EE/ME/AE324:Dynamical Systems

Chapter 7: Transform Solutions of Linear Models

The Laplace Transform• Converts systems or signals from the real timeConverts systems or signals from the real time

domain, e.g., functions of the real variable t, to the complex frequency domain e g functions of thecomplex frequency domain, e.g., functions of the complex variable s, where

s jσ ω= +

• Can be used to convert linear differential eqns. into l b i h i l

s jσ ω= +

algebraic eqns. that are easier to solve

• Given a time domain signal f(t), the Laplace transform is defined as

[ ] ( ) ( )( ) stF f t dtf t∞

−∫L[ ]0

( ) ( )( ) stF s f t e dtf t = ∫L

Transform Pairs: Step Function• ( )Suppose for 0, then by definition :

( )t

st st

f t A tAF A dt

→∞∞− −

= >

∫00

( ) st st

t

F s Ae dt es

A A A=

= = − ∫

h l k f

[ ]0 0 1A A Ae es s s

−∞ = − − = − − = • These signals are known as transform pairs:

[ ]( ) ( ) for 0AF f tA A t >L

• A = 1 is a special case known as the Unit Step fcn

[ ]( ) ( ) , for 0F s f tA A ts

>== =L

A = 1 is a special case known as the Unit Step fcn., denoted U(t) = 1 for all t > 0

Transform Pairs: Exp. Function( ) t• ( )

( )

Suppose for 0, then by definition :atf t Ae t−

∞ ∞

= >

∫ ∫ ( )

0 0

( ) s a tat at stF s Ae Ae e dt Ae dt− +− − −= = = ∫ ∫L

( ) 0 t

s a tA Ae e es a s

→∞− + −∞ = − = − − +

[ ]0

0 1

ts a sA A

=+

= − − =[ ] 0 1s s a+

Transform Pairs: Exp. FunctionPl t b l ill t t th f f th f• Plots below illustrate the exp. fcn. for three cases of :a

• The exp. fcn. is equivalent to U(t) when 1, 0A a= =

Transform Pairs: Ramp Function( )S f 0 h b d fi i if A• ( )

[ ]

Suppose for 0, then by definition :

t

f t At t∞

= >

∫[ ]0

( ) stF s At Ate dt−= = ∫L

1 tst

stAte Ae dt→∞ ∞−

−= − + ∫00

Step Fcn =0 1ts s=

=

∫

2

Step Fcn. =

s

A= 2s

Transform Pairs: Trig. Functions( )S i ( ) f 0 th b d ff t A t t• ( )

[ ]

Suppose sin( ) for 0, then by def. :

( ) i ( ) i ( ) st

f t A t t

F A A d

ω∞

= >

∫L[ ]0

( ) sin( ) sin( ) st

t

F s A t A t e dtω ω −

→∞

= = ∫L

( )[ ]2 2

sin( ) cos( )t

stAe s t ts

ω ω ωω

→∞−

= − −+( )

0ts

A

ω

ω=

− +

=

[ ]

2 2

Similarly cos( )

sAsA t

ω

ω

=+

L[ ] 2 2Similarly, cos( )A ts

ωω

=+

L

Transform Pairs: Rectangular Pulse• ( )Suppose is a pulse of width and height :f t L A( )

( ) ( )( ) 1LL

st st sLA AF s Ae dt e es s

− − −= = = −∫ ( ) ( )0 0

s s−∫

, 1L A→∞ =

1 0L = →

Height of arrowis area under curve0L

A= →

1L A⇒ ⋅ =

Transform Pairs: Unit Impulse• The Laplace transform of the unit impulse ( )tδp p ( )

can be derived from the pulse transform1using the assumption that 0 :L = →

( )1using the assumption that 0 :

1 0( ) li i d t i t fsL

L Ae

δ−

= →

−( )0

( ) lim , an indeterminant form0

Applying L'Hospital's Rule:L

ssL

δ→

= =

Applying LHospital s Rule:( )1 sL

sLd e−

−

−

m( ) liL

sδ→

= ( )0 0li 1 (m )

sL

LtsedL

d sL sδ

→= =

dL

Transform Pairs: Unit Impulse• 1Assuming 0, the definition of the unitL A= →Assuming 0, the definition of the unit

impulse ( ) can be expressed heuristically as:

L Atδ

→

( ) 0, for 0 only exists at the origint tδ = ≠ ⇒

( ) 1, for 0 unity area under curve t dtε

εδ ε

−= > ⇒∫

• Assuming a continuous function ( ), a formaldefinition of the unit impulse ( ) is given by:

f ttδdefinition of the unit impulse ( ) is given by:

( ) ( ) (0), for 0, 0b

t

f t t dt f a b

δ

δ = > >∫This definition will be used in EE321 (not EE3 4 !2 )

a−∫

Transform Properties• Tables E 1 and E 2 in Appendix E contain a list ofTables E.1 and E.2 in Appendix E contain a list of

helpful Laplace transform pairs and properties; key transform properties will now be discussedtransform properties will now be discussed

• Multiplication by a Constant:

[ ]0 0

( ) ( ) ( ) ( )st staf t af t e dt a f t e dt aF s∞ ∞

− −= = =∫ ∫L

•[ ]

0 0

Superposition:

[ ](

Thi i li th t th L l t f i li

) ( ) ( ) ( )af t bg t aF s bG s+ = +L

This implies that the Laplace transform is a linearoperator, like integration and differentiation!

Multiplication Properties• [ ] [ ] [ ]Generally ( ) ( ) ( ) ( )f t g t f t g t⋅⋅ ≠L L L••

[ ] [ ] [ ]Multiplication by an ExpoGenerally, ( ) ( ) ( ) ( )

nential:f t g t f t g t⋅⋅ ≠L L L

( ) ( )( ) ( ) s a tate f t f t e dt F s a∞

− +− = ⋅ +=∫L

•( )

0

2Common cases: 1 ,atte− = L( )

( )

2

( )at t

s as a−

++

L( )

( )2 2cos( ) ,ate t

s aω

ωω

= + +L

( )2 2sin( )ate t

s aω

ωω− =

+ +L

Multiplication by Time• Signal multiplication by time occurs commonly:• Signal multiplication by time occurs commonly:

[ ] { ( )}( ) d F st f t = −⋅L

•

[ ] { ( )}

Special Case:

( ) F sds

t f t =L

• Special Case:

!n n L 1

!n

nt ns +

= L

[ ] [ ] 3 43

221 1 2 61 , , , t

st t

s s s ⇒ = = = = L L L L

Differentiation by Time2• 23

2For example, if ( ) ( ) ,f t t F ss

= =

then ( ) ( ) 2( )( )

g t f t tdf tG

= =

L( )( )

( ) (0)

df

F f

tG sdt

= L

2 2

( ) (0)sF s f

s

=

=

−

= 3 2

This can be verified from Table

E.1

!

ss s

== ⋅

This can be verified from Table E.1!

Differentiation by Time1Gi U( ) 1 f 0 ( )t t U>• Given U( ) 1, for 0 ( ) ,

( )if ( ) th

t t U ss

dU tf

= > =

( )if ( ) , then

( )

f tdtdU t

=

( )( ) dU tF sdt

= L

( ) (0)sU s U= −

1 1 1 )(s

s sδ= ⋅ = =

( )This implies that ( !) dUd

t tt

δ =

Integration by Time• Signal integration by time occurs commonly: g g y y

1 ( )( )t

F sf dλ λ

= ∫L0

1 1 1S ifi Ct

s

dλ λ

∫

∫L• 30

2Specific Cases: s s s

dλ λ = =

∫L

( ) ( )1

tat a A

s s aA e Ae da

λ λ− −− =+∫( ) ( )0

(0) ( )t

s s a

g

a

F s

+∫

∫0

(0)( ) ( ( )(0) ) ( ) gg F sf d G ss

t gs

λ λ == + +∫

Application to Systems Analysis• You will learn e g in EE321 that•

[ ] [ ] [ ]You will learn, e.g., in EE321, that

( ) ( ) ( ) ( )f t g t f t g t= ∗⋅L L L

where is the convolution operator defined as:∞

∗

0

( )* ( ) ( ) ( )h t u t h u t dλ λ λ⋅ −∫

• In general, multiplication in one domain implies

convo tilu

[ ] [ ]( ) ( ) (on in the

) ( ) other domain

( ) ( )hh YL L[ ] [ ]( ) ( )

( ) ( ) ( ) ( )( ) ( )U s H s

t t u t h ty t u h Y s⇒ = ∗ ⋅ =L L

Thi f t i f l h l i t 'Application to Systems Analysis

• This fact is useful when analyzing a system'sresponse where h(t) is the system impulseresponse H(s) the system transfer fcn.,u(t) is the system input and y(t) is the systemu(t) is the system input and y(t) is the systemoutput:

( ) ( ) ICs 0( ) ( )

u t th t y t

δ=

System( )u t ( ) ( ) ( )y t h t u t= ∗

( ) ( ), ICs 0( ) ( )

u t ty

δ= =

System( )U s ( ) ( ) ( )Y s H s U s= ⋅

( ) ( )H Y( ) 1, ICs 0

( ) ( )U s

H s Y s= =

=

Finding System Responses Using Laplace Transforms

• A system response (regardless of its order) can be solved using the Laplace transform in three steps:

Laplace Transforms

g p p1. Write and immediately transform the intergral‐

differential eqns. describing the system for t>0,evaluating all ICs and inputs, e.g., unit step fcns.

2. Solve the resulting algebraic eqns., which are now fcns. of the complex frequency domain variable s, for the transform of the desired outputs.

3. Evaluate the inverse transform to obtain the output as a fcn. of time. This is done using Tables E.1 and E.2 along with some techniques to be discussed in Sectionalong with some techniques to be discussed in Section 7.6 of the text.

Unit Step and Impulse Responses• If the procedure on the prior slide is applied using

a unit step fcn., U(t), as the system input and zero p , ( ), y pICs, then the system output is called the Step Response and is denoted in the text as yU(t)p yU( )

• If the procedure is applied using a unit impluse( ) as the system input and zero ICs, then the

system output is called Impulse Responsthe etδ

system output is called Impulse Responsthe and is denoted

e( )h t

•Unit Step and Impulse Responses

Knowing a system's Impulse Response• Knowing a system s Impulse Response( ) ( ) is important since this can beh t H s

used to calculate the system's response to anyother input ( ) ( ), since:u t U sp ( ) ( ), ( ) ( ) ( ) ( ) y t Y s H s U s= ⋅

( ) ( )h t y t( )u t ( ) ( ) ( )y t h t u t= ∗

( ) ( ), ICs 0( ) ( )

u t th t y t

δ= ==

System( )U s ( ) ( ) ( )Y s H s U s= ⋅

( )H s

Unit Step and Impulse Responses• Based on the Laplace operator properties,

it can be shown that:it can be shown that:( ) ( ) ( )Udy th t H sdt

=

A system's impulse response can bedt

⇒ obtained indirectly by differentiating its step response!p p

• Consider a general first‐order system of the form:General First‐Order Models

Consider a general first order system of the form:1 ( )y y f tτ

+ =

where is a real, non-zero constant called thesystem time constant

τ

•system time constantIf the fcn. ( ) is a constant for all 0,the Laplace transform of the eqn is given by

f t A t >the Laplace transform of the eqn. is given by

1 ( ) (0) ( ) AsY s y Y s− + =( ) ( ) ( )

(0) ( ) 1 1

ys

y AY s

τ

⇒ = +

( ) 1 1s s sτ τ

+ +

•General First‐Order Models

The inverse transforms of the terms on the right-hand sideThe inverse transforms of the terms on the right hand sideof the eqn. can be found directly using Tables E.1 and E.2in Appendix E as shown:in Appendix E, as shown:

(0) ( ) 1 1y AY s = +

( ) 1 1s s s

τ τ + +

( )

( ) (0)( )

t t

zi zsy ty t y e A Ae

y tτ ττ τ

− −= + −

( ) ( )

where ( ) is the zero-input response and ( ) is thezero state response

zi zs

zi zsy y

y t y tzero-state response

•General First‐Order Models

If the input is zero i e ( ) 0 then the stability ofy t =If the input is zero, i.e., ( ) 0, then the stability ofthe system can be determined from the limit of ( ) as

thi i d t i d b th l f h

zs

zi

y ty t

t ; this is determined by the value of as shown:t τ→∞

0τ >

unst0

ableτ <⇒

stable⇒

0τ =marginally st0

ableτ⇒

•General First‐Order Models

The system response can also be expressed in termsThe system response can also be expressed in termsof its transient ( ) and ( ) steady-state responses:tr ss

t t t

y t y t− − −

[ ]( )( )

( ) (0) (0)( ) ( ) 0 as

t t t

sszi trzs y ty ty t y e A Ae A y A e

y t y t t

τ τ ττ τ τ τ= + − = + −

→ →∞( )try

is the time for the τresponse to reach 63%of its total transition;;alternately, the timefor the initial slope

The response settles to within 2% of in 4ssy τ for the initial slope

to reach as shownssy

ssy

•Ex. 7.2: Rotational Mass‐Damper

Find the system's step response assuming the system isFind the system s step response assuming the system isinitially at rest with no stored energy and the input is

( ) for 0 i e a scaled step fcn :t A tω = >( ) for 0, i.e., a scaled step fcn.:a t A tω = >

Th E M i•

( ) ( )1 2 1

The system EoM is:

( ) ( )B B BJ B B B t tω ω ω ω ω ω+

+ + = ⇒ + =( )1 1 2 1 1 1 1( ) ( )a aJ B B B t tJ J

ω ω ω ω ω ω+ + = ⇒ + =

•Ex. 7.2: Rotational Mass‐Damper

Substituting the input ( ) yields:a t Aω =

( )1 2 11 1

B B B AJ J

ω ω+

+ =

( )1 2 11 1 1( ) (0) ( )

J JB B B As s sJ Js

ω ω ω+

− + =

( )1 2 11 1 1( ) ( ) (0)

J JsB B B As s sJ Js

ω ω ω+

⇒ + = +J Js

( )1

11 ( ) , where B A Js s

J B Bω τ ⇒ + = =

( )

( )1

1 2

1 1( ) ( ) 1t

Js B B

B A B As t e τ

τ

ω ω −

+

⇒ = = ( )1 1

1 2

( ) ( ) 11

s t eB BJs s

τω ω

τ

⇒ = = − + +

•Ex. 7.3: RL Circuit

Find for 0 assuming the system input ( )ie t e t A> =Find for 0 assuming the system input ( )for 0, with zero ICs:

o ie t e t At

>

>

KCL i ld•

( )

KCL yields:

1 1(0) 0t

e e i e dλ− + + =∫( )0

(0) 0i o oe e i e dR L

λ+ + =∫

•Ex. 7.3: RL Circuit

Substituting (0) 0, , taking the Laplaceii e A= =Substituting (0) 0, , taking the Laplacetransform and solving system eqn. for ( ) yields:1 1

i

o

i e Ae s

A A 1 1( ) ( ) ( )o o oA Ae s e s e s

RR s Ls sL

− = ⇒ = +

• Taking the inverse Laplace transform of ( ) yields: o

L

e s

g p ( ) y

( )

o

oAe s = ( ) for 0

RtL

oe t Ae tR

−= >

RsL

+

•Ex. 7.4: Switched Circuit

Assume the switch is open with no energy stored in theAssume the switch is open with no energy stored in thecapacitor for 0 and the switch closes at 0.Find ( ) for 0:

t te t t

< =>Find ( ) for 0:oe t t >

• KCL at nodes and for 0:A O t >

( )1 2 63 o A A o A Ae e e e e e− = ⇒ = +

( ) ( )1 1 1 124 66 3 6 2o o A o o Ae e e e e e+ − = − ⇒ = +

•Ex. 7.4: Switched Circuit

Transforming the eqns. term by term yields:g q y y6

( ) 6 ( ) (0)o A Ae e e= +

( )( ) 6 1 ( ) ( ) 6 ( ) (0)o A Ae s se s e= − ( )( ) 6 1 ( )

1 6 16 ( ) ( )

A Ae s s e s

e e e s e s

+ = +

= + = +

•

6 ( ) ( )2 2

Combining the eqns to solve for ( ) yields:

o A o Ae e e s e ss

e s

= + = +

• Combining the eqns. to solve for ( ) yields:6 1 6 1( )

1 1 1

o

o

e sse s +

= = + 1 1 1

12 12 12s s s s s + + +

( )12 12 12( ) 6 12 1 12 6t t t

oe t e e e− − −

= + − = −

Ex. 7.4: Switched Circuit12Given ( ) 12 6t

e t e−

• 12Given ( ) 12 6

(0 ) 6 V and lim ( ) 12 Vo

o oSS ot

e t e

e e e t+

→∞

= −

⇒ = = =

The system has a time constant 12 secsystem settles to SS within 5 60 sec

ττ=

⇒ =

•system settles to SS within 5 60 secτ⇒

The system can be written in state-space form as shown:1 11 1 1 and 6

12 2A A o Ae e e e= − + = +

Ex. 7.4: Simulink Model and Plot

• Electrical circuits (Chapter 6):Exam 2 Review

– How to model resistors, capacitors, inductors, open/closed switches, op‐amps, etc.M d l i N d V l M h d b i– Model using Node Voltage Method to obtain:• Input‐Output models

St t d l• State‐space models

• First‐order system responses (through Section 7.5):A li ti f th L l t f t b th– Application of the Laplace transform to both mechanical and electrical systems, use of Tables E.1‐2

– How to handle common inputs e g unit step and unitHow to handle common inputs, e.g., unit step and unit impulse, and initial conditions

– Understand the concepts of zero state and zero input p presponses, transient and steady‐state responses, system time constant and stability

•Section 7.6: Transform Inversion

We now discuss a method for computing an inverse Laplacetransform that allows us to compute system output ( )from ( ) assuming that it can be written as a strictly proper

y tY s( ) g y p p

rational function, i.e., m n< , where:( ) mb s bN s + +

•

01

1 0

( ) ( )( )

In addition assume that ( ) can be factored into roots:

mn n

n

b s bN sY sD s s a s a

D s n

−−

+ += =

+ + +

•1 2

In addition, assume that ( ) can be factored into roots: ( ) ( )( ) ( )

h h ll d ln

D s nD s s s s s s s= − − −

f ( ) d h i l iYwhere the terms are called polis es of ( ) and their locationsdetermine the stability and shape of the system response

Y s

• Note, the poles are generally complex valued i i is jσ ω⇒ = +

•Partial‐Fraction Expansion Method

Partial Fraction Exapnsion (PFE) allows us to writePartial Fraction Exapnsion (PFE) allows us to writehigher-order s-domain expressions in terms of a sum oflower order expressions having known inverses

•

lower-order expressions having known inverses

There are two basic cases when computing PFEs:Dist (Real or Complex) aninct Poles Repeatedd Poles

•Distinct Poles (Real or Complex)

Let's examine the case of Distinct Poles i e ;s s s≠ ≠1 2Let s examine the case of Distinct Poles, i.e., ;assume ( ) can be written as:

n

n

s s sY s

A AA A

≠ ≠

1 2

11 2

( )( ) ( ) ( ) ( )

K id h i d f ll

nn i

in i

A AA AY ss s s s s s s s

A=

= + + + =− − − −∑

• Known as a residue, the term is computed as follows

iA ( ) ( ) , for 1, ,i i is sA s s Y s i n== − =

• The inverse Laplace transform can then be computed asn n

i

A 1 1

1 1[( ) ( )] , for 0

( )i

n ns ti

ii ii

Ay t Y s Ae ts s= =

− − = = = > −

∑ ∑L L

•Distinct Poles (Real or Complex)

Complex poles are treated the same as real poles otherComplex poles are treated the same as real poles otherthan the need to combine the resulting complex conjugateterms so that ( ) is a real fcn of timey tterms so that ( ) is a real fcn. of time

For example, assume ( ) contains complex poles a

y t

Y s t

1,2

1 2

s so that:

( )

jA AY

σ ω= − ±

+ +1 2

1

( )( ) ( )

where ( ) ( ) ,jj

Y ss j s j

A s j Y s Ae φ

σ ω σ ωσ ω

= + ++ − + +

= + − =1

*2 1

where ( ) ( ) ,

( ) ( ) j

s j

s j

A s j Y s Ae

A s j Y s A Ae φ

σ ω

σ ω

σ ω

σ ω −

=− +

=− −

+

= + + = =

( )2 2 1with and tanA ωσ ω φ σ−= + =

•Distinct Poles (Real or Complex)

Transforming the complex poles back into the time domainTransforming the complex poles back into the time domainusing the same transform pair as in the real poles case andapplying Euler's formula yields:applying Euler s formula yields:

( )( ) ( )

j jAe AeY sj j

φ φ−

= + ++ + +( ) ( )

( ) jj

s j s j

y t Ae e σφ

σ ω σ ω− +

+ − + +

= ( ) ( )t j tjAe eω σ ωφ − −−+ +( ) ( )

22

j t j tt e eAe

ω φ ω φσ

+ − +− +

=

( ) 2 cos , for 0h d ll l

tAe t tt A

σ ω φ

φ

−

= + >

where , , , , and are all realy t Aσ ω φ

•Ex. 7.9: PFE w/District Real Poles

Find the inverse transform of•

( )( )1 2

Find the inverse transform of5 5 ( ) ,

2 5 4 1 4 ( 1) ( 4)A As sY s

s s s s s s− + − +

= = = ++ + + + + +( )( )

1 2

2 5 4 1 4 ( 1) ( 4)where s 1, s 4 are the system poles

s s s s s s+ + + + + +

= − = −

1 1 1 1( ) ( ) ( 1) ( )

5 1 5 6

s s sA s s Y s s Y s

s

= =−⇒ = − = +

+ +( ) ( )

5 1 5 6 24 1 4 31

ss s

− + += = = =

+ − +=−

2 2 2( ) ( ) s sA s s Y s =⇒ = − 4( 4) ( )

5 4 5 9

ss Y s

s

=−= +

− + +( ) ( )

5 4 5 9 31 4 1 34

ss s

+ += = = = −

+ − + −=−

•Ex. 7.9: PFE w/District Real Poles

Given the PFE for ( ) we can now write:Y s

4

Given the PFE for ( ), we can now write: 2 3( ) ( ) 2 3 , for 0

( 1) ( 4)t t

Y s

Y s y t e e t− −= − = − >

• Plotting this fcn.

( ) ( ) ,( 1) ( 4)

ys s+ +

over 5τ using Matlab:– Do the initial and

final values of the figure make sense, e.g., y(0)and y ?and ySS?

•Ex. 7.11 PFE: w/District Complex Poles

Find ( ) given:y t

( )( )2

Find ( ) given:4 8 4 8( )

2 5 1 2 1 2

y ts sY s

s s s j s j+ +

= =+ + + − + +( )( )

1 2

2 5 1 2 1 2

,( 1 2) ( 1 2)

s s s j s jA A

s j s j

+ + + + +

= ++ + +

*1 2 1

( 1 2) ( 1 2)where s 1 2, s 1 2 are the system poles

s j s jj j s

+ − + +

= − + = − − =

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

4 1s s s jA s s Y s s j Y s

j= =− +⇒ = − = + −

− +( ) 0 4642 8 4 8 jj+ +4 1

j+=

( ) 0.4642 8 4 8 2 5( 1 2 1 2) 4( ) ( ) ( 1 2) ( )

jj j ej j j

A s s Y s s j Y s

−+ += = − =

− + + +

⇒ = = + +2 2 2

0.464 *1

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

2 5 j

s s s jA s s Y s s j Y s

j e A

= =− −⇒ = − = + +

= + = =

•Ex. 7.11: PFE w/District Complex PolesGiven the PFE for ( ):Y s

0.464 0.464

Given the PFE for ( ):

5 5( )j j

Y s

e eY s−

= +

( )

( )1 2 1 2

( ) 2 5 cos 2 0.464 , for 0t

Y ss j s j

y t e t t−

++ − + +

= − >

Pl tti thi f

( )( ) 2 5 cos 2 0.464 , for 0y t e t t >

• Plotting this fcn. over 5τ using Matlab yields:

– Do the initial and final values of the figure

k (0)make sense, e.g., y(0)and ySS?

•Repeated Poles (Real or Complex)

Poles are said to be repeated if several roots of ( ) 0 D s =

1 2are the same; assuming repeated poles ,we represent the PFE of ( ) as:

rr s s sY s

= = =

1, 1, 1 112

1 1 1

p ( )

( )( ) ( ) ( )

r rr

A A AY ss s s s s s

−= + + + +− − −1 1 1

1,

( ) ( ) ( )

( ) r

s s s s s s

y t A= +( )

( ) 111, 1 11

11 !

r s trA t A t e−−

+ + +

1,r ( )

( )

1, 1 11

1

1 !

1h ( ) ( )

r

pr

r

dA Y−

−

( ) ( )1, 11

1

where ( ) ( )1 !

rp p

s sA s s Y s

p ds −

=

= − −

11 1 12 1

11

( ) ( ) , ( ) ( ) ,r r

s ss s

dA s s Y s A s s Y sds=

=

⇒ = − = −

•Ex. 7.10: Repeated Real Poles

Find the inverse Laplace transform of:

( ) ( )23 2

5 16 5 16 ( )9 24 20 2 5s sY s

s s s s s+ +

= =+ + + + + Matlab plot

( )

12 11 22

2 5

( 2) ( 2) ( 5)

t t

A A As s s

= + ++ + +

( ) 2 512 11 2

211 2

( )

5 16where ( 2) ( ) 25

t t

s

y t A A t e A e

sA s Y s

− −

=

= + +

+= + = =11

12

2 25

( 2)

s ss

dA s

=− =−+

= + 22

5 16 9( ) 1d sY s + = = = 12 ( )

ds ( )22 2 2

>>eval(limit(simple(di

( )5

ff('(5

5

*s+16)/(s+5)',s)),s, 2)- )s s s

ds s s=− =− =− + +

2 5( 5) ( ) 1sA s Y s =−= + = −

•Long Division

When , PFE can't be applied directly to ( );m n Y s=When , PFE can t be applied directly to ( );performing long division fixes the problem as shown:

m n Y s

1( ) ( )( ) ( ) ( )( ) ( )

N s N sY s A y t A tD s D s

δ −′ ′ = + = +

L

where ( ) has order PFE can be appliedN s p n

′ < ⇒( )N s′to (Y ′ ( )) as shown previously( )

N ssD s

=

Ex. 7.13: Long Division22 7 8 4s s s+ + +

• 2 2

2 7 8 4For example, if ( ) 23 2 3 2

s s sY ss s s s

+ + += = +

+ + + +

12

4 ( ) 2 ( )3 2sy t t

s sδ − + = + + +

L3 2

2

s s+ +

2 2

2

3 2 2 7 82 6 4

s s s ss s

⇒ + + + +

+ +2 6 44

s ss

+ ++

• There are a few additional Laplace transformAdditional Transform Properties

There are a few additional Laplace transform properties that are helpful to know when analyzing system responses including: modeling Time Delayssystem responses, including: modeling Time Delays, the Initial and Final Value Theorems

T d l ti d l id th it t fTo model time delay, consider the unit step fcn.,0, for 0

( )t

U t≤

= ( )1, for 0

and a unit step delayed by T seconds

U tt

= >p y y

0, for ( )

1 ft T

U t Tt T≤

− = >1, for t T>

•Time Delay

If ( ) ( ) ( )f t f t U t=

[ ] ( ) ( ) ,

If ( ) ( ) ( )

( ) ( ) ( ) stf t U t e dt

f t f t U t

F s f t U t∞

−== ∫L[ ]0

( ) ( )( ) ( ) ( )

then ( ) ( ) ( )

ff

f t T f t T U t T− = − −

∫

[ ] ( ) ( ) ( ( )) stf t T U t T e dtf t T U t T∞

−= − −− − ∫L

( )

0

( ) ( ) s t Tst sTf t T e dt e f t T e dt∞ ∞

− −− −= − = −∫ ∫( ) ( )

( ) ( ) ( ) )(

a a

sT s sT sT

f f

f d F f t T Fλλ λ∞

− − − −

∫ ∫

∫ ( ) ( ) ( ) )(sT s sT sT

a

e f e d e F s f t T e F sλλ λ= = ⇒ −∫

•Example of Time Delay

Find the Laplace transform of ( ) ( ) ( 2)f t U t U t= − −

( ) 1( )( ) ( 1 ( ))sT

sT sT ee U sF s U e U ss

s−

− − −= − == −

s

•Initial and Final Value Theorems

The Initial Value Theorem (IVT) and• The Initial Value Theorem (IVT) andFinal Value Theorem (FVT) allow computation of ( )y t

0

at zero and steady-state from ( ) as shown:(0 ) lim ( ) lim ( ), when the limit in exists

t

Y sy y t sY s s+ = =

0

lim ( )t s

SS ty y t

→ →∞

→∞=

0lim ( ), if all poles in left-half s-planessY s

→=

For example, find (0 ) and for ( ) when4 8( ) h 1 2 t bl l

SSy y Y ssY j

+

+±1,22( ) , where 1 2 stable poles

2 5Y s s j

s s= = − ± ⇒

+ +

Initial and Final Value Theorems24 8s s+• 2

4 8From IVT, (0 ) lim ( ) lim 42 5s s

s sy sY ss s

+

→∞ →∞

+= = =

+ +2

20 0

4 8From FVT, lim ( ) lim 02 5SS s s

s sy sY ss s→ →

+= = =

+ +

These results agree with those of Ex. 7.11 (see graph)!

Questions?