Laplace Transforms – recap for ccts
Transcript of Laplace Transforms – recap for ccts
MAE140 Linear Circuits107
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
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 (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
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 δ(t)
0if1
)()(
0
)(
000
>=+
!=!===
"+!"!"
!
!"
!
!## $
%$
%$
sj
e
s
edtedtetusF
tjststst
≥
<=
0for,10for,0
)(tt
tu
! !" "
"+#
+###>
+=
+#===
0 0 0
)()( if
1)( $%
$$
$$$
ss
edtedteesF
tstsstt
sdtetsFst allfor1)()(
0
== !"
#
#$
MAE140 Linear Circuits111
Laplace Transform Pair Tables
damped cosine
damped sine
cosine
sine
damped ramp
exponential
ramp
step
impulse
TransformWaveformSignal)(t!
22)( !"
"
++
+
s
s
22)( !"
!
++s
22 !
!
+s
22 !+s
s
1
s
1
2
1
s
!+s
1
2)(
1
!+s
)(tu
)(ttu
)(tuet!"
)(tut
te!"
( ) )(sin tut!
( ) )(cos tut!
( ) )(sin tutt
e !"#
( ) )(cos tutt
e !"#
MAE140 Linear Circuits112
Laplace Transform Properties
Linearity – absolutely critical propertyFollows from the integral definition
Example
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL
[ ] ( ) ( )
22
1
2
1
2
222))cos((
!
!!
! !!!!
+=
++
"=
+=#$
%&'
(+= ""
s
As
js
A
js
A
eA
eA
eeA
tAtjtjtjtj
LLLL
MAE140 Linear Circuits113
Laplace Transform Properties
Integration property
Proof
Denote
so
Integrate by parts
( )s
sFdf
t
=!"#
$%&'0
)( ((L
dtstet
dft
df !"#
"=" $%
&'(
)
*+,
-./
0 0
)(
0
)( 0000L
)(and,
0
)(and,
tfdt
dye
dt
dx
tdfy
s
stex
st==
!=
""=
"
##
!!!"
#"
#+
$$%
&
''(
)#=
$$%
&
''(
)
0000
)(1
)()( dtetfs
dfs
edf st
tstt
****L
MAE140 Linear Circuits114
Laplace Transform Properties
Differentiation Property
Proof via integration by parts again
Second derivative
)0()()(
!!="#$
%&'
fssFdt
tdfL
)0()(
0
)(0
)(
0
)()(
!!=
"#
!
!+
#
!$%
&'(
) !="
#
!
!=
*+,
-./
fssF
dtstetfsstedt
tdfdtste
dt
tdf
dt
tdfL
)0()0()(2
)0()()(
2
)(2
!"!!!=
!!#$%
&'(
=#$%
&'(
)*
+,-
.=
/#
/$%
/&
/'(
fsfsFs
dt
df
dt
tdfs
dt
tdf
dt
d
dt
tfdLLL
MAE140 Linear Circuits115
Laplace Transform PropertiesGeneral derivative formula
Translation propertiess-domain translation
t-domain translation
)0()0()0()()( )(21 !!!!"!!!=#$
#%&
#'
#() !! mmm
m
m
ffsfssFdt
tfdL
msL
)()}({ !!+=
" sFtfe tL
{ } 0for)()()( >=!!! asFeatuatf as
L
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 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
pspsps
zszszsK
asasasa
bsbsbsbsF
!!!
!!!=
++++
++++=
!!
!!
L
L
L
L
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-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
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 2πj times the sum ofresidues at the poles inside
)()(lim sFas
as
!"
[ ])()(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-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
MAE140 Linear Circuits122
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
!"
!#$
<
%=
+=
&
'+
'&(
0for0
0for
1)(
t
te
dseas
tf
at
j
j
st)
)
assF
+= 1)(
C1 C2!"<<+= r
jres ,
2
3
2,
#$
#$%
0foras0
0cos)Re(
)Im()Re(>!""=
<+=
treee
rs
tsjtsst
#$
MAE140 Linear Circuits123
Inverting Laplace Transforms
Compute residues at the poles
Example
)()(lim sFas
as
!"
!"#
$%& '
'
'
(')()(
1
1lim
)!1(
1sF
mas
mds
md
asm
( ) ( ) ( ) ( )31
3
21
1
1
2
31
3)1(2)1(2
31
522
+!
++
+=
+
!+++=
+
+
ssss
ss
s
ss
3)1(
)52()1(lim 3
23
1−=
+
++
−→ ssss
s1
)1()52()1(lim 3
23
1=
+
++
−→ ssss
dsd
s
2)1(
)52()1(lim!2
13
23
2
2
1=
+
++
−→ ssss
dsd
s
( ) )(32)1(52 23
21 tutte
sss t −+=
+
+ −−L
MAE140 Linear Circuits124
Inverting Laplace Transforms
Compute residues at the poles
Bundle complex conjugate pole pairs into second-order terms if you want
but you will need to be careful
Inverse Laplace Transform is a sum of complexexponentials
For circuits the answers will be real
( )[ ]222 2))(( !""!"!" ++#=+### ssjsjs
)()(lim sFas
as
!"
!
1
(m "1)!lims#a
dm"1
dsm"1
(s" a)mF(s)[ ]
MAE140 Linear Circuits125
Inverting Laplace Transforms in Practice
We have a table of inverse LTsWrite F(s) as a partial fraction expansion
Now appeal to linearity to invert via the tableSurprise!Nastiness: computing the partial fraction expansion is best
done by calculating the residues
!
F(s) =bms
m + bm"1sm"1 +L+ b1s+ b0
ansn + an"1s
n"1 +L+ a1s+ a0
= K(s" z1)(s" z2)L(s" zm)
(s" p1)(s" p2)L(s" pn )
=#1
s" p1( )+
#2
s" p2( )+
#31
(s" p3)+
#32
s" p3( )2
+#33
s" p3( )3
+ ...+#q
s" pq( )
MAE140 Linear Circuits126
Example 9-12
Find the inverse LT of)52)(1(
)3(20)( 2 +++
+=
sssssF
21211)(
*221
js
k
js
k
s
ksF
+++
!++
+=
!4
5
2555
21)21)(1(
)3(20)()21(
21lim2
10
1522
)3(20)()1(
1lim1
jej
jsjss
ssFjs
jsk
sss
ssFs
sk
=""=
+"=+++
+="+
+"#=
=
"=++
+=+
"#=
)()4
52cos(21010
)(252510)( 4
5)21(
4
5)21(
tutee
tueeetf
tt
jtjjtjt
!"
#$%
&++=
!!
"
#
$$
%
&
++=
''
'''++''
(
((
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
!"
#$%
&+++=
''((
MAE140 Linear Circuits128
Multiple Poles
Look for partial fraction decomposition
Equate like powers of s to find coefficients
Solve
)())(()(
)())((
)()(
12221212
211
22
22
2
21
1
12
21
1
pskpspskpskKzKs
ps
k
ps
k
ps
k
psps
zsKsF
!+!!+!=!
!+
!+
!=
!!
!=
112221122
1
22212121
211
2
)(22
0
Kzpkppkpk
Kkppkpk
kk
=!+
=++!!
=+
MAE140 Linear Circuits129
Introductory s-Domain Cct Analysis
First-order RC cctKVL
instantaneous for each tSubstitute element relations
Ordinary differential equation in terms of capacitor voltage
Laplace transform
Solve
Invert LT
+_
R
VA
i(t)t=0
CvcvS
vR+ ++
-
-
-0)()()( =!! tvtvtv CRS
dt
tdvCtitRitvtuVtv
CRAS
)()(),()(),()( ===
)()()(
tuVtvdt
tdvRC AC
C =+
ACCC Vs
sVvssVRC1
)()]0()([ =+!
RCs
v
RCss
RCVsV
CAC
/1
)0(
)/1(
/)(
++
+=
Volts)()0(1)( tueveVtv RCt
CRCt
AC !"
#$%
&+''(
)**+
,-=
--
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
CvcvS
vR+ ++
-
-
-
+_
R
VA
I(s)
Vc(s)
+
-
sC
1
s
1
+_s
Cv )0(
)0()()(
)0()(
1)(
CCC
CCC
CvssCVsI
s
vsI
CssV
!=
+= Impedance + source
Admittance + source
ACCC Vs
sVCRvssCRV1
)()0()( =+!
MAE140 Linear Circuits131
Time-varying inputsSuppose vS(t)=Vacos(βt), what happens?
KVL as before
Solve
+_
R
VA
i(t)t=0
CvcvS
vR+ ++
-
-
-
+_
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 !
"
#$%
& '+
'
+'+
+
=(
)((