Signals and Systems Analysis

Laplace Transform
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OUTLINE

• Introduction

• Laplace Transform

• Properties of Laplace Transform

• Inverse Laplace Transform

• Applications of Laplace Transform

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INTRODUCTION
• Why Laplace transform?
o Easier to solve circuit problems containing capacitors and inductors.
o Convolution in time domain  Multiplication in frequency domain.
o Provides the total response containing natural response and forced
response.

• It is used to transform circuits from time domain to frequency domain.

• After obtaining the result, apply inverse Laplace transform to find the
result in time domain
• It is a tool that allows us to represent a signal x(t) as a
continuous sum of exponentials of the form e(st) .
• Such a representation is quite valuable in the analysis
and processing of signals
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OUTLINE

• Introduction

• Laplace Transform

• Properties of Laplace Transform

• Inverse Lapalace Transform

• Applications of Fourier Transform

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LAPLACE TRANSFORM

– Is an integral transform of a function from time domain x(t) to complex

frequency domain X(S)

– Integral transformation maps (converts) a function from its original

domain (time domain) to another domain (S-domain) via integration

– General form of integral transformation

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LAPLACE TRANSFORM: BILATERAL LAPLACE TRANSFORM

• Bilateral Laplace transform (two-sided Laplace transform)


X (s)   x(t) e(st)dt, s    j


– s    j is a complex variable
– s is often called the complex frequency
– Notations: X (s)  L[x(t)]
x(t)  X (s)
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LAPLACE TRANSFORM: BILATERAL LAPLACE TRANSFORM

• Bilateral Laplace transform (two-sided Laplace transform)


X (s)   x(t) e(st)dt, s    j


• It is clear that a

• x(t) can be synthesized with growing exponentials along the path   j

with  varying from -oo to oo.

• The value of  is flexible.

• Notice, there are infinite choices for .

• This means the spectrum of x(t) is not unique, and there are infinite possible ways of
synthesizing x(t).

•  has a certain minimum value for a given x(t). The region in the complex plane is
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LAPLACE TRANSFORM

• Time domain v.s. s-domain

o x(t) : a function of time t  x(t) is called the time domain signal

o X(s) : a function of s  X (s) is called the s-domain signal

o S-domain is also called the complex frequency domain

– By converting the time domain signal into the s-domain, we can usually
greatly simplifies the analysis of the LTI system.

– S-domain system analysis:

1. Convert the time domain signals to the s-domain with the Laplace
transform

2. Perform system analysis in the s-domain

3. Convert the s-domain results back to the time-domain

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LAPLACE TRANSFORM: BILATERAL LAPLACE TRANSFORM

• Region of Convergence (ROC)

– The range of s that the Laplace transform of a signal converges.
– The Laplace transform always contains two components
• The mathematical expression of Laplace transform
• ROC.

• Example
• For a signal x(t)  e(at)u(t), find the Laplace transform X(s) and its ROC.
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LAPLACE TRANSFORM: BILATERAL LAPLACE TRANSFORM

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LAPLACE TRANSFORM: BILATERAL LAPLACE TRANSFORM

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LAPLACE TRANSFORM: BILATERAL LAPLACE TRANSFORM

• Region of Convergence (ROC)

• The ROC of X(s) is Re s > -a, as shown in the shaded area.
• This fact means that the integral defining X(s) exists only for the values of s
in the shaded region.
• For other values of s, the integral does not converge.
• For this reason, the shaded region is called the ROC (or the region of
existence) for X(s).
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LAPLACE TRANSFORM: UNILATERAL LAPLACE TRANSFORM

• Unilateral Laplace transform (one-sided Laplace transform)



X (s)  0  x(t) exp(st)dt
estdt

– 0 :The value of x(t) at t = 0 is considered.

– Useful when dealing with causal signals or causal systems.

• Causal signal: x(t) = 0, t < 0.

• Causal system: h(t) = 0, t < 0.

– We are going to simply call unilateral Laplace transform as Laplace

transform.
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INVERSE LAPLACE TRANSFORM

  j
L-1[F(S)] = f(t)   (F(S) estdt
  j

– Requires knowledge about complex analysis and contour

integration.(Beyond the scope of this course )

– We will see simpler methods to find the inverse Laplace transform.

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LAPLACE TRANSFORM: UNILATERAL LAPLACE TRANSFORM

• Example: find the unilateral Laplace transform of the

following signals.
– 1. x(t)  U(t)
– 2. x(t)   (t)
– 3. x(t)  e( -at)u(t)

– 4. x(t)  cos(at) u(t)

– 5. x(t)  sin( at) u(t)

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LAPLACE TRANSFORM: UNILATERAL LAPLACE TRANSFORM

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LAPLACE TRANSFORM: UNILATERAL LAPLACE TRANSFORM

• By direct integration, find the Laplace transform X(s) and the
region of convergence of X(s) for the gate functions shown
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LAPLACE TRANSFORM: UNILATERAL LAPLACE TRANSFORM
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OUTLINE

• Introduction

• Laplace Transform

• Properties of Laplace Transform

• Inverse Lapalace Transform

• Applications of Fourier Transform

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PROPERTIES: LINEARITY
• Linearity
– If f1 (t)  F 1 (s) f2 (t)  F 2 (s)
– Then af1 (t)  bf2 (t)  aF 1 (s)  bF 2 (s)

The ROC is the intersection between the two original signals

• Example
– Find the Laplace transfrom of A  B e (bt)u(t)
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PROPERTIES: TIME SHIFTING

• Time shifting
– If f(t)  F (s) and a = to  0
– Then f(t  a)u(t  a)  F (s) esa
The ROC remain unchanged
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PROPERTIES: TIME SHIFTING

• Time shifting
• Find the Laplace transform of x(t) depicted
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PROPERTIES: TIME SHIFTING

• Time shifting
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PROPERTIES: TIME SHIFTING

• Time shifting
• Find the Laplace transform of the signal illustrated
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PROPERTIES: SHIFTING IN THE s DOMAIN

• Shifting in the s domain
– If f(t)  F(s)
– Then y(t)  f(t) e-sot  F (s  so)

• Example
– Find the Laplace transform of x(t)  Aeat cos(bt)u(t)
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PROPERTIES: TIME SCALING

• Time scaling
– If f(t)  F (s) Re{s}   1
– Then
F 
1 s
f(at)  Re{s}  a 1
a a

• Example
– Find the Laplace transform of x(t)  u(at)
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PROPERTIES: DIFFERENTIATION IN TIME DOMAIN

• Differentiation in time domain

– If f(t)  F(s)
– Then df(t)
 sF(s)  g(0  )
dt
d 2 f(t)  2
2
s F(s)  sf(0 
)  f '(0 
)
dt
d n f(t)  n
n
s F(s)  s n1
f(0 
) …  sf (n2)
(0  )  f (n1)
(0  )
dt
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PROPERTIES: DIFFERENTIATION IN TIME DOMAIN

• Find the Laplace transform of the signal x(t) the time-

differentiation and time-shifting properties of the Laplace
transform.
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PROPERTIES: DIFFERENTIATION IN TIME DOMAIN

• Differentiation in time domain

• Example
– Find the Laplace transform of g(t)  sin 2 t u(t), g (0  )  0
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PROPERTIES: DIFFERENTIATION IN S DOMAIN

• Differentiation in s domain
– If f(t)  F (s)
– Then
t f (t)  -d F(s)
ds

d n F(s)
(t) f(t) 
n

dsn

• Example
– Find the Laplace transform of t u(t)
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PROPERTIES: CONVOLUTION

• Convolution
– If f(t)  F(s) h(t)  H (s)
– Then f(t)  h(t)  F(s)H (s)

The ROC of X (s)H (s) is the intersection of the ROCs of X(s)

and H(s)
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PROPERTIES: INTEGRATION IN TIME DOMAIN

• Integration in time domain

– If f(t)  F (s)
– Then
0
t 1
f( )d   F(s)
s

• Example
– Find the Laplace transform of r(t)  tu(t)
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PROPERTIES
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OUTLINE

• Introduction

• Laplace Transform

• Properties of Laplace Transform

• Inverse Lapalace Transform

• Applications of Fourier Transform

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INVERSE LAPLACE TRANSFORM

• Inverse Laplace transform
1   j
x(t) 
2j   j
X (s) exp(st)ds

– Evaluation of the above integral requires the use of contour

integration in the complex plan  difficult.
• Inverse Laplace transform: special case
– In many cases, the Laplace transform can be expressed as a
rational function of s
bm s m  bm1 s m1   b s1  b
X (s)  0
an s n  a n1 s n1   a s1  a 0

– Procedure of Inverse Laplace Transform

• 1. Partial fraction expansion of X(s)
• 2. Find the inverse Laplace transform through Laplace
transform table.
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INVERSE LAPLACE TRANSFORM

• Review: Partial Fraction Expansion with non-repeated
linear factors
A B C
X (s)   
s  a1 s  a 2 s  a3

A  (s  a1) X (s) sa B  (s  a2 ) X (s) sa C  (s  a3 ) X (s) sa

1 2 3

• Example 2s 1
– Find the inverse Laplace transform of X (s) 
s 3  3s 2  4s
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INVERSE LAPLACE TRANSFORM

• Example
– Find the Inverse Laplace transform of 2s2
X (s)  2
s  3s  2

• If the numerator polynomial has order higher than or equal to the order
of denominator polynomial, we need to rearrange it such that the
denominator polynomial has a higher order.
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INVERSE LAPLACE TRANSFORM

• Partial Fraction Expansion with repeated linear factors

1 A2 A1 B
X (s)    
(s  a) 2 (s  b) s  a 2 s  a s  b

A2  s  a  X (s)
2
sa
A1 
d
ds

s  a  X (s)
2
 B  s  b X (s) sb
sa
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INVERSE LAPLACE TRANSFORM

• High-order repeated linear factors

1 A1 A2 AN B
X (s)      
(s  a) (s  b) s  a (s  a)
N 2
(s  a) N
s b

 
N k
Ak 
1 d
s  a N
X (s) k  1, , N
(N  k)! ds N k
sa

B  s  b X (s)
sb
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INVERSE LAPLACE TRANSFORM

• Find the inverse unilateral Laplace transforms of
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INVERSE LAPLACE TRANSFORM

• Find the inverse unilateral Laplace transforms of
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INVERSE LAPLACE TRANSFORM

• Find the inverse unilateral Laplace transforms of
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INVERSE LAPLACE TRANSFORM

• Find the inverse unilateral Laplace transforms of
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PROPERTIES: INITIAL VALUE THEOREM

• Initial value theorem
– if x(t) and its derivative dx/dt are both Laplace transformable
then
f(0  )  lim sF (s)
s

– The behavior of x(t) for small t is determined by the behavior of

F(s) for large s.

• The initial value theorem applies only if X(s) is strictly proper

(M < N),
• because for M > N, lim. s X(s) does not exist, and the
theorem does not apply.
• In such a case, we can still find the answer by using long
division to express as a polynomial in s plus a strictly proper
fraction, where M < N.
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PROPERTIES: INITIAL VALUE THEOREM

• Example cs  d
– The Laplace transform of f(t) is F (s) 
(s  a)(s  b)
Find the value of f(0  )
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PROPERTIES: FINAL VALUE THEOREM

• Final value theorem
– If x(t)  X(s)
– Then: lim x(t)  lim sX (s)
t s0

• The final value theorem applies only if the poles of X(s) are in the
LHP (including s = o).
• If X(s) has a pole in the RHP, x(t) contains an exponentially
growing term and x () does not exist.
• If there is a pole on the imaginary axis, then x(t) contains an
oscillating term and x () does not exist.
• However, if there is a pole at the origin, then x(t) contains a
constant term, and hence, x () exists and is a constant.
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PROPERTIES: FINAL VALUE THEOREM

• Determine the initial and final values of y(t) if its Laplace
transform Y(s) is given by
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OUTLINE

• Introduction

• Laplace Transform

• Properties of Laplace Transform

• Inverse Lapalace Transform

• Applications of Laplace Transform

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APPLICATION: LTI SYSTEM REPRESENTATION

• LTI system
– System equation: a differential equation describes the input output
relationship of the system.
y ( N ) (t)  a y ( N 1) (t)   a y (1) (t)  a y(t)  b x (M ) (t)   b x (1) (t)  b x(t)
N 1 1 0 M 1 0
N 1 M
y ( N ) (t)   an y (n) (t)   bm x (m) (t)
n0 m0

– S-domain representation
 N N 1 n M m
 s   a n s  Y (s)    m  X (s)
b s
 n0   m0 

– Transfer function M

Y (s) b s m
m

H (s)   m0
N 1
s   a ns n
X (s) N

n 0
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APPLICATION: LTI SYSTEM REPRESENTATION

• Example: Solve the second-order linear differential

equation
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APPLICATION: LTI SYSTEM REPRESENTATION

• Example: Solve the second-order linear differential

equation
 42

APPLICATION: LTI SYSTEM REPRESENTATION

• Example: Solve the second-order linear differential

equation
 42

APPLICATION: LTI SYSTEM REPRESENTATION

• Example: Solve the second-order linear differential

equation
 42

APPLICATION: LTI SYSTEM REPRESENTATION

• Zero-Input and Zero-State Components of

Response
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APPLICATION: LTI SYSTEM REPRESENTATION

• Solve
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APPLICATION: LTI SYSTEM REPRESENTATION

• Laplace Transform to Solve an Electric Circuit

• In the circuit, the switch is in the closed position for a long time
before t = 0 when it is opened instantaneously. Find the inductor
current y(t) for t > 0.
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APPLICATION: LTI SYSTEM REPRESENTATION

• Laplace Transform to Solve an Electric Circuit

• When the switch is in the closed position (for a long time), the
inductor current is 2 amperes and the capacitor voltage is 10 volts.
When the switch is opened, the circuit is equivalent to that depicted
in Fig., with the initial inductor current y/(0“) = 2 and the initial
capacitor voltage vc(0“) =10v. The input voltage is 10 volts, starting
at t = 0, and, therefore, can be represented by 10u(t).
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APPLICATION: LTI SYSTEM REPRESENTATION

• Laplace Transform to Solve an Electric Circuit

• The loop equation of the circuit
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APPLICATION: LTI SYSTEM REPRESENTATION

• Laplace Transform to Solve an Electric Circuit

• The loop equation of the circuit
 42

APPLICATION: LTI SYSTEM REPRESENTATION

• Laplace Transform to Solve an Electric Circuit

• The loop equation of the circuit
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APPLICATION: LTI SYSTEM REPRESENTATION

• Zero-State Response
• Consider an Nth-order LTIC system specified by the equation
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APPLICATION: LTI SYSTEM REPRESENTATION

• Zero-State Response
• Consider an Nth-order LTIC system specified by the equation
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APPLICATION: LTI SYSTEM REPRESENTATION

• Zero-State Response
• the Laplace transform of the zero-state response y(t), is the product
of X(s) and H(s), where X(s) is the Laplace transform of the input
x(t) and H(s) is the system transfer function
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APPLICATION: LTI SYSTEM REPRESENTATION

• Find the response y(t) of an LTIC system described by the

equation
 42

APPLICATION: LTI SYSTEM REPRESENTATION

• Find the response y(f) of an LTIC system described by the

equation
 42

APPLICATION: LTI SYSTEM REPRESENTATION

• Show that the transfer function of:

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APPLICATION: LTI SYSTEM REPRESENTATION

• Show that the transfer function of:

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APPLICATION: COMBINATIONS OF SYSTEMS

• Combination of systems
– Cascade of systems

H (S )  H 1 (s)H 2 (s)

– Parallel systems

H (S )  H 1 (s)  H 2 (s)
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APPLICATION: LTI SYSTEM REPRESENTATION

• Example
– Represent the system to the cascade of subsystems.
s 2  3s  2
H (S )  3
s 6s 2 11s  6
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APPLICATION: LTI SYSTEM REPRESENTATION

• Example:
– Find the transfer function of the system

LTI system
 46

APPLICATION: LTI SYSTEM REPRESENTATION

• Poles and zeros

(s  z M )(s  z M 1 ) (s  z1 )
H (s) 
(s  p N )(s  p N 1 ) (s  p1 )

– Zeros: z1 , z 2 , , z M
– Poles: p1 , p2 , , p N
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APPLICATION: STABILITY
• Review: BIBO Stable
– Bounded input always leads to bounded output



| h(t) | dt  

• The positions of poles of H(s) in the s-domain

determine if a system is BIBO stable.
A1 A2 AN
H (s)    
s  s1 (s  s 2 ) m s  sN

– Simple poles: the order of the pole is 1, e.g. s1 sN

– Multiple-order poles: the poles with higher order. E.g. s2
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APPLICATION: STABILITY
• Case 1: simple poles in the left half plane
1 1
k  0

s   k   
2 2
k
(s   k  j k )(s   k  j k )

p1   k  jk p2   k  jk

1
hk (t)  exp( kt) sin(  kt)u(t)
k

Impulse response

• If all the poles of the system are on the left half plane,
then the system is stable.
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APPLICATION: STABILITY
• Case 2: Simple poles on the right half plane
1 1
k  0

s   k   
2 2
k
(s   k  j k )(s   k  j k )

p1   k  jk p2   k  jk

1
hk (t)  exp( kt) sin(  kt)u(t)
k

Impulse response

• If at least one pole of the system is on the right half

plane, then the system is unstable.
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APPLICATION: STABILITY
• Case 3: Simple poles on the imaginary axis
1 1
k  0

s   k   
2 2
k
(s   k  jk )(s   k  jk )

1
hk (t)  sin( kt)u(t)
k

• If the pole of the system is on the imaginary axis, it’s

unstable.
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APPLICATION: STABILITY
• Case 4: multiple-order poles in the left half plane
1 m
hk (t)  t exp( kt) sin(  kt)u(t) k  0 stable
k
• Case 5: multiple-order poles in the right half plane
1 m
hk (t)  t exp( kt) sin(  kt)u(t) k  0 unstable
k
• Case 6: multiple-order poles on the imaginary axis
1 m
hk (t)  t sin( kt)u(t) unstable
k

k  0 k  0
 52

APPLICATION: STABILITY
• Example:
– Check the stability of the following system.
3s  2
H (s) 
s2  6s  13
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The Bilateral Laplace Transform

• The bilateral Laplace transform of x(t) is given by

2
 52
The Bilateral Laplace Transform

• Example:
• The bilateral Laplace transform of x(t) is given by

• To summarize, the bilateral transform X(s) can be computed from

the unilateral transforms in two steps:
• Split x(t) into its causal and anticausal components, x1(t) and
x2(t), respectively.
• Since the signals x1(t) and x2(-t) are both causal, take the
(unilateral) Laplace transform of x1(t) and add to it the
(unilateral) Laplace transform of x2(-t), with s replaced by -s.
This procedure gives the (bilateral) Laplace transform of x(t).
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The Bilateral Laplace Transform

• Example:

•
 52

The Bilateral Laplace Transform

• Example: Let us find the bilateral Laplace transform of
 52

• Example: Let us find the bilateral Laplace transform of

 52

• Find the inverse bilateral Laplace transform of

 52

• Find the inverse bilateral Laplace transform of