Why Laplace transforms? - University of California, San Diegocarmenere.ucsd.edu/jorge/teaching/mae40/w20/lectures/5laplace.pdf · Our Laplace Transforms will consist of rational functions

133MAE40 Linear Circuits 133

Why Laplace transforms?First-order RC cct

KVL

instantaneous for each tSubstitute element relations

Ordinary differential equation in terms of capacitor voltage

Laplace transform

Solve

Invert LT

+_

R

VA

i(t)t=0

CvcvSvR+ +

+-

-

-0)()()( =−− tvtvtv CRS

dttdvCtitRitvtuVtv C

RAS)()(),()(),()( ===

)()()( tuVtvdttdvRC AC

C =+

ACCC Vs

sVvssVRC 1)()]0()([ =+−

RCsv

RCssRCVsV CA

C /1)0(

)/1(/)(

++

+=

Volts)()0(1)( tueveVtv RCt

CRCt

AC !"

#$%

&+''(

)**+

,−=

−−


Laplace transforms

The diagram commutesSame answer whichever way you go

Linearcct

Differentialequation

Classicaltechniques

Responsesignal

Laplacetransform L

Inverse Laplacetransform L-1

Algebraicequation

Algebraictechniques

Responsetransform

Tim

e do

mai

n (t

dom

ain) Complex frequency domain

(s domain)


Laplace Transform - definition

Function f(t) of timePiecewise continuous and exponential order

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

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

Units of s are sec-1=HzA frequency

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

btKetf <)(

ò¥

-=0-

)()( dtetfsF st


Laplace transform examplesStep function – unit Heavyside Function

After Oliver Heavyside (1850-1925)

Exponential functionAfter Oliver Exponential (1176 BC- 1066 BC)

Delta (impulse) function d(t)

îíì

³<

=0for,10for,0

)(tt

tu






0if1)()(0

)(

000>=

+−=−===

∞+−∞−∞

−

−∞

−

−∫∫ σ

ωσ

ωσ

sje

sedtedtetusF

tjststst

îíì

³<

=0for,10for,0

)(tt

tu






0if1)()(0

)(

000>=

+−=−===

∞+−∞−∞

−

−∞

−

−∫∫ σ

ωσ

ωσ

sje

sedtedtetusF

tjststst

îíì

³<

=0for,10for,0

)(tt

tu

€

F(s) = e−αte−stdt = e−(s+α )tdt = −e−(s+α )t

s+α 0

∞

=1

s+αif σ > −α

0

∞

∫0

∞

∫






0if1)()(0

)(

000>=

+−=−===

∞+−∞−∞

−

−∞

−

−∫∫ σ

ωσ

ωσ

sje

sedtedtetusF

tjststst

îíì

³<

=0for,10for,0

)(tt

tu

€

F(s) = e−αte−stdt = e−(s+α )tdt = −e−(s+α )t

s+α 0

∞

=1

s+αif σ > −α

0

∞

∫0

∞

∫

sdtetsF st allfor1)()(0

== ∫∞

−

−δ


Laplace Transform Pair TablesSignal Waveform Transformimpulsestep

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−


Laplace Transform Properties

Linearity – absolutely critical propertyFollows from the integral definition

Example

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

€

L(Acos(βt)) =



Linearity – absolutely critical propertyFollows from the integral definition

Example

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

€

L(Acos(βt)) = LA2e jβt + e − jβt( )

$ % &

' ( ) =

A2

L e jβt( ) +A2

L e − jβt( )

=A2

1s− jβ

+A2

1s+ jβ

=As

s 2 +β 2



Integration property

Proof

( )ssFdf

t=

!"#

$%&∫0

)( ττL

dtstet

dft

df −∫∞

∫=∫ $%

&'(

)

*+,

-./

0 0)(

0)( ττττL



Integration property

Proof

Denote

so

Integrate by parts

( )ssFdf

t=

!"#

$%&∫0

)( ττL

dtstet

dft

df −∫∞

∫=∫ $%

&'(

)

*+,

-./

0 0)(

0)( ττττL

)(and,

0)(and,

tfdtdy

edtdx

tdfy

s

stex

st ==

∫=−−

=

−

ττ

∫∫∫∞

−∞

−+

$$%

&

''(

)−=

$$%

&

''(

)

0000)(

1)()( dtetf

sdf

sedf st

tsttττττL



Differentiation Property

Proof via integration by parts again

Second derivative

)0()()(−−=

"#$

%&' fssFdttdf

L

€

Ldf (t)dt

" # $

% & '

=df (t)dt

e−stdt0−

∞∫ = f (t)e−st[ ]0−

∞+ s f (t)e−stdt0−

∞∫

= sF(s)− f (0−)

)0()0()(2

)0()()(2)(2

−"−−−=

−−#$%

&'(=

#$%

&'(

)*+

,-.=

/#

/$%

/&

/'(

fsfsFs

dtdf

dttdfs

dttdf

dtd

dt

tfdLLL


Laplace Transform PropertiesGeneral derivative formula

Translation propertiess-domain translation

t-domain translation

€

Ldm f (t)

dtm

" # $

% & '

= s m F(s) − sm−1 f (0−) − sm−2 ) f (0−) −− f (m−1)(0−)

)()}({ αα +=− sFtfe tL

{ } 0for)()()( >=−− − asFeatuatf asL



Initial Value Property

Final Value Property

Caveats:Laplace transform pairs do not always handle

discontinuities properlyOften get the average value

Initial value property no good with impulsesFinal value property no good with cos, sin etc

)(lim)(lim0

ssFtfst ∞→+→

=

)(lim)(lim0

ssFtfst →∞→

=


Rational FunctionsWe shall mostly be dealing with LTs which are

rational functions – ratios of polynomials in s

pi are the poles and zi are the zeros of the functionK is the scale factor or (sometimes) gain

A proper rational function has n³mA strictly proper rational function has n>mAn improper rational function has n<m

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

)(

21

21

011

1

011

1

n

m

nn

nn

mm

mm

pspspszszszsK

asasasabsbsbsbsF

−−−−−−

=

++++

++++=

−−

−−


A Little Complex Analysis

We are dealing with linear cctsOur Laplace Transforms will consist of rational functions

(ratios of polynomials in s) and exponentials like e-stThese arise from • discrete component relations of capacitors and inductors• the kinds of input signals we apply

– Steps, impulses, exponentials, sinusoids, delayed versions of functions

Rational functions have a finite set of discrete polese-st is an entire function and has no poles anywhere

To understand linear cct responses you need to look at the poles – they determine the exponential modes in the response circuit variables.Two sources of poles: the cct – seen in the response to Ics

the input signal LT poles – seen in the forced response


Residues at poles

Functions of a complex variable with isolated, finite order poles have residues at the polesSimple pole: residue =

Multiple pole: residue =

The residue is the c-1 term in the Laurent Series

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 exponentialsFor circuits the answers will be real

)()(lim sFasas

−→

€

1(m−1)!

lims→ a

d m−1

dsm−1(s− a)m F(s)( )

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

MAE40 Linear Circuits


Inverting Laplace Transforms in Practice

We have a table of inverse LTsWrite F(s) as a partial fraction expansion

Now appeal to linearity to invert via the tableSurprise!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( )


Inverting Laplace TransformsCompute residues at the poles

Example

)()(lim sFasas

−→

!"#

$%& −

−

−

→−)()(1

1lim

)!1(1 sFmasmds

mdasm

( ) ( ) ( ) ( )313

21

112

31

3)1(2)1(231

522

+−

++

+=

+

−+++=

+

+

ssss

ss

s

ss


Inverting Laplace TransformsCompute residues at the poles

Example

)()(lim sFasas

−→

!"#

$%& −

−

−

→−)()(1

1lim

)!1(1 sFmasmds

mdasm

( ) ( ) ( ) ( )313

21

112

31

3)1(2)1(231

522

+−

++

+=

+

−+++=

+

+

ssss

ss

s

ss

3)1(

)52()1(lim 3

23

1-=

+

++

-® ssss

s1

)1()52()1(lim 3

23

1=

úúû

ù

êêë

é

+

++

-® ssss

dsd

s

2)1(

)52()1(lim!2

13

23

2

2

1=

úúû

ù

êêë

é

+

++

-® ssss

dsd

s

( ) )(32)1(52 23

21 tutte

sss t -+=úúû

ù

êêë

é

+

+ --L



T&R, 5th ed, Example 9-12

Find the inverse LT of )52)(1(

)3(20)( 2 +++

+=sss

ssF


T&R, 5th ed, Example 9-12

Find the inverse LT of )52)(1(

)3(20)( 2 +++

+=sss

ssF

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



Not Strictly Proper Laplace Transforms

Find the inverse LT of34

8126)( 2

23

++

+++=ssssssF


Not Strictly Proper Laplace Transforms

Find the inverse LT of

Convert to polynomial plus strictly proper rational functionUse polynomial division

Invert as normal

348126)( 2

23

++

+++=ssssssF

35.0

15.02

3422)( 2

++

+++=

++

+++=

sss

sssssF

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

!"#

$%& +++= −−δδ



Multiple Poles

Look for partial fraction decomposition

Equate like powers of s to find coefficients

Solve

)())(()(

)())(()()(

12221212

211

22

22

2

21

1

12

21

1

pskpspskpskKzKs

psk

psk

psk

pspszsKsF

−+−−+−=−

−+

−+

−=

−−

−=

112221122

1

22212121

211

2

)(220

Kzpkppkpk

Kkppkpkkk

=−+

=++−−

=+


Recall motivating example for LTFirst-order RC cct

KVL

instantaneous for each tSubstitute element relations

Ordinary differential equation in terms of capacitor voltage

Laplace transform

Solve

Invert LT

+_

R

VA

i(t)t=0

CvcvSvR+ +

+-

-

-0)()()( =−− tvtvtv CRS

dttdvCtitRitvtuVtv C

RAS)()(),()(),()( ===

)()()( tuVtvdttdvRC AC

C =+

ACCC Vs

sVvssVRC 1)()]0()([ =+−

RCsv

RCssRCVsV CA

C /1)0(

)/1(/)(

++

+=

Volts)()0(1)( tueveVtv RCt

CRCt

AC !"

#$%

&+''(

)**+

,−=

−−


An Alternative s-Domain Approach

Transform the cct element relationsWork in s-domain directly OK since L is linear

KVL in s-Domain

+_

R

VA

i(t)t=0

CvcvSvR+ +

+-

-

-

+_

R

VA

I(s)

Vc(s)

+

-

sC1

s1

+_ sCv )0(

)0()()(

)0()(1)(

CCC

CCC

CvssCVsIs

vsICs

sV

−=

+= Impedance + source

Admittance + source

ACCC Vs

sVCRvssCRV 1)()0()( =+−


Time-varying inputsSuppose vS(t)=VAcos(bt), what happens?

KVL as before

Solve

+_

R i(t)t=0

CvcvSvR+ +

+-

-

-

+_

R I(s)

Vc(s)

+

-

sC1

22 β+sAsV

+_ sCv )0(

RCsv

RCssRC

sVsV

ssVRCvsVRCs

CA

C

ACC

1)0(

)1)(()(

)0()()1(

22

22

++

++=

+=−+

β

β

)()0(2)(1)cos(

2)(1)( tuRC

teCvRC

te

RCAVt

RCAVtCv !

"

#$%

& −+

−

+−+

+=

βθβ

β

€

VAcos(βt)

Documents

Why Laplace transforms? - University of California, San Diegocarmenere.ucsd.edu/jorge/teaching/mae40/w20/lectures/5laplace.pdf · Our Laplace Transforms will consist of rational functions