307A-LaplaceTransformTechniques

8/8/2019 307A-LaplaceTransformTechniques

1/61

EGGN 307Introduction to Feedback Control Systems

Laplace Transform Techniques

(Lectures 8-12)

Professor Kevin L. Moore

Spring 2010

http://engineering.mines.edu/course/eggn307a


2/61

2

Colorado School of Mines

3.0 Laplace Transform Modeling

3.1 Review of Complex Numbers

3.2 Laplace Transforms

Due to Katie Johnson or Tyrone Vincent or someone


3/61

3


Laplace Transform Motivation

Differential equations model dynamic systems

Control system design requires simple methods forsolving these equations!

Laplace Transforms allow us to systematically solve linear time invariant (LTI) differential

equations for arbitrary inputs.

easily combine coupled differential equations into oneequation.

use with block diagrams to find representations for systemsthat are made up of smaller subsystems.

uxbxm !



4/61

4


The Laplace Transform Definition

Laplace Transform exists if integral converges for any

value ofs Region of convergence is not as important for inverting one-

sided transforms



5/61

5


Laplace Transform Example (1)

Example:

Show that

Notation for unit step



6/61

6


Laplace Transform Example (2)



7/61

7


Laplace Transform of a Unit Step

Find the Laplace Transform for the following function

gee

!otherwise0

01)(

ttus

? A

s

s

es

dtesFstst

1

101

11)(

00

!

!

!!

g

g



8/61

8


Exercise

Find the Laplace Transform for the following function

ee

!otherwise0

103)(

ttf

? A? A

tttF

!

!

!!

13

13

33)(

1

0

1

0



9/61

9


The Laplace Transform Definition (Review)

Recall:

The easiest way to use the Laplace Transform is by

creating a table of Laplace Transform pairs. We can

use several Laplace Transform properties to build the

table.



10/61

10


The function with the simplest Laplace Transform (1)

A special input (class) has a very simple LaplaceTransform

The impulse function:

Has unit energy

Is zero except at t=0

Think of pulse in the limit



11/61

11


The function with the simplest Laplace Transform (2)

1t

sFtf

H



12/61

12


LT Properties: Scaling and Linearity

Proof: Both properties inherited from linearity ofintegration and the Laplace Transform definition

sFsFtftf

saFtaf

sFtf

2121



13/61

13


Example 1

Find the following Laplace Transforms

Hint: Use Eulers Formula



14/61

14


Example 1 (2)

tjtj eejt[[[

! 2

1sin

? A

22

11

2

1sin

[

[

[[[

!

!

s

jsjsjtL

tjtj

eet[[

[

! 2

1

cos

? A

22

11

2

1cos

[

[[[

!

!

s

s

jsjstL

22

22

cos

sin

[[

[

[[

s

st

st

sFtf



15/61


16/61

16


Example 2

Find the following Laplace Transforms



17/61

17


Example 2 (2)

? A

22

22

)(

cos

[

[[

!

!

!

as

as

s

ste

ass

atL ? A

22

22

)(

sin

[

[

[

[[

!

!

!

as

ste

ass

atL

22

22

cos

sin

[[

[[[

as

aste

aste

stf

at

at



18/61

18


LT Properties: Integration & Differentiation

Proof of Differentiation Theorem: Integration by parts

! vduuvudv



19/61

19


LT Properties: Integration & Differentiation (2)

gg

01

0

XXXX dfss

sFdf

fssFtfdt

d

sFtf

t



20/61

20


Example 3

Find Laplace Transform for

What is the Laplace Transform of

Derivative of a step?

Derivative of sine?



21/61

21


Example 3 (2)

? A ? A 2111)( ssstdtut !!! LL

1)0(1)(

!!

-

u

ss

dt

tduL Impulse!

10

11)sin(

22 !

!

-

ss

ss

dttdL Cosine!

? A 22

11

asste

ass

at

!!

!

L

1

sin

1

1

1

2

2

2

s

st

dt

d

tudt

d

aste

st

stf

at



22/61

22


Exercise

What is the Laplace Transform of

tdt

dcos

1

1

1

1

11

1

)cos(22

2

2

2

2

!

!

!

-

ss

s

s

s

s

ss

dt

tdL -Sine!

1

1cos

2 s

tdt

d

stf



23/61

23


Initial Value Theorem


24/61


25/61

25


3.0 Laplace Transform Modeling

3.1 Review of Complex Numbers

3.2 Laplace Transforms

3.3 Inverse Laplace and LODE Solutions



26/61

26


Inverse Laplace Transform


27/61

27


Partial Fraction Idea -1


28/61

28




29/61

29




30/61

30


The differentiation theorem

Higher order derivatives

Recall: Laplace differentiation theorem (1)



31/61

31


Differentiation Theorem (revisited)

Differentiation Theorem when initial conditions are

zero

m 0001

121 fdtdf

dtdsfssFstf

dtd

n

n

nnn

n

n

-



32/61

32


Solving differential equations: a simple example (1)

Consider

0,1 u! tdt

dx



33/61

33


Solving differential equations: a simple example (2)

Solution Summary

Use differentiation theorem to take Laplace Transform of

differential equation

Solve for the unknown Laplace Transform Function

Find the inverse Laplace Transform

0,0 u! txttx



34/61

34


Example 1

Find the Laplace Transform for the solution to

? A 120300

00...0

2

121

!

!

sXxssXxsxsXs

fsffssFstfdt

d

L

nnnn

n

n

Notation:

30,100,123 !!u! xxtxxx



35/61

35


- Partial Fraction Expansions

In general, LODEs can be transformed into a function that isexpressed 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)


36/61

36


- 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

!!


37/61

37


Example 1, Part 2

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

This Laplace Transform function is not immediatelyfamiliar, but it is made up of parts that are.

Factor denominator, then use partial fraction

expansion:



38/61


39/61

39


Final Step

Example 1 completed:

Since

By inspection,

E

E

mu

ste t 0,

0,2

52

2

1 2 u! teetx tt



40/61

40


Residue Formula (1)

The residue formula allows us to find one coefficientat a time by multiplying both sides of the equation by

the appropriate factor.

Returning to Example 1:



41/61

41


Residue Formula (2)

For Laplace Transform with non-repeating roots,

The general residue formula is:



42/61

42


Example 2

Find the solution to the following differential equation:



43/61

43


Example 2 (2)

)2(

1

)1(

2)(

!

sssX

0,2)( 2 u! teetx tt

0)(21)(3)(

0)(20)(300)(

2

2

!

!

sXssXssXs

sXxssXxsxsXs

3)()23( 2 ! ssXss

)2()1()2)(1(3)(

!! s

BsA

ssssX

221

311

1!

!!

!ssXsA 1

12

322

2!

!!

!ssXsB



44/61

44


Inverse Laplace Transform with Repeated Roots

We have discussed taking the inverse Laplacetransform of functions with non-repeated, real roots

using partial fraction expansion.

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(

sssssss

s

!



45/61


46/61


47/61

47


Example with repeated roots (2)

Terms with repeated roots:



48/61

48


Example with repeated roots (3)

C = 1B = 2



49/61

49


Exercise 1

Find the solution to the following differential equation0)0(1)0(044 !!! xxxxx

? A

? A

tt

ss

teetx

Ass

sA

s

s

sssB

s

B

s

A

s

ss

ssss

ssssss

sxssxsxss

22

22

22

2

22

2

2

2

2

1:2

22

2

4

242

222

4

444

04440

040400

!!

!

!

!

!!!

!

!

!

!

!



50/61

Above

3.3 Inverse Laplace and LODE solutions

- Partial fraction expansions

- LODE solution examples

* Real roots

* Real, repeated roots

Next:* Complex roots


51/61

NOTE:

A complex conjugate pair is actually two distinct,simple first order poles, so can find residues and

combine in the usual way:


52/61


53/61

53


Inverse Laplace Transform with Complex Roots

To simplify your algebra, dont use first-orderdenominators such as

Instead, rename variables

So that

21 KKB ! 21 11 KjKjC !



54/61

54


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(

)()(

[W

[[

[W

W

[

W

W

ste

s

s

te

stf

t

t



55/61

Return to example from above:


56/61

E l i h l


57/61

57


Example with complex roots

Example: findx

(t)

Laplace Transform



58/61

E l ith l t (3)


59/61

59


Example with complex roots (3)


E l ith l t (5)


60/61

60


Example with complex roots (5)


E i 2


61/61


Exercise 2

Find solution to the following differential equation

0)0(1)0(084 !!! xxxxx

? A

tttx

xxx

tt2si2cos

22

2

22

2

42

4

84

4

0844

080400

22

2222

2

2

2

2

!

!

!

!

!

!

Documents

307A-LaplaceTransformTechniques