control system lecture notes on laplace transform

1. Chapter two
Laplace Transform
2.1 Introduction
The method of transforming a function from time domain to s domain is known
as Laplace transform, where s is a complex operator denoted by s=α+jβ.
Use the Laplace transformation to transform
the circuit from the time domain to the frequency domain, obtain the solution, and
apply the inverse Laplace transform to the result to transform it back to the time domain.
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2. The Laplace transform is significant for a number of reasons.
1. it can be applied to a wider variety of inputs than phasor
analysis.
2. it provides an easy way to solve circuit problems involving
initial conditions, because it allows us to work with algebraic
equations instead of differential equations.
3. the Laplace transform is capable of providing us, in one single
operation, the total response of the circuit comprising both
the natural and forced responses
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3. 2.2 Laplace Transformation Theorems
Given a function f(t), its Laplace transform, denoted by F(s) or is defined by
where s is a complex variable given by s=α+jβ.
The Laplace transform is an integral transformation of a function f (t) from
the time domain into the complex frequency domain, giving F (s).


0
st-
dt(t)ef=(t))L(f
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4. 

0
st-
dt(t)ef=L(f(t))
Laplace Transform
Example 1:
-st st
0
0
-bt -bt -st -(b+s)t ( s)t
0
0 0
-st
0
a a a
L(a)= ae dt e 0
s s s
1 1
L(e )= e e dt e dt -e
b+s s+b
df df
L(f ) L e dt
dt dt
b



 

 

 
          
    
     
 

 
 f(0)sL(f)
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5.  
n
n
dt
fd
foretc,(0)fsf(0)F(s)s=
(0)ff(0)-sF(s)s=
φ(0)-φ(s)
dt
df
=φwhere
dt
dφ
L=
dt
fd
L
2
2
2















s
22
2222
ωs
s
=
ωs
jωs
ωs
jωs
2
1
=
jωs
1
jωs
1
2
1
=





















  
 
 
 

j t j t
2 2
e - e
L(sin ωt) =L
2j
ω
=
s ω





  
2
ee
L=ωt)L(cos
tjt-j
Note:
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6. Laplace Transforms of Common Functions
Name f(t) F(s)
Impulse
Step
Ramp
Exponential
Sine
1
s
1
2
1
s
as 
1
22
1
s
1)( tf
ttf )(
at
etf )(
)sin()( ttf 






00
01
)(
t
t
tf
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7. Laplace Transform Properties
   
)(lim)(limtheoremvalueFinal
)(lim)0(theoremvalueInitial
)()()nConvolutio
)(
1)(
)(nIntegratio
)0()()(ationDifferenti
)()()]()([calingAddition/S
0
0
2121
0
2121
ssFtf-
ssFf-
sFsFdτ(ττ)f(tf
dttf
ss
sF
dttfL
fssFtf
dt
d
L
sbFsaFtbftafL
st
s
t
t














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8. Table of Laplace transform
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9. Example
Find the Laplace transform f(t)=δ(t) + 2u(t) - 3𝑒−2𝑡
u(t)
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10. 2.3 Inverse Laplace Transformation
• Inversion of the Laplace transform to find the signal x ( t ) from its Laplace
transform X(s) is called the inverse Laplace transform
• symbolically denoted as
• Inverse Laplace transform permits to go back from S domain to time function
Methods to find Inverse Laplace transform
1. Inversion Formula:
There is a procedure that is applicable to all classes of transform functions
that involves the evaluation of a line integral in complex s-plane; that is,
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11. 2. Using Tables of Laplace Transform Pairs:
• In the second method for the inversion of X(s), we attempt to express X(s) as a
sum,
where X,(s),. .., Xn(s)are functions with known inverse transforms X1(t),.. ., Xn(t).
3. Partial-Fraction Expansion:
• If X(s) is a rational function, that is, of the form,
• a simple technique based on partial-fraction expansion can be used for the
inversion of X(s)
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12. Example
• Find the inverse Laplace transform of
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13. 2.4 Solving Linear time-invariant Differential Equation
we considered a continuous-time LTI system for which input x ( t ) and output y(t)
satisfy the general linear constant-coefficient differential equation of the form,
• Applying the Laplace transform and using the differentiation property of the
Laplace transform, we obtain
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14. Example
Solve the ODE,
First, take L of both sides of above equation
    
2
5 1 4sY s Y s
s
  
Rearrange,
Take L-1,
 
 
1 5 2
5 4
s
y t
s s
  
  
 
L
By using partial fraction
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15. Example:
system at rest (initial value is zero)
Step 1 Take L.T. (note zero initial conditions)
3 2
3 2
6 11 6 4
0 0 0 0
d y d y dy
y
dt dt dt
y( )= y ( )= y ( )=
   
 
3 2 4
6 11 6 ( )s Y(s)+ s Y(s)+ sY(s) Y s =
s

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16. Rearranging,
Step 2a. Factor denominator of Y(s)
Step 2b. Use partial fraction decomposition
Multiply by s, set s = 0
3 2
4
( 6 11 6)
Y(s)=
s s s s  
))(s+)(s+)=s(s+s++s+s(s 3216116 23
31 2 44
1 2 3 1 2 3
αα α α
s(s+ )(s+ )(s+ ) s s s s
   
  
32 4
1
00
1
4
1 2 3 1 2 3
4 2
1 2 3 3
ss
αα α
α s
(s+ )(s+ )(s+ ) s s s
α

 
       
 
 
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17. For a2, multiply by (s+1), set s=-1 (same procedure
for a3, a4)
2 3 4
2
2 2
3
α , α , α    
2 32 2
2 2
3 3
2
0 (0) 0.
3
t t t
y(t)= e e e
t y(t) t y
  
  
    
Step 3. Take inverse of L.T.
You can use this method on any order of ODE,
limited only by factoring of denominator polynomial
(characteristic equation)
2 2 2 2/3
( + )
3 1 2 3
Y(s)=
s s s s
 
  
(check original ODE)
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18. Transform Circuits
• Signal Sources:
where u ( t )and i ( t ) are the voltage and current source signals,
respectively.
• Resistance R:
• Inductance L:
Or
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19. • Capacitance C:
Or
The output y ( t ) of a continuous-time LTI system is found to be 2𝑒−3𝑡u(t) when the input
x ( t )is u(t ).
1. Find the impulse response h(t) of the system.
2. Find the output y(t) when the input x ( t ) is 𝑒−𝑡u(t)
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Example