# Laplace Transforms – recap for ccts

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• MAE140 Linear Circuits107

Laplace Transforms recap for ccts

Whats 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 Transforms2. Look at forced response of ccts due to input ICSs and IVSs

as 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

• MAE140 Linear Circuits108

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)

• MAE140 Linear Circuits109

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 (+j)

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

Units of s are sec-1=HzA frequency

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

btKetf

• MAE140 Linear Circuits110

Laplace transform examplesStep function unit Heavyside Function

After Oliver Heavyside (1850-1925)

Exponential functionAfter Oliver Exponential (1176 BC- 1066 BC)

Delta (impulse) function (t)

0if1

)()(

0

)(

000

>=+

!=!===

"+!"!"

!

!"

!

!## \$

%\$

%\$

sj

e

s

edtedtetusF

tjststst

=!! ! asFeatuatf asL

• MAE140 Linear Circuits116

Laplace Transform Properties

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 !"!

=

• MAE140 Linear Circuits117

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 nmA strictly proper rational function has n>mAn improper rational function has n

• MAE140 Linear Circuits118

A Little Complex Analysis

We are dealing with linear cctsOur Laplace Transforms will consist of rational function (ratios

of polynomials in s) and exponentials like e-sThese 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-s 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

• MAE140 Linear Circuits119

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)(!2

1)()()()( +!!"+!"+= afazafazafzf

! !"

"=

#

=

"+"=1

0

)()()(

Nn n

nn

nn azcazczf

• MAE140 Linear Circuits120

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 2j times the sum ofresidues at the poles inside

)()(lim sFasas

!"

[ ])()(lim)!1(

11

1

sFas

ds

d

m

m

m

m

as

!! !

!

"

• MAE140 Linear Circuits121

Inverse Laplace Transforms the Bromwich Integral

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

Re(s)=It yields f(t) for t0

The inverse Laplace transform is always a causal functionFor t

• MAE140 Linear Circuits122

Inverse Laplace Transform Examples

Bromwich integral of

On curve C1

For given there is r such that

Integral disappears on C1 for positive t

x

R

s-plane

pole a

t0 t

• MAE140 Linear Circuits123

Inverting Laplace Transforms

Compute residues at the poles

Example

)()(lim sFasas

!"

!"#

\$%& '

'

'

(')()(

1

1lim

)!1(

1sF

mas

mds

md

asm

( ) ( ) ( ) ( )31

3

21

1

1

2

31

3)1(2)1(2

31

522

+!

++

+=

+

!+++=

+

+

ssss

ss

s

ss

3)1(

)52()1(lim 323

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

• MAE140 Linear Circuits124

Inverting Laplace Transforms

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 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)[ ]

• MAE140 Linear Circuits125

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!Nastiness: 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 )

=#1

s" p1( )+

#2

s" p2( )+

#31

(s" p3)+

#32

s" p3( )2

+#33

s" p3( )3

+ ...+#q

s" pq( )

• MAE140 Linear Circuits126

Example 9-12

Find the inverse LT of)52)(1(

)3(20)( 2 ++++

=sss

ssF

21211)(

*221

js

k

js

k

s

ksF

+++

!++

+=

!4

5

2555

21)21)(1(

)3(20)()21(

21lim2

10

1522

)3(20)()1(

1lim1

jej

jsjss

ssFjs

jsk

sss

ssFs

sk

=""=

+"=+++

+="+

+"#=

=

"=++

+=+

"#=

)()4

52cos(21010

)(252510)( 4

5)21(

4

5)21(

tutee

tueeetf

tt

jtjjtjt

!"

#\$%

&++=

!!

"

#

\$\$

%

&

++=

''

'''++''

(

((

• MAE140 Linear Circuits127

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

3

5.0

1

5.02

34

22)(

2

++

+++=

++

+++=

sss

ss

sssF

)(5.05.0)(2)(

)( 3 tueetdt

tdtf tt !"

#\$%

&+++=

''((

• MAE140 Linear Circuits128

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

ps

k

ps

k

ps

k

psps

zsKsF

!+!!+!=!

!+

!+

!=

!!

!=

112221122

1

22212121

211

2

)(22

0

Kzpkppkpk

Kkppkpk

kk

=!+

=++!!

=+

• MAE140 Linear Circuits129

Introductory s-Domain Cct AnalysisFirst-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

dt

tdvCtitRitvtuVtv

CRAS

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

)()()(

tuVtvdt

tdvRC AC

C =+

ACCC Vs

sVvssVRC1

)()]0()([ =+!

RCs

v

RCss

RCVsV

CAC

/1

)0(

)/1(

/)(

++

+=

Volts)()0(1)( tueveVtv RCt

CRCt

AC !"

#\$%

&+''(

)**+

,-=

--

• MAE140 Linear Circuits130

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)

+

-

sC

1

s

1

+_s

Cv )0(

)0()()(

)0()(

1)(

CCC

CCC

CvssCVsI

s

vsI

CssV

!=

+= Impedance + source

ACCC Vs

sVCRvssCRV1

)()0()( =+!

• MAE140 Linear Circuits131

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

KVL as before

Solve

+_

R

VA

i(t)t=0

CvcvSvR+ + +

-

-

-

+_

R I(s)

Vc(s)

+

-

sC

1

22 !+s

AsV

+_s

Cv )0(

RCs

v

RCss

RC

sV

sV

s

sVRCvsVRCs

C

A

C

ACC

1

)0(

)1)(()(

)0()()1(

22

22

++

++=

+=!+

"

"

)()0(2)(1

)cos(2)(1

)( tuRCt

eCvRCt

e

RC

AVt

RC

AVtCv !

"

#\$%

& '+

'

+'+

+

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