Department of Electronics & Communication Engineering

MODULE II

Gibbs phenomenon
Gibbs discovered that for a periodic signal with discontinuities, if the signal is reconstructed by
adding the Fourier series, overshoot appears around the edges. These overshoots decay outward in a damped
oscillatory manner away from the edges. This is called Gibbs phenomenon and is illustrated in figure.

Figure: Illustration of Gibbs phenomenon

The overshoots at the discontinuity according to Gibbs are found to be around 9 percent of the height
of discontinuity irrespective of the number of terms in the Fourier series. As more number of terms in the
series are added, the frequency increases, overshoots gets sharper, but the adjoining oscillation amplitude
reduces. That is, the error between the original signal x(t) and the truncated signal xn(t) reduces except at the
edges as n increases. Thus the truncated Fourier series approaches x(t) as the number of terms the
approximation increases.

EC202 Signals & Systems 1 Prepared by: Dr. Giby Jose

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MODULE II
FOURIER REPRESENTATION OF CONTINUOUS TIME SIGNALS
Continuous Time Fourier series
1. Find the trigonometric Fourier series for the periodic signal x(t) shown in figure below:

2 2
From figure, we find that T = 2 and 0     . For convenience take integration
T 2
interval from t = -1 to t = 1
During this interval x(t) = t.
We have
 
x t   a 0   a k cos k 0 t    b k sin k 0 t 
k 1 k 1
 
 a 0   a k cos k  t    b k sin k  t 
k 1 k 1
1
1 t2 
1
1 1 1
a 0   x t  dt   t dt   
1 1
    0
TT 2 1 2  2  2 2 2
1
1
a k   x t  cos k 0 t  dt   t cos k  t  dt
2 2
TT 2 1

 t sin k  t   cos k  t 
1 1
    2 2  000
 k  1  k   1

1
x t sin k 0 t  dt   t sin k  t  dt
2 2
bk 
TT 2 1

  t cos k  t    cos k  t    2 cos k 

1 1
   2 2   0
 k  1  k   1 k

2   1k 
  
  k 
Therefore, the trigonometric Fourier series,
 
x t   a 0   a k cos k 0 t    b k sin k 0 t 
k 1 k 1

2   1k 
   sin k 0 t 
k 1  
 k 
2 
 sin  t   sin 2  t   sin 3  t   sin 4  t     
1 1 1
 2 3 4 

ECT 204 Signals and Systems 1 Prepared by: Dr. Giby Jose
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2. Determine the FS coefficients for the signal x(t) depicted in Fig.

T =; x(t) = e-2t
2 2
0   
T 2
2

 xt  e
1  j0 k t 1
ck  dt  e
2t
e  j0 k t dt
T T
2 0

 e  2  j0 k  t 
2 2
1  2  j0 k  t 1
 e dt   
2 0
2   2  j0 k   0



e  2  j0 k  2  e  2  j0 k  0 e  4  2 j 0 k  1


2   2  j 0 k   4  2 j 0 k 


e  4  2 j k
 
1 e  4  2 j  k 1


 4  2 j  k    4  2 j  k 



 1 e  4  2 j k

 
1 e  4  e  2 j k 
  4  2 j  k  4  2 j  k


1 e 4
cos 2  k  jsin 2  k  1 e  4 1 j 0
 4  2 j  k  4  2 j  k
1 e  4

4  2 j k
3. Determine the FS coefficients for the signal defined by

The fundamental period is T = 4, and each period of this signal contains an impulse.
2 2 
T = 4; 0   
T 4 2
x(t) = δ(t)
4 
j kt
 xt  e dt   t  e 2 dt
1  j0 k t 1
ck 
T T
4 0

Using sampling or sifting property of impulse function,

4
t  dt  1 
1 1 1
ck  
4 0 4 4

Since integral over any period of δ(t) = 1.

ECT 204 Signals and Systems 2 Prepared by: Dr. Giby Jose
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In this case, the magnitude spectrum is constant and the phase spectrum is zero.
4. Determine the FS representation of the signal x(t) = 3 cos(πt/2 + π/4). Also plot magnitude
and phase spectra.
x(t) = 3 cos(πt/2 + π/4)
Whenever x(t) is given as a sinusoid, always try to express it in terms of complex
exponential.
 j  t     
 j t   
  t   e  2 4   e  2 4  
x t   3 cos    3 
 2 4  2 
 

   
3 j t j j t j 
x t   e 2  e 4  e 2  e 4  (1)
2 
 
Generalized Fourier series representation is

x t    C k e jk t 0
(2)
k  

Compare equation (1) and (2), we get,

 
3 j j 0 k t
j t
C k  e 4 ; k  1; e  e 2
2
 
3 j j 0 k t
j t
C k  e 4 ; k   1; e  e 2
2
3 j
 e 4 ; ; k 1
2
 
3 j 4
Ck   e ; ; k 1
2
0; otherwise




Figure: Amplitude and phase spectra.

5. Determine the FS representation of x(t) = 2sin(2πt - 3) + sin(6πt).
x(t) = 2sin(2πt - 3) + sin(6πt)
 e j 2  t  3   e  j 2  t  3   e j 6  t  e  j 6  t
 2 
 2j  2j

ECT 204 Signals and Systems 3 Prepared by: Dr. Giby Jose
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1
j
  
1
 e j 2  t  e  3 j  e  j 2  t  e 3 j  e j6  t  e  j6  t
2j

1 1 j j j
   2   j
j j j j 1


x t    j e j 2  t  e  3 j  e  j 2  t  e 3 j   2j e j6  t
 e  j6  t 
x t    j e j 2  t  e  3 j  j e  j 2  t  e 3 j  e j 6  t  e  j 6  t
j j
(1)
   2
signal with   2   2 
0
signal with 0  6 

2  2 2 2  1
T1   1 T2   
 0 2  0 6 3
Combined time period, T = LCM of 1, 1/3 = 1
1 LCM of numerator LCM 1,1 1
LCM of 1,    1
3 HCF of deno min ator HCF 1, 3 1
2 2
0    2
T 1

x t    C k e jk t 0

k  

x t    C k e jk 2  t (2)
k  

Comparing equation (1) and (2), we get

For k 1; Ck   je3 j
k  1; C k  j e 3 j
j
k  3; Ck 
2
j
k   3; Ck 
2
j
2 ; k3
 3j
 je ; k  1
Ck  
3 j
 j e ; k 1
 j
 ; k 3
2

6. Find the FS coefficients of the full-wave rectified cosine depicted in Fig.

ECT 204 Signals and Systems 4 Prepared by: Dr. Giby Jose
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Here T = 1
2 2
0    2
T 1

x t   cos  t ;   t
1 1
2 2

x t  e  j 0 k t dt
1
Ck 
TT
1
cos  t e  j 2  k t dt
1 T


d
d  ea x 
e
ax
cosb x    2 a cos b x  b sin b x
 a  b
2
c  c
1
 e  j2  k t 2
Ck    j 2  k cos  t   sin  t 
  j 2k   
2 2
  1
2

1   jk     jk        
  e   j 2  k cos   sin   e 
  j 2  k cos      sin    
 4 2k 2  2   2 2   2  2  

 e 0    e j k  0   
1  jk 

 14k
2
 2

  e 
1  jk 
   e jk 
 14k
2
 2

 e    
  jk   2  jk 
  e jk    e jk 

2 1  4 k 2 
2 1  4 k 2 2
e

2 cos  k 2 1k


 14k2    1  4 k  2

Another method

x t  e  j 0 k t dt
1
Ck 
TT
1
2
1
  cos  t e  j 2  k t dt
1 1

2
1 1
 e j t  e  j t   jk 2  t
 
2
 e 1 2 j t  jk 2  t
  dt   e e  e  j  t e  j k 2  t dt
 2  2 1
1   
2 2
1 1
1  e j   k 2  t e  j   k 2  t 
 
2 2
e j   k 2  t  e  j   k 2  t dt  
1
  
2 2  j   k 2   j   k 2 
1 1
2 2

ECT 204 Signals and Systems 5 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

 j   k 2   2  j   k 2    j   k 2   j   k 2   
1 e e 2 e 2 e 2

   
2 j   k 2   j   k 2  j   k 2   j   k 2 

 
 j   k 2   2  j   k 2    j   k 2   j   k 2   
 
  
1 e e 2 e 2 e 2

2 j   k 2   j   k 2  
 

 j   k 2   2  j   k 2   
j   k 2    j   k 2   e  e 2 
e  
e 2 2
 
 
2 j   k 2   2 j   k 2 
sin(90 – θ) = cosθ;
sin(90 + θ) = cosθ
sin   2  k  sin   2  k  cos 2  k cos 2  k
 2 2 2 2  cos  k  cos  k
2k 2k 2k 2k 2k 2k


  2  k  cos  k    2  k  cos  k 
 cos  k  2  k cos  k   cos  k  2  k cos  k
  2  k   2  k    2  k   2  k 
2  cos  k 2  cos  k 2 cos  k 2  1k
   
2  4 2 k 2 
2 1  4 k 2   1  4 k   1  4 k 
2 2

k jk 
1
7. Find the time-domain signal x(t) corresponding to the FS coefficients C k    e 20 .
2
Assume that the fundamental period is T = 2.
This problem is an example for inverse continuous time Fourier series problem
2 2
Here T = 2; 0   
T 2
jk  k
1
C k    e 20
2
  jk  k
1
x t    Cke j k 0 t
    e 20 e j k 0 t
k   k    2 

 1 
 jk   t 
k
1
    e  20 
k    2 
 k; k  0
k 
 k; k 0
k  1   1 
1 jk   t  
 1  jk   t 
k
1
 x t      e  20 
    e  20 
k    2  k 0 2 

 1   1 
 1  jm  t    1  jk   t 
1 m k
    e  20      e  20 
m  2  k 0 2 

ECT 204 Signals and Systems 6 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

 1   1 

 1   jm 20  t    1  jk  20  t 
m k
   e    e
m 1 2  k 0 2 

a
Sum of a G.P. = ; a  first term and r  common differece .
1 r
First geometric series is evaluated by summing from m = 0 to m = ∞ and subtracting m = 0
term. The result of summing both infinite geometric series is

x t  
1 1
 1 
  1 
1  j t   1 j t  
1 e  20 
1 e  20 
2 2

1 j   t  20   1 j   t  20    1  j   t  20  
 1   1   1   1 
1  j   t  20 
1 e 1 e  1 e 1 e
2 2  2  2 
   
 1  j   t  201   1 j   t  201  
1  e   
1 e  
 2  2 
  

1  j   t  20   j t     1  j   t  20  1 j   t  20  1 
 1   1   1   1 

2  e e  20  
  1 e  e 
2   4 
   2 2
 1   1 
1  j t   1 j t   1
1 e  20   e  20  
2 2 4
3 3
 4  4
  1 j t   
 1  5  1 
5 1  j   t  20   20    cos   t  
 e  e 4  20 
4 2 
 
3

 
5  4 cos   t  
 20 
8. Using Parseval’s theorem find the total power and plot power spectrum of the periodic
signal, x(t) = 4 + 2cos3t+3sin4t.
x(t) = 4 + 2cos3t+3sin4t
2 2
1  3;
  T1 
1 3
2  2
 2  4;  T2  
2 4
Fundamental time period, T = LCM ( T1, T2) = LCM (2π/3, 2π/4) = 2π
2 2
 0   1
T 2
 
x t    C k e j k 0 t   C k e j k t (1)
k   k  

x(t) = 4 + 2cos3t+3sin4t

ECT 204 Signals and Systems 7 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

 e j3t  e  j3t   e j4 t  e  j4 t 
 4  2   3  (2)
 2   2j 
   
Comparing equations (1) and (2), we get
k  0; Ck  4
k  3; C k 1
k   3 ; C k 1
3
k  4; Ck   j
2
3
k   4; Ck  j
2

3
 2 j; k   4

1 ; k   3

C k  4 ; k  0
1 ; k  3

 3
 2 j ; k  4


P   Ck
2



2 2
3 2 2 2 3
P  j  1  4  1   j
2 2

a  jb  a 2  b 2

2 2
P   3   1  16  1   3    22.5 watts
90
 2   2  4

9
4 ; k4

1 ; k  3
2 
C k  16 ; k 0
1 ; k 3

9
 4 ; k4

Figure: Power spectrum

ECT 204 Signals and Systems 8 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

Continuous Time Fourier Transform

9. Find Fourier transform of the following functions.
(i) x t   e 2 t u t  (ii) x t   e (iii) x t   e  2 t u t 1
t

(i) x t   e 2 t u t 
 
X j   xt e dt 
 j t
e
2t
u t  e  j t dt
 

 
 e 2t
e  j t
dt   e   2  j t dt
0 0


 e    2  j  t 
  
1
e  e0  
   2  j 0  2  j
1
 0 1  1
 2  j  j  2
(ii) xt   e
t

|t| means –t and +t, that is, ‘-t’ from ‘-∞’ to ‘0’ and ‘t’ from ‘0’ to ‘∞’.
 
X j   xt e
 j t t
dt  e e  j t dt
 
0  0 
  t   j t   t   j t t  j t  t  j t
 e e dt  e e dt  e dt  e dt
 0  0


 e  1 j t   e   1 j t 
0  0
  e  1 j t dt  e
1  j t
dt      
 0  1  j      1  j 0


1
1 j

e0  e 
1
1 j
e  e0   

1
1 0  1 0  1  1  1
1 j 1 j 1 j 1 j
1  j  1  j 2
  2 2 2
1  j1  j 1  j 
2

1 2

(iii) x t   e  2 t u t 1
 
X j   xt e dt 
 j t
e
2 t
ut 1 e  j t dt
 

 
 e 2 t
e  j t
dt   e  2 j t dt
1 1

ECT 204 Signals and Systems 9 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering


 e   2  j  t 
  
1

e   e  2 j 
  2  j1 2  j

e  2 j

1
2  j

0  e  2 j  
2  j
10. x(t) is shown in the figure. Find its Fourier transform.

 1 ; 0  t  2
x t   
1 ;  2  t  0
 0 2
X j   xt e dt  1e dt    1e dt
 j t  j t  j t

 2 0

0 2
 e  j t   e  j t  e 0  e 2 j e 2 j  e 0
     
  j   2   j  0  j j

 e 0  e 2 j e 2 j  e 0  1  e 2 j  e 2 j  1
  
j j j

 2
2

2 2 j
e  e 2 j 
 2  2 cos 2 j
 
j j j
 j  2  2 cos 2 j 2  2 cos 2
 
 
2 j 1  cos 2


11. Consider a rectangular pulse defined as
1 ;  T0  t  T0
x t   
0 ; t  T0
Find Fourier transform. Give an idea about its magnitude spectrum.

 T0
X j   xt e dt 
 j t
 1e
 j t
dt
 T0
T
 e  j t  0 e  j T0  e j T0
  
  j   T  j
0


j
e 
 1  j T0
 e j T0 
j
e 
1 j T0
 e  j T0 
2
2

ECT 204 Signals and Systems 10 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

2 sin  T0 


sin  x x
S a x   and sin c x   and S a x   sin c  
sin x
x x 
sin  T0
  T0 
 Sa  T0  
 sin c  
 T0
  
2 sin  T0 
Similarly here X  j 

Multiply numerator and denominator by T0, we get
2 T sin  T0    T0 
X  j  0  2 T0 sin c  
T0    
For ω = 0, using L’Hospital’s rule, we get
Lt 2 Lt 2 T0 cos  T0 
sin  T0    2 T0 cos 0  2 T0
   1
For ω = 0, X(jω) = 2T0
As T0 increases, x(t) becomes less concentrated about time origin, while X(jω)
becomes more concentrated about frequency origin and vice versa. That is, width of x(t) is
inversely related to width of X(jω). Hence signal concentrated in one domain are spread out
in other domain.
12. Find inverse Fourier transform of the following spectra.
2 cos   
X  j   
0 ,   ,

x t   X j  e j t d
1

2  

 e j  e  j  j  j
2 cos   2  e e
 2 
 
1  
  
x t  
1
2 
e j
e  j j t
e d  
2 
 e e
j j t
d    e 
 j j t
e d  

  

1 
   1  e j  1 t     e j  t  t   


2 
 e j  1

t  
d   e j   
t  1 d   


   
   
 2   j 1  t     j t  1    
    
1  e j 1 t    e  j 1 t    e j t 1   e  j t 1  
   
  2 j 1  t   2 j t  1 
sin 1  t   sin t  1 
   sin c 1  t   sin ct 1
 1  t  t 1
13. Given X j   3    4, find x(t).
 
x t   X j  e j t d  3    4 e j t d
1 1

2   
2  

   4 e j  t d
3
 
2  

ECT 204 Signals and Systems 11 Prepared by: Dr. Giby Jose
Downloaded from Ktunotes.in
 Dept. of Electronics & Communication Engineering

We know that  xt  t  t o dt  xt 0 

x t  
3 j4 t
e
2
14. Find the inverse FT of the rectangular spectrum depicted in Fig. and plot x(t).


x t   X j  e j t d
1

2  
 
1  e j t 
x t  
1 j t
2 
1  e d    
2  j  



1
2 jt

e j  e  j  
sin  t   sin  t
t   t


   t 
 Sa t   Sinc  
    
x
 Sax   Sinc  

At t = 0, x(t) becomes
Lt sin  t Lt  cos  t  cos 0 
  
t 0 t t 0   

15. x(t) = δ(t). Find X(jω).


X j    xt  e
 j t
dt



  t  e
 j t
 dt



 t e d t  x0
 j t
We know that 


X j   e 0 1 s
16. Find inverse FT of X(j ω) = 2 π δ(ω)

x t   X j  e j t d
1
2 
 
2   e j  t d    e j  t d
1
 
2   

ECT 204 Signals and Systems 12 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

 t e d  x 0
j t
We know that


x t   e 0 1
Hence FT when x(t) = 1 is X(jω) = 2π δ(ω)
17. Find x(t) for the given X(jω) shown below.


x t   X j  e j t d
1

2  

1 
1 2  1  e j  t    e j  t  2 
      
2 2
  1 e j t
d   1 e j t
d 
 2   j t   jt  
 1   2  1 


1
2 j t

e jt  e 2 jt  e2 jt  e jt 
1
2 j t
 
e2 jt  e 2 jt   e jt  e jt  

1
sin 2 t  sin t 
t
18. Find X(jω) for the time signal given below.

t , t 1
Here x t   
0, t 1

X j    xt  e
 j t
dt

 1
 j t  e  j t  1  j t
e1
  t e d t  t   
  j   1 1  j 
1 dt



1
j
  1

1 e  j    1 e  j   2 2 e  j   e j  
j 
 1  j
j
e  1

 e  j  2 e j  e  j

  
j2  e  j  e  j   1  e j  e  j 
  2 2  2 j
j   2    2j 
2j 2j
 cos   2 sin 
 
Laplace Transform
1. Determine LT of x(t) = eat u(t) and depict ROC and pole and zero locations in S – plane.
Assume ‘a’ is real.
Solution:

ECT 204 Signals and Systems 13 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

 
X s    x t e  s t dt   e a t ut e  s t dt u(t) = 0 for t < 0 and u(t) = 1 for t > o
 


 e  s  a t 
  s  a t
  1 e dt    (1)
0   s  a 0
To evaluate e-(s – a) at the limits, substitute s = σ + jω

 1 
X s    e    a  t e  j t 
   j  a  0
If (σ – a) > 0 i.e., σ > a then e    a  t goes to zero as t approaches ∞ and if σ < a then
e    a  t becomes ∞.

X s   
1
0 1 ,   a
   j  a  
 1 
 X s     , Re s a
 s  a 
LT, X(s) does not exist for σ ≤ a, since integral is unbounded. ROC for this signal is
thus σ > ‘a’ or Re s a which is shown by the shaded portion of s – plane.
Pole is located at s = a. There is only one pole and no zeros.

2. Determine LT of y(t) = -eat u(- t) and depict ROC and pole and zero locations in S – plane.
Assume ‘a’ is real.
Solution:
 
Y s    yt e dt    e u t e dt
st at st
u(t) = 0 for t > 0 and u(t) = -1 for t < o
 
o o
0
  s  a t  e  s  a t   e  s  a t 
   1 e dt       (1)
   s  a     s  a    
To evaluate e-(s – a) at the limits, substitute s = σ + jω
 e    a  t e  j t 
0

Y s    
   j  a    
If (σ – a) < 0 i.e., σ < a then e    a  t goes to zero as ‘t’ approaches -∞ and if σ > a then
e    a  t becomes ∞.

Y s   
1
1  0 ,  a
   j  a 
 1 
Y s    , Re s a
s a 

ECT 204 Signals and Systems 14 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

Pole is located at s = a. There is only one pole and no zeros. Shaded region is the ROC.

In problem (1) and (2) we are having two different signals. But their LT are identical
and ROCs are different i.e., two different signals may have identical LTs but different ROCs.
Hence expression of LT does not uniquely correspond to a signal x(t) if ROC is not specified.
In first problem, the signal is causal, i.e., x(t) > 0, for t > 0 and its ROC will be in the
form Re s  max where σmax represents maximum real part of any of the poles of X(s).

3. Determine the Laplace transform of the signal x(t) = sin(ω0t) u(t) and plot the ROC and the
locations of poles and zeros in the s-plane.
Solution:
e j0 t u t   e  j0 t u t 
xt   sin  0 t u t  
2j

Take Laplace transform on both sides, we get

Lxt 
1
2j
  
L e j0 t u t   L e  j0 t u t  
1  1 1  1  s  j o  s  j o 
X s       
2 j  s  j  o s  j o  2 j  s  j o 
o

s  o
2 2

The set of values of Re{s} for which the Laplace transforms of both terms converge is
Re{s} > 0.
o
sin  0 t u t  
LT
, Res 0
s 2  o
2

Here it may be noted that ROC: s > jω0 means,

ROC: σ + jω > 0 + jω0
Normally, we consider only real part for ROC.
Hence, above equation can be written as
ROC: σ > 0. i.e., Re{s} > 0.
Similarly, for the second term s > jω0, we can write,
ROC: σ + jω > 0 - jω0
ROC: σ > 0. i.e., Re{s} > 0.
Thus, the ROC for both the terms is Re{s} > 0.

ECT 204 Signals and Systems 15 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

Laplace transform of impulse function x(t) = δ(t)

1 for t  0
We know that t   
0 for t  0

Lt  Xs    t  e st dt  e st t 0 1 (for any s)
0
That is, ROC is the entire s-plane. The ROC is shown in figure.
t  
LT
 ; for all s

Figure: ROC of δ(t)

Laplace transform of step function x(t) = u(t)
1 for t  0
We know that u t   
0 for t  0

Lu t  Lx t  Xs    u t  e st dt
0

 
st  e  st 
  1e dt   
0   s  0
1

  e   e 0 
s
1
s

This integral converges when Re (s) > 0
Lu t 
1
s
The ROC is Re (s) > 0, that is, the entire right half of the s-plane as shown in figure.

ECT 204 Signals and Systems 16 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

Figure: ROC of L[u(t)]

Laplace transform of ramp function x(t) = t u(t)
 
Lt u t  Xs    t u t  e st
dt   t e st dt
0 0

 e st  
 st  e st 

e
 t   1dt  0   2 
  s  0 0  s  s  0
 1  1
 00 2   2
 s  s
The above integral converges if Re (s) > 0, that is, ROC is Re (s) > 0 as shown in figure.

Figure: ROC of L[t u(t)]

Laplace transform of parabolic function x(t) = t2 u(t)

 
 
L t u t   Xs    t u t  e
2 2 st
dt   t 2 e st dt
0 0

 2e st  
 st 
e 2
 t   2 t dt  0   t e st
  s  0 0  s s0
   st  
2  e st  e  2  1 st 
 
 s  s  s 0
  dt   0  e dt
s  s  
 0 0 

2  e st  1
 2   3
s   s  0 s
The above integral converges if Re (s) > 0, that is the ROC id Re (s) > 0.

ECT 204 Signals and Systems 17 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

Laplace transform of cosine function x(t) = cos ωt u(t)

 e jt  e  jt 
Xs   L cos t u t  L  u t 
 2 
 
 L e jt u t   L e  jt u t 
1
2
1
2
 
1  1  1  1  1  s  j   s  j 
    
2  s  j  2  s  j  2  s  js  j 
s
 (for Re (s)  0
s  2
2

4. Find the Laplace transform of x(t) = e-t u(t) + e-4t u(t) and find ROC.
Given x(t) = e-t u(t) + e-4t u(t)

L x t  Xs   L e  t u t   e 4 t u t  
 e ut   e ut e dt

t 4 t st


 

 e ut e u t  e st dt
t st 4 t
 dt  e
 
   
s 1 t
 e t
e st
dt   e 4 t
e st
dt   e dt   e s  4  t dt
0 0 0 
 0 

Converges if Converges if
Re s   1 Re s    4

Converges if Re s   1

t  t 
 e s 1 t   e s  4  t  0 1  0 1
Xs        
  s  1  t  0   s  4  t  0  s  1  s  4 


1

1

s  4  s  1
s  1 s  4 s  1s  4 
2s  5
 2 ; ROC; Re s   1
s  5s  4
The ROC is shown in figure. This example shows that the ROC of the sum of two signals is
equal to the intersection of the ROCs of the two signals.

Figure: ROC of x(t) = e-t u(t) + e-4t u(t)

ECT 204 Signals and Systems 18 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

5. Find the Laplace transform of x t   cos2 2t u t 

 1  cos 4t  
Given x t   cos2 2t u t     u t 
 2 
 1  cos 4t  
L x t  Xs   L  u t 
 2 
 L u t  l cos 4t u t 
1
2
1 1 s  1  2s 2  16 
    

2  s s 2  42  2  s s 2  16 


Inverse Laplace transform
1. Find the inverse Laplace transform of the following
3s 2  8s  6
(i) Xs   (ii) Xs   (iii) Xs  
s s
s  2 s  (iv)
s  5s  6
2 2
 2s  1 s  22
2s  1 s 3 1
Xs   (v) Xs   ; Res    2 (vi) Xs  
s
(vii)
s  22  1 s2 s s 1s  2
 5s  7
Xs   with ROC -1 < Re(s) < 1
s 1s 1s  2
Solution
(i) Given Xs  
s s

s  5s  6
2 s  2s  3
By using partial fraction
s A

B

s  3A  s  2 B
s 2  5s  6 s  2  s  3 s  2s  3
Put s = -3
 3   3  3 A   3  2  B
 3  B
 B3
Put s = -2
 2   2  3 A   2  2  B
 2A
 A 2
We know that
 1  2 t
L1    e u t 
s  2 
 1  3 t
L1    e u t 
 s  3 
Therefore, inverse Laplace transform of X(s) is
 1  1  1 
x t    2 L1    3L     2 e u t   3 e u t 
2 t 3t
s  2  s  3

ECT 204 Signals and Systems 19 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

3s 2  8s  6 3s 2  8s  6
(ii) Given Xs   
s  2 s 2  2s 1 s  2s 12
By using partial fraction
3s 2  8s  6

A

B

C

s  12 A  s  2s  1 B  s  2 C
s  2s  12 s  2 s  1 s  12 s  2s  12
Put s = -2
3 22  8 2  6   2  12 A   2  2 2  1 B   2  2 C
12 16  6  A  0  0
 A2
Put s = -1
3 12  8 1  6   1  12 A   1  2 1  1 B   1  2 C
38 6  0  0  C
 C 1
3s 2  8s  6  s  12 A  s  2 s  1 B  s  2  C
 s  12 2   s  2 s  1 B  s  2 1
   
 s 2  2s  1 2   s 2  3s  2 B  s  2
 2s 2  4s  2  Bs 2  3Bs  2B  s  2
 2  Bs 2  5  3Bs  2B  4
Compare the coefficient of s2, s, or constant terms on both sides, (compare constant terms),
we get
6  2B  4
2
 B 
2
Xs  
2 1 1
  
s  2 s  1 s  12
Taking inverse Laplace transform, we get
 2  1  1  1  1 
L1 Xs  L1    L  s  1  L  2
s  2     s  1 
x t   2 e 2 t u t   e t u t   t e t u t 

(iii) Given Xs  

s
s  22
By using partial fraction
s

A

B

s  2 A  B
s  22 s  2 s  22 s  22
 sA  2A  B
Compare the coefficients of s in both side, we get
1 A
Compare constant terms, we get
0  2A  B   2B  A 1
Therefore, inverse Laplace transform of X(s) is

ECT 204 Signals and Systems 20 Prepared by: Dr. Giby Jose
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 Dept. of Electronics & Communication Engineering

 1   1 
x t   L1    2 L1 
2
s2  s  2 
x t   e 2 t u t   2 t e 2 t u t   e 2 t 1  2t  u t 

  1  
at 
 L1    t e 
 s  a  
2

 


(iv) Given Xs  

s
s  22  1
  1  s  2  2 
x t   L1 Xs  L1 
s
L  
 s  2  1  s  2  1
2 2

 s2   1 
 L1    2 L 1
 
 s  2  1  s  2  1
2 2

We know that
 sa  at    at
L1    e cos t and L1    e sin t
 s  a      s  a    
2 2 2 2

  
x t   e 2 t cos t  2 e 2 t sin t u t 
2s  1
(v) Given Xs   ; Res    2
s2
2s  1 2s  4  3 2s  2 3 3
   2
s2 s2 s2 s2 s2
Taking inverse Laplace transform on both sides, we have
x t   2 t   3 e 2 t u t 
 L1 1 t 
s 3 1
(vi) Given Xs  
s s 1s  2
 s 3 1  1  s 3  1 
L1    L 
 s s  1s  2  s 3  3s 2  2s 
 
Since order of numerator and denominator are equal, partial fraction cannot be
obtained directly.
 
s 3  3s 2  2s s 3  11
s 3  3s 2  2s
 3s 2  2s  1
s 3 1 3s 2  2s 1
Xs   1 
s 3  3s 2  2s s 3  3s 2  2s
3s 2  2s 1 A B C s 1s  2 A  s s  2 B  s s 1 C
   
s s 1s  2 s s 1 s  2 s s 1s  2
Put s = 0

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30   20  1  0  10  2  A  0  0

 1  2 A
1
 A
2
Put s = -1
3 12  2 1 1   1  1 1  2 A   1 1  2 B   1 1  1
 3  2 1   B
 B0
Put s= -2
3 22  2 2 1   2  1 2  2 A   2 2  2 B   2 2  1 C
 12  4 1  0  0  2C
7
 C
2
 1 7 
 Xs  1   2  2  1  1  7 1
 s s  2  2s 2 s  2
 
Take inverse Laplace transform
1 7  1 
L1 Xs  L1 1 L1    L1  
 2s  2 s  2 
x t   t   u t   e 2 t u t 
1 7

2 2
 5s  7
(vii) Given Xs  
s 1s 1s  2
By using partial fraction
 5s  7

A

B

C

s 1s  2 A  s 1s  2 B  s 1s 1C
s 1s 1s  2 s 1 s 1 s  2 s 1s 1s  2
Put s = 1
  5 1  7  1 11  2A  1  11  2B  1  11 1C
 12  0  6B  0
 12
 B  2
6
Put s = -2
  5  2  7   2 1 2  2A   2  1 2  2B   2  1 2 1C
10  7  0  0   3C
3
 C  1
3
Put s = -1
  5  1  7   1 1 1  2A   1  1 1  2 B   1  1 1 1C
5  7   2A  0  0
2
 A 1
2
Xs  
1 2 1
  
s 1 s 1 s  2

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The ROC and the locations of the poles are depicted in figure. The ROC, -1 < Re(s) < 1, is a
strip.

The pole of the first term is at s = -1. The ROC lies to the right of this pole, so this pole
corresponds to a causal (right sided) signal. Therefore,
e t u t  
1
s 1
The second term has a pole at s = 1. Here the ROC lies to the left of this pole, so this pole
corresponds to an anti-causal (left sided) signal. Therefore,
 2 e t u  t  
2
s 1
The third term has a pole at s = -2. Here the ROC lies to the right of this pole, so this pole
corresponds to a causal (right sided) signal. Therefore,
e 2 t u t  
1
s2
There the inverse Laplace transform of X(s),
x t   e  t u t   2 e t u  t   e 2 t u t 

ECT 204 Signals and Systems 23 Prepared by: Dr. Giby Jose
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MODULE II
Fourier representation of continuous time signals – continuous time Fourier series
Any input signal can be represented as a weighted superposition of time shifted impulses. i.e.,
 
for discrete case, xn   xk  n  k  and x(t )   x() (t  ) d for continuous time case. So we can
k   
create a signal by adding the product of a constant and time shifted impulses and output is obtained if
we know the impulse response of the system.
Here we represent a signal as a weighted superposition of complex sinusoids instead of time
shifted impulses. Hence, exponential, ramp signals etc. may be used instead of time shifted impulses.
A signal is expressed in terms of lot of component signals to make the analysis simpler. i.e.,
instead of analyzing output of a system to an input by directly calculating its output, we split the input
and analyze the output for each of this small input component and get the final output due to linearity
property. Thus if a signal which is the weighted superposition of complex sinusoids is applied to a
linear system, then system output is a weighted superposition of system response to each complex
sinusoids.
The study of signals and systems using sinusoidal representations is termed Fourier analysis,
after Joseph Fourier (1768-1830) for his development of the theory. This is a frequency domain
approach.
Let impulse response of a continuous time system be h(t) and input be x(t) = ejωt. Convolution
 
integral gives output as y( t )   x() h(t  )d   h() x(t  )d due to commutative property.
 
 
j( t )
 y( t )   h()x(t  )d   h()e d
 

 e jt  h()e
 j
d Since x(t – τ) = ejω(t – τ).


 j
= ejωtH(jω) where H( j)   h()e d

i.e., y(t) = x(t) H(jω).
Output of the system is a complex sinusoidal of same frequency as input multiplied by complex
constant H(jω). H(jω) is a function of only frequency ‘ω’, and not time ‘t’. Hence H(jω) is termed
frequency response of continuous time systems.
Thus complex sinusoidal input to a LTI system generates an output equal to sinusoidal input
multiplied by the system frequency response.
These type of inputs in which output can be expressed as input multiplied by some constant is
called Eigen function and the constant is called Eigen value of the system. So the complex sinusoidal
t   e jt is an Eigen function of the system H associated with Eigen value   H j .

By representing arbitrary signals as weighted superposition of Eigen functions, we transform

the operation of convolution to multiplication. Consider expressing the input to an LTI system as the
weighted sum of ‘M’ complex sinusoids

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M
x t    a k e jk t
k 1
j k t
If e is an Eigen function of the system with Eigen value H (jωk), then each term in the input,
a k e j k t produces an output term a k H j k  e j k t . Hence, we express the output of the system as
M jω t
y t    a k Hjω k e k
k 1
The output is a weighted sum of ‘M’ complex sinusoids. Here weight becomes a k H  j k  . Also
operation of convolution becomes multiplication because x(t) is expressed as a sum of Eigen functions.
An analogous relationship holds in the discrete-time case. There are four distinct Fourier
representations, each applicable to a different class of signals. The four classes are defined by the
periodicity properties of a signal and whether the signal is continuous or discrete in time. The Fourier
series (FS) applies to continuous-time periodic signals, and the discrete-time Fourier series (DTFS)
applies to discrete-time periodic signals. Non-periodic signals have Fourier transform representations.
The Fourier transform (FT) applies to a signal that is continuous in time and non-periodic. The
discrete- time Fourier transform (DTFT) applies to a signal that is discrete in time and non-periodic.
DTFS is often referred to as Discrete Fourier Transform (DFT). Table illustrates the relationship
between the temporal properties of a signal and the appropriate Fourier representation.
Time
Periodic Non-periodic
Property
Continuous Fourier Series (FS) Fourier Transform (FT)
Discrete-Time Fourier Series Discrete-Time Fourier Transform
Discrete
(DTFS) (DTFT)
Exponential Fourier series

x t    C k x k t 
k 1

A continuous time complex exponential signal is expressed as e j0 t  cos0 t  j sin 0 t .

Let xk(t) be the complex exponential signal representation.


 x k t   .....,e  j20 t , e  j0 t ,1, e j0 t , e j20 t ,......
2
Where 0 
T0
2
 T0  , the fundamental time period condition for orthogonality is
0

 x m t x k t dt  0 ; m  k


T0

If exponential signal e j k0 t is to be orthogonal to e jm0 t , where e j k 0 t  x k t  and

e j m 0 t  x m t 

  x m t x k t dt   e jm 0 t e  jk 0 t dt   e j0 m  k t dt
T0 T0 T0

e  jk 0 t is the complex conjugate of e jk 0 t

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Consider a constant time period ‘t’, which is possible since it is a continuous signal.
t1  T0
t1  T0
j0 m  k t  e j0 m  k t  e j0 m  k t1  T0   e j0 m  k t1
  x m t  x k t dt   e dt    
T0 t1  j 0 m  k   j 0 m  k 
t 1

 2 
e j0 mk t1 j0 mk T01 e j0 mk t1  j0 mk  0

1
j0 m  k 
e  e 
j0 m k t1 j0 m k T0
1 
j0 m  k 
e  1 
j0 m  k  
e  1


 

e j0 mk t1 j2mk 


j0 m  k 
e 
1 
Put m – k = n
e j0 nt1 j2n
 e j0 nt1

j0 nt1
  x m t x k t dt  e 1  cos 2n  j sin 2n  1  e 1  0  1
T0
j0 n j0 n j0 n

  x m t x k t dt 0
T0

That is complex exponential signals xm(t) and xk(t) are orthogonal over the interval T0.

Generalized Fourier series is given by x t    C k x k t  .
k 1
Hence for exponential Fourier series, it is given as

x t    C k e jk t 0

k  
[Ck can also be denoted as x(k), ak, bk etc. followed in text books of different authors]

To find expression for coefficient Ck:

 xt x k t dt  xt x k t dt  xt e

   j0 kt
dt
T0 T0 T0
Ck   
 t dt  x k t  t dt  e
x kj kt  j kt
x 2k dt 0 0

T0 T0 T0

 xt e  xt e
 j0 kt  j0 kt
dt dt
T0 T0

e
0
dt t T0 0

T0

 xt e
1  j0 kt
C k  dt is the expression for which error is minimum and the series is
T0 T0

x t    C k e jk t 0

k  
A signal can be expressed as a series of complex exponential signals multiplied by some constant.
Trigonometric Fourier series
The trigonometric form of the Fourier series of a periodic signal, x(t), with period T is defined as

EC202 Signals & Systems 3 Prepared by: Dr. Giby Jose

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 
x t   a 0   a k cos k 0 t    b k sin k 0 t 
k 1 k 1

Where ak, bk are real valued coefficients.

x t  dt
1
T T
a0 

x t  cos k 0 t dt
2
T T
ak 

x t sin k 0 t  dt
2
T T
bk 

Periodicity property of Fourier series


x t    C k e jk t 0

k  
Condition for periodicity is x(t) = x(t + T).
 
 x t  T    C k e jk 0 t T    C k e jk t e jk T
0 0

k   k  
2
T
0
2
 jk 0 
 xt  T    C k e jk0t e 0
  C k e jk t e j2k
0

k  k 
j2 k
e  cos 2k  j sin 2k  1  j0  1

 
 x t  T    C k e jk t .1   C k e jk t  xt 
0 0

k   k  

That is Fourier series representation is periodic.

Existence of Fourier integral or convergence of Fourier series
A periodic signal x(t) has a Fourier series representation if it satisfies the following Dirichlet
conditions:
1. x(t) is absolutely integrable over any period, that is,
 xt  dt  
T
2. x(t) has a finite number of maxima and minima within any finite interval of t .
3. x(t) has a finite number of discontinuities within any finite interval of t, and each of these
discontinuities is finite.
Dirichlet conditions are sufficient, but not necessary conditions for Fourier series representation.
Fourier series theorems/properties
Let us consider x(t) and y(t) which denotes two periodic signals with period ‘T’ which have
Fourier series coefficients denoted by Ck and dk respectively. The important properties of Fourier
series are the following.

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I. Linearity
If x t  
 C k and yt   d k , then
FS FS

zt   Ax t   Byt  

FS
 Z k  AC k  Bd k .
2
Where x(t), y(t), and z(t) are periodic signals with fundamental time period, T 
0
Proof
x t e  j0 kt dt ; d k   yt e  j0 kt dt
1 1
Ck 
TT TT

zt e  j0 kt dt   Ax t   Byt e  j0 kt dt  A  zt e  j0 kt dt  B  zt e  j0 kt dt
1 1 1
z k 
TT TT TT
T

 AC k  Bd k [A periodic waveform x(t) and its Fourier coefficient Ck can be represented

symbolically as x t  
FS
 C k ], hence proved.
II. Translation or time shift
If x t  
 C k then zt   xt  t 0    z k  e  j0kt 0 C k
FS FS

Where z(t) is a signal time shifted from x(t) by t0

Proof
z k   zt e  j0kt dt   x t  t 0 e  j0kt dt
1 1
TT TT
Put t – t0 = λ, then t = λ + t0; dt = dλ

x  e  j0 k   t 0 d  e  j0 kt 0  x  e  j0 k d ;

1 1
z k 
TT TT
λ being a dummy variable, we get
z k  e  j0 kt 0 C k , hence proved.

III. Frequency shift

If x t  
 C k , then zt   e j0kt 0 xt  
FS FS
 z k  Ck  k 0

Proof

zt e  j0 kt dt   e j0 k 0 t x t e  j0 kt dt   x t e  j0 k k 0 t dt

1 1 1
zk 
TT TT TT

x t e  j0 kt dt , we get
1
T T
Ck 

z k  C k  k 0 , hence proved.
IV. Time scaling
If x t  
 C k , then zt   x at  
FS FS
z k  Ck ; a  0
Proof
If x(t) is periodic, then z(t) = x(at) is also periodic. If x(t) has a fundamental period ‘T’, then z(t)
T
= x(at) has fundamental period . Consequently, if x(t) has a fundamental frequency equal to ω 0,
a
then fundamental frequency of z(t) is aω0.

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x t e  j0 kt dt
1
We have C k 
TT

 zt e  xat e
1  j0 akt 1  j0 akt
z k  dt  dt
T T
T T
a a a a

Put at = λ,  t  a ; dt  d
1
a

x  e  j0 kt d   x  e  j0kt d
a 1 1
zk 
TT a TT
z k  C k , hence proved.

So FS coefficient of x(t) and x(at) are identical, even though the harmonic spacing changes from ω 0
to aω0.

That is x t    C k e j kt 0

k  

While x at    C k e ja  kt 0

k  
Here when time scaling is performed, Ck remains same while ω0 changes to aω0
V. Time differentiation

If x t  
 C k , then zt   x t   z k  jk 0 C k
FS d FS
dt
Proof

We know that x t    C k e jk t 0
(1)
k  
Differentiating with respect to ‘t’, we get
 
x t    C k j0 ke j0 kt   C k j0 k e j0 kt
d
(2)
dt k   k  
Comparing equation (1) and (2), we get

x t    z k j 0 ke j0 kt
d
dt k  
 z k  jk 0 C k , hence proved.

VI. Time domain convolution

If x t  
 C k and yt   d k , then
FS FS

2
zt   x t   yt   z k  TC k d k , where T 
FS
0
Proof
 
We have yt   x t   h t    xht  d and yn   xn   hn    xk hn  k 
  k  

2
It is assumed that x(t), y(t), and z(t) have same fundamental period T 
0

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zt e  j0 kt dt   x t   yt e  j0 kt dt

1 1
zk 
TT TT

From definition we have for a period T, x t   yt    xht  d

T
1
xyt  de  j0kt dt
T t T T
zk 

Changing order of integration, we get

x  yt  e  j0kt dtd

1
T T
zk 
t T
Put t    ;  t    ; dt  d

x  y e  j0k  dd  xe  j0k d  y e  j0k d  TC k Td k

1 1 1
zk  
T  T 
T  T T
t T t T
 z k  TC k d k , hence proved.

VII. Modulation or multiplication theorem

If x t  
 C k and yt   d k then,
FS FS

zt   x t yt   z k  C k  d k

FS

Proof
zt e  j0 kt dt   x t yt e  j0 kt dt
1 1
zk 
TT TT

We have x t    C k e jk t0

k  

Replacing ‘k’ by ‘m’, we have x t    C m e jm t 0

m  

C m e jm0 t yt e  j0 kt dt
1
z k   
T T m
Changing order of summation and integration, we get,
1  1 
zk   C m e jm 0 t
yt e  j0 kt
dt   C m  yt e  j0 k m t dt
T m T T m T
 
 j0 k  m t
 Cm
1
   C m d k m
T T
zk  y t e dt 
m   m  

yt e  j0 k m t dt
1
 d k m 
TT

 z k  C k  d k , xn   hn    xk hn  k  , hence proved.
k  
VIII. Parseval’s theorem
Let x(t) be a power signal with Fourier coefficient Ck, then average power in the waveform is
given by

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
1
   Ck
T T
2 2
P x t dt 
k 
Proof
Let us assume that x(t) is complex. Then xt   xt   x  t 
2

x t  dt   x t   x  t dt
1 1

2
P (1)
TT TT

We know that, x t    C k e jk t 0

k  
Taking conjugates on both sides, we get

x  t    C k e  jk  t 0
(2)
k  
Put equation (2) in (1), we get

P   x t   C k e  jk 0 t dt
1
TT k 
Changing order of summation and integration, we get
1     
 k    jk 0 t  1
 k T    jk 0 t
 k k  Ck

T k  T
2
P C x t e dt  C x t e dt  C C  , hence proved.
k  T k  k 

Continuous time Fourier transform (CTFT)

Any periodic signal x(t) expressed as complex exponential signals gives the Fourier series. When a
non-periodic signal (or aperiodic signal) is expressed as complex exponential signal then that
representation is called the Fourier transform.

Consider an arbitrary signal x(t) that is of finite duration is given below. It is an aperiodic signal.
From this aperiodic signal, we can construct a periodic signal xT0 t  for which x(t) is one period as
shown below. T0 is arbitrary taken such that there is no overlapping. We construct xT0 t  from x(t).
Where xT0 t  is periodic with fundamental period T0. Then we can express xT0 t  in terms of complex
exponential.

x T0 t    C k e jk t 0
(1)
k  

 x T t e
1  jk 0 t
Ck  dt (2)
T0 T0
0

2
0 
T0

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If we choose T0 to be larger that is, if to repeat a pulse infinite time is required, then as T0   ,
x T0 is identical to x(t) over a longer interval. That is,
x T0 is equal to x(t) for any finite value of ‘t’. That is,
Lt x T0  x t 
T0 
Since x T0  x t  for t 
T0
and also since x(t) = 0. Outside this interval, we
2
can rewrite equation of Ck as:
T0
2 

 x T t e  x T t e
1  jk 0 t 1  jk 0 t
Ck  dt  dt (3)
T0 T0
0
T0 
0

Define another signal on the envelop X(jω).


X j   xt e
 jt
dt


 X j0 k    xt e
 j0 k
dt (4)

Comparing equation (3) an (4), we get
X jk 0 
X jk 0  
1
Ck 
T0 T0
Substituting Ck in equation (1), we get

X jk 0  jk 0 t
x T0 t    e
k   T0

2 1 
T0  ,  0
0 T0 2

0 1 
 x T0 t    X jk  0 e jk 0 t   X jk 0 e jk0t 0 (5)
k   2 2 k 

Let T0   , then 0  0 , therefore 0   , an arbitrary small frequency.

1 
Then, Lt x T0 t   Lt  X jk e jkt   xt  (6)
T0  T0  2 k  

This is called the limiting form of x T0 t  . Here k can take values ±1, ±2, ±3, ….., resulting in
k  ,2,3    

k can be considered as a continuous variation of ω. That is, k sweeps out all frequencies from -∞
to ∞. Since it covers all frequencies, we can consider k as a continuous frequency variable ω.
That is, Lt k   and Lt   d
0 0

Substituting in equation (6), we get,

1  1  1 
x t   Lt  X  jk     e jk t
  X  j   d  e jt
  X j  e jt d
0 2 k  2 k  2 k 

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Changing summation to integration, we get,


x t   X j  e jt d
1

2 

This is known as inverse Fourier transform equation.


X j   xt   e
 jt
dt


This is known as Fourier transform or Fourier integral of x(t).

Fourier transform theorems/properties
I. Linearity

If x t  
 X j and yt  
 Y j , then
FT FT

zt   ax t   byt  

 Z j  aX j  bY j
FT

Proof
   
Z j   zt e dt   axt   byt e dt  a  xt e dt  b  yt e dt
 jt  jt  jt  jt

   
 aX j  bY  j , hence proved.

II. Time shifting

If x t  
 X j , then
FT

yt   xt  t 0 

 Y j  e  jt 0 X j
FT

Proof
 
Y j   yt e
 jt
dt   xt  t 0 e
 jt
dt
 

Put t – t0 = λ, then dt = dλ
 
 j  t 0 
 Y j   x e d  e  jt 0  x e
 j
d
 

 e  jt0 X  j  , hence proved.

III. Frequency shifting

If x t  
 X j , then
FT

yt   e jt x t  

 Y j  X j  
FT

Proof
  
 j t
Y j   yt e dt 
 jt
 e xt e dt 
j  jt
 xt e dt
  

Y j  X j    , hence proved.

IV. Time scaling

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If x t  
 X j , then
FT

1  j 
yt   x at  
 Y j 
FT
X 
a  a
Proof
 
Y j   yt e
 jt
dt   xat e
 jt
dt
 
d
Put at = λ, then dt 
a
d
Since ‘a’ can be +ve or -ve, instead of ‘a’ we can use dt 
a
   
 j d 1  j
Y j   X e a   X e a d


a a 
1  j 
 X  , hence proved.
a  a

V. Folding or conjugation property

If x t  
 X j , then
FT

x  t  
 X j  X *  j
FT

Proof
By substituting a = -1 in time scaling property, we can obtain this.
VI. Frequency differentiation

If x t  
 X j , then
FT

 jtx t   X j

FT d

d
Proof
Differentiating with respect to ‘ω’, we get,

dX j d 
     
d  jt   
   xt e  jt
dt    xt  e dt    x t  jt e  jt
dt     jtx t e  jt dt =
d d    d     
F. T. of [  jtx t  ], hence proved.

VII. Time differentiation

If x t  
 X j , then x t  
 jX j
FT d FT
dt
Proof

x t   X j  e jt d
1

2 
Differentiating both sides with respect to ‘t’, we get,
  
d
x t  
1
 X j e jt d 
d 1
 X j je jt d 
1
jX je jt d
dt 2   dt 2  2 

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= inverse F.T. of jX  j , hence proved.

VIII. Convolution

If x t  
 X j and yt  
 Y j , then
FT FT

zt   x t   yt  

 Z j  X jY j
FT

Proof
 
Z j   zt e  xt   yt e
 jt  jt
dt  dt
 
We have,

x t   yt    xyt  d
 
   
 Z j    xyt  e ddt 
 jt
 x  yt  e ddt
 jt

     
Put t     , then dt =dλ
   
 j   
 Z j   x  y e dd   xe d
 j
 y e
 j
d
       
 X jY j , hence proved.

IX. Integration or accumulation

If x t  
 X j , then
FT

 xd  X j  X j0

FT 1


j
Proof

We know that  x yt  d  x t   yt 
 

Similarly we can write x t   u t    xu t  d
 



But u t     1, fort    0 ie. t   or   t

0, otherwise
t
 x t   u t    x d (1)
 

By convolution property, we have x t   yt   X jY j

 x t   u t   X jU j (2)

Comparing equation (1) and (2), we get,

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 x d 
 X jU j
FT
(3)
 

At a later stage, we will show that

u t  
 U j   
FT 1
(4)
j

Substituting equation (4) in (3), we get

t
 1  FT
 xd 
 X j   
 X j  X j
FT 1
   j  j

X j  X j0

FT 1


j
Since X j  X j0 using property of impulse function, hence proved.

X. Modulation or multiplication

If x t  
 X j and yt  
 Y j , then
FT FT

zt   x t yt  

FT
 Z j 
1
X j  Y j
2

Proof
 
Z j   zt e
 jt
dt   xt yt e
 jt
dt
 

We know that x t   X j  e jt d
1

2 

This can be written as x t   X j   e jt d
1
2 
(2)

‘λ’ is used instead of ‘ω’ because if we substitute this in equation (1) with ‘ω’ itself, ejωt will vanish.
Substitute equation (2) in (1), we
   
 j t
Get, Z j   2  X j e yt e ddt  2  X j   yt e
1 j t  jt 1
ddt
t         t  


X j Y j   d
1
2  
 (3)


We know that,  x yt  d  x t   yt  (4)
 
Comparing equation (3) and (4), we get,
Z j 
1
X j  Y j , hence proved.
2
So multiplication in time domain is equivalent to convolution in frequency domain. That is,
convolution and multiplication properties are duals of each other. Multiplication of one signal by
another can be thought of as using one signal to scale or modulate amplitude of another and

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consequently multiplication of two signals is often referred to as amplitude modulation. Hence this
property is referred to as modulation property also.
XI. Parseval’s theorem or Rayleigh’s theorem

If x t  
 X j , then
FT

 

 xt  X j d
1

2 2
E dt 

2  

Where ‘E’ is the total energy content of the signal x(t). Also X  j  is defined as the energy
2

density spectrum of the signal x(t).

Proof
 

 xt   xt x t dt

2 
E dt  (1)
 t  


We have x t   X j  e jt d
1
2 
Taking conjugates on both sides, we get,

x t   X   j  e jt d
 1

2 
(2)

Substituting equation (2) in (1), we get,

   
E   x t  X   je  jt ddt  X   j  x t e  jt ddt
1 1

2  
2 
t   t 

 
X   jX jd  X j d , hence proved.
1 1
 
2

2  2 

XII. Symmetry

Consider a real signal x(t) and let X j  FTx t   A  jB and x(t) = xe(t) + x0(t), where
xe(t) and x0(t) are even and odd components of x(t) respectively. Then,
x e t  
 A
FT

x o t  
 jB 
FT

Proof

x e t  
1
xt   x t 
2
x 0 t   x t   x  t 
1
2
Given, x t  
 X j  A  jB 
FT

Since x(t) is real, we have,

x  t  
 X j  X   j  A  jB 
FT

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x e t  
FT

1
2
 
X j  X   j  A  jB   A  jB   jB  , hence proved.
1
2
Thus we have shown that F. T. of a real even signal is a real function of ‘ω’ and that of a real odd signal
is an imaginary function of ‘ω’.
XIII. Duality or similarity theorem

If x t  
 X j , then
FT

X jt  
 2x  
FT

Here, X(jt) implies that it is X  j  itself, but ‘ω’ replaced by ‘t’. For example,

Figure:
Proof

x t   X j  e jt d
1

2 

Interchanging ‘t’ and ‘ω’, we get,


x   X jt   e jt dt
1
2 

Replacing ‘ω’ by ‘-ω’, we get,


x    X jt   e  jt dt  F.T.of x  jt  , hence proved.
1

2 

Two important functions are

I. Sa function
It is defined as: Sax  
sin x
x
‘Sa’ stands for sine over argument. It has a maximum value of unity at x = 0 and approaches zero
in an oscillatory manner as ‘x’ goes from -∞ to ∞. Since sinx = 0 for x  n , Sa(x) crosses x-axis
whenever x is an integral multiple of π. It is an even function.
[ Lt
sin x
x0 x
 
is of the form 0 0 . Apply L-Hospital’s rule gives Lt
cos x
x 0 1
 1]

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Figure: Sa function

II. Sinc function

Figure: Sinc function

sinπx 
It is defined as sinc(x) 
πx
x
Thus sinc(x)  Sa πx  and Sax   sinc  . It has same shape as Sa(x).
π
Laplace transform
We can generalize the complex sinusoidal representation of a continuous time signal offered by
Fourier transform to a representation in terms of complex exponential signals, which is termed the
Laplace transform. Fourier transform does not exist for signals that are not absolutely integrable, such
as impulse response of an unstable system. Hence Laplace transform may be used since FT based
methods cannot be used in this class of problems.
Let est be a complex exponential with complex frequency s = σ + jω.
est = e(σ + jω)t = eσ t . ejωt = eσ t (cosωt + jsinωt)
= eσ t cosωt + j eσ t sinωt.

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i.e., the real part of est is an exponentially damped cosine, and the imaginary part is an exponentially
damped sine. The real part of ‘s’ is the exponential damping factor ‘σ’, and the imaginary part of ‘s’ is
the frequency of the cosine and sine factor, ‘ω’.
When σ = 0, s = jω, then FT is a special case of LT.
Consider applying an input of the form x(t) = est to an LTI system with impulse response h(t).
The system output y(t) is given by
y(t) = H{x(t)} = H{est}
= h(t) * x(t)
  
  h( )x(t  τ)dτ   h e s(t  τ)dτ  est  h( )e sτ dτ
  
Transfer function, H(s) is defined as

H(s)   h( )e sτ dτ

Where H(s) is the LT of impulse response.
So, y(t) = estH(s)
H{est} = H(s)est.
The action of the system on an input est is multiplication by the transfer function H(s). Hence,
we identify est as an Eigen function of the LTI system and H(s) as the corresponding Eigen value.
We have the transfer function defined by

Hs    hτ e st dτ

Replace ‘τ’ by ’t’

Hs    ht e st dt (1)

Put s = σ + jω
 
Hσ  jω   ht e σ jωt dt   ht e σt e
j t
dt
 
FT
 ht e σt   H σ  jω
  j t
Since X jω   x t e dt

σ t 1  j t
 ht e   Hσ  jωe dω
2π 
1  j t
Since x t    X jωe dω
2π 
σt 
e j t 1  σjωt dω
 ht    Hσ  jωe dω   H σ  jωe (2)
2π  2π 
Substitute s = σ + jω
 ω     s  σ  jω
ω    s  σ  jω
ds
 ds  jd  or dω 
j
Substitute these in equation (2), we get

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1 σ j st ds  1
σ j
st
ht    Hs  e  Hs e ds
2π σj j 2 j σj

1 σ j st
 ht    Hs e ds (3)
2  j σj
Hence we say that H(s) is the LT of h(t) and h(t) is the inverse LT of H(s).
Laplace transform comes in two varieties:
1. Unilateral or one sided and
2. Bi lateral or two sided.
Bilateral LT offers insight into nature of systems characteristics such as stability, causality and
frequency response. Equation (1) indicates bi-lateral LT since integration limit is from - ∞ to ∞ that is,
two sided. Hence for unilateral LT, equation (1) becomes,
 st
Hs    ht e dt
0 
The lower limit of 0+ implies that we do not include the point t = 0 in the integral. Hence H(s)
depends only on h(t) for t > 0. Discontinuities and impulses at t = 0 are excluded that is, it is one sided.
Unilateral LT is a convenient tool for solving differential equations with initial conditions.
For any arbitrary signal, x(t), the LT is
  st
X s    x t e dt and inverse LT of X(s) is

1 σ j st
xt    X s e ds .
2  j σj
L
We express this relationship with the notation x t   X s 

Convergence or existence of Laplace transform

LT exists if X s   
  
We have X s    xt e  st dt   xt e  σ jωt dt   x t e  σ t e j ω t dt
  
  σ t j ω t
X s    xt e e dt

  σ t j ω t
X s    x t e e dt

[ a  b  a  b . For example: a = 2 and b = 3 it becomes 2  3  2  3 , that is 1  5 ]

Thus LT will exist if

 σ t
 xt e dt   ; for existence of Fourier integral it is  xt  dt   and in LT x(t) is given by
 T
σ t
xt e .
σ t
Hence LT is FT of xt e . So the necessary condition for convergence of LT is absolute
σ t
integrability of xt e .
The necessary and sufficient conditions for the existence of the Laplace transform are:

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1. x(t) should be continuous or piece-wise conditions in the given closed integral.

2. x(t)e–σt must be absolutely integrable.


 xt e
t
That is, X(s) exists only if dt  


Or only if, Lt e st x t   0

t 

By limiting ourselves to certain range of ‘σ’, we may ensure that xt e

 σ t is absolutely
integrable even though x(t) is not absolutely integrable by itself. The range of ‘σ’ for which the LT
converges is termed the region of convergence (ROC).
Generally ROC is defined as the range of values of complex variable ‘s’ for which LT converges.
As s = σ + jω, it is ‘σ’ that determines ROC and not ‘ω’.
It is convenient to graphically represent complex frequency ‘s’ in terms of a complex plain
termed the s-plane. Horizontal axis represents real part of ‘s’, that is exponential damping factor ‘σ’
and vertical axis represents imaginary part of ‘s’, or the sinusoidal frequency ‘ω’.

Fourier transform is given by LT evaluated along imaginary axis since in s-plane σ = 0

corresponds to imaginary axis. The jω axis divides s-plane in half. Region of s-plane to left of jω axis is
termed left half of the s-plane while region to right of jω axis is termed right half of s-plane. Real part of
‘s’ is –ve in left half of s-plane and +ve in right half of s-plane.
The most commonly encountered form of the Laplace transform in engineering is a ratio of two
polynomials in s; that is,

bMsM  bM1sM1      b1s  b0

Xs 
sN  a N1sN1    a1s  a0

If N > M, X(s) is proper rational function and if M > N is improper rational function.
Since these are polynomials, we can factorize them. That is, we rewrite X(s) as a product of
terms involving the roots of the denominator and numerator polynomials, given by:
bM M s  c k 
Xs  k 1
k 1 s  d k 
N

The ck are the roots of the numerator polynomial and are termed the zeros of X(s). The dk are
the roots of the denominator polynomial and are termed the poles of X(s). We denote the locations of
zeros in the S-plane with the ‘0’ symbol and the locations of poles with the ‘X’ symbol. The locations of
poles and zeros in the S-plane completely specify X(s), except for the constant gain factor bM.
Advantages and limitations of Laplace transform
Advantages
1. Signals which are not convergent in Fourier transform are convergent in Laplace transform.
2. Convolution in time domain can be obtained by multiplication in s-domain.

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3. Intego differential equations of a system can be converted into simple algebraic equations. So
LTI systems can be analyzed easily using Laplace transform.
Limitations
1. Frequency response of the system cannot be drawn or estimated. Instead only the pole-zero
plot can be drawn.
2. s = jω is used only for sinusoidal steady state analysis.
Properties of unilateral Laplace transform
In the properties given below, assume that
Lu Lu
xt  
 X s  and y t  
 Ys 

I. Linearity
Lu
axt   by t  
 aX s   bYs 
Linearity of LT follows from its definition as an integral and the fact that integration is a linear
operation. The linearity property states that LT of weighted sum of two or more signals is equal to
similar weighted sum of LTs of the individual signals.
Proof
By definition of LT,
 st
X s   Lxt    xt e dt

  st
Ys   Ly t    y t e dt

  st  st   st
Lax t   by t    ax t   by t e dt   ax t e dt   by t e dt
  
  st   st
 a  xt e dt  b  y t e dt  aX s   bYs 
 
II. Scaling

Lu 1 s
xat  
 X 
a a
Scaling in time introduces inverse scaling in S.
Proof
By definition of LT,
  st
X s   Lxt    xt e dt

  st
 Lx at    x at e dt

Put at  τ
τ
t 
a
dτ
dt 
a

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  s τ dτ 1  sτ 1 s
 Lxat    xτ e a   xτ e a dτ  X 
 a a  a a
The above transform is applicable for positive values of ‘a’. If ‘a’ happens to be negative, it can be
proved that
1 s
x at    X 
a a
Lu 1 s
Hence in general xat  
 X 
a a

III. Time shift

Lu  sτ
xt  τ  
 e Xs  for all ‘τ’ such that xt  τ  u t   x t  τ  u t  τ 
A shift of ‘τ’ in time corresponds to multiplication of the Laplace transform by the complex
exponential e-sτ. The restriction on the shift arises because the unilateral transform is defined solely in
terms of the positive-time portions of the signal. Hence, this property applies only if the shift does not
move a nonzero +ve time component of the signal to –ve time or move a nonzero –ve time portion of
the signal to +ve time. The time-shift property is most commonly applied to causal signals x(t) with
shifts τ > 0, in which case the shift restriction is always satisfied.
Proof
By definition of LT,

Xs   Lx t    xt e
 st
dt


 Lx t     xt  e
 st
dt

Let t    z ; t  z  
dt  dz
  sτz    sz  sτ  sτ   sz
 Lxt  τ    x z e dz   xz e e dz e  x z e dz
  
Since ‘z’ is a dummy variable, for integration we can replace ‘z’ by ‘t’.

 Lx t    e  s  xt e
 st
dt  e  s Xs 


IV. S-domain shift

s0t Lu
e xt  
 X s  s 0 
Multiplication by a complex exponential in time introduces a shift in complex frequency s into
the Laplace transform.
Proof
s0t   s t  st  s  s0 t
Lx t  e    x t  e 0 e dt   x t e dt  X s  s 0 
   
V. Convolution
Lu
xt   y t  
 Xs Ys 

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Convolution in time corresponds to multiplication of Laplace transforms. This property only

applies when x(t) = 0 and y(t) = 0 for t < 0.
Proof

Let x(t) and y(t) be two time domain signals. By definition of LT, Ys   Ly t    y t e  st dt

The convolution of two
continuous time signals x(t) and y(t) is 
xt   y t    xλ  y t  λ dλ
defined as 

Where ‘λ’ is the dummy variable used for integration.

    st
Lxt   y t      xλ  y t  λ dλ  e dt
  
 st  sλ  st  sλ s t λ   sλ  sM
Let e  esλ  e e e e e e
Where M  t  λ  dM  dt
   sλ  sM   sλ   sM
 Lxt   y t     xλ  y Me e dλ   xλ e dλ   y Me dM
   
Here ‘λ’ and ‘M’ are dummy variables used for integration and so they can be replaced by ‘t’.
  st   st
 Lxt   y t    x t e dt   y t e dt  X s  Ys 
 
VI. Differentiation in the S-domain
Lu dXs
 tx t  

ds
Differentiation in the s-domain corresponds to multiplication by -t in the time domain.
Proof
By definition of LT,
  st
X s   Lxt    xt e dt

On differentiating the above equation with respect to ‘t’, we get,
d
ds
X s  
d 

ds 
x t  e
 st
dt 


 x t  
d  st
ds
e dt 



 xt   te
 stdt

  st
   t x t e dt  L t xt 

Lu dXs
  t x t   
ds
VII. Differentiation in time domain
dxt  Lu
dt
 SX s   x 0 
  
Proof
  st Lu
Since X s    xt e dt and xt  
 X s 
0
Differentiating with respect to time ‘t’, we get

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dxt  Lu dXt 
 
dt dt
dxt  Lu   dxt    st
   e dt
dt 0  dt 
Integrating by parts, we obtain
dxt  Lu
dt

 e
 st  
 
xt  0   s  xt e

 st
dt
0
dxt  Lu
dt
 e 
 st 
 
xt  0   s  xt e

 st
dt
0
 st
e xt  0
t 
dxt  Lu

dt
  
 0  x 0   sX s 

dxt  Lu

dt
  sX s   x 0   
VIII. Integration

t Lu x
 xτ dτ 

1 0 Xs

 
where x 1   0
 
0   xτ dτ in the area under x(t) from
 s s 
t = -∞ to t = 0 .
+

Proof
By definition of LT, the LT of a causal signal is given by

X s   Lxt    xt e
 st
 
dt  uv  u  v   dv  v 

   st    st
 st e  e
L xt     xt e dt   xt dt    xt  dt
0   s  s
0
 
 e  e0 1   st 1 1  st
  xt dt t      x t  dt t 0  s s  x t e dt   x t dt t  0 s  xt e
 dt
  s  0 s 0


 
x 1 0  X s 

s s
IX. Initial value theorems
The initial and final-value theorems allow us to determine the initial value of x(t), x(0+), and the
final value, x(∞) of x(t) directly from X(s). These theorems are most often used to evaluate either the
initial or final values of a system output without explicitly determining the entire time response of the
system.
The initial-value theorem states that Lt sX s  Lt xt 
s  t 0
That is, initial value of signal, x0  Lt sX s   Lt xt 
s t 0
Proof

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 dxt 
We know that L   SX s   x0
 dt 
On taking limit s   on both sides of the above equation, we get
 dxt 
Lt L   Lt SX s  x0
s  dt  s
By definition of LT
 dxt   dxt   st
L   e dt
 dt  0 dt
 dxt   st
 Lt  e dt  Lt SX s   x0
s 0 dt s
 dxt    st
  Lt e
 st 
dt  Lt SX s   x0  Lt e  0
0 dt  s  s s 
dxt 
0  Lt SX s   x 0 Here and x(0) are not functions of ‘s’.
s dt
x0  Lt SX s 
s
Lt x t   Lt SX s  x0   Lt xt 
t 0 s t 0 

X. Final value theorems

The initial-value theorem does not apply to rational functions X(s) in which the order of the
numerator polynomial is greater than or equal to that of the denominator polynomial order, which is
improper rational function.
The final-value theorem states that Lt sX s   Lt x t 
s 0 t 
That is, final value of signal, x   Lt sX s   Lt xt 
s 0 t 0

Proof
On taking limit s  0 on both sides of the above equation, we get
 dxt 
Lt L   Lt SX s   x0
s  0  dt  s  0
By definition of LT
 dxt   dxt   st
L   e dt
 dt  0 dt
 dxt   st
 Lt  e dt  Lt SX s   x0
s  0 0 dt s 0
 dxt     st 
  Lt e  st dt  Lt SX s   x0  Lt e  1
0 dt  s  0  s 0 s  0 
xt 0  Lt SX s  x0 Here dxt  and x(0) are not functions of ‘s’.
s 0 dt
x   x0  Lt SX s   x 0
s
 
 x    Lt SX s  x    Lt xt 
s 0 t   

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 Lt x t   Lt SX s 
t  s 0
The final-value theorem applies only if all the poles of X(s) are in the left half of the s-plane, with at
most a single pole at s = 0.
Basic Laplace transforms

x(t) X(s)
δ(t) 1
1 or u(t) 1
s
1
t
s2
1
e-at
sa
1
eat
sa
1
te-at
s  a 2
ω
Sin ωt
s  ω2
2
s
Cos ωt
s2  ω2
ω
e-at Sin ωt
s  a 2  ω2
sa
e-at Cos ωt
s  a 2  ω2
n!
tn
s n1
δ(t-τ), τ>0 e-sτ

Inversions of unilateral Laplace transform

The direct method of finding the inverse Laplace transform is time consuming. So the inverse
Laplace transform of X(s) is usually obtained by using partial fraction expansion. The partial fraction
expansion can be used only for proper rational functions. i.e., for functions in which the order of the
numerator is smaller than that of the denominator.

Ns  b M s M  b M 1s M 1        b1s  b 0

Let Xs   
Ds  s N  a N 1s N 1        a 1s  a 0

If M = N, that is, degree of N(s) = degree of D(s), then divide the numerator by the denominator
N1 s 
Xs   C 
Ds 
Where C is a constant and degree of N1(s) < degree of D(s).
If degree of N(s) > degree of D(s), divide the numerator by the denominator and write X(s) as
N1 s 
Xs   Terms with powers of s greater than or equal to zero 
Ds 

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Where degree of N1(s) < degree of D(s)

If the given X(s) is of any of the standard forms, we straight away write down its inverse
transform. If the given X(s) is not of any of the standard form, D(s) is factorized. The roots of the
characteristic equation, D(s) = 0 are called the poles of X(s).
X(s) can be expressed as:
Ns  b M s M  b M 1s M 1        b1s  b 0
Xs   
Ds  N
 s  Pk 
k 1

Now, consider the two possible cases:

I. Distinct Poles
If all the poles Pk are distinct, then X(s) can be written as a sum of single pole terms given by
K K2 K Kn
Xs   1   i 
s  P1 s  P2 s  Pi s  Pn
To determine the value of Ki, multiply both sides by (s – Pi) and then put s = Pi. Then we have,

 K1 K K K 
s  Pi  Xs s  P  s  Pi   s  Pi  2       s  Pi  i  s  Pi  n 
i
 s  P1 s  P2 s  Pi s  Pn  s  P
i

In the RHS of the above equation, all the terms except Ki vanish. Hence we get,
K i  s  Pi  Xs  s  P
i

II. Multiple Poles

In this case, all the poles are not distinct. Some of the poles may repeat. Let the pole Pi, repeat l
times. Then, if all the other poles are distinct, we write,
K K K i1 K i2 K il Kn
Xs   1  2              
s  P1 s  P2 s  Pi  s  Pi 2 s  Pi l s  Pn
The coefficient Kil is obtained by multiplying both sides of the above equation by (s – Pi)l and
evaluating it at s = Pi.
K il  s  Pi l Xs  s P
i

The coefficients Ki(l-b) are evaluated by multiplying both sides of the preceding equation by
(s – Pi)l
differentiating (l – b) times and then evaluating the resultant equation at s = Pi. Thus,
K il 1  s  Pi l Xs  s P
d
ds i

1d 
K il  b    s  Pi l Xs 
b!  ds  s P i

Complex Roots
If X(s) have complex poles, then the partial fraction expansion of the same can be expressed as:
K1 K*
Xs    1*
s  P1 s  P1

Where K1* is complex conjugate of K1 and P1* is complex conjugate of P1.

Alternate method
When the denominator has complex roots, the complex pair can be expressed as a single
quadratic factor instead of having first order partial fractions. That is, if s = - a ± jb, we can retain the
quadratic factor.

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Ds   s  a  jb s  a  jb   s  a 2  b 2

Relationship between Laplace transform and Fourier transform

Consider a continuous time signal x(t). Its Laplace transform is defined as

LTxt  Xs    xt e
 st
dt

Substitute s = σ + jω in the above equation

 xt  e e
 
  j t
LTx t   xt  e dt   t  j t
dt
 


 FT x t  e t

Therefore, Laplace transform of x(t) is the Fourier transform of x t  e   t .
If σ = 0, then
LTx t  FTx t 

Properties of ROC
1. The ROC of X(s), the Laplace transform of x(t) is bounded by poles or extends up to infinity.
2. The ROC does not contain any pole.
3. If x(t) is a right sided signal, the ROC of X(s) extends to the right of the right most pole and no
pole is located inside the ROC.
4. If x(t) is a left sided signal, the ROC of X(s) extends to the left of the left most pole and no pole is
located inside the ROC.
5. If x(t) is a two sided signal, the ROC of X(s) is a strip in the s-plane bounded by poles and no
pole is located inside the ROC.
6. Impulse function is the only function for which the ROC is the entire s-plane.
7. The ROC must be a connected region.
8. The ROC of an LTI stable system contains the imaginary axis of s-plane.
9. The ROC of the sum of two or more signals is equal to the intersection of the ROCs of those
signals.

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