Inverse Laplace Transform Lecture-3

8/22/2019 Inverse Laplace Transform Lecture-3

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By Mr S M WijewardanaPhD student QMUL12-04-2013

Learning Objectives:

1. Introduction to Inverse Laplace Transform

2. Three Methods used to find Inverse Laplace Transform3. Initial value theorem and the Final value theorem4. Examples on Inverse Laplace Transform5. Eigenvalues and Eigenvectors.


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Given the Laplace transform F(s), the reverse process offinding f(t) is called Inverse Laplace Transform.

f(t) = -1 [F(s)]

Inverse Laplace transform contains no information aboutthe initial condition. Because Inverse Laplace transformyields the response for t>0, and the Laplace transformmethod is capable of solving problems in which there is a

discontinuity at the origin. f(0-) f(0+) Hence, f(0+) is not the initial value f(0).

Three main Methods

1. Partial Fraction Method 2.Method of Residues3.Convplution integral method.


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Initiaal and Final Value TheoremsFinal Value Theorem- determine the steady-state valueof the system response without finding the inverse

Laplace transform.

Procedure: 1) find the transfer function X (s)

2) multiply X(s) by s

3) take the limit of s X(s) as s goes to zero

4) result is the value of x(t) when t = infinity

Initial Value Theorem- determine the value of the timeFunction when t =0 without finding the inverse transform

Procedure: 1) find the transfer function X(s)

2) multiply X(s)b y s

3) take the limit of sX(s) as Sgoes to infinity

4) Result is the value of x(t) when t =0

lim x(t) = lim s.X(s)

t goes to 0 then s reaches to infinity


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Partial Fraction Method

)2)(1(

2

)( sssFExample: Given

Find the time domain function using partial fraction method.

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

2)(

s

B

s

A

sssFLet A & B are constants

Multiplying by(s+1)(s+2) both sides of the equationgives:

2 = A(s+2) + B(s+1)When s=-2, B= -2: When s=-1, A=2

)2(

2

)1(

2)(

sssF

By using Inverse Laplace Transform table we can write:f(t) = 2 e-t - 2 e-2t


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G(s)

Input

Output = C(s)R(s)

Plant

Block diagram represents a simple open-loop transfer function. Find the timeresponse of the system if:

)6)(4(

)2()(

ss

ssG

Output = Input . G(s)

Hence C(s) = R(s).G(s)

)6)(4(

)2(.

1)(

ss

s

ssC

Applying the Initial-value theorem: )(limlim )(0

ssFst

tf

)(limlim )(0

ssFst

tf

e.g.

Solution:


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)(limlim )]([0

ssCst

tC

Applying the Final value theorem:Final value theorem can be applied only if the Laplace transform of f(t) is F(s)

and if sF(s) is analytic on the imaginary axis and in the right half of s-plane.

01

0

24101

21

lim

)6)(4(

)2(

lim

2

2

valueInitial

ss

ss

sss

ss

s

s

)(limlim0

)]([ ssFs

tft

Analytic: take it as convergent function for the time being.


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The function G(s):

Is an analytic function at all points in the s-plane except s=-4 and s=-6

Since sF(s) is analytic on the imaginary axis and in the right half of s-plane, the final-value theorem can be applied.Therefore:

)6)(4(

)2()(

ss

ssG

)(limlim0

)]([ ssCs

tCt

)6)(4(

)2(lim

)6)(4(

)2(lim

0

0

ss

s

sss

ss

s

s

Hence the final value(F.V) = 1/12


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)6()4()6)(4(

)2()(

s

C

s

B

s

A

sss

ssC

By obtaining the partial fractions of C(s) : (A,B,C are constants.

(s+2) =A(s+4)(s+6) + Bs(s+6) + Cs(s+4)

From which A=1/12, B= =1/4, C= -1/3

)6()4()( 3

141

121

ssssC

By taking Inverse Laplace Transform:

tt

eetc

64

3

1

4

1

12

1)(

Since we know the initial value andthe final value of c(t) now we can plotthe graph of c(t) vs t


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t

C(t)

1/12

0


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kSS

stn

kn

n

kesFss

ds

d

nRs

).()(

)1(!

1

1

1

Method of Residues:

Method of residues gives the inverse Laplace Transform of functions with

coefficients. This method is useful to find the Inverse Laplace transformwhen the Laplace transform functions are not available in the tables.The numbers associated with the poles of a function are known as functionresidues

The Residue theorem states that if F(s) is a rational function theninverse Laplace transform of a function is given by:

Polesall

stesFofsidues ]).([Re -1 [F(s)] =

Where the residue, RsK of nth order pole at S= -Sk is obtained by:

(Rational function has a numerator and a denominator and it is like a ratio of

two polynomial functions: rational number means it can be written as a fraction:

Sqrt of 2 is an irrational number)


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Use method of residues to determine thetime function

-1 [F(s)] = -1 ]

)4)(3)(2(

66[

2

sss

ss

The function F(s) has poles at s=-2, s=-3 and s=-4:assume the residues are R-2, R-3 and R-4

ttst

eess

ss

eR s

222

2 .2

2

)4)(3(

66

. 2

ttst eess

sseR

s33

2

3 3.1

3

)4)(3(

66.

3

tste

ss

sseR

s

42

4 .1)4)(3(

66.

4

Hence the required time function

f(t) = -[e-2t

-3e-3t

+e-4t

]


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Convolution Theorem

Theorem says that if [f1(t)] = F1(s) and [f2(t)]= F2(s) and f1(t) =0,f2(t)= 0,

for t


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By substituting t= (t-) and t= in f1 and f2 respectively

t

tdeetftheoremnConvolutiobyNow

0

2)(..2)(

From tables: f1(t)= 2e-t , f2(t)= e

-2t

f1(t-) =2 e-(t-), f2=e

-2

dee

t

t.2

0

tt

tt

t

t

t

ee

eee

tee

dee

2

0

0

22

)]/1([2

0][2

.2


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Eigenvalues and Eigenvectors

IfAis an byn matrix andXis a column vector of order n, we can generate a newcolumn vector Yby multiplication.

Y= AXIf we consider that Y also in the same direction as of X, then we should be able torepresent Y, as a scalar quantity(say) multiplied by the vector X such that Y is

linearly related to X.

Hence we should be able to write that:

Y = A X = XThe values ofthat satisfy the above equation are the characteristic values of A.The vector solution xi 0, are called the eigenvectors of A.

(A- I)X= 0 or (I-A)X=0


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Example: (Eigenvalues)

If we have Y = AX and AX = X for non zero Xs if and only ifdet(In-A) = 0 then is called the eigenvalues of A.

34

21A

Find the eigenvalues of A

Solution:

For nontrivial null space (I-A):

0340

201det

034

21

0

0det

0)34

21

10

01det(

(-1)(-3)-8 =0


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We call the equation that we get is the Characteristic Polynomial:

2 -4-5 =0

(-5)(+1)=0

Which gives us =5 and = -1 which are the eigenvalues of A

22

25A=

Example: Compute eigenvalues and eigenvectors for the matrix A:

(Definition: (you may need to revise)Homogeneous: a collection of linear equations or a set oflinear equations):Singular matrix: a square matrix whose determinant is zero.


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-----------------------------------------------------------------------------------------------

34

21A

Another method:

Solution: it can be shown that eigenvalues give the addition along the diagonalmatrix of the singular matrix: 1 + 2 = 4: Other relationship is theDeterminant is equal to the multiple of the two eigenvalues: 1 . 2 = 3-4x2 = -5By solving the two equations we get the same answers : Eigenvalues, 5 and -1.

Solution:

We know that:Ax = x for non zerox, iff det(In-A) =0;

22

25A =

Example:

(A-I) x = 0x is the eigenvector and it must be a columnvector with x1 and x2 etc.

Compute the eigenvalues andeigenvectors.


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Solution: part-1 :eigenvalueWe know that:Ax = x for non zerox, iff det(In-A) =0; is an eigenvalue.To find eigenvalue det (In-A) =0

Or we can write: det (A-In) =0 : A=

To compute correspondingeigenvectors: (A-I) x = 0Take = 6 first:

2225

1,6

0)1)(6(

22

250

IAor

0

22

25

010

01

22

25


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0

0

42

21

6

022

25

0

12

11

12

11

x

x

x

x

xIA

From which: -x11 + 2x12 =0 and2x11 - 4x12 =0(same as above)

Therefore x11= 2x12, choose x12=1 then x11=2Hence eigenvector

1

2*x


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When = 1

2221

2221

22

21

22

21

2

024

0

0

12

24

0

0

22

25

xx

xx

x

x

x

x

Choose x22=-2 then x21 = 1

Hence the other eigenvector

1

2**x


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Pictorial representation of Eigenvectors


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Learning Outcomes 1. Method of residues

2.Partial fraction method

3. Application of Initial Value theorem & Final Valuetheorem

4. Convolution theorem and the limitations of InverseLaplace transform

5.Use of mathematical function in control theory. 6.Eigenvalues and Eigenvectors

Inverse Laplace Transform Lecture-3

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