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
MAE140 Linear Circuits 108
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)
MAE140 Linear Circuits 109
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
MAE140 Linear Circuits 110
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
MAE140 Linear Circuits 111
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)
0if1)()(0
)(
000>=
+−=−===
∞+−∞−∞
−
−∞
−
−∫∫ σ
ωσ
ωσ
sje
sedtedtetusF
tjststst
≥ <
= 0 for , 1 0 for , 0
) ( t t
t u
MAE140 Linear Circuits 112
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)
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
MAE140 Linear Circuits 113
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)
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
MAE140 Linear Circuits 114
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−
MAE140 Linear Circuits 115
Laplace Transform Properties
Linearity – absolutely critical property Follows from the integral definition
Example
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL
€
L(Acos(βt)) =
MAE140 Linear Circuits 116
Laplace Transform Properties
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
MAE140 Linear Circuits 117
Laplace Transform Properties
Integration property
Proof
( )ssFdf
t=
∫0
)( ττL
dtstet
dft
df −∫∞
∫=∫
0 0)(
0)( ττττL
MAE140 Linear Circuits 118
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,
tfdtdye
dtdx
tdfy
s
stex
st ==
∫=−−
=
−
ττ
∫∫∫∞
−∞
−+
−=
0000)(
1)()( dtetf
sdf
sedf st
tsttττττL
MAE140 Linear Circuits 119
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
MAE140 Linear Circuits 120
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
MAE140 Linear Circuits 121
Laplace Transform Properties
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 →∞→
=
MAE140 Linear Circuits 122
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
−−−−−−
=
++++
++++=
−−
−−
MAE140 Linear Circuits 123
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
MAE140 Linear Circuits 124
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
MAE140 Linear Circuits 125
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−
− −
−
→
MAE140 Linear Circuits 126
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
MAE140 Linear Circuits 127
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
MAE140 Linear Circuits 128
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
MAE140 Linear Circuits 129
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)[ ]
MAE140 Linear Circuits 130
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( )
MAE140 Linear Circuits 131
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
MAE140 Linear Circuits 132
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
MAE140 Linear Circuits 133
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
=−+
=++−−
=+
MAE140 Linear Circuits 134
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
+
−=
−−
MAE140 Linear Circuits 135
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()( =+−
MAE140 Linear Circuits 136
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