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Classical Control – Prof. Eugenio Schuster – Lehigh University 1

Control Systems

Lecture 2 Laplace Transform

Classical Control – Prof. Eugenio Schuster – Lehigh University 2

Laplace Transform

Function f(t) of time Piecewise continuous and exponential order

0- limit is used to capture transients and discontinuities at t=0 s is a complex variable (σ+jω)

There is a need to worry about regions of convergence of the integral

Units of s are sec-1=Hz A frequency

If f(t) is volts (amps) then F(s) is volt-seconds (amp-seconds)

btKetf <)(

∫ ∞

� = 0-

) ( ) ( dt e t f s F st ( ) [ ] ∫ ∞ +

∞ � � = =

j

j st ds e s F

j t f s F

α

α π ) ( 2 1 ) ( 1 L

2

Classical Control – Prof. Eugenio Schuster – Lehigh University 3

Laplace Transform Examples Step function – unit Heavyside Function

After Oliver Heavyside (1850-1925)

Exponential function After Oliver Exponential (1176 BC- 1066 BC)

Delta (impulse) function δ(t)

0if1)()(0

)(

000>=

+−=−===

∞+−∞−∞

−

−∞

−

−∫∫ σ

ωσ

ωσ

sje

sedtedtetusF

tjststst

% & '

≥ <

= 0 for , 1 0 for , 0

) ( t t

t u

∫ ∫∞ ∞ ∞+−

+−−− >+

=+

−===0 0 0

)()( if1)( ασ

αα

ααα

ssedtedteesF

tstsstt

sdtetsF st allfor1)()(0

== ∫∞

−

−δ

Classical Control – Prof. Eugenio Schuster – Lehigh University 4

Laplace Transform Table Signal Waveform Transform impulse step

ramp

exponential

damped ramp

sine

cosine

damped sine

damped cosine

)(tδ

22)( βα

α

++

+

s

s

22)( βα

β

++s

22 β

β

+s

22 β+s

s

1

s1

21

s

α+s1

2)(

1

α+s

)(tu

)(ttu

)(tue tα−

)(tutte α−

( ) )(sin tutβ

( ) )(cos tutβ

( ) )(sin tutte βα−

( ) )(cos tutte βα−

3

Classical Control – Prof. Eugenio Schuster – Lehigh University 5

Laplace Transform Properties

{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL

Linearity: (absolutely critical property)

( )ssFdf

t=

!"#

$%&∫0

)( ττLIntegration property:

)0()()(−−=

"#$

%&' fssFdttdf

LDifferentiation property:

)0()0()(22)(2

−"−−−=#$

#%&

#'

#()

fsfsFsdt

tfdL

)0()0()0()()( )(21 −−−−"−−−=#$

#%&

#'

#() −− mmm

m

mffsfssF

dttfd

msL

Classical Control – Prof. Eugenio Schuster – Lehigh University 6

Laplace Transform Properties

)()}({ αα +=− sFtfe tL

{ } 0)()()( >=−− − asFeatuatf as forL

Translation properties:

s-domain translation:

t-domain translation:

Initial Value Property: )(lim)(lim0

ssFtfst ∞→+→

=

Final Value Property: )(lim)(lim0

ssFtfst →∞→

=

If all poles of F(s) are in the LHP

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Classical Control – Prof. Eugenio Schuster – Lehigh University 7

Laplace Transform Properties )(1)}({asF

aatf =LTime Scaling:

Multiplication by time:

Convolution:

dssdFttf )()}({ −=L

)()(})()({0

sGsFdtgft

=−∫ τττL

Time product: λλπ

ωσ

ωσdsGsF

jtgtf

j

j∫+

−−= )()(

21)}()({L

Classical Control – Prof. Eugenio Schuster – Lehigh University 8

Laplace Transform

Exercise: Find the Laplace transform of the following waveform

[ ] )()2cos(2)2sin(22)( tutttf −+=( )( )4

24)( 2 ++

=ssssF

Exercise: Find the Laplace transform of the following waveform

( )

[ ] ( )tudttedtuetf

dxxtuetft

t

tt

4040

0

4

5)(5)(

4sin5)()(−

−

−

+=

+= ∫ ( )( )

( )2

2

3

4020010)(

1648036)(

+

+=

++++

=

sssF

ssssssF

Exercise: Find the Laplace transform of the following waveform

)2()(2)()( TtAuTtAutAutf −+−−=( )seAsFTs 21)(−−

=

5

Classical Control – Prof. Eugenio Schuster – Lehigh University 9

Laplace Transform

The diagram commutes Same answer whichever way you go

Linear system

Differential equation

Classical techniques

Response signal

Laplace transform L

Inverse Laplace transform L-1

Algebraic equation

Algebraic techniques

Response transform

Tim

e do

mai

n (t

dom

ain)

Complex frequency domain (s domain)

Classical Control – Prof. Eugenio Schuster – Lehigh University 10

Solving LTI ODE�s via Laplace Transform

011

1

1

0

)1(

0

1

0

)1(1

0

011

1

011

1

)0()0()()(

asasas

subsyasU

asasasbsbsbsbsY n

nn

ji

j

jim

ii

ji

j

jin

ii

nn

n

mm

mm

++++

−

+++++++++

= −−

−

=

−−

=

−

=

−−−

=−

−

−−

∑∑∑∑

( ) ( ) ( ) ( ) ubububyayay mm

mm

nn

n0

110

11 +++=+++ −

−−

−

Initial Conditions:

!"

#$%

&−=!

"

#$%

&−+− ∑∑∑∑∑

−

=

−−

=

−

=

−−−

=

−

=

−− ji

j

jiim

ii

ji

j

jiin

ii

n

j

jjnn susUsbsysYsasysYs1

0

)1(

0

1

0

)1(1

0

1

0

)1( )0()()0()()0()(

( )( ) ( ) ( )( ) ( )0,,0,0,,0 11 uuyy mn …… −−

Recall jk

j

jkkk

k

sfsFdttfd

∑−

=

−−−=#$%

&'( 1

0

)1( )0()()(sL

For a given rational U(s) we get Y(s)=Q(s)/P(s)

6

Classical Control – Prof. Eugenio Schuster – Lehigh University 11

Laplace Transform

Exercise: Find the Laplace transform V(s)

3)0(

)(4)(6)(

−=−

=+

v

tutvdttdv

( ) 63

64)(

+−

+=

ssssV

Exercise: Find the Laplace transform V(s)

( )( )( ) 12

3215)(

+−

+++=

sssssV

2)0(',2)0(

5)(3)(4)( 22

2

=−−=−

=++ −

vv

etvdttdv

dttvd t

What about v(t)?

Classical Control – Prof. Eugenio Schuster – Lehigh University 12

Computing Transfer Functions via Laplace Transform

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

)()()(

)()()()()(

21

21

011

1

011

1

011

1

011

1

n

m

nn

n

mm

nn

n

mm

pspspszszszsK

asasasbsbsb

sUsYsH

sUsAsBsU

asasasbsbsbsY

−−−−−−

=

+++++++

==

=++++

+++=

−−

−−

−−

−−

( ) ( ) ( ) ububyayay mm

nn

n0

110

11 ++=+++ −

−−

−

Assume all Initial Conditions Zero:

( ) ( ) )()( 011

1011

1 sUbsbsbsYasasas mm

nn

n +++=++++ −−

−−

Input Output

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Classical Control – Prof. Eugenio Schuster – Lehigh University 13

Rational Functions

We shall mostly be dealing with TFs which are rational functions – ratios of polynomials in s pi are the poles and zi are the zeros of the function K is the scale factor or (sometimes) gain

A proper rational function has n≥m A strictly proper rational function has n>m An improper rational function has n<m

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

)(

21

21

011

1

011

1

n

m

nn

nn

mm

mm

pspspszszszsK

asasasabsbsbsbsF

−−−−−−

=

++++++++

= −−

−−

Classical Control – Prof. Eugenio Schuster – Lehigh University 14

Partial Fraction Expansion - Residues at Simple Poles

)()(lim sFpsk ipsii

−=→

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

2

2

1

1

21

21

n

n

n

m

psk

psk

psk

pspspszszszsKsF

−++

−+

−=

−−−−−−

=

( ))()(

)()(

)()()(

2

2

1

1

n

ini

iii ps

pskkpspsk

pspsksFps

−−

++++−−

+−−

=−

Functions of a complex variable with isolated, finite order poles have residues at the poles

Residue at a simple pole:

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Classical Control – Prof. Eugenio Schuster – Lehigh University 15

Partial Fraction Expansion - Residues at multiple poles

rjsFrpsjrds

jrdpsjr

k ii

j 1,)()(lim)!(

1=!"

#$%& −

−

−

→−=

( )31522

+

+

s

ssExample:

( ) ( ) ( )( ) )(32

13

11

12

152 2

321

3

21 tutte

sssL

sssL t −+=""

#

$%%&

'

+−

++

+=""#

$%%&

'

+

+ −−−

( ) ( )31211321

++

++

+=

s

k

s

ksk

rr

rm

psk

psk

psk

pszszszsKsF

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

12

1

2

1

1

1

21

−++

−+

−=

−−−−

=

Residue at a multiple pole:

2 ) 1 (

) 5 2 ( ) 1 ( lim ! 2 1

3 2 3

2 2

1 = ) ) * +

, , - .

+ + +

� → s s s s

ds d

s 1k

1 ) 1 (

) 5 2 ( ) 1 ( lim 3 2 3

1 = ) ) * +

, , - .

+ + +

� → s s s s

ds d

s 2k

3 ) 1 (

) 5 2 ( ) 1 ( lim 3 2 3

3 � =

+ + +

� → s s s s

s 3k

Classical Control – Prof. Eugenio Schuster – Lehigh University 16

Partial Fraction Expansion - Residues at Complex Poles

Compute residues at the poles Bundle complex conjugate pole pairs into second-order terms if you want … but you will need to be careful!

Inverse Laplace Transform is a sum of complex exponentials. But the answer will be real.

( )[ ]222 2))(( βααβαβα ++−=+−−− ssjsjs

)()(lim sFasas

−→

9

Classical Control – Prof. Eugenio Schuster – Lehigh University 17

Inverting Laplace Transforms in Practice

We have a table of inverse LTs Write F(s) as a partial fraction expansion Now appeal to linearity to invert via the table

Surprise! Nastiness: computing the partial fraction expansion is best done by calculating the residues

€

F(s) =bms

m + bm−1sm−1 ++ b1s+ b0

ansn + an−1s

n−1 ++ a1s+ a0= K (s− z1)(s− z2)(s− zm)

(s− p1)(s− p2)(s− pn )

=α1s− p1( )

+α2s− p2( )

+α31(s− p3)

+α32s− p3( )2

+α33s− p3( )3

+ ...+αqs− pq( )

Classical Control – Prof. Eugenio Schuster – Lehigh University 18

Inverse Laplace Transform

) 5 2 )( 1 ( ) 3 ( 20 ) ( 2 + + +

+ =

s s s s s F

21211)(

*221js

kjs

ksksF

+++

−++

+=

π45

255521)21)(1(

)3(20)()21(21

lim2

101522

)3(20)()1(1

lim1

jej

jsjssssFjs

jsk

sss

ssFss

k

=−−=+−=+++

+=−+

+−→=

=−=++

+=+

−→=

)()452cos(21010

)(252510)( 45)21(

45)21(

tutee

tueeetf

tt

jtjjtjt

!"#

$%& ++=

!!

"

#

$$

%

&++=

−−

−−−++−−

π

ππ

Exercise: Find the Inverse Laplace transform of

10

Classical Control – Prof. Eugenio Schuster – Lehigh University 19

Inverse Laplace Transform

Exercise: Find v(t)

3)0(

)(4)(6)(

−=−

=+

v

tutvdttdv

( ) 63

64)(

+−

+=

ssssV

Exercise: Find v(t)

( )( )( ) 12

3215)(

+−

+++=

sssssV

2)0(',2)0(

5)(3)(4)( 22

2

=−−=−

=++ −

vv

etvdttdv

dttvd t

What about v(t)?

)(311)(

32)( 6 tuetutv t−−=

)(255

21)( 32 tueeetv ttt !

"

#$%

& +−= −−−

Classical Control – Prof. Eugenio Schuster – Lehigh University 20

Not Strictly Proper Laplace Transforms

Find the inverse LT of Convert to polynomial plus strictly proper rational function

Use polynomial division Invert as normal

3 4 8 12 6 ) ( 2

2 3

+ + + + +

= s s

s s s s F

35.0

15.02

3422)( 2

++

+++=

++

+++=

sss

sssssF

)(5.05.0)(2)()( 3 tueetdttdtf tt

!"#

$%& +++= −−δδ