Physics for informatics
Lecture 3
Laplace transform
Ing. Jaroslav Jíra, CSc.
The Laplace transform
What is it good for?
•
Solving of differential equations
•
System modeling
•
System response analysis
•
Process control application
The Laplace transform
Solving of differential equations procedure
A system analysis can be done by several simple steps
1. Finding differential equations describing the
system
2. Obtaining the Laplace transform of these
equations
3. Performing simple algebra to solve for output
or variable of interest
4. Applying inverse transform to find solution
The Laplace transform
The definition
1. The Laplace transform is an operator that
switches a function of real variable f(t) to the
function of complex variable F(s).
2. We are transforming a function of time - real
argument t to a function of complex angular
frequency s.
3. The Laplace transform creates an image F(s)
of the original function f(t)

F ( s)  L f (t )   f (t )e  st dt
0
where s= σ+ iω
The Laplace transform
Restrictions
1. The function f(t) must be at least piecewise
continuous for t ≥ 0.
2. |f(t)| ≤ Meαt where M and α are constants. The
function f(t) must be bounded, otherwise the
Laplace integral will not converge.
3. We assume that the function f(t) = 0 for all t < 0
The Laplace transform
Inverse Laplace transform
  j
1
st
f (t )  L1 F ( s ) 
F
(
s
)
e
ds

2j   j
1. Inverse transform requires complex analysis to
solve
2. If there exists a unique function F(s)=L[f(t)], then
there is also a unique function f(t)=L-1[F(s)]
3. Using the previous statement, we can simply
create a set of transform pairs and calculate the
inverse transform by comparing our image with
known results in time scope
The Laplace transform
Basic properties
Linearity
L[c1 f1 (t )  c2 f 2 (t )]  c1F1 (s)  c2 F2 (s)
Scaling in time
1
s
L[ f (at )]  F ( )
a a
 st 0
Time shift
L[ f (t  t0 )u(t  t0 )]  e
Frequency shift
L[eat f (t )]  F (s  a)
F ( s)
The Laplace transform
Another properties
Original
f (t )
f (t )e at
f (at)
f (t )
Image
F (s )
F ( s  a)
1 s
F( )
a a
sF ( s)  f (0)
f (t )
s 2 F (s)  sf (0)  f (0)
f ( n ) (t )
s n F (s)  s n1 f (0)  s n2 f (0)  ... f ( n1) (0)
tf (t )
t n f (t )
t

0
f (t )
 F (s)
(1) n F ( n) (s)
1
F (s)
s
The Laplace transform
The most commonly used transform pairs
Original
Image
Original
Image
sin(t )

s2   2
1
s2
cos(t )
s
s2   2
2
s3
sinh(t )

s 2
cosh(t )
s
s2   2
t sin(t )
2 s
(s 2   2 )2
te at
1
( s  a) 2
t cos(t )
2 at
2
( s  a)3
eat sin(t )
n!
( s  a) n 1
e at cos(t )
s2  
(s 2   2 )2

( s  a) 2   2
sa
( s  a) 2   2
a
a
s
t
t
2
t , n N
n
e
at
t e
t neat , n  N
n!
s n 1
1
sa
2
The Laplace transform
Transform pair deduction
Unit step
u (t )  0
for t  0
The unit step u(t)
is defined by
u (t )  1
for t  0
Shifted unit step

1
  1  st 
L[u (t )]   1e dt   e  
 s
0 s
0
 st
f (t )  0 for t  a
f (t )  1 for t  a

e  as
  1  st 
L[ f (t )]   1e dt   e  
s
 s
a
a
 st
L[u (t )] 
1
s
u(t)
1
0
The Laplace image

t
0

The Laplace image
u(t)
1
a
e  as
L[ f (t )] 
s
t
The Laplace transform
Transform pair deduction
f(t)
Unit impulse
The unit impulse f(t)
is characterized by unit
area under its function
f (t ) 
1
,
t1
f (t )  0
for 0  t  t1
for t  0 and t  t1
0
The Laplace image
t1
1  st
1 e 
1  e  st1 1 
L[ f (t )]   e dt  
 
  
t
t1   s  0 t1   s s 
0 1
 st
t1
1/t1
t1
t
1  e  st1
L[ f (t )]
s t1
f(t)
 (t )  , for t  0
 (t )  0, for t  0
Dirac delta
It is a unit impulse
for t1 → 0
 st1
1 e
t1 0
s t1
L[ (t )] lim
 st1
se
t1 0
s
 lim
0
1
L[ (t )]  1
t
The Laplace transform
Transform pair deduction
Exponential function
f (t )  eat
e at ˆ
1
sa

  e ( s  a)t 
  1
L[ f (t )u (t )]   e at e  st dt   e ( s a )t dt  
sa
 ( s  a) 
0
0

0


f (t )  t
Linear function

L[ f (t )u (t )]   te  st dt
 uv  uv   uv
0

t e
L[ f (t )u (t )]   te dt  
 s
0
 st
Per partes integration
 st


u  t; u '  1
 e  st
v'  e ; v 
s
 st


 e 
1  ( e )
1
dt  0   2   2
 
s
0 0
 s 0 s
 st
 st
1
t ˆ 2
s
The Laplace transform
Transform pair deduction
Square function

L[ f (t )u (t )]   t 2e  st dt
0

f (t )  t 2
Per partes integration
 uv  uv   uv
u  t 2 ; u '  2t
 e  st
v'  e ; v 
s
 st



2  st
 st



t
e

2
te
2
2 1
 st
L[ f (t )u (t )]   t 2 e  st dt  

dt

0

te
dt


2


s
s
s
s
s

0 0
0
0
t 2 ˆ
The n-th power function
f (t )  t n
n!
t ˆ n 1
s
n
2
s3
The Laplace transform
Transform pair deduction
Cosine function

f (t )  cos(t )

eit  e it  st
1 1
1 
L[ f (t )u (t )]   cos( t )e dt  
e dt  


2
2
s

i

s

i



0
0
 st
1 ( s  i )  ( s  i )
s
L[ f (t )u (t )] 
 2
2
2
2
s 
s 2
Sine function

s
cos(  t ) ˆ 2
s 2
f (t )  sin(t )

it
 it
e

e
1 1
1 
L[ f (t )u (t )]   sin(wt )e  st dt  
e  st dt  


2
i
2
i
s

i

s

i



0
0
1 ( s  i )  ( s  i ) 1 2i
L[ f (t )u (t )] 

2
2
2i
s 
2i s 2   2

sin( t ) ˆ 2
s 2
The Laplace transform
Transform pair deduction
Time shift
f(t)
f(t-a)
The original function f(t) is shifted in time to f(t-a)

L[ f (t  a)u (t  a)]   f (t  a)e  st dt
0
0
Let x  t  a, then dx  dt and t  x  a
As t  a, x  0 and as t  , x  .


0
0
a
f (t  a) ˆ eas F (s)
L[ f (t  a)]  f ( x)e  s ( x  a ) dx  e as  f ( x)e  sx dx
Frequency shift


0
0
L[e at f (t )]   [e at f (t )]e  st dt   f (t )e ( s  a )t dt  F ( s  a )
L[eat f (t )] F (s  a)
t
Inverse Laplace transform
The algorithm of inverse Laplace transform
Since the F(s) is mostly fractional function, then the most important step is to
perform partial fraction decomposition of it.
Depending on roots in denominator, we are looking for the following functions,
where A and B are real numbers:
A
sa
for a single real root s= a
A
( s  a) 2
for a double real root s= a
A
( s  a) n
for a real root of multiplicity n
As  B
s2   2
for a pair of pure imaginary roots s= ± iω
As  B
( s  a) 2   2
for a pair of complex conjugated roots s= a ± iω
Inverse Laplace transform
Basic examples of partial fraction decomposition to find the original f(t)
Two distinct real roots
F ( s) 
2s  3
s 2  4s  3
The equation s2 + 4s + 3= 0 has two distinct real roots s1= -3 and s2= -1
2s  3
A
B


s 2  4s  3 s  3 s  1
We have to find coefficients
A and B for
F ( s) 
Multiplying the equation
by its denominator
2s  3  A(s  1)  B(s  3)
s  3 :  3  2 A  A 
Now we can substitute
s  1 : 1  2 B  B 
Decomposed F(s)
F (s) 
so the original function
f (t ) 
1
2
3 1
1 1

2 s  3 2 s 1
3  3t 1  t
e  e ,t 0
2
2
3
2
Inverse Laplace transform
One real root
and one real double root
3s 2  5
F ( s) 
( s  1)(s  3) 2
The denominator has a single root s1= -1 and a double root s23= -3
We are looking for coefficients
A, B and C
Multiplying the equation
by its denominator
Now we can substitute to get A,B;
the C coefficient can be obtained
by comparison of s2 factors
Decomposed F(s)
so the original function
3s 2  5
A
B
C
F ( s) 



(s  1)(s  3) 2 s  1 (s  3) 2 s  3
3s 2  5  A(s  3)2  B(s  1)  C(s  1)(s  3)
s  1 : 8  4 A  A  2
s  3 : 32  2B  B  16
s2 : 3  A  C  C  3  A  1
F ( s) 
2
16
1


s  1 ( s  3) 2 s  3
f (t )  2et 16 t e3t  e3t , t  0
Inverse Laplace transform
Two pure imaginary roots
F ( p) 
4s  7
s 2  16

ˆ sin  t ,
2
2
s 
Since we know, that
s
ˆ cos  t
2
2
s 
it will be helpful to rearrange
the original formula
4s  7
s
7 4

4

s 2  16
s 2  16 4 s 2  16
Now we can directly
write the result
7
f (t )  4 cos 4t  sin 4t , t  0
4
and
Inverse Laplace transform
One real root
and two pure imaginary roots
We are looking for coefficients
A, B and C
Multiplying the equation
by its denominator
F ( s) 
2s  7
( s  6)(s 2  4)
2s  7
A
Bs  C


( s  6)(s 2  4) s  6 s 2  4
2s  7  A(s 2  4)  ( Bs  C)(s  6)
19
40
19
s2 : 0  A  B  B   A 
40
1
19
51
s 0 :  7  4 A  6C  C  (7  )  
6
10
60
s  6 :  19  40A  A  
Now we can substitute to get A;
B,C coefficients can be obtained
by comparison of s0,s2 factors
19 1
19 p
51 2


40 s  6 40 s 2  4 120 s 2  4
Decomposed F(s)
F ( s)  
so the original function
f (t )  
19 6t 19
51
e  cos 2t 
sin 2t , t  0
40
40
120
Inverse Laplace transform
Two complex conjugated roots
F (s) 
4s  5
s 2  4 s  13
We have to rearrange the
denominator in the first step
s 2  4s  13  s 2  4s  4  9  (s  2)2  9
Decomposed F(s)
4s  5
4( s  2)
58


s 2  4s  13 ( s  2) 2  9 ( s  2) 2  9
Now we have to assemble all
necessary relations
so the original function
F ( s  2) ˆ f (t )e 2t
3
s
ˆ sin 3t ; 2
ˆ cos3t
2
s 9
s 9
f (t )  e2t (4 cos3t  sin 3t ), t  0
Solving of differential equations by the Laplace transform
Example 1
Find the x(t) on the interval <0,∞)
x  2 x  3, x(0)  0
The image of desired function is
X (s) ˆ x(t )
From the former definitions we
know, that
Then we can write
3
A
B
 
s( s  2) s s  2
x(t ) ˆ s X (s)  x(0)
3
s X ( s)  0  2 X ( s) 

s
3  A(s  2)  Bs
3
X ( s) 
s( s  2)
s  0 : 3  2A  A 
3
2
s  2 : 3  2 B  B  
X ( s) 
3
3

2s 2( s  2)
The original function
x(t ) 
3
(1  e  2t ), t  0
2
3
2
Solving of differential equations by the Laplace transform
x  4 x  sin 2t , x(0)  3
Example 2
Function on the right side
Necessary relations
x(t ) ˆ s X (s)  x(0)
sin  t ˆ
Equation in the Laplace form
s X (s)  3  4 X (s) 
2
s2  4
2
X ( s)(s  4)  3  2

s 4
2
A
Bs  C

 2
2
( s  4)(s  4) s  4 s  4
s  4 : 2  20A  A 
1
10
s2 : 0  A  B  B   A  
p2   2
3
2
X ( s) 
 2
s  4 ( s  4)(s  4)
2  A(s 2  4)  ( Bs  C)(s  4)
s 0 : 2  4 A  4C  C 
1
10

2  4A 2

4
5
Solving of differential equations by the Laplace transform
Example 2 - continued
A formula for the X(s) after the
partial fraction decomposition
1
1
2
 s
3
5
X ( s) 
 10  102
s4 s4
s 4
after some small arrangements
31
1
2
 s
5
X (s)  10  102
s4
s 4
31
1 s
1 2
X (s)  10 

s  4 10 s 2  4 5 s 2  4
The original function
x(t ) 
31  4t 1
1
e  cos 2t  sin 2t , t  0
10
10
5
Solving of differential equations by the Laplace transform
Example 3
Homogeneous second order LDR
Necessary relations
x  6 x  9 x  0, x(0)  2; x(0)  4
x(t ) ˆ s X (s)  x(0)
x(t ) ˆ s 2 X (s)  s x(0)  x(0)
Equation in the Laplace form
s 2 X (s)  2s  4  6(s X (s)  2)  9 X (s)  0
X ( s)( s  6s  9)  2s  16 
2
2s  16
X ( s)  2
s  6s  9
2s  16 2( s  3)  10
2
10
X ( s) 



2
2
( s  3)
( s  3)
s  3 ( s  3) 2
The original function
x(t )  (2 10t )e3t , t  0
Solving of differential equations by the Laplace transform
Example 4
Inhomogeneous second order LDR
x  4 x  2 cos2t , x(0)  0; x(0)  4
x(t ) ˆ s X (s)  x(0)
Necessary relations
x(t ) ˆ s 2 X (s)  s x(0)  x(0)
Equation in the Laplace form
2s
s X (s)  2  0  4  4 X ( s)  2
s 4
2
2s
X ( s)(s  4)  4  2

s 4
2
knowing that
2s
t sin t ˆ 2
(s   2 )2
The original function
4
2s
X ( s)  2
 2
s  4 ( s  4) 2
1
f (t )  2 sin 2t  t sin 2t
2
1
x(t )  (4  t ) sin 2t , t  0
2
Solving of differential equations by the Laplace transform
t
Example 5
Integro-differential equation
Necessary relations
x   x( )d  1, x(0)  1
0
t
 x( )d ˆ
0
1
X ( s );
s
x(t ) ˆ s X (s)  x(0)
Equation in the Laplace form
1
1
sX ( s )  1  X ( s ) 

s
s
X ( s )( s  1)  1  s 
2
The original function
1
1
X ( s )( s  )  1 
s
s
s 1
s
1
X (s)  2
 2
 2
s 1 s 1 s 1
x(t )  cost  sin t , t  0