S. Awad, Ph.D. M. Corless, M.S.E.E. E.C.E. Department University of Michigan-Dearborn Laplace Transform Math Review with Matlab: Calculating the Inverse.

S. Awad, Ph.D.

M. Corless, M.S.E.E.

E.C.E. Department

University of Michigan-Dearborn

LaplaceTransform

Math Review with Matlab:

Calculating the Inverse Laplace Transform

U of M-Dearborn ECE DepartmentMath Review with Matlab

2

Laplace Transform:X(s) Inverse Laplace Transform

Inverse Laplace Transform

Laplace Nomenclature Table Look Up Method Simple Table Look Up Example Inverse Laplace Transform General Form Distinct Pole Example Repeated Poles Example


3


Laplace Nomenclature The Laplace Transform of a time domain function

x(t), will be a complex domain function X(s)

)()( sXtx LT

)()(

)()(1 sXLTtx

txLTsX

This relationship is also denoted as:


4


Inverse Laplace Transform Inverse Laplace Transform is used to compute x(t) from X(s)

The Inverse Laplace Transform is strictly defined as:

Strict computation is complicated and rarely used in engineering

jc

jc

stdsesXj

sXLTtx )(2

1)()( 1

Practically, the Inverse Laplace Transform of a rational function is calculated using a method of table look-up


5


Table Look Up Method If X(s) can be written as a sum of terms with known

Inverse Laplace Transforms, x(t) will be the sum of these Inverse Laplace Transforms

)()()()(

)()()()(

21

21

txtxtxtx

sXsXsXsX

n

n

Requires knowledge or reference of Laplace Transform pairs, but is much simpler than directly calculating the Inverse Laplace Transform


6


Simple Table Look Up Example

Given:

3

410)(

ss

sX

Use superposition and the Table Look-Up Method determine the Inverse Laplace Transform of X(s)

Verify the result using Matlab


7


Tabular Solution X(s) is written as a Linear Summation of terms with Known

Inverse Laplace Transforms

)(1

)()(

sXs

txetu t

3

410)(

ss

sX

)(4)(10)( 3 tuetutx t

LT-1 LT-1LT-1

x(t) can be directly determined using the Laplace Transform Pairs shown below to the right


8


Matlab Verification

3

410)(

ss

sX

» syms X s» X=((10/s)+(4/(s+3)));» x=ilaplace(X)

x =10+4*exp(-3*t)

)(410)( 3 tuetx t

The ilaplace Matlab command can be used to quickly verify the solution

NOTE: The output of ilaplace must be interpreted

as a Causal Solution


9


Inverse Laplace Transform General Form

In general X(s) can be written as a rational function

))...()((

))...()(()(

21

21

nn

mm

pspspsa

zszszsbsX

z1, z2, … , zm are the zeros of X(s)

p1, p2, … , pn are the poles of X(s)

If X(s) is written as a strictly rational function, a method using partial-fraction expansion can be used to determine the inverse Laplace transform


10


Strictly Rational Function A function X(s) is strictly rational if the Degree of its

Numerator Polynomial N(s) is Less than the Degree of its Denominator Polynomial D(s)

If N(s) D(s), perform Long Division until the remainder polynomial R(s) is of lesser order than A(s)

)(

)()(

)(

)()(

sD

sRsQ

sD

sNsX

N(s) = Numerator

D(s) = Denominator

Q(s) = Quotient

R(s) = Remainder


11


Partial-Fraction Expansion based on Poles

Once X(s) is written in a strictly rational form, Partial Fraction Expansion can be performed

Partial-Fraction Expansion of of X(s) can be classified into two categories based on the Poles of X(s)

CASE I: All Poles are Distinct CASE II: All or Some Poles are Repeated


12


CASE1: Distinct Poles Assume that all Poles are Different ( pipj if i j )

Assume numerator order is less than denominator order, then the Partial Fraction Expansion of X(s) is given by:

n

n

ps

c

ps

c

ps

csX

...)(

2

2

1

1

where: nisXpscipsii ,...,2,1,)(

))...()((

))...()(()(

21

21

nn

mm

pspspsa

zszszsbsX


13


Building from previous work, x(t) is the summation of unit steps multiplied by exponentials

LT-1 of Distinct Terms X(s) can be written as the sum of terms Due to the Linearity Property, x(t) will be the sum of the

Inverse Laplace Transform of the terms

n

i

tpi tuectx i

1

)()(

n

i i

iLTn

i i

i txps

cLT

ps

csX

1

1

1

)()(


14


CASE2: Repeated Pole

))...(()(

)(

)(

)()(

11 nrr pspsps

sN

sD

sNsX

n

n

r

rr

r

ps

c

ps

c

ps

c

ps

c

ps

csX

...)()(

...)(

)(1

1

12

1

2

1

1

Assume that Some or All Poles (Roots) are Repeated For the case below, Pole p1 is Repeated r Times

Partial Fraction Expansion shows Repeated and Distinct poles

Repeated Pole Distinct Poles


15


Determine Coefficients Find Distinct Poles Coefficients, i = r+1, r+2, ...,n

nrrisXpscipsii ,...,2,1,)()(

ipsr

rr sXpsc )()(

Find General Repeated Poles, i = 1, 2, ... , r-1

1

)( )!(

11)(

)(

ps

r

ir

ir

i sXpsds

d

irc

Find First Repeated Pole Coefficient, i = r


16


Partial-Fraction Expansion Method for Determining Inverse Laplace Transform

Put Rational Function into Strictly Rational Form where the degree of the numerator polynomial less than that of the denominator polynomial

Factor the Denominator Polynomial

Perform Partial-Fraction Expansion

Use Laplace Transform Pair Table to obtain the inverse Laplace transform


17


Distinct Pole Example Given:

Use the Partial-Fraction Expansion and the Table Look-Up Method determine the Inverse Laplace Transform of X(s)

Verify the result using Matlab

)3)(2)(1(

52)(

2

sss

sssX

Notice all Poles are Distinct

Order of Numerator (2) < Order of Denominator (3)


18


Partial Fraction Expansion

)3)(2)(1(

52)(

2

sss

sssX

Start by finding Partial Fraction Expansion of X(s)

321)( 321

s

c

s

c

s

csX

Poles of

p1=-1, p2=-2, p3=-3


Find ci coefficients


19


c1 is explicitly evaluated as:

Find Coefficient C1

C1 = -2

2)31)(21(

5)1()1(2

)3)(2(

52

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

52)1)((

2

1

2

1

2

11

S

Ss

ss

ss

ssss

ssssXc

Each coefficient is determined by evaluating:

nisXpscipsii ,...,2,1,)(


20


Find Coefficient C2 and C3

3)32)(12(

5)2()2(2

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

52

)2)((

2

2

2

2

2

22

c

ssss

ssc

ssXc

s

s

C2 = 3

1)23)(13(

5)3()3(2

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

52

)3)((

2

3

3

2

3

33

c

ssss

ssc

ssXc

s

s

C3 = 1


21


)3)(2)(1(

52)(

2

sss

sssX

Partial Fraction Expanded X(s)

Original Expression

3

1

2

3

1

2)(

sss

sX

321)( 321

s

c

s

c

s

csX

Expansion

Replace with Coefficients


22


)()(3)(2)( 32 tuetuetuetx ttt

3

1

2

3

1

2)(

sss

sX

Inverse Laplace Transform x(t) = Sum of the Inverse Laplace Transform of

the individual terms of X(s)

)(32)( 32 tueeetx ttt

LT-1LT-1 LT-1LT-1


23


» syms Xnum Xden s» Xnum = 2*s^2 -s +5; % Numerator X(s)» Xden = (s+1)*(s+2)*(s-3); % Denominator» x = ilaplace( Xnum/Xden ) x =-2*exp(-t)+3*exp(-2*t)+exp(3*t)

Distinct Pole Matlab Verification This can easily be verified in

Matlab using ilaplace

)(32)( 32 tueeetx ttt

)3)(2)(1(

52)(

2

sss

sssX


24


Repeated Pole Example Given:

Use the Partial-Fraction Expansion and the Table Look-Up Method determine the Inverse Laplace Transform of X(s)

Verify each step using Matlab

)1()2(

22)(

3

34

ss

ssssX

Order of Numerator (4) = Order of Denominator (4) therefore X(s) is NOT Strictly Rational

Notice Pole s = -2 is Repeated 3 times


25


Solution Steps Convert X(s) to a strictly rational function

Perform long division by hand Verify long division using Matlab deconv

Perform partial fraction expansion on rational part of X(s) Calculate coefficients by hand Calculate coefficients in Matlab using diff and subs

Verify conversion to strictly rational function using combination of int and diff commands

Use table method to determine Inverse Laplace Transform

Verify entire process using ilaplace


26


Repeated Pole Example X(s) must be decomposed into a constant plus a Strictly Rational

Function

)1()2(

)()(

)1()2(

22)(

33

34

ss

sRsQ

ss

ssssX

)1()2(

)()(

3

ss

sNcsX o )(lim sXc

so

Q(s) will just be a constant since the order of the numerator and the denominator are the same


27


Convert to Rational Functions

Use Long Division to Normalize the Transfer Function such that the Highest Order of the Numerator is Less Than the Highest Order of the Denominator

ssssssssX

sNumsDensX

22820187)(

)()()(

34234

820187

22

)1()2(

22)(

234

34

3

34

ssss

sss

ss

ssssX

Must Decompose X(s) into Strictly Rational Functions of s


28


Long Division Perform long division to decompose X(s) into a

constant plus a purely rational function

Thus:

)1()2(

164236132)(

3

23

ss

ssssX

2

16423613

164036142

0 2s 0s s2s820187

22

234

23 4234

sss

ssss

ssss


29


Matlab Polynomial Division: deconv

The Matlab command deconv can be used to perform polynomial division

deconv operates on arrays of polynomial coefficients, NOT symbolic variables

A

RQ

A

B

[Q, R] = deconv( B, A)B = Numerator polynomial coefficients

A = Denominator polynomial coefficients

Q = Quotient of B/A

R = Remainder of B/A


30


Long Division Verification Use Matlab to verify the polynomial long division:

820187

820187

164236132

820187

22

234

234

23

234

34

ssss

RQ

ssss

sss

ssss

sss

» polynum = [2 1 0 -2 0]; » polyden = [1 7 18 20 8];» [Q, R]=deconv(polynum, polyden)Q = 2R = 0 -13 -36 -42 -16


31


Expand Rational Part of X(s) The rational part of X(s) will be referred to as Xo(s)

)1()2(

16423613)(

3

23

ss

ssssX o

1)2()2(2)( 4

33

221

s

c

s

c

s

c

s

csX o

Xo(s) has 3 repeated roots and one distinct root


)(2)1()2(

164236132)(

3

23

sXss

ssssX o

Partial Fraction Expansion must be performed on Xo(s)


32


Evaluating Coefficient C4 C4 is evaluated using the distinct pole expression as

shown in the previous example

14 )()1( so sXsc

3

23

1

3

23

1

3

23

4

)21(

16142136113

)2(

16423613

)2)(1(

16423613)1(

s

s

s

sss

ss

ssssc

34 c


33


Matlab C4 Verification c4 can verified in Matlab by creating the symbolic

expression Xo and evaluating for c4

» Xo =( -13* s^3 -36*s^2 -42*s -16)/((s+1)*(s+2)^3);

» c4 = (Xo)*(s+1)c4 =(-13*s^3-36*s^2-42*s-16)/(s+2)^3

» c4=subs(c4,s,-1)c4 = 3


34


Evaluating Coefficient C3 C3 is evaluated similar to C4

28

)21(

16242236213

)2)(1(

16423613)2(

)()2(

3

23

2

3

233

23

3

s

so

ss

ssss

sXsc

283 c

» c3 = (Xo)*((s+2)^3);» c3 = subs(c3,s,-2)c3 = -28

C3 is easily verified in Matlab


35


Repeated Pole Coefficients To find C2 and C1, the following expression must be evaluated

for each case ( r = 3, p1 = -2, i = 2, 1 )

2

23

)3(

)3(

23

233

)3(

)3(

1)(

)(

)1(

16423613

)!3(

1

)2)(1(

164236132

)!3(

1

)( )!(

1

1

si

i

si

i

ps

r

ir

ir

i

s

sss

ds

d

i

ss

ssss

ds

d

i

sXpsds

d

irc

)1(

16423613)(

23

s

ssssY

For simplicity, let Y(s) be the expression to be differentiated


36


Repeated Pole Coefficients

Before differentiating, Y(s) can be rewritten as:

222)23(

)23(

2 )( )( !1

1)(

)!23(

1

sss

sYsYds

dsY

ds

dc

12323

)1)(16423613()1(

16423613)(

sssss

ssssY

21

222

2

2)13(

)13(

1

2

)(

)(2

1 )(

!2

1)(

)!13(

1

s

sss

sYc

sYds

dsY

ds

dsY

ds

dc

Similarly, C1 equals the second derivative of Y(s) / 2

C2 equals the first derivative of Y(s)


37


Calculating First Derivative Calculating First Derivative by Hand

123 )1)(16423613()( sssssY

2

232

21223

)1(

16423613

)1(

427239)(

)427239()1()1)(1)(16423613()(

s

sss

s

sssY

ssssssssY

» Y =(Xo)*((s+2)^3);» dY = diff(Y); pretty(dY) 2 3 2 -39 s - 72 s - 42 -13 s - 36 s - 42 s - 16 ------------------ - -------------------------- s + 1 2 (s + 1)

Matlab Verification of First Derivative


38


Simplify Y’(s) Since Y’(s) will eventually have to be calculated, it will be helpful to simplify its terms

» dY=simplify(dY); pretty(dY) 3 2 26 s + 75 s + 72 s + 26 - ------------------------- 2 (s + 1)

2

23

2

232

)1(

26727526

)1(

16423613

)1)(1(

)1)(427239(

s

sss

s

sss

ss

sssY

Matlab can also be used to perform this simplification


39


Evaluate C2 C2 can now be directly evaluated from Y’(s)

» c2 = subs(dY,s,-2)c2 = 26

Matlab Verification of C2

26)12(

26)2(72)2(75)2(26

)1(

26727526)(

2

23

2

2

23

22

s

s s

ssssYc

262 c


40


Calculate Second Derivative The second derivative of Y(s) must be calculated to find C1

3

23

2

2

22323

2232

23

)1(

267275262

)1(

7215078

)7215078()1()1)(2)(26727526()(

)1)(26727526()1(

26727526)(

s

sss

s

ss

ssssssssY

sssss

ssssY

» ddY = diff(dY); pretty(ddY) 2 3 2 78 s + 150 s + 72 26 s + 75 s + 72 s + 26 - ------------------ -+ 2 ------------------------- 2 3 (s + 1) (s + 1)

Matlab Verification of Y’’(s)


41


Calculate C1

The following expression can be calculated by hand

16

)1(

267275262

)1(

7215078

2

1

2

)(

2

3

23

2

2

21

s

s

s

sss

s

ss

sYc

Matlab Verification

» c1 = subs( ddY/2 ,s ,-2)c1 = -16

161 c


42


Result of Expansion The Rational Part of X(s) is expanded to:

1

3

)2(

28

)2(

26

2

16

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

16423613)(

32

43

32

213

23

ssss

s

c

s

c

s

c

s

c

ss

ssssX o

2

16

)2(

26

)2(

28

1

32)(2)(

230

sssssXsX

Thus X(s) can be rewritten as:


43


Matlab Partial Fraction Expansion

As of Matlab 5.3.x, there is currently no function to directly convert a symbolic expression to a strictly rational function, the following “trick” can be performed

Integrate the function to be expanded Differentiate the result of the integration The result of the symbolic differentiation will be expressed in

strictly rational form Thus the result of the entire process is the original function

expanded into strictly rational form

Remember that the integration process MUST occur first so that no constant data is lost

Vector expressions can be evaluated using residues


44


Expansion Verification By hand, X(s) was converted to a strictly rational function

» num = 2*s^4 + s^3 -2*s;» den = (s+1)*( (s+2)^3 );

2

16

)2(

26

)2(

28

1

32

)2)(1(

22)(

233

34

ssssss

ssssX

This can be verified in Matlab

» X = diff(int( num/den ));» pretty(X) 3 28 26 16 2 + ----- - -------- + -------- - ----- s + 1 3 2 s + 2 (s + 2) (s + 2)


45


Table Method Solution

)(1626143)(2)( 2222 tueteetettx tttt

2

16

)2(

26

)2(

28

1

32)(

23

sssssX

The following transform pairs can be used to evaluate x(t)

)(1

)()(

sX

txet t

LT-1

The Inverse Laplace Transform can be calculated directly


46


Verification Using Matlab

»x = ilaplace( X )

x =

2*Dirac(t)+3*exp(-t)-14*t^2*exp(-2*t)

+26*t*exp(-2*t)-16*exp(-2*t)

»pretty(x)

2

2 Dirac(t) + 3 exp(-t) - 14 t exp(-2 t) +

26 t exp(-2 t) - 16 exp(-2 t)

)(1626143)(2)( 2222 tueteetettx tttt

Of course all of our hard work can be easily done in one step by using the ilaplace command in Matlab


47


Summary Direct calculation of Inverse Laplace Transform is

difficult

Practically, the Inverse Laplace Transform of a rational function is calculated using a table look-up method

Use long division and partial fraction expansion to put X(s) in strictly rational form

Two general types of poles: distinct and repeated

Matlab can be used to verify each step by hand or quickly perform the entire Inverse Laplace Transformation using ilaplace

S. Awad, Ph.D. M. Corless, M.S.E.E. E.C.E. Department University of Michigan-Dearborn Laplace Transform Math Review with Matlab: Calculating the Inverse.

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