Post on 18-Jul-2018
133MAE140 Linear Circuits 133
Why Laplace transforms?First-order RC cct
KVLinstantaneous 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
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 !"
#$%
&+''(
)**+
,−=
−−
134MAE140 Linear Circuits 134
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)
135MAE140 Linear Circuits 135
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
136MAE140 Linear Circuits 136
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
137MAE140 Linear Circuits 137
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)
0if1)()(0
)(
000>=
+−=−===
∞+−∞−∞
−
−∞
−
−∫∫ σ
ωσ
ωσ
sje
sedtedtetusF
tjststst
îíì
³<
=0for,10for,0
)(tt
tu
138MAE140 Linear Circuits 138
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)
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
∞
∫
139MAE140 Linear Circuits 139
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)
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
140MAE140 Linear Circuits 140
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−
141MAE140 Linear Circuits 141
Laplace Transform Properties
Linearity – absolutely critical propertyFollows from the integral definition
Example
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL
€
L(Acos(βt)) =
142MAE140 Linear Circuits 142
Laplace Transform Properties
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
143MAE140 Linear Circuits 143
Laplace Transform Properties
Integration property
Proof
( )ssFdf
t=
!"#
$%&∫0
)( ττL
dtstet
dft
df −∫∞
∫=∫ $%
&'(
)
*+,
-./
0 0)(
0)( ττττL
144MAE140 Linear Circuits 144
Laplace Transform Properties
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
145MAE140 Linear Circuits 145
Laplace Transform Properties
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
146MAE140 Linear Circuits 146
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
147MAE140 Linear Circuits 147
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 →∞→
=
148MAE140 Linear Circuits 148
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
−−−−−−
=
++++
++++=
−−
−−
149MAE140 Linear Circuits 149
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
150MAE140 Linear Circuits 150
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
MAE140 Linear Circuits
151MAE140 Linear Circuits 151
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( )
152MAE140 Linear Circuits 152
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
153MAE140 Linear Circuits 153
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-=
+
++-® s
ssss
1)1(
)52()1(lim 3
23
1=
úúû
ù
êêë
é
+
++-® s
sssdsd
s
2)1(
)52()1(lim!2
13
23
2
2
1=
úúû
ù
êêë
é
+
++-® s
sssdsd
s
( ) )(32)1(52 23
21 tutte
sss t -+=úúû
ù
êêë
é
+
+ --L
MAE140 Linear Circuits
154MAE140 Linear Circuits 154
T&R, 5th ed, Example 9-12
Find the inverse LT of )52)(1(
)3(20)( 2 +++
+=
sssssF
155MAE140 Linear Circuits 155
T&R, 5th ed, Example 9-12
Find the inverse LT of )52)(1(
)3(20)( 2 +++
+=
sssssF
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
MAE140 Linear Circuits
156MAE140 Linear Circuits 156
Not Strictly Proper Laplace Transforms
Find the inverse LT of34
8126)( 2
23
++
+++=
ssssssF
157MAE140 Linear Circuits 157
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
!"#
$%& +++= −−δδ
MAE140 Linear Circuits
158MAE140 Linear Circuits 158
Multiple Poles
Look for partial fraction decomposition
Equate like powers of s to find coefficients
Solve
)())(()(
)())(()()(
12221212
211
22
22
2
21
1
12
21
1
pskpspskpskKzKs
psk
psk
psk
pspszsKsF
−+−−+−=−
−+
−+
−=
−−
−=
112221122
1
22212121
211
2
)(220
Kzpkppkpk
Kkppkpkkk
=−+
=++−−
=+
159MAE140 Linear Circuits 159
Recall motivating example for LTFirst-order RC cct
KVLinstantaneous for each t
Substitute element relations
Ordinary differential equation in terms of capacitor voltage
Laplace transform
Solve
Invert LT
+_
R
VA
i(t)t=0
CvcvSvR+ +
+-
-
-0)()()( =−− tvtvtv CRS
dttdvCtitRitvtuVtv C
RAS)()(),()(),()( ===
)()()( tuVtvdttdvRC AC
C =+
ACCC Vs
sVvssVRC 1)()]0()([ =+−
RCsv
RCssRCVsV CA
C /1)0(
)/1(/)(
++
+=
Volts)()0(1)( tueveVtv RCt
CRCt
AC !"
#$%
&+''(
)**+
,−=
−−
160MAE140 Linear Circuits 160
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()( =+−
161MAE140 Linear Circuits 161
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)
162MAE140 Linear Circuits 162
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 Transforms2. Look at forced response of ccts due to input ICSs and IVSs
as functions of timeInput 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 2The same as with ODEs