Laplace Transform
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ME 467 Systems Dynamics & Automatic Control
Chapter 2: LAPLACE Transforms
1
Transformation Definition
Transforms -- a mathematical conversion from one domain (form) to another to make a problem easier to solve
problem in srcinal form
solution in srcinal form transform
solution in transform way of thinking
inverse transform
Laplace Transform problem in time domain
solution in time domain
Laplace transform
•
inverse solution in s domain
Other transforms • Fourier • z-transform • wavelets
Laplace transform
Nature of the Laplace Transform
Why study Laplace Transforms
The Laplace Transform provides a tool for the rapid solution of ordinary constant coefficient linear differential equations It gives us a straight forward procedure to solve and analyze dynamic systems.
Nature of the Laplace Transform
L(f(t)) Laplace Domain Differential Equation f(t)
Algebraic Equation f(s)
Time Domain Non Differential Equation f(t)
L-1(f(s))
Algebraic Manipulation
Algebraic Equation f(s)
Laplace Transform L[
f (t )] F ( s)
0
f (t )e st dt Complex Number
Convert time-domain functions and operations into frequency-domain
f(t) F(s) (tR, sC) Linear differential equations (LDE) algebraic expression in Complex plane
Graphical solution for LDE characteristics
The Complex Plane (review) Imaginary axis (j)
u x jy
y
y
u tan 1 y
r
x
x
Real axis
r
| u | r | u |
u x jy (complex) conjugate
x2 y2
Basic Theorems of Linearity L[ Kf()]t KL [ ()]f t ( )KF s
L[ f1( )t 2 f( )]t1 [ L(2)]f t [ (L)] f t
F1 ( s) F(2) s
The Laplace transform of a product is not the product of the transforms.
L[ f1( )t2 f( )]t 1 2(F) s( )F s
Unit step function. F (s)
0
()f t e st dt
f(t) = u(t) = 1
F ( s) (1) e st dt
e0 F ( s) 0 s 0 s e
st
0
1
s
Exponential Function (t ) e at F (s )
0
e
at st
e
e
( s a )t
dt
0
0 (s ) a 0 ( )
1
s a
ast (
e
)
dt e 0 s
a
Exponential function multiplied by another function
g ()t e at()f t
at
G )s( e f)( t e dt 0
F )(s Example
s s a
at
st
)(f t e
( F) s a
L [1*e ]
1
s
s s a
dtt sa
0
1
s a
(
)
Derive the Laplace transform of the sinusoidal function 1 L [ cos t )] L e j 2
t
e j t
1 L [e j ]t [ 1 L] e j t
2 1
1
2 1 1
2 sj
s s 2 2
2 s j
Complex Differential Theorem L [t f()]t
In general
d ds
( )F s
L [t f ( )]t (1) n
n
dn (n) F s ds
example
L [t *1] 1
d 1 1 2 ds s s
Laplace Transforms of Common Functions Name Impulse
f(t)
1 f (t ) 0
F(s) t0 t0
1
Step
f (t ) 1
1
Ramp
f (t ) t
1
Exponential
f (t ) e at
Sine
f (t ) sin(t )
s
s2 1
sa 1
2 s2
a is +ve
Example
(t ) 50 u( )t
F ( s)
50
s
Example
vt ( ) e5
2t
V (s ) L[v (t )] 5
20
sin t 4 4
( s 2) 2 (4) 20
2
2 s 4s 4 16 s 4 20 s 2
Example
p(t) 5cos2 t 3 e4t s 3 s 2 (2) 2
P()s [ (L)] p5 t
5s
s 4 2
3
s4
1 s 4
Most mathematical handbooks have tables of Laplace transforms
Inverse Laplace Transforms by Identification When a differential equation is solved by
Laplace transforms, the solution is obtained as a function of the variable s. The inverse transform must be formed in order to determine the time response. The simplest forms are those that can be recognized within the tables and a few of those will now be considered.
Example F ( s)
5 12
s
2
s
8
s3
f (t) 5 12 t 8 e3t
Example V ( s)
200
s 2 100
2 2 s (10)
V ( s) 20
10
v(t) 20sin10 t
Determine the inverse transform of the function below. 8s 4 V ( s) 2 s 6s 13
When the denominator contains a quadratic, check the roots. If they are real, a partial fraction expansion will be required. If they are complex, the table may be used. In this case, the roots are
s1,2 3 2i
Example 8. Continuation.
s 6s 13 2
2
2
s 2 6 s (3) 13 ( 3) s 6s 9 4
( s 3) 2 (2)
2
2
Example, Cont. V ( s)
8( s 3)
4 24
2 2 ( s 3) (2) ( s3) 2
2
(2)
8( s 3) 2 10(2) 2 2 ( s 3) (2) ( s3)2 (2)
tv( )e 8
3t
tcose2 10 t
3t
sin 2
LAPLACE TRANSFORMS PARTIAL FRACTION EXPANSION
Definition
Definition -- Partial fractions are several fractions whose sum equals a given fraction
Purpose -- Working with transforms requires breaking
complex
fractions
into
simpler
fractions to allow use of tables of transforms
Partial Fraction Expansions s 1 A B ( s 2) ( s 3) s2 s3 s 1 A( s 3) Bs 2) ( s 2) ( s 3) ( s 2) ( s 3)
A B 1
3 A 2 B 1
s 1 1 2 ( s 2) ( s 3) s 2 s3
Expand into a term for each factor in the denominator. Recombine RHS Equate terms in s and constant terms. Solve. Each term is in a form so that inverse Laplace transforms can be applied.
Example F ( s)
s6 s6 s 2 3s 2 ( s 1)( s2)
F (s)
s6 A A 1 2 ( s 1)( s 2) 1s 2s
A1 ()1) s(
F s
s 1
1 6 s 6 5 s 2 s 1 1 2
Example, cont. A2 ()s2) (
F s s 2
F ( s)
5
s 1
s 6 2 6 4 s 1 s 2 2 1
4
s2
f ()t 5 e4t e2t
Example 50( s 3)
F ( s)
( s 1)( s 2)( s 22 s5)
F1 ( s)
A1 s 1
A2 s2
Example, Cont. A1
( s 2 )( s 2 2 s 5) 50( s 3)
(50)(2)
(1)(4)
25
s 1
A2
(50)(1) 2 ( s 1)( s 2 s 5) s 2 ( 1)(5) 50( s 3)
f1 (t) 25 et 10 e2t
10
Example, Cont. To get A and B, substitute by any two convenient numbers, (i.e. s=0 and s=1)
50( s 3)
25
( s 1)( s2)( s22 5)s
50(3) (1)(2)(5) 50(4) (2)(3)(8)
25 1 0
25 10
1 2
F ( s)
B 5
2
3
10
2 1 s 2 s 2 5s
A B 8
25
10
s 1
s2
B 25
A 15 15 s25 2 2s 5
s
As B s
Example, cont. F2 ( s)
15s 25 s 2 2s 5
s 2 2s25 s 2 s15122 ( 1) s(2) F2 ( s)
s1)2 2 5(2) 15 s2 25 2 15( 2 2 ( s 1) (2) ( s1) (2) ( 1)s (2)
f (t) (f1) t ()f 2 t t 25et 10 e
t2
t e 15 t ecos 2t 5
sin 2
Example, Repeated Roots F (s)
F (s)
60
2
s( s 2)
A )sF ( ss 0
C1 s( 2) )F ( s2
60
s( s 2)2
C1 A s ( 2) s(
2
C2 2) s
60 60 2 15 ( s 2) s 0 (0 2)
s 2
2
60 60 30 s s 2 2
Example, Cont. F ( s)
60
s( s 2)
15
2
C2 2 2 s
30
s( 2)s
Let s=1, and solve for C2
60
15
30
(1)(1 2) 2 1 (1 2)
F (s)
60
s(s 2)
2
2
15
C 2 (1 2)
C2 15
30
s( 2)s
f (t) 15 30 te2t15 e2t1515
15 2 s
2
(1 e22t )
t
Repeated Roots –General Form Given
) s zm B( s) K s z)1 s ) z 2 ,for n A( s) s p) It can be expressed as )F ( s
)F ( s B(s) A( s)
b2 2 b1 1 ... s) p )s p )s p
n B( s ) s p bn ) A ( s ) s p
1
n !1ds )
mn
bn
n
d n B( s ) s p bn1 ) ds As ( ) s p
d n 1
s p) n 1
)A s(
n
B (s )
b1 s p
Laplace of Differentiated Functions
L[ f '( )] t sF ( ) s (0) f L[ f "( )] t s2 F ( ) s (0) sf n 1 L [f n( )] t n s n(F) s s (0) f
n
f '(0)
2 s '(0).... f
(0) f 1 d f (t )
Note: s is called Differentiation Operator
(t )
dt
s
Laplace of Integrated Functions t F ( s) L f (t )dt 0 s
Note: 1/s is called Integrator operator (t )
1
s
f (t ) dt
Solve This Differential Equation dy y (0) 10 2 y 12 dt dy L 2 L y L 12 dt sY ( s) 10 2 Y ( ) s 12 s
s 2 ) Y( s) 10 Y ( s)
10
s2
12
12
s( s 2)
s
Example Solve the following differential equation
dy 2 y 12sin4 t dt
y (0) 10 12(4)
sY ( s) 10 2 Y ( ) s s 2 16
Y ( s)
10
48
s 2 ( s 2)( s 2 16)
48 ( s 2)( s 2 16)
A s2
B1s B2 2 s16
Example, cont. 48 2.4 2 s 16 s 2 20 48
A
48
( s 2)( s 2 16) 48 (2)(16) 48 (1)(17)
2.4 2
2.4 1
2.4
s2
B2 16
B1s B2 2 s16
B2 4.8
B1 B2
B1 2.4
17
Example, Cont.
Y ( s)
10
2.4
s 2 s2
2.4 s 2
s16
4.8 2
16 s
y(t) 12.4 e2t 2.4cos4 t 1.2sin 4 t
Example d 2y dy 3 2 2y 4 2 dt dt y(0) 10 and y'(0) 0
(sY) 1s0 2 ( )
s 2Y ( s) 1 0 s0 3
Y s
10 s 30 3s2 2 s s ( s 2 3 s2) 24 10 s30 s ( s 1)(s 2) ( s1)( s2)
Y ( s)
24
24 s
Example, Cont. 24
12
s(s 1)( s2) 10s 30
24
s 1s 20
12
s
4
s 1
2 s
10
( s 1)( s 2) 1s F ( s)
12
2s 2
s2
f (t) 12 4 et2 e2t
Example d 2y dy 2 5 2y 0 2 dt dt y(0) 0 and y'(0) 10
s 2Y ( s) 010 2 Y ( s)
(sY) 0s 20
s( s 2 2 s5)
5( ) Y s 10
2 2s 5 s
20
s
Example, Cont. 20
s( s 2 2s5)
As B 2 (s 2s 5) s 4
A B (1)(1 2 5) 1 (1 2 5) 20
20 (1)(1 2 5)
A 4
4
A B 1 (1 2 5) 4
B 8
Example, Cont. Y ( s) 2
4 8s
4
s 2 2s5
s 2
10
4 4 2
2s 25 s
s
2s5 s 2
22
s 2s 5 s 2 s151 ( 1) s(2) Y ( s)
4
s
4( s1) 3(2) 2 ( s 1)2 (2) ( 2s1)2 (2)
t t y (t) 4 4 ecos 2 t3 esin 2
t
s
Initial- and Final-Value Theorems
For the given differential equations find the initial and final value of y(t)
y (t) 6 (y )t 8 ( y) t 2( )u t
* Apply Initial- and Final-Value Theorems
y (t) 6 (y )t 8 ( y) t 2( ) u t Y ( s)
lim t f (t )
lim t 0 f (t )
2
s ( s 2) ( s 4) 2 (0)
(0) (0 2) (0 4) 2 () () ( 2) ( 4)
1
4
0
Laplace transform of the function. Apply final-value theorem Apply initialvalue theorem
Step functions Steps functions are used to form piecewise continuous functions.
Unit step function(Heaviside function):
A plot of uc(t) is :
Step functions For a given function f(t), if it is multiplied with uc(t), then
uc picks up the interval [c;).
Rectangular pulse. The plot of the function looks like
for 0 · a < b < 1. We see it can be expressed as
and it picks up the interval [a; b).
Example For the function
In terms of the unit step function, f(t) can rewritten as:
Laplace transform of Step functions uc(t)
Shift of a function Given f(t), t > 0, then
is the shift of f by c units.
Shift of a function Let y = t - c, so t = y + c, and dt = dy,
which is equivalent to
Note now we are only considering the domain t 0. So u0(t) = 1 for all t 0.
Example For the function
In terms of the unit step function, f(t) can rewritten as:
The Laplace transform is
Example For the function
In terms of the unit step function, f(t) can rewritten as:
The Laplace transform is
Summary Steps for Solving an ODE d2y dy 6 8y 2 2 dt dt
y (0) y ' (0) 0 ODE w/initial conditions
s 2 Y ( s) 6s Y ( s) 8 Y ( s) 2 / s
Y ( s)
Y ( s)
1 4s
2
s ( s 2) ( s 4)
1 1 2 ( s 2) 4 ( s 4)
y (t )
1 4
e 2t e 4t 2 4
Apply Laplace transform
to each term Solve for Y(s)
Apply partial fraction expansion Apply inverse Laplace transform to each term
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