Laplace Transforms – recap for cctscarmenere.ucsd.edu/jorge/teaching/mae140/f11/lectures/5laplace.pdfInverse Laplace Transforms – the Bromwich Integral This is a contour integral

132MAE140 Linear Circuits

132

Laplace Transforms – recap for ccts

What’s the big idea?1. Look at initial condition responses of ccts due to capacitor

voltages and inductor currents at time t=0Mesh or nodal analysis with s-domain impedances

(resistances) or admittances (conductances)Solution of ODEs driven by their initial conditions

Done in the s-domain using Laplace Transforms

2. Look at forced response of ccts due to input ICSs and IVSsas functions of time

Input and output signals IO(s)=Y(s)VS(s) or VO(s)=Z(s)IS(s)The cct is a system which converts input signal to

output signal3. Linearity says we add up parts 1 and 2

The same as with ODEs


133

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)


134

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 (σ+jω)

There is a need to worry about regions of convergence ofthe 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-

)()( dtetfsF st


135

Laplace transform examplesStep function – unit Heavyside Function

After Oliver Heavyside (1850-1925)

Exponential functionAfter Oliver Exponential (1176 BC- 1066 BC)

Delta (impulse) function δ(t)

⎩⎨⎧

≥

<=

0for,10for,0

)(tt

tu


136





0if1)()(0

)(

000>=

+!=!===

"+!"!"

!

!"

!

!## $

%$

%$

sje

sedtedtetusF

tjststst

⎩⎨⎧

≥

<=

0for,10for,0

)(tt

tu


137





0if1)()(0

)(

000>=

+!=!===

"+!"!"

!

!"

!

!## $

%$

%$

sje

sedtedtetusF

tjststst

⎩⎨⎧

≥

<=

0for,10for,0

)(tt

tu

!

F(s) = e"#te"stdt = e"(s+# )tdt = "e"(s+# )t

s+# 0

$

=1

s+#if % > "#

0

$

&0

$

&


138





0if1)()(0

)(

000>=

+!=!===

"+!"!"

!

!"

!

!## $

%$

%$

sje

sedtedtetusF

tjststst

⎩⎨⎧

≥

<=

0for,10for,0

)(tt

tu

!

F(s) = e"#te"stdt = e"(s+# )tdt = "e"(s+# )t

s+# 0

$

=1

s+#if % > "#

0

$

&0

$

&

sdtetsF st allfor1)()(0

== !"

#

#$


139

Laplace Transform Pair TablesSignal Waveform Transform

impulse

step

ramp

exponential

damped ramp

sine

cosine

damped sine

damped cosine

)(t!

22)( !"

"

++

+

s

s

22)( !"

!

++s

22 !

!

+s

22 !+s

s

1

s1

21

s

!+s1

2)(

1

!+s

)(tu

)(ttu

)(tue t!"

)(tutte !"

( ) )(sin tut!

( ) )(cos tut!

( ) )(sin tutte !"#

( ) )(cos tutte !"#


140

Laplace Transform Properties

Linearity – absolutely critical propertyFollows from the integral definition

Example

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

!

L(Acos("t)) =


141


Linearity – absolutely critical propertyFollows from the integral definition

Example

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

!

L(Acos("t)) = LA2e j"t + e # j"t( )

$ % &

' ( ) =

A2

L e j"t( ) +A2

L e # j"t( )

=A2

1s# j"

+A2

1s+ j"

=As

s 2 +" 2


142


Integration property

Proof

( )ssFdf

t=

!"#

$%&'0

)( ((L

dtstet

dft

df !"#

"=" $%

&'(

)

*+,

-./

0 0)(

0)( 0000L


143


Integration property

Proof

Denote

so

Integrate by parts

( )ssFdf

t=

!"#

$%&'0

)( ((L

dtstet

dft

df !"#

"=" $%

&'(

)

*+,

-./

0 0)(

0)( 0000L

)(and,

0)(and,

tfdtdy

edtdx

tdfy

s

stex

st ==

!=""

=

"

##

!!!"

#"

#+

$$%

&

''(

)#=

$$%

&

''(

)

0000)(

1)()( dtetf

sdf

sedf st

tstt****L


144


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


145

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() (L( f (m(1)(0()

)()}({ !! +=" sFtfe tL

{ } 0for)()()( >=!! ! asFeatuatf asL


146


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 !"!

=


147

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 function

K is the scale factor or (sometimes) gain

A proper rational function has n≥mA 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

!!!!!!

=

++++

++++=

!!

!!

L

L

L

L


148

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-sτ

These arise from• discrete component relations of capacitors and inductors• the kinds of input signals we apply

– Steps, impulses, exponentials, sinusoids, delayedversions of functions

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

To understand linear cct responses you need to look atthe poles – they determine the exponential modes inthe 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


149

A Little More Complex Analysis

A complex function is analytic in regions where it hasno polesRational functions are analytic everywhere except at a finite

number of isolated points, where they have poles of finiteorderRational functions can be expanded in a Taylor Series

about a point of analyticity

They can also be expanded in a Laurent Series about anisolated pole

General functions do not have N necessarily finite

...)(2)(!21)()()()( +!!"+!"+= afazafazafzf

! !"

"=

#

="+"=

1

0)()()(

Nn n

nn

nn azcazczf


150

Residues at poles

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

Multiple pole: residue =

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

Cauchy Residue TheoremThe integral around a simple closed rectifiable positively

oriented curve (scroc) is given by 2πj times the sum ofresidues at the poles inside

)()(lim sFasas

!"

!

1(m"1)!

lims# a

d m"1

dsm"1(s" a)m F(s)( )


151

Inverse Laplace Transforms– the Bromwich Integral

This is a contour integral in the complex s-planeα is chosen so that all singularities of F(s) are to the left of

Re(s)=αIt yields f(t) for t≥0

The inverse Laplace transform is always a causal functionFor t<0 f(t)=0

Remember Cauchy’s Integral FormulaCounterclockwise contour integral =

2πj×(sum of residues inside contour)

( )[ ] ∫∞+

∞−

− ==j

j

stdsesFj

tfsFα

απ )(21)(1L


152

Inverse Laplace Transform Examples

Bromwich integral of

On curve C1

For given θ there is r→∞ such that

Integral disappears on C1 for positive t

x

R→∞

s-plane

pole a

t≥0 t<0

!"

!#$

<

%=

+=

&

'+

'&(

0for00for

1)(

tte

dseas

tf

at

j

j

st)

)

assF

+= 1)(

C1 C2!"<<+= rjres ,

23

2, #

$#$%

0foras0

0cos)Re()Im()Re( >!""=

<+=

treee

rstsjtsst

#$


153

Inverting Laplace Transforms

Compute residues at the poles

Example

)()(lim sFasas

!"

!"#

$%& '

'

'

(')()(1

1lim

)!1(1 sFmasmds

mdasm

( ) ( ) ( ) ( )313

21

112

31

3)1(2)1(231

522

+!

++

+=

+

!+++=

+

+

ssssss

sss


154



Example

)()(lim sFasas

!"

!"#

$%& '

'

'

(')()(1

1lim

)!1(1 sFmasmds

mdasm

( ) ( ) ( ) ( )313

21

112

31

3)1(2)1(231

522

+!

++

+=

+

!+++=

+

+

ssssss

sss

3)1(

)52()1(lim 3

23

1−=

+

++

−→ ssss

s1

)1()52()1(lim 3

23

1=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+

++

−→ ssss

dsd

s

2)1(

)52()1(lim!2

13

23

2

2

1=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

+

++

−→ ssss

dsd

s

( ) )(32)1(52 23

21 tutte

sss t −+=⎥⎥⎦

⎤

⎢⎢⎣

⎡

+

+ −−L


155



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 complexexponentials

For circuits the answers will be real

( )[ ]222 2))(( !""!"!" ++#=+### ssjsjs

)()(lim sFasas

!"

!

1(m "1)!

lims#a

dm"1

dsm"1 (s" a)mF(s)[ ]


156

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 +L+ b1s+ b0

ansn + an"1s

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

(s" p1)(s" p2)L(s" pn )

=#1s" p1( )

+#2s" p2( )

+#31(s" p3)

+#32s" p3( )2

+#33s" p3( )3

+ ...+#qs" pq( )


157

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

Find the inverse LT of)52)(1(

)3(20)( 2 +++

+=

sssssF


158

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

Find the inverse LT of)52)(1(

)3(20)( 2 +++

+=

sssssF

21211)(

*221js

kjs

ksksF

+++

!++

+=

!45

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

!"#

$%& ++=

!!

"

#

$$

%

&++=

''

'''++''

(

((


159

Not Strictly Proper Laplace Transforms

Find the inverse LT of34

8126)( 2

23

++

+++=

ssssssF


160

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

!"#

$%& +++= ''((


161

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

=!+

=++!!

=+


162

Introductory s-Domain Cct Analysis

First-order RC cctKVL

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

CvcvS

vR+ ++

-

-

-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 !"

#$%

&+''(

)**+

,-=

--


163

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

CvcvS

vR+ ++

-

-

-

+_

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()( =+!


164

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

KVL as before

Solve

+_

R

i(t)t=0

CvcvS

vR+ ++

-

-

-

+_

R I(s)

Vc(s)

+

-

sC1

22 !+sAsV

+_sCv )0(

RCsv

RCssRC

sVsV

ssVRCvsVRCs

CA

C

ACC

1)0(

)1)(()(

)0()()1(

22

22

++

++=

+=!+

"

"

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

2)(1)( tuRC

teCvRC

te

RCAVt

RCAVtCv !

"

#$%

& '+

'

+'+

+=

()(

(

!

VAcos("t)

Documents

Laplace Transforms – recap for cctscarmenere.ucsd.edu/jorge/teaching/mae140/f11/lectures/5laplace.pdfInverse Laplace Transforms – the Bromwich Integral This is a contour integral