Laplace Transforms – recap for cctsnodes.ucsd.edu/sonia/teaching/140_LinearCircuits/data/5laplace.pdf · MAE140 Linear Circuits 109 Laplace Transform - definition Function f(t)

MAE140 Linear Circuits 107

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=0 Mesh 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 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 signal 3.  Linearity says we add up parts 1 and 2

The same as with ODEs


Laplace transforms

The diagram commutes Same answer whichever way you go

Linear cct

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)


Laplace Transform - definition

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


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)

≥ <

= 0 for , 1 0 for , 0

) ( t t

t u






0if1)()(0

)(

000>=

+−=−===

∞+−∞−∞

−

−∞

−

−∫∫ σ

ωσ

ωσ

sje

sedtedtetusF

tjststst

≥ <

= 0 for , 1 0 for , 0

) ( t t

t u






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






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


Laplace Transform Pair Tables 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−


Laplace Transform Properties

Linearity – absolutely critical property Follows from the integral definition

Example

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

€

L(Acos(βt)) =



Linearity – absolutely critical property Follows from the integral definition

Example

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

[ ] ( ) ( )

22

12

12

222))cos((

β

ββ

β ββββ

+=

++

−=

+=

+= −−

sAs

jsA

jsA

eAeAeeAtA tjtjtjtj LLLL



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,

tfdtdye

dtdx

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 Properties General derivative formula

Translation properties s-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 properly Often get the average value

Initial value property no good with impulses Final value property no good with cos, sin etc

)(lim)(lim0

ssFtfst ∞→+→

=

)(lim)(lim0

ssFtfst →∞→

=


Rational Functions We 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≥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

−−−−−−

=

++++

++++=

−−

−−


A Little Complex Analysis

We are dealing with linear ccts Our Laplace Transforms will consist of rational function (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, delayed versions of functions

Rational functions have a finite set of discrete poles e-sτ 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


A Little More Complex Analysis

A complex function is analytic in regions where it has no poles Rational functions are analytic everywhere except at a finite

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

about a point of analyticity

They can also be expanded in a Laurent Series about an isolated pole

General functions do not have N necessarily finite

...)(2)(!21)()()()( +′′−+′−+= afazafazafzf

∑ ∑−

−=

∞

=−+−=

1

0)()()(

Nn n

nn

nn azcazczf


Residues at poles

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

Multiple pole: residue =

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

Cauchy Residue Theorem The integral around a simple closed rectifiable positively

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

)()(lim sFasas

−→

[ ])()(lim)!1(

11

1sFas

dsd

mm

m

m

as−

− −

−

→


Inverse Laplace Transforms – the Bromwich Integral

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

Re(s)=α It yields f(t) for t≥0

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

Remember Cauchy’s Integral Formula Counterclockwise contour integral =

2πj×(sum of residues inside contour)

( ) [ ] ∫ ∞ +

∞ - - = =

j

j st ds e s F

j t f s F

α

α π ) ( 2 1 ) ( 1 L


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

stα

α

a s s F

+ = 1 ) (

C1 C2 ∞→<<+= rjres ,

23

2, π

θπθα

0foras0

0cos)Re()Im()Re( >∞→→=

<+=

treee

rstsjtsst

θα


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

3 ) 1 (

) 5 2 ( ) 1 ( lim 3 2 3

1 - =

+ + +

- → s s s s

s 1

) 1 ( ) 5 2 ( ) 1 ( lim 3

2 3 1

=

+

+ + - → s

s s s ds d

s

2 ) 1 (

) 5 2 ( ) 1 ( lim ! 2

1 3 2 3

2 2

1 =

+

+ + - → s

s s s ds d

s

( ) ) ( 3 2 ) 1 ( 5 2 2 3

2 1 t u t t e

s s s t - + =

+ + - -

L


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 complex exponentials

For circuits the answers will be real

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

)()(lim sFasas

−→

€

1(m −1)!

lims→a

dm−1

dsm−1 (s− a)mF(s)[ ]


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( )


Example 9-12

Find the inverse LT of ) 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


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


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

=−+

=++−−

=+


Introductory s-Domain Cct Analysis First-order RC cct

KVL

instantaneous for each t Substitute element relations

Ordinary differential equation in terms of capacitor voltage

Laplace transform

Solve

Invert LT

+ _

R

VA

i(t) t=0

C vc vS 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

+

−=

−−


An Alternative s-Domain Approach

Transform the cct element relations

Work in s-domain directly OK since L is linear

KVL in s-Domain

+ _

R

VA

i(t) t=0

C vc vS 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()( =+−


Time-varying inputs Suppose vS(t)=Vacos(βt), what happens?

KVL as before

Solve

+ _

R

VA

i(t) t=0

C vc vS vR

+ + +

-

-

-

+ _

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

Documents

Laplace Transforms – recap for cctsnodes.ucsd.edu/sonia/teaching/140_LinearCircuits/data/5laplace.pdf · MAE140 Linear Circuits 109 Laplace Transform - definition Function f(t)