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© Copyright F.L. Lewis 1999All rights reserved

EE 4314 - Control SystemsLECTURE 3

INVERSE LAPLACE TRANSFORM

INVERSE LAPLACE TRANSFORM

Given a time function f(t), its unilateral Laplace transform is given by

∫∞

−

−

=0

)()( dtetfsF st ,

where ωσ js += is a complex variable. The inverse Laplace transform is a complexintegral given by

∫+

−

=ωσ

ωσπ

j

j

st dsesFj

tf )(2

1)( ,

where the integration is performed along a contour in the complex plane. Since this istedious to deal with, one usually uses the Cauchy theorem to evaluate the inversetransform using

stesFofresiduesenclosedtf )()( Σ= .

In this course we shall use lookup tables to evaluate the inverse Laplacetransform. An abbreviated table of Laplace transforms was given in the previous lecture.The text has a more detailed table. Given a realistic Laplace transform with several polesand zeros, it is not likely to be contained in the table. However, it is easy to break atransform down as into sum of simpler transforms that are in the table by using thePartial Fraction Expansion (PFE).

PARTIAL FRACTION EXPANSION (PFE)

The PFE is simply a technique for splitting a rational transform up into a sum ofsimpler terms.

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Case 1. Non-Repeated Poles

Let a rational polynomial function of s be given as

)()(

)()(

1 sdps

snsY

+=

where n(s) is the numerator and d(s) is a polynomial which does not have any poles at

1ps −= . The relative degree, defined as the degree of the denominator (e.g.

)()( 1 sdps + ) minus the degree of the numerator n(s), should be at least 1.

One may split out a term with divisor )( 1ps + and write

)()()()(

)()( 1

1

1

1

sYps

K

sdps

snsY +

+=

+=

where K1 is known as the residue of the pole at 1ps −= , and )(1 sY is a polynomial

fraction that does not have any poles at 1ps −= . To determine the residue K1 one maywrite

)()()(

)()()( 1111 sYpsK

sd

snsYps ++==+ .

Now evaluating all terms at 1ps −= , the last term becomes zero and one obtains

1)()( 11 ps

sYpsK−=

+=

Example 1- Non-Repeated Poles

Find the inverse Laplace transform of

6116)(

23 +++=

sss

ssY .

First note that this function is not in the transform table. To write it as the sum of termsthat are in the table, factor the denominator to obtain

)3)(2)(1()(

+++=

sss

ssY .

Now write

321)( 321

++

++

+=

s

K

s

K

s

KsY

and evaluate the residues by

3

2

1

)31)(21(

)1(

)3)(2(1

1 −=+−+−

−=

++=

−=sss

sK

2)32)(12(

)2(

)3)(1(2

2 =+−+−

−=

++=

−=sss

sK

2

3

)23)(13(

)3(

)2)(1(3

3 −=+−+−

−=

++=

−=sss

sK

Therefore

32

3

2

2

12

1)(

+

−+

++

+

−=

ssssY .

Check this now by combining the terms back into one fraction to get the original Y(s).

Each term is now in the table, whence one may write down the inverse Laplacetransform as

)()2

32

2

1()( 1

32 tueeety ttt−

−−− −+−= .

We multiplied by the unit step to make this function causal.

Case 2- Repeated Real Poles

Let a rational polynomial function of s with relative degree of at least 1 be givenas

)()(

)()(

1 sdps

snsY

n+=

where n(s) is the numerator and d(s) is a polynomial which does not have any poles at

1ps −= . This has a pole repeated n times at 1ps −= .

One may show that

)()()()()()(

)()( 1

1

11

1

12

1

11

1

sYps

K

ps

K

ps

K

sdps

snsY n

nnn+

+++

++

+=

+= − L

where K11 is the residue of the pole at 1ps −= , and )(1 sY is a polynomial fraction that

does not have any poles at 1ps −= . Note that, for a pole of order n, one requires in thePFE all terms of order n and below.

To determine the residue K11 one may proceed as above and use

4

1

)()( 111 ps

n sYpsK−=

+= .

The other numerator terms Kij can be determined using some derivative formulas given inthe text. An alternative technique is given in the next example.

Example 2- Repeated Real Pole

Find the inverse Laplace transform of

254)(

23 +++=

sss

ssY .

First, factor the denominator to obtain

)2()1()( 2 ++

=ss

ssY .

This may be written as the sum of terms

21)1()2()1()( 212

211

2 ++

++

+=

++=

s

K

s

K

s

K

ss

ssY . (1)

The residues K11 and K2 are determined using the technique in Example 1 to obtain

2

2

1)1(

1)( 12

2 +−

++

++−

=ss

K

ssY .

To evaluate the term K12, one notes that the equality (1) must hold for all valuesof s. Select therefore a convenient value of s, evaluate (1), and solve for K12. A veryeasy value is s=0, for which

20

2

10)10(

1

)20()10(

0)0( 12

22 +−

++

++−

=++

=K

Y ,

yielding K12= 2.

Therefore, the PFE is

2

2

1

2

)1(

1)(

2 +−

++

++−

=sss

sY ,

whence the Laplace transform table gives the time function)()22()( 1

2 tueteety ttt−

−−− −−= .

Note that one may not select a value s corresponding to a pole in using (1) todetermine K12, since at the pole values Y(s) blows up to infinity.

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Case 3- Complex Pole Pair

Let a rational polynomial function of s with relative degree of at least 1 be givenas

)())((

)()(

22 sds

snsY

βα ++=

where n(s) is the numerator and d(s) is a polynomial which does not have the factor))(( 22 βα ++s .

We have seen in the previous lecture that( )( ))()())(( 22 βαβαβα jsjss −+++=++ ,

so this factor corresponds to a complex pole pair at βα js ±−= . In this course we shallnot write factors using 'j'. We shall keep complex pole pairs together in the quadraticform.

One may split out a term with divisor ))(( 22 βα ++s and write

)())()()))((

)()( 12222

sYs

bas

sds

snsY +

+++

=++

=βαβα

where )(1 sY is a polynomial fraction that does not contain the factor ))(( 22 βα ++s .Note that the term above the quadratic factor has degree one less, namely, in this case,one.

The next example shows how to solve for the constants a and b.

Example 3- Imaginary Pole Pair

Find the inverse Laplace transform of

842

1)(

23 ++++

=sss

ssY .

Factor the denominator and write the PFE to obtain

428

1

42)4)(2(

1)( 22

12 +

++

+

−=

++

++

=++

+=

s

bas

ss

bas

s

K

ss

ssY , (2)

where the residue at s=-2 was determined as in Example 1.

To find a and b, select two values of s and evaluate both sides of (2). From theresulting algebraic equations, one may solve for a and b. If s=0 is not a pole, one mayselect s=0 and find b first, then find a using another value of s. If s=0 is a pole, one must

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select two nonzero values of s, which generally give two simultaneous equations for aand b.

Since zero is not a pole in this example, evaluate (2) at s=0 to obtain

40

0.

208

1

)40)(20(

10)0( 22 +

++

+

−=

+++

=ba

Y .

Solving yields b=3/4.

Selecting now s=1 gives

414

3

218

1

)41)(21(

11)1( 22 +

++

+

−=

+++

=a

Y ,

which yields a=1/8.

The PFE is thus given by

44

3

48

1

28

1

44

38

1

28

1

)4)(2(

1)( 2222 +

++

++

−=

+

++

+

−=

+++

=ss

s

ss

s

sss

ssY .

Now the transform table says that

)(2sin8

32cos

8

1

8

1)( 1

2 tuttety t−

−

++−= .

Note that the poles are at 2,2 js ±−= . However, we did not use any terms in 'j' in

this evaluation. We kept the imaginary pole pair together as 22 β+s .

Example 4- Complex Pole Pair

Find the inverse transform of

)74)(1(

2)( 2 +++

−=

sss

ssY .

Using the techniques in Example 3 one may see that

744

134

3

14

3)(

2 ++

++

+

−=

ss

s

ssY .

To find the inverse transform, write the denominator in the form 22)( βα ++s to get

7

3)2(4

134

3

14

3)( 2 ++

++

+

−=

s

s

ssY .

Now use the table to see that

)(3sin34

73cos

4

3

4

3)( 1

2 tutteety tt−

−−

++

−= .

Note that the complex poles are at 32 js ±−= , yet we never use 'j' in thisevaluation. We are careful to keep the complex pair together in the quadratic term

22)( βα ++s . Note further that the exponential decay term of the complex pair is te α− ,

while the frequency of oscillation is given by 3=β .

Case 4-Relative Degree Equal to Zero

In control theory we do not use Laplace transforms with relative degree less thanzero. If the relative degree of Y(s) is equal to zero, then prior to using the above PFEtechniques one must do one step of long division to write

)()( 1 sYconstsY +=

where const is a constant and the relative degree of Y1(s) is one or more.

Example 5- Relative Degree Zero

Find the inverse Laplace transform of

6116

1834183)(

23

23

++++++

=sss

ssssY .

This transform has relative degree of zero, so the PFE does not give the correctanswer. To find the time function, perform one step of long division to write

61163)(

23 ++++=

sss

ssY .

The second term now has relative degree of 1 and one can proceed using the techniquesgiven above.

In fact, using the results of Example 1, one sees that

)()2

32

2

1()()( 1

320 tueeetuty ttt

−−−− −+−+= .

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If the transform relative degree is zero, it means that the time function contains theimpulse function u0(t).