L03-Rev2-LCCDE-LaplaceTransform.pdf

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

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

LAPLACE TRANSFORM PAIRS

tωsin

)(tδ)(tf

tωcos

as −1ates1

22 ωω+s

22 ω+ss

LAPLACE TRANSFORM PAIRS (FIRST-SHIFTING THEOREM)

teat ωsin

assat tfLtfeL −→= )]([)]([

teat ωcos

2)(1as −teat

22)( ωω+− as

22)()(ω+−

OTHER LAPLACE TRANSFORM PAIRS

))(( bsasab++

−btat ee −− −

2 assbsa

ate−−1

)( assa+

ateat −+−1

))(( bsassab

++btat e

bab −−

−−

ba −

−+−

LAPLACE TRANSFORM PAIRS (INTEGRAL & DERIVATIVES)

dttgd )(

dttdg )(

ssG )(∫

)(sGsn

Note: All Initial Conditions are assumed zero.

COVER-UP METHOD

sFasR=

−= )()(

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

use Residue Theorem

COVER-UP METHOD FOR MULTIPLE ROOTS

k sFasdsd

−−

= )()()!(

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

Example 1:

sYssYsYs =++ )()()(

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

Solution: Converting the DE into LaplaceTransform, we have

tydtdy

Example 1:Solving for Y(s)

ssssY )(

simplifying

))(()(

Example 1:

Taking the Partial Fraction Expansion

)()())(()(

Using cover-up method

+−−=

−= )()())(()(

sssssC

+−−=

−= )()())(()(

sssssD

Example 1:

))(())(( sssssA

Using cover-up method for multiple roots

ssssds

)())(()(

554252452 2

22 −=−=+++−= −

− )()()()(s

Example 1:

)()()(

+=ssss

substituting

tt eetty −− +−−=3

taking the Inverse Laplace Transform

Example 2:

+−=−+

ssYssYsYs )()()(

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

Solution: Converting the DE into LaplaceTransform, we have

ttydtdy

240222

cos−=−+

Example 2:Solving for Y(s)

sssY )(

simplifying

))(()(

8240222

−++++−

Example 2:Taking the Partial Fraction Expansion, we have

−−=

−++++−

sssssssY

))(()(

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

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

Example 2:

221411

812140

824022

−=++++−

=++++−

== ))(()(

)()())(( ssss

822240

824022

23 −=

−−+−−+−+−−

=−+++−

=−= ))(()(

)()())(( ssss

solving for the residues using cover-up method

1102040

802040

−=−++++−

=−++++−

== ))()((

)()())()(( ssss

Example 2:jmmartinezjr 07092003

−+++++−

−++++−

ssssss

))()((!

23232234

443482404120842

−++++++++−−+−−+++

sssssssssssssssD

)())(())((

−−−

))(())((D

Example 2:

822240

sssssFEs

−−=+−

−−+−−+−+−−

=−+++−

=+−=

))(()()()(

822240

sssssFEs

+−=+

−+++−

=−+++−

))(()()()(

Example 2:

substituting into the expression shown in slide 14

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

ttteety tt 22262

7 2 sincos)( −+−−−−= −

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

−−=

−++++−

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.

422 sFEDCBA )( +++++−+

322244 sFEDCBA )( +−++++

Expanding each term, and then grouping similar terms we have

244284 sFDCBA )( −+++−+

523 8240 sEDBAss )( +++=++−

)()( CsDC 884 −+−+

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 =−

−−

−−−

000800

008400

404284

222144

211121

011011

FEDCBA

In matrix form

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

−=−=

• 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

7=++=−−−= DBAE

−+−−−−−=

−−−−=

)()(EDCBAF

Example 3:Determine the Inverse Laplace Transform of the

function

Solution: By completing the square of the expression

)()()(

7236422

+−−+

ss-ssY

2562 +− ss we have

32536256

−++−=+−

Example 3:Taking the Partial Fraction Expansion of

Y(s), we obtain

22222 43

11 +−+

+−−

)()()(

225161

17213614

723642

=+−+

== )()(

)()()()(

ss-sAs

612172136147217211225161

6272364727212256

−+−++−=

+−−+−++−

sssss-ss-sss

ssss-s

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

)())(())((

Example 3:Expanding each term of the numerator

256125672364

)())((

))(()(

−+−−+

+−−++−=+

sssBssAss-s

)()()()(DCBAsDCBAsDCBAsCBss-s

43252587316

45772364 2323

+−−+−++−++−−++=+

and then grouping similar terms we have

Example 3:equating coefficients of s3 to solve for C

=−=−=

equating coefficients of s2 to solve for D

36457 −=+−− DCBA

35172364

++−−=

tDetCeBeAtety tttt 44 33 sincos)( +++=

By taking the Inverse Laplace Transform of

we have

22222 43

11 +−+

+−−

)()()(

teteetety tttt 44432 33 sincos)( −++=

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

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)

>> 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

MATLAB IMPLEMENTATION

Example 1: Find the partial fraction expansion of

2544)( 23 +++

Y(s) can be written as

Multiple poles should be written in increasing order

>> 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

MATLAB IMPLEMENTATIONExample 2: Find the partial fraction expansion of

))(()(

8240222

−++++−

K =[ ]

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

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

-2.0000

>> 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

MATLAB IMPLEMENTATIONExample 3: Find the partial fraction expansion of

)()()(

7236422

+−−+

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

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)

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

−=−+

tttf 2cos402)( −=

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

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)

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

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 .

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

−=−+

Solution:

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 −=−=−

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!

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

L03-Rev2-LCCDE-LaplaceTransform.pdf

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