system in frequency domain can also be obtained by taking
Fourier Transform of both sides of eq7.1-1. Using convolution theorem, we get

R   E    H                                                    7.1-2

Where              R   F r t 

E    F et 

H    F ht 

H   is said to be the transfer function or frequency response of the system.

Of course, the transfer function H   is, in general, a complex quantity and can be

written in polar form

H    H  e jh  

Where H   is known as amplitude-frequently response and  h   is known as

203
phase-frequently response. Since ht  is a real function of time, so that H  f  is an even

function of frequency and  h   is an odd function of frequency.

We can, also writing 7.1-2 as follow

R  e jr    E  e je    H   e jh  

or                 R   E    H                                             7.1-3

r    e    h                                         7.1-4

eq7.1-3, 7.1-4 implied that output spectrum, both Amplitude-frequently and phase-frequency,
has been modified by transfer function of the system. The input to system spectrum has

changed into output spectrum. And the out Amplitude spectrum is weighted by H   . The

out phase spectrum is weighted by  n   .

2. Zero-state Response
As we discussed in chapter 6, a particular important and useful class of continuous-time
LTI (Linear-Time Invariant) systems are those for which the input and output satisfy a linear
constant-coefficient differential equation of the form
n
d k r t  m    d k et 

k 0
ak
dt k

 bk
k 0  dt k
7.1-5

From               R   E   H                                               7.1-6

or, equivalently

R 
H                                                             7.1-7
E  
Here, of course, we are assuming that these three Fourier transforms all exist. Next,
consider applying the Fourier transforms to both sides of eq7.1-5 to obtain

n    d k r t       m d k et 
F  ak              F  bk
k 0  dt k         k 0 dt k  
From the linearly properly this becomes
n
 d k r t  m     d k et 
 ak F  dt k    bk F  dt k 
k 0               k 0           
and from the differentiation property
204
n                           m


k 0
ak  j  R  
k
 b  j  E  
k 0
k
k

or, equivalently,

 n        k         m         k
    
R 　 ak  j    E  　 bk  j  
 k 0      

 k 0      

Thus, from eq7.1-7 get
m

R 
 b  j  k
k

H                k 0
7.1-8
E           n

 a  j 
k 0
k
k

From eq7.1-8 we observe that H   is a rational function, that is, it is a ratio of

polynomials in  j  . The coefficients of the numerator polynomial are the same coefficients

as those that appear on the right hand side of differential equation, and the coefficients of
denominator polynomial are the same coefficients as appear on the left side of differential
equation. Thus, we see that the frequency response given in eq7.1-8 for the LTI system
characterized by eq7.1-5 can be written down directly by inspection.
Example 7.1-1
Consider the LTI system that is initially at rest and that is characterized by
drt 
 a r t   et 
dt
with a  0 , from eq7.1-8, the frequency response is

H   
1
j  a

inspection of H   ，we see that is the Fourier transform of e -atu t  . Thus, the impulse

response of the system is recognized as

ht   e -atu t 

Example 7.1-2
Consider an LTI system initially at rest that is characterized by the differential equation
d 2 r t     dr t              det 
2
4          3r t           2et 
dt           dt                  dt
From eq7.1-8 the frequency response is

205
H   
 j   2
 j 2  4 j   3
To determine the corresponding impulse response, we use the method of partial fraction

to obtain the following expressions for H  

j  2
H   
A1     A2
      
 j  1 j  3 j  1 j  3
j  2                      1
and                A1                             
j  3        j  1
2

j  2                      1
A2                            
j  1        j  3
2

The inverse transform of each term can be recognized by inspection with the result that

1               
ht    e  t  e  3t  u t 
1
2       2       

In addition, if the Fourier transform E   of input to such a system is also of

polynomials in j , then so is R   E   H   . In this case we can use the same

technique to solve the differential equation, in other word, we can use the same technique to

find the r t  the system response for arbitrary input et  . That is, to find the response r t 

to the input et  , we shall illustrate it in the next examples.

Example 7.1-3
Consider the system that as same as Example r  4r  3r  e  2e and suppose
that the input is

et   e t u t 

Then, we have that
j  2
R   H    E   
1

 j  1 j  3 j  1
In this case we seek a partial fraction expansion of form

R  
A11       A12        A2
           
j  1  j  12
j  3

206
A11 

d  j  1 R 
2
                  
j  3  j  2

1
dj                             j  1
 j  32                 j  1
4

j  2
A12   j  1 R 
1
                            
2
j 1              j  3      j 1
2

j  2                         1
A2   j  3R  j  3                                                              
 j  12       j  3
4

R  
1                   1                      1
So that                                 4
           2
           4
j  1              j  1      2
j  3

The inverse transform is then found to be

 t        1
4
e t  1 te t  1 e 3t u t 
2         4

Example 7.1-4: The RC lower pass filter is illustrated shown in figure 7-1 and input is

et   Eut -ut-  . Find the vc t  on the capacitor C, where the circuit is at rest.

Figure 7-1

Solution: (1) find the E  

E    F E u t   u t   
           1                 1   j 
 E     
                   e      
          j  
               j 
       

E
j
1  e j              
(2) find the H  

The H   can derived directly from the frequently model as illustrated as follow

Figure 7-2
207
      1 
we see:              E    I    R 
         
     jc 


R   I   
1
jc
1     1
R     jc                          
H   
1       RC
by using                                                     
E   R   1    jRc  1 j   1   j  
jc                 RC
1
where                
RC
(3) find the Vc  

Vc    E   H    E


1
j   j

1  e  j      
        A   A2
here                              1 
 j    j j j  

A1                             1
j       j  0

                      
A2                                  1
j   j  
a
so that
 1      1 
Vc    E 
 j  j    1  e

 j
             
             
 1                                 
E            
1  e j 
1

1  e j                     
 j              j              
(4) find the uc t 

from above we see

E u t   u t   — E                             
F      1
1  e  j
j

so that             E
1
j  

 E u t  e t  u t    e  t  
                                                 
therefore


uc t   E u t   u t     E u t e  at  u t    e  t     
208
                               
 E 1  e at u t   1  e  t   u t                        
3. Response of System encouraged by Periodic Signals

Assuming et  is a periodic signal with T1 period. We represent (one of) first period

of et  as


et 
T1
t 
e1 t                              2
 0
                      otherwise

and               E1    F e1 t 

then, et  expressed as exponential Fourier series


et     c e
n  
n
jn 1t

so that
             
E    F   cn e jn1t 
 n         

 2            c    n 
n  
n                     1

E1 n1  thus
1
because cn 
T1


2
E   
T1            E n     n 
n  
1         1                   1


 1        E n     n 
n  
1         1                   1

by using of R   H   E   get

R   H   E  
                                                            7.1-9
 1        H n  E n    n 
n  
1       1       1               1

The response of the system, then, let we take the inverse of Fourier Transform to both
sides of eq7.1-9
209
 t   F 1R 

 1  H n1  E1 n1  F 1   n1 
n  

1 
        H n1  E1 n1  e jn1t
T1 n  
Example 7.1-5 Find the response of the lower pass filter that the input Signal is periodic

triangular pulse with T1 as the period. The input et  and low pass filter is shown in figure

7-3 (a) and (b) respectivilly.

Figure 7-3

Solution: (1) from former example the frequency response of the low pass filter is

H   
j  
1
where      
RC
ET1 2  T1 
(2) E1    F e1 t             Sa      
2     4 
1 
(3)  t          
T1 n  
H n1  E1 n1  e jn1t

1                 ET 2  n T 
                       1 S a  1 1  e jn1t
T1 n   jn 1   2       4 
                                              n 1  
E                                n  j  n1t  artg
2                       

1                                    
                                 Sa      e
2   n       n1   
2    2     2 
7.2 Distortion-less Transmission
In communication systems a distortion-less channel is often desired. This implies that the
channel output is just proportional to a delayed version of the input

r t   Ket  td                                              7.2-1

210
where K is the gain (which may be less than one) and t is the delay.

The corresponding requirement in the frequency domain specification is obtained by
taking the Fourier transform of both sides of (7.2-1)

R   KE   e  j t d                                         7.2-2

Thus, for distortion less transmission, we require that the transfer function of the
channel be given by

R 
H               Ke  j t d
E  
This implies that for no distortion at the output of a linear time-invariant system, two
requirements must be satisfied:
(1) The amplitude response is flat. That is

H    constamt  K                                             7.2-3(a)

(2) The phase response is a linear function of frequency. That is,

     td                                                      7.2-3(b)

The two requirements can figured as bellow

Figure 7-4

When the first condition is satisfied, it is said that there is no amplitude distortion. When
the second condition is satisfied, there is no phase distortion. For distortion-less transmission
both conditions must be satisfied.
The second requirement is often specified in an equivalent way using the time delay.
Define the time delay of the system by
d  
Td                                                                  7.2-4
d
for distortion-less transmission, the time delay is
d  
Td               t  constant
d

211
If Td  is not a constants as a function of frequency, there is phase distortion because the

phase response    is not a linear function of frequency.

7.3 Ideal Low pass Filters
An ideal filter is a system that transmits without distortion all the frequencies in a
certain band. The amplitude spectrum over the band is a constant and the phase spectrum
over the band is linear. Figure 7-5 depicts the filter characteristics for the ideal low pass and
band pass filters

Figure 7-5

1. Ideal low pass filter and impulse Response

The ideal low-pass filter is define by the transfer function H   given by

e  j t 0 ;    m

H                                                                 7.3-1
 0;
               m

The frequency  m is often called the cutoff frequency of the fitter. We can find the

impulse response of this system by taking the inverse Fourier transform of 7.3-1. Thus

ht   F 1 H  

H  e j t d
1

2        


e j t t 0 d
1

2        

m
cos t  t0  d
1


0

sin  t  t0  0 m
1   1                    

 t  t0

212
sin m t  t0 
2   1

 t  t0
m
      s  t  t0                                                   7.3-2
 a m
A sketch of the impulse response is shown in Figure 7-6. We conclude from the

impulse-response function that the peak value of the response,  m  , is proportional to the

cut-off frequency. Notice that

Figure 7-6

as  m   the filter becomes an all-pass filter and the output response peak   . In

other words, the output response approaches the input, an impulse. That is
m
lim             s  t  t0    t  t0                                     7.3-3
 m          a m
    m
so that                       s  t  t0  dt  1
    a m

or                         sa x  dx                                                            7.3-4


Again, the impulse response is not causal, because of ht   0 , t  0 . Why are the ideal

low pass filter non causal？They have finite bandwidths and linear phase characteristics.
2. Constraint Condition for physically realizable filter.

It can be shown that, for all H   satisfying                       H   d   , a necessary and
2


sufficient condition on the magnitude spectrum for a filter to be physically realizable is that

    ln H  
        1 2
d  

This is called the Paley-Wiener criterion. From this result, we can conclude that for a

realizable filter, the magnitude of H   may not fall off toward zero faster than a function

213
k                      k 2
of simple exponential order (for example, e                   , but not as e            , and the attenuation may

not be infinite over any band of frequencies of nonzero width.

For example, let H    e  , then
2

 ln H             2             11
2

  1   2 d    1   2 d     2  1 d
         1 
  1  2         d  

   1

So that H    e 
2
cannot realize. Therefore we often attempt to built realizable

filters to approximate the ideal filter characteristic as closely as possible.
Example 7.3-1, A realizable Low pass filter can be built with RLC Network depict in Figure
7-7.

Figure 7-7
Solution:
Applied model of frequency in above circuit, we have
1
R         j c  1
H                       R
E   jL     1
1
R
 j c
1

 j 2
LC  j
L
1
R
L      L   L
Assume R          then        LC
C      R    L
C

1
Let  c        then
LC

H   
1                        1
                                                 7.3-5
 j 
2
                   2      

   1 j 
                       1      j
      c
 c          c                c 


214
H   
1
7.3-6
2                       2
      2 
 
1       
     2   
     c    c


c
     tg1              2
7.3-7
 
1  
 
 c
The amplitude-frequency and phase-frequency response are depicted in Figure 7-8

Figure 7-8

The impulse response take can inverse of Fourier transform to Eq. 7.3-5. We first

change H   slightly.

H   
1                              1
2
           2
                              S

S
1
j  j
        1
c                    c      2
c
 c                                                           j  s

c     2
c      2
                  
S  c  c                     2 2 
2                             2

2

 S  c   c    
     2        4 

3                        3
2
c 2             2 c
c
              2                       2
2                           2
c   3                   c   3 
2                           2
3                        3 
S        c           S        c 
   2   2

             2   2 



then the

215
                                                           
                                                3
c       
1  2c                                                       
ht   F                                                 2
2
   3  
2
 3                                  
j  c    c  


                                            2   2


 
2c  2c t
ct u t 
3
       e sin
3         2
The impulse response of the RLC network is depicted in Figure 7-9

Figure 7-9

3. Step Response of the Ideal low pass filter
The concepts of impulse function and impulse response are valuable analytically, but
more practical system test consists in applying a unit step to the input of a system and
observing its response. Using the results obtained above, the step response follows using the
convolution property:
g t   ht * ut 

 h  d
t
                                                                  7.3 - 8


Combining Eq 7.3-2, and 7.3-8, we have
m
g t                               Sa  m   t0  d
t

              

Changing variable, let x   m   t0  so that

 m t  t 0 
g t                                       Sa x  dx
1
          

Using the fact that

Sa x  dx 
0
                                      2
we have
 m t  t 0 
g t                                                Sa x  dx
1 1

2                     0
7.3-9

216
S a  x  dx is a tabulated function known as the sine-integral function
y
The integral        0

and denoted by

Si  y              S a  x  dx
y
   0

Form the properties of the Sa x  function, we can deduce the fallowing:

(1) Si  y  is an odd function: that is, Si  y    Si  y  ;

(2) Si 0  0 ;

                              
(3) Si           , Si     
2                              2
A sketch of Si  y  is shown in Figure 7-10 (a) Using the Si  y  the step response

g t  is shown in Figure 7-10 (b)

Figure 7-10

Several important observations can be made from Figure 7-10 (b). Note that as the

bandwidth W of LPF is decreased, the filter output g t  rises to its peak value more slowly.

A measure of the time it takes for this rise is called the “rise time” the filter. A convenient
of
choice for the LPF is the time required for the output ware form to rise from its minimum to

2       1
its maximum. From Fig 7-10 (b), tr                            
m       fm

217
Example 7.3-2 H1   and H 2   are distinct system they are all the LPF and expressed,

individually, as

H1    u  1   u  1 

H 2    u   2   u   2 

Where  2  1

find the impulse response of cascade system with H1   and H 2   as follow form as

Figure 7-11 shown

Figure 7-11

Solution: from Figure 7-11 we have

E   E H1   H 2    R 

or                 H    H 2    H 2  H1  

because  2  1 , so that H 2  H1    H1   thus

H    H 2    H1  

therefore          ht   F 1H 2    H1  

2             
      Sa  2t   1 Sa 1t 
              
Example 7.3-3 The H   and input et  given as shown as Figure 7-12 find the response

of the system.

218
Figure 7-12

Solution:

The et  can be expanded to Fourier-Series form

4                                     
et  
1            1
sin 1t  3 sin 31t  5 sin 51t  
                                     

2                                                8       2
Where 1         consider the bandwidth of the system m          4       41 . So that
T                                                T        T
only 1th and 3th harmonic can going through the LPF. Thus, the response of the LPF is

4                                      
r t      sin 1 t  t0   3 sin 31 t  t0 
1
                                      

219
Problems

7-1 Find the zero state response by use of Fourier analysis method. The system function and
encourage signal is given as following.

H              　　　　 et   e  3t u t 
1
j  2
7-2 Find the frequency response and the impulse response of the systems having the output

r t  for the input et  .

(a) et   e u t 　　　　 r t   e
t
   2 t

 e 3t u t 

(b) et   e u t 　　　　 r t   e                      u t  2
3t                             3t  2 

7-3 Determine the frequency response and the impulse response for the systems described by
the following differential equation.
dr t 
(a)            3r t   et 
dt
d 2 r t     dr t               det 
(b)         2
5          6r t  
dt            dt                   dt
7-4 In order to distortion-less transmission for given system (or circuit) determine the relation

of R1 , R2 and C1　 C2 .
,

220
7-5 The system function of ideal low-pass filler is

      2

1　　  
                               
H              and the E    Sa    . Determine the response r t  by
0 　　  2

2

        

using the time convolution theorem. r t   et   ht  .

7-6 The system is depicted in Fig 7-1.

Fig 7-1

e
Where H i   is ideal Low-pass filter let　t   u t  determine the r t  .

e  jt 0 　　　  1
             
H i    
0　　　　  1
            
7-7 Find the system function and the impulse response for given systems.

7-8 Find the response of the system that depicted in Fig 7-2. the input

et                                             S t   c o 1 0 0 t
sin 2t

cos1000t　　　  t                            s     0
2t

Fig 7-2

221

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