L03-Rev2-LCCDE-LaplaceTransform.pdf

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Control Systems Engineering

ECE 563

07092003

SOLUTION TO LINEAR DIFFERENTIAL EQUATIONS WITH

CONSTANT-COEFFICIENTS (LCCDE) BY LAPLACE TRANSFORM

J.M.Martinez, Jr.School of EE-ECE-CoE

Mapúa Institute of Technology

• Laplace Transform of Common Functions

• First-Shifting Theorem

• Other Laplace Transform Pairs

•Laplace Transform of Derivatives and Integral of Functions

• Cover-up Method (Residue Theorem)

• Inverse Laplace Transform by Partial Fraction Expansion

• Determination of Residues by Cover-up Method

• Determination of Residues by Equating Coefficients of the Numerator

• Using Matlab to solve LCCDE

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LAPLACE TRANSFORM

Steps in Solving LCCDE using Laplace Transform

1. Convert the differential equation to Laplace Transform (time domain s-domain).

2. Solve for Y(s) (isolate Y(s)).3. Expand Y(s) into partial fractions.4. Take the Inverse Laplace Transform of Y(s).

(s-domain time domain).

jmmartinezjr 07092003

3

LAPLACE TRANSFORM PAIRS

)(sF

tωsin

1

)(tu

)(tδ)(tf

tωcos

2

1st

as −1ates1

22 ωω+s

22 ω+ss


4

LAPLACE TRANSFORM PAIRS (FIRST-SHIFTING THEOREM)

)(sF

teat ωsin

assat tfLtfeL −→= )]([)]([

)(tf

teat ωcos

2)(1as −teat

22)( ωω+− as

22)()(ω+−

−as

as


5

OTHER LAPLACE TRANSFORM PAIRS

))(( bsasab++

−btat ee −− −

)()(

2 assbsa

++

)(sF

ate−−1

)(tf

)(2

2

assa+

nt

)( assa+

ateat −+−1

1

!+ns

n

))(( bsassab

++btat e

baae

bab −−

−−

−+1

atea

babta

ba −

−

−+−


6

LAPLACE TRANSFORM PAIRS (INTEGRAL & DERIVATIVES)

n

n

dttgd )(

)(sF

dttdg )(

)(tf

ssG )(∫

τ

ττ0

)( dg

)(ssG

)(sGsn

Note: All Initial Conditions are assumed zero.


7

COVER-UP METHOD

[ ]as

sFasR=

−= )()(

To find the constants (residues) of the partial fraction expansion of rational functions we can

use Residue Theorem


8

COVER-UP METHOD FOR MULTIPLE ROOTS

[ ]as

nkn

kn

k sFasdsd

knR

=−

−

−−

= )()()!(

1

where:

a = multiple roots

n = multiplicity of the roots

k = order of root = n, n-1, … 3, 2, 1


For multiple roots, Residue Theorem can be expressed in the form

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Example 1:

22 2

45s

sYssYsYs =++ )()()(

Find the Total Solution to the Differential Equation (Assume all initial conditions=0)

Solution: Converting the DE into LaplaceTransform, we have

tydtdy

dtyd

2452

2

=++


10

Example 1:Solving for Y(s)

45

2

2

2

++=

ssssY )(

simplifying

))(()(

14

22 ++

=sss

sY


11

Example 1:

Taking the Partial Fraction Expansion

)()())(()(

1414

222 +

++

++=++

=sD

sC

sB

sA

ssssY

Using cover-up method

24

1

144

2

14

24

24

2

−=

+−−=

+++=

−= )()())(()(

sssssC

3

2

411

2

14

21

21

2=

+−−=

+++=

−= )()())(()(

sssssD


12

Example 1:

2

1

1040

2

14

2

02

2 =++

=++

==

))(())(( sssssA

Using cover-up method for multiple roots

02

02

2

45

2

14

2

1

1

==

++=

++=

ssssds

dsss

sdsdB

)())(()(

!

[ ]8

554252452 2

0

22 −=−=+++−= −

=

− )()()()(s

sssB


13

Example 1:

)()()(

13

2

424

1

8

5

2

1

2 ++

+

−

+

−

+=ssss

sY

substituting

tt eetty −− +−−=3

2

24

1

8

5

2

1 4)(

taking the Inverse Laplace Transform


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Example 2:

440

22

222

+−=−+

ss

ssYssYsYs )()()(

Find the Total Solution to the Differential Equation (Assume all initial conditions=0)

Solution: Converting the DE into LaplaceTransform, we have

ttydtdy

dtyd

240222

2

cos−=−+


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Example 2:Solving for Y(s)

24

402

2

22

−++

−=

sss

sssY )(

simplifying

))(()(

24

8240222

23

−++++−

=ssss

sssY


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Example 2:Taking the Partial Fraction Expansion, we have

4

2

412

24

8240

222

222

23

++

++++

−+

−−=

−++++−

=

sF

sEs

sD

sC

sB

sA

sssssssY

)()(

))(()(


tFtEDCtBeAety tt 222 sincos)( +++++= −

By taking the Inverse Laplace Transform, y(t) is of the form

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Example 2:

221411

812140

24

824022

23

122

23

−=++++−

=++++−

== ))(()(

)()())(( ssss

ssB

2

7

12422

822240

14

824022

23

222

23 −=

−−+−−+−+−−

=−+++−

=−= ))(()(

)()())(( ssss

ssA


solving for the residues using cover-up method

1102040

802040

124

82402

23

02

23

−=−++++−

=−++++−

== ))()((

)()())()(( ssss

ssC

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Example 2:jmmartinezjr 07092003

0234

23

02

23

842

8240

124

8240

1

1

=

=

−+++++−

=

−++++−

=

s

s

ssssss

dsdD

sssss

dsdD

)(

))()((!

02234

23232234

842

443482404120842

=

−++++++++−−+−−+++

=s

sssssssssssssssD

)())(())((

2

1

8

48082

−=

−−−

=)(

))(())((D


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Example 2:

12422

12222

822240

12

82402

2

23

22

23

jFjE

jjjjj

sssssFEs

js

−−=+−

−−+−−+−+−−

=−+++−

=+−=

)(

))(()()()(

))((

12422

12222

822240

12

82402

2

23

22

23

jFjE

jjjjj

sssssFEs

js

+−=+

−+++−

=−+++−

=+=

)(

))(()()()(

))((



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Example 2:

substituting into the expression shown in slide 14


244

84

jEjF

=−=

Taking the sum and difference of the two expressions will result into

6

2

=−=

EF

ttteety tt 22262

12

2

7 2 sincos)( −+−−−−= −

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(Alternative Solution to Example 2)Determination of Residues by Equating Coefficients


4

2

412

24

8240

222

222

23

++

++++

−+

−−=

−++++−

=

sF

sEs

sD

sC

sB

sA

sssssssY

)()(

))(()(

• When the denominator contains complex roots, determining the residues is more difficult because we have to evaluate complex expressions.

• We can eliminate the need for complex number operations by equating coefficients of each term in the numerator and then solving the resulting system of linear equations.

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422 sFEDCBA )( +++++−+

322244 sFEDCBA )( +−++++

Expanding each term, and then grouping similar terms we have

244284 sFDCBA )( −+++−+

523 8240 sEDBAss )( +++=++−

)()( CsDC 884 −+−+


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By equating coefficients of similar terms we have the following set of linear equations

).( 10 eqEDBA =+++

).( 2022 eqFEDCBA =+++++−

).( 34022244 eqFEDCBA −=+−+++

).( 4244284 eqFDCBA =−+++−

).( 5084 eqDC =−

).( 688 eqC =−


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−=

−−

−−−

−

8

0

2

40

0

0

000800

008400

404284

222144

211121

011011

FEDCBA

In matrix form


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Using Cramer’s Rule or other methods for system of linear equations

2

62

11

22

7

−==

−=

−=−=

−=

FE

DCB

A


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• Solving a large number simultaneous linear equations is difficult!

• If we already knew A, B, C, D using cover-up method, we only need two of the six equations to solve for E and F. Using eq.(1) and eq.(2) we have


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72

2

7=++=−−−= DBAE

22

62

1122

2

7

2

2−=

−+−−−−−=

−−−−=

)()(EDCBAF

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Example 3:Determine the Inverse Laplace Transform of the

function

Solution: By completing the square of the expression

)()()(

2561

7236422

23

+−−+

=sss

ss-ssY


2562 +− ss we have

22

2222

43

32536256

+−=

−++−=+−

)(

)()(

s

ssss

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Example 3:Taking the Partial Fraction Expansion of

Y(s), we obtain


22222 43

4

43

3

11 +−+

+−−

+−

+−

=)(

)()(

)()()(

)(s

Ds

sCs

Bs

AsY


225161

17213614

256

723642

23

12

23

=+−+

=+−+

== )()(

)()()()(

-ss

ss-sAs

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1

25161

612172136147217211225161

256

6272364727212256

256

72364

1

1

22

2322

122

2322

12

23

=+−

−+−++−=

+−−+−++−

=

+−+

=

=

=

B

--

sssss-ss-sss

ssss-s

dsdB

s

s

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

)())(())((

)(!

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Example 3:Expanding each term of the numerator


22

2223

1413

256125672364

)())((

))(()(

−+−−+

+−−++−=+

sDssC

sssBssAss-s

)()()()(DCBAsDCBAsDCBAsCBss-s

43252587316

45772364 2323

+−−+−++−++−−++=+

and then grouping similar terms we have

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Example 3:equating coefficients of s3 to solve for C


4=+CB

3

14

4

=−=−=

CC

BC

equating coefficients of s2 to solve for D

36457 −=+−− DCBA

44

35172364

5736

−=

++−−=

++−−=

D

D

CBAD

)()(

32


tDetCeBeAtety tttt 44 33 sincos)( +++=

By taking the Inverse Laplace Transform of

we have

22222 43

4

43

3

11 +−+

+−−

+−

+−

=)(

)()(

)()()(

)(s

Ds

sCs

Bs

AsY

teteetety tttt 44432 33 sincos)( −++=

Substituting the values of A,B,C, and D

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MATLAB IMPLEMENTATIONPartial Fraction Expansion of polynomial

given the Coefficients of Numerator num and Denominator den

[R,P,K]=residue(num,den)

where:

R=residues (numerator)

P=poles i.e (s-p1)(s-p2)…

K=direct term (null if order of num<den)


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>> num=[4];>> den=[1 4 5 2];>> [R,P,K]=residue(num,den)R =

4.0000-4.00004.0000

P =-2.0000-1.0000-1.0000

K =[]

MATLAB IMPLEMENTATION

2)1(4

)1(4

24)(

++

+−

++

=sss

sY

Example 1: Find the partial fraction expansion of

2544)( 23 +++

=sss

sY

Y(s) can be written as

Multiple poles should be written in increasing order


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>> num=[-40 2 0 8];>> den= poly([0 0 2i -2i -2 1]);>> [R,P,K]=residue(num,den)R =

3.0000 + 1.0000i3.0000 - 1.0000i

-3.5000 -2.0000 -0.5000 -1.0000

P =0.0000 + 2.0000i0.0000 - 2.0000i

-2.0000 1.0000

0 0

MATLAB IMPLEMENTATIONExample 2: Find the partial fraction expansion of

))(()(

24

8240222

23

−++++−

=ssss

sssY


K =[ ]

>> E=R(1)+R(2) %E in Ex.2E =

6.0000>> F=i*(R(1)-R(2)) %F in Ex.2F =

-2.0000

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>> num=[4 -36 72 0];>> den=conv([1 -6 25],poly([1 1]));>> [R,P,K]=residue(num,den)R =

1.5000 + 2.0000i1.5000 - 2.0000i1.0000 2.0000

P =3.0000 + 4.0000i3.0000 - 4.0000i1.0000 1.0000

K =[]

MATLAB IMPLEMENTATIONExample 3: Find the partial fraction expansion of

)()()(

2561

7236422

23

+−−+

=sss

ss-ssY


>> C=R(1)+R(2) % C in Example 3C =

3.0000>> D=i*(R(1)-R(2))% D in Example 3D =

-4.0000

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MATLAB IMPLEMENTATIONUsing MatLab Symbolic Toolbox we can find the

solution to LCCDE by

syms s tFirst creating symbolic variables s and t using the command

Then, converting time functions into LaplaceTransform using the command

laplace(F)And finally, taking the Inverse Laplace Transform

using the command

ilaplace(F)


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MATLAB IMPLEMENTATION Example 4:

Find the Total Solution of the Differential Equation (Assume all initial conditions=0)

Solution: First, take the Laplace Transform of

ttydtdy

dtyd 2cos40222

2

−=−+

tttf 2cos402)( −=

>>F=laplace(2*t-40*cos(2*t))F =2/s^2-40*s/(s^2+4)


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Then solve for Y(s) by dividing F(s) with the auxiliary polynomial derived from the left side of

the D.E.

>>Y=F/(s^2+s-2)Y =(2/s^2-40*s/(s^2+4))/(s^2+s-2)


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Finally, convert Y(s) to y(t)

>> y=ilaplace(Y)y =-t-1/2-7/2*exp(-2*t)- 2*exp(t)+ 6*cos(2*t)-2*sin(2*t)

Note: Matlab variables are case sensitive, so y is different from Y


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MATLAB IMPLEMENTATION(Other useful Matlab Functions )

dsolve(‘D.E.’,’initial cond.’,’ind.var.’)•Function dsolve computes symbolic solutions to ordinary differential equations.

•The equations are specified by symbolic expressions containing the letter D to denote differentiation.

•The symbols D2, D3, ... DN, correspond to the second, third, ..., Nth derivative, respectively. Thus, D2y is the Symbolic Math Toolbox equivalent of .

2

2

dtyd


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Example 5:

>> dsolve('D2y=-Dy+2*y+2*t-40*cos(2*t)','y(0)=0','Dy(0)=0', 't')ans =-t-13/2+12*cos(t)^2-4*cos(t)*sin(t)-2*exp(t)-7/2*exp(-2*t)

Find the Total Solution of the Differential Equation (All initial conditions=0)

ttydtdy

dtyd 2cos40222

2

−=−+

Solution:


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Example 5:

tt eettttty 22

272sincos4cos12

213)( −−−−+−−=

tt eetttty 2

2722sin22cos6

21)( −−−−+−−=

ttt 2cos662

)2cos(112cos12 2 +=+

=

The Total Solution

can also be written as

by considering that

tttt 2sin22

)2sin(4sincos4 −=−=−


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pretty(F)The pretty function prints symbolic output in a format that resembles typeset mathematics.>> syms s t>> y=ilaplace(4/(s^3+4*s^2+5*s+2))y =

4*exp(-2*t)+4*t*exp(-t)-4*exp(-t)>> pretty(y)

4 exp(-2 t) + 4 t exp(-t) - 4 exp(-t) Difficult to read? Better!


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REFERENCES:

• Elementary Differential Equations,7thedition,by Rainville E.D.,and Bedient P.E.• Schaum’s Outline Series “Feedback

and Control Systems”, 2nd edition, by DiStefano III, J.J., Stubberud A.R., and Williams I.J.

• Control Systems Engineering, 3rd

edition, by Nise N.S.


2nd Revision: 08202007

Documents

L03-Rev2-LCCDE-LaplaceTransform.pdf