Differential equations model dynamic systems Laplace transform …ncbs.knu.ac.kr/Teaching/ACS_Files/ACS_Week3a.pdf · 2017-03-05 · Laplace Transform of a Unit Step Find the Laplace

Lecture 4

Laplace Transform: Motivation

Differential equations model dynamic systems

Control system design requires simple methods for solving these equations!

Laplace Transforms allow us to

– systematically solve linear time invariant (LTI) differential equations for arbitrary inputs.

– easily combine coupled differential equations into one equation.

– use with block diagrams to find representations for systems that are made up of smaller subsystems.

uxbxm =+⇔ &&&

Lecture 4

Laplace and Inverse Laplace

Dr. Kalyana Veluvolu

1 2

Laplace Transforms:

Laplace transform of a time domain function f(t) is

(Doubled-sided Laplace transform)

where s = σ + jω is the complex variable.

In most engineering problems, f(t) is causal, i.e.

Thus, single-sided Laplace transform is used, i.e.

{ } ∫∞

∞−

−== ττ τdefsFtf

s)()()(L

0for 0)( <= ttf

{ } ∫∞

−==

0

)()()( ττ τdefsFtf

sL

Lecture 4



Laplace Transform Example 1

Example:

Show that

Notation for “unit step”

Lecture 4



3

Laplace Transform Example 1 contd..

Lecture 4



4

Laplace Transform of a Unit Step

Find the Laplace Transform for the following function

∞≤≤

=otherwise0

01)(

ttu s

[ ]

s

s

es

dtesFstst

1

101

11)(

00

=

−−

=

−==

∞

−

∞

−

−−

∫

Lecture 4



5

Example 2

Find the Laplace Transform for the following function

≤≤

=otherwise0

103)(

ttf

[ ]

[ ]s

s

stst

es

es

es

dtesF

−

−

−−

−=

−−

=

−==

−−

∫

13

13

33)(

1

0

1

0

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6

The function with the simplest Laplace Transform (1)

A special input (class) has a very simple Laplace Transform

The impulse function:

– Has unit “energy”

– Is zero except at t=0

Think of pulse in the limit

Lecture 4



( ) ( )( ) 1t

sFtf

δ

7

LT Properties: Scaling and Linearity

Proof: Both properties inherited from linearity of integration and

the Laplace Transform definition

( ) ( )( ) ( )

( ) ( ) ( ) ( )sFsFtftf

saFtaf

sFtf

2121 ++

Lecture 4



8

Example 3

Find the following Laplace Transforms

Use Euler’s Formula

Lecture 4



( )tjtjee

jt

ωωω −−=2

1sin

[ ]

22

11

2

1sin

ω

ω

ωωω

+=

+−

−=

s

jsjsjtL [ ]

22

11

2

1cos

ω

ωωω

+=

++

−=

s

s

jsjstL

( )tjtjeet

ωωω −+=2

1cos

( ) ( )

( )

( )22

22

cos

sin

ωω

ω

ωω

+

+

s

st

st

sFtf

9

LT Properties: Time and Frequency Shift

Proof of frequency shift: Combine exponentials

( ) ( )( ) ( ) ( )

( ) ( )σ

ττσ

τ

+

−−−

−

sFtfe

sFetutf

sFtf

t

s

Lecture 4



10

Example 4

[ ]

22

22

)(

cos

ω

ωω

++

+=

+=

+=

−

as

as

s

ste

ass

atL

[ ]

22

22

)(

sin

ω

ω

ω

ωω

++=

+=

+=

−

as

ste

ass

atL

( ) ( )

( )( )

( )( ) 22

22

cos

sin

ωω

ω

ωω

++

+++

−

−

as

aste

aste

sFtf

at

at

Find the following Laplace Transforms

Lecture 4



11

LT Properties: Integration & Differentiation

Proof of Differentiation Theorem: Integration by parts

∫∫ −= vduuvudv

Lecture 4



12

LT Properties: Integration & Differentiation

( ) ( )

( ) ( ) ( )

( )( )

( )∫∫ ∞−∞−

−

+

−

01

0

ττττ dfss

sFdf

fssFtfdt

d

sFtf

t

Lecture 4



13

Some Examples:

[ ] [ ]2

111)(

ssstdtut === ∫LL

1)0(1)(

=−=

−u

ss

dt

tduL Impulse!

10

1

1)sin(22 +

=−+

=

s

s

ss

dt

tdL Cosine!

[ ]( )22

11

asste

ass

at

+==

+=

−L

( ) ( )

( )

( )

( )1

sin

1

1

1

2

2

2

+

+

−

s

st

dt

d

tudt

d

aste

st

sFtf

at

Derivative of a step?

Derivative of sine?

Lecture 4



14

15

Properties of the Laplace Transforms

Let F(s) = LLLL {f(t)}.

(1) Time Differentiation

(2) Time Integration

(3) Complex Translation (Shifting in the s-domain)

in

inn

i

in

n

n

dt

fdssFstf

dt

d

−−

−−−

=∑−=

1

11

0

)0()()(L

n

t t t

ns

sFddf

n )()(

1 2

0 0 0

11 =

∫ ∫ ∫ τττ LLL

{ } )()( asFtfe at ±=mL

Lecture 4



16

(4) Real Translation (Shifting in the time domain)

where u(t – T) is a shifted unit-step function.

(5) Real Convolution (Complex Multiplication)

(6) Initial Value Theorem

{ } )()()( sFeTtuTtfsT−=−−L

∫∫ −=−=∗

tt

dtffdtfftftf0

12

0

2121 )()( )()()()( ττττττ

{ } )()()()( 2121 sFsFtftf =∗L

)(lim)0()(lim0

ssFftfst ∞→→

==

Lecture 4



17

(7) Final Value Theorem

provided sF(s) is analytic in the closed right-half of the s–plane

(i.e. the denominator of sF(s) has no roots on the jω−axis or in the

right-half of the s–plane).

Example: Consider

i.e.

and hence So, yss = 1.

Alternatively, by final value theorem, we have

)(lim)(lim0

ssFtfst →∞→

=

ssR

sssT

sR

sY 1)( ;

)2)(1(

2)(

)(

)(=

++==

2( ) 1 2 .

t ty t e e

− −= + −

1)(lim)(lim0

===→∞→

ssYtyyst

ss

2 1 2 1( )

( 1)( 2) 1 2=Y s

s s s s s s= − +

+ + + +

Lecture 4



18

Example:

In this case, denominator of F(s) has 2 roots on the jω−axis, so the

f.v.t. cannot be used. (If we applied the f.v.t., it gives a final value of

zero which is wrong!)

ttfs

sF ωω

ωsin)( i.e )(

22=

+=

Example:

Clearly, sF(s) is analytic in the closed right-half of the s–plane, so

the f.v.t. can be applied to give

)2(

5)(

2 ++=

ssssF

2

5)(lim)(lim

0===

→∞→ssFtff

stss

Lecture 4



Basic Laplace Transform Pairs

Lecture 4



19

Inverse Laplace Transform

1σ• is such that the integral is taken over a line in the region

of convergence

• Very difficult to apply directly

• Instead, we convert X(s) to a form such that we can easily

find the inverse

Lecture 4



20

Example

Lecture 4



21

The differentiation theorem

Higher order derivatives

Laplace Differentiation Theorem

Lecture 4



22

Differentiation Theorem

Differentiation Theorem when initial conditions are zero

( ) ( ) ( ) ( ) ( )−

−

−−−−− −−−−↔ 000

1

121 f

dt

df

dt

dsfssFstf

dt

dn

nnnn

n

n

K

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23

Solving Differential Equations: an Example

Consider0,1 ≥= t

dt

dx

Lecture 4



24

Solving Differential Equations: an Example

Solution Summary

– Use differentiation theorem to take Laplace Transform of the

differential equation

– Solve for the unknown Laplace Transform Function

– Find the inverse Laplace Transform

( ) ( ) 0,0 ≥+= − txttx

Lecture 4



25

Example 5

Find the Laplace Transform for the solution to

( ) ( ) ( ) ( )( ) ( )( )

( ) ( ) ( ) ( ) ( )[ ] ( ) 120300

00...0

2

121

=+−+−−⇒

−−−−=

−−−

sXxssXxsxsXs

fsffssFstfdt

dL

nnnn

n

n

&

Notation:

( ) ( ) 30,100,123 −==≥=++ xxtxxx &&&&

Lecture 4



26

Partial Fraction Expansions

In general, ODEs can be transformed into a function that is expressed

as a ratio of polynomials

In a partial fraction expansion we try to break it into its parts, so we can

use a table to go back to the time domain:

Three ways of finding coefficients

– Put partial fraction expansion over common denominator and

equate coefficients of s (Example 1)

– Residue formula

– Equate both sides for several values of s (not covered)

Lecture 4



27

Partial Fraction Expansions

Have to consider that in general we can encounter:

– Real, distinct roots

– Real repeated roots

– Complex conjugate pair roots (2nd order terms)

– Repeated complex conjugate roots

222

2

222

11 ))(()()()(

)()(

bas

GFsEs

bas

DCs

ps

B

ps

AK

sD

sNsX

++

+++

++

++

++

++==


Example 6

Given X(s), find x(t).

Step 1:

Factorize the denominator, then use partial fraction expansion:

Lecture 4



29

Step 2: Finding A, B, and C

To solve, re-combine RHS and equate numerator coefficients (“Equate

coefficients” method)

Example 6 contd..Lecture 4



30

Finally,

Since

By inspection,

α

α

+↔≥−

s

KtKe

t 0,

( ) 0,2

52

2

1 2 ≥+−= −−teetx

tt

Example 6 contd..

Lecture 4



31

Residue Formula (1)

The residue formula allows us to find one coefficient at a time by

multiplying both sides of the equation by the appropriate factor.

Returning to Example 1:

Lecture 4



32

Residue Formula (2)

For Laplace Transform with non-repeating roots,

The general residue formula is:

Lecture 4



33

)2(

1

)1(

2)(

+

−+

+=

sssX 0,2)( 2 ≥−= −− teetx tt

( ) ( ) ( )( )

( ) 0)(21)(3)(

0)(20)(300)(

2

2

=+−+−

=+−+−−

sXssXssXs

sXxssXxsxsXs &

3)()23( 2 +=++ ssXss

)2()1()2)(1(

3)(

++

+=

++

+=

s

B

s

A

ss

ssX

( ) ( ) 221

311

1=

+−

+−=+=

−=ssXsA ( ) ( ) 1

12

322

2−=

+−

+−=+=

−=ssXsB

Example 7: Find the solution to the following differential equation:

0)0( 1(0) 023 ===++ xxxxx &&&&

Solution:

Lecture 4



34

Inverse Laplace Transform with Repeated Roots

Now we will consider partial fraction expansion rules for functions

with repeated (real) roots:

– # of constants = order of repeated roots

Example:

23223

4

)3()3(3)3(

)1(

s

E

s

D

s

C

s

B

s

A

ss

s++

++

++

+=

+

+

Lecture 4



35

1)3()3(3)1()3(

)1(23223

4

−+++

++

++

+=

−+

+

s

F

s

E

s

D

s

C

s

B

s

A

sss

s

The easiest way to take an inverse Laplace transform is to use a table

of Laplace transform pairs.

Repeated real roots in Laplace transform table

( )

( )222

222

1

2

)(

)(2)sin(

2)sin(

)(

1

!

)(

1

)()(

ω

ωω

ω

ωω

++

++

+

+

−

+

−

−

as

astte

s

stt

asn

et

aste

sFtf

at

n

atn

at

Repeated Real Roots

Repeated Imaginary Roots

(also use cosine term)

Repeated Complex Roots

(also use cosine term)

}

Lecture 4



36

Example with repeated roots

Find x(t)

Take Laplace Transform of both sides:

tef

2−=

21

1x

( ) ( ) 10,00

2 2

==

=++ −

xx

exxx t

&

&&&

Lecture 4



37


Terms with repeated roots:

Lecture 4



38


C = 1B = 2

Lecture 4



39

Example

Find the solution to the following differential equation

0)0(1)0(044 ===++ xxxxx &&&&

( ) ( ) ( ) ( ) ( )[ ] ( )

( ) ( ) ( )

( )[ ]

( )( ) ( )

( ) ( ) ( )

( )( )( )

( ) tt

ss

teetx

Ass

sA

s

s

ssXsB

s

B

s

A

s

ssX

ssssX

sXssXssXs

sXxssXxsxsXs

22

22

22

2

22

2

2

2

2

1:2

22

2

4

242

222

4

444

04440

040400

−−

−=−=

+=

=⇒+

++=

+

+

=+=+=

++

+=

+

+=

+=++

=+−+−−

=+−+−− &

Lecture 4



40

Inverse Laplace Transform with Complex Roots

To simplify your algebra, don’t use first-order denominators such as

Instead, rename variables

So that

21 KKB += ( ) ( )( )21 11 KjKjC −++=

Lecture 4



41

More Laplace transform pairs (complex roots):

Also, see the table in your textbook and most other control systems

textbooks.

Laplace Transform Pairs for Complex Roots

22

22

)()sin(

)()cos(

)()(

ωσ

ωω

ωσ

σω

σ

σ

++

++

+

−

−

ste

s

ste

sFtf

t

t

Lecture 4



42

Example with complex roots

Example: find x(t)

Laplace Transform

Lecture 4



43

Example with complex roots (2)

Lecture 4



44


Lecture 4



45


Lecture 4



46

Example

Find solution to the following differential equation

0)0(1)0(084 ===++ xxxxx &&&&

( ) ( ) ( ) ( ) ( )[ ] ( )

( ) ( ) ( )

( )

( )

( ) ( )

( ) ( ) ( )tetetx

ss

s

s

s

ss

ssX

sXssXssXs

sXxssXxsxsXs

tt 2sin2cos

22

2

22

2

42

4

84

4

0844

080400

22

2222

2

2

2

2

−− +=

+++

++

+=

++

+=

++

+=

=+−−−

=+−+−− &

Lecture 4



47

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Differential equations model dynamic systems Laplace transform …ncbs.knu.ac.kr/Teaching/ACS_Files/ACS_Week3a.pdf · 2017-03-05 · Laplace Transform of a Unit Step Find the Laplace