133MAE40 Linear Circuits 133
Why Laplace transforms?
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
CvcvS
vR+ + +
-
-
-
0)()()( =−− tvtvtv CRS
dt
tdvCtitRitvtuVtv CRAS
)()(),()(),()( ===
)()()( tuVtv
dt
tdvRC ACC =+
ACCC Vs
sVvssVRC 1)()]0()([ =+−
RCs
v
RCss
RCVsV CAC /1
)0(
)/1(
/)(
+
+
+
=
Volts)()0(1)( tueveVtv RC
t
CRC
t
AC !
"
#
$
%
&
+''
(
)
**
+
,
−=
−−
134MAE40 Linear Circuits 134
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
Ti
m
e
do
m
ai
n
(t
do
m
ai
n) Complex frequency domain(s domain)
135MAE40 Linear Circuits 135
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 (s+jw)
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
136MAE40 Linear Circuits 136
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 d(t)
î
í
ì
³
<
=
0for,1
0for,0
)(
t
t
tu
137MAE40 Linear Circuits 137
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 d(t)
0if1)()(
0
)(
000
>=
+
−=−===
∞+−∞−∞
−
−
∞
−
−
∫∫ σωσ
ωσ
sj
e
s
edtedtetusF
tjst
stst
î
í
ì
³
<
=
0for,1
0for,0
)(
t
t
tu
138MAE40 Linear Circuits 138
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 d(t)
0if1)()(
0
)(
000
>=
+
−=−===
∞+−∞−∞
−
−
∞
−
−
∫∫ σωσ
ωσ
sj
e
s
edtedtetusF
tjst
stst
î
í
ì
³
<
=
0for,1
0for,0
)(
t
t
tu
€
F(s) = e−αte−stdt = e−(s+α )tdt = −e
−(s+α )t
s+α 0
∞
=
1
s+α
if σ > −α
0
∞
∫
0
∞
∫
139MAE40 Linear Circuits 139
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 d(t)
0if1)()(
0
)(
000
>=
+
−=−===
∞+−∞−∞
−
−
∞
−
−
∫∫ σωσ
ωσ
sj
e
s
edtedtetusF
tjst
stst
î
í
ì
³
<
=
0for,1
0for,0
)(
t
t
tu
€
F(s) = e−αte−stdt = e−(s+α )tdt = −e
−(s+α )t
s+α 0
∞
=
1
s+α
if σ > −α
0
∞
∫
0
∞
∫
sdtetsF st allfor1)()(
0
== ∫
∞
−
−δ
140MAE40 Linear Circuits 140
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
s
1
2
1
s
α+s
1
2)(
1
α+s
)(tu
)(ttu
)(tue tα−
)(tutte α−
( ) )(sin tutβ
( ) )(cos tutβ
( ) )(sin tutte βα−
( ) )(cos tutte βα−
141MAE40 Linear Circuits 141
Laplace Transform Properties
Linearity – absolutely critical property
Follows from the integral definition
Example
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL
€
L(Acos(βt)) =
142MAE40 Linear Circuits 142
Laplace Transform Properties
Linearity – absolutely critical property
Follows from the integral definition
Example
{ } { } { } )()()()()()( 2121 sBFsAFtBtAtBftAf +=+=+ 21 fLfLL
€
L(Acos(βt)) = L A
2
e jβt + e − jβt( )
$
%
&
'
(
) =
A
2
L e jβt( ) +
A
2
L e − jβt( )
=
A
2
1
s− jβ
+
A
2
1
s+ jβ
=
As
s 2 +β 2
143MAE40 Linear Circuits 143
Laplace Transform Properties
Integration property
Proof
( )
s
sFdf
t
=
!
"
#
$
%
&
∫
0
)( ττL
dtste
t
df
t
df −∫
∞
∫=∫ $
%
&
'
(
)
*
+
,
-
.
/
0 0
)(
0
)( ττττL
144MAE40 Linear Circuits 144
Laplace Transform Properties
Integration property
Proof
Denote
so
Integrate by parts
( )
s
sFdf
t
=
!
"
#
$
%
&
∫
0
)( ττL
dtste
t
df
t
df −∫
∞
∫=∫ $
%
&
'
(
)
*
+
,
-
.
/
0 0
)(
0
)( ττττL
)(and,
0
)(and,
tf
dt
dy
e
dt
dx
t
dfy
s
ste
x
st ==
∫=
−−
=
−
ττ
∫∫∫
∞
−
∞
−
+
$
$
%
&
'
'
(
)−
=
$
$
%
&
'
'
(
)
0000
)(
1
)()( dtetf
s
df
s
edf st
tstt
ττττL
145MAE40 Linear Circuits 145
Laplace Transform Properties
Differentiation Property
Proof via integration by parts again
Second derivative
)0()()( −−=
"
#
$
%
&
' fssF
dt
tdf
L
€
L
df (t)
dt
"
#
$
%
&
'
=
df (t)
dt
e−stdt
0−
∞
∫ = f (t)e−st[ ]0−
∞
+ s f (t)e−stdt
0−
∞
∫
= sF(s)− f (0−)
)0()0()(2
)0()()(2
)(2
−"−−−=
−−
#
$
%
&
'
(=
#
$
%
&
'
(
)*
+
,-
.=
/#
/
$
%
/&
/
'
(
fsfsFs
dt
df
dt
tdfs
dt
tdf
dt
d
dt
tfd
LLL
146MAE40 Linear Circuits 146
Laplace Transform Properties
General derivative formula
Translation properties
s-domain translation
t-domain translation
€
L
dm 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
147MAE40 Linear Circuits 147
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)(lim
0
ssFtf
st ∞→+→
=
)(lim)(lim
0
ssFtf
st →∞→
=
148MAE40 Linear Circuits 148
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 n³m
A strictly proper rational function has n>m
An improper rational function has n
149MAE40 Linear Circuits 149
A Little Complex Analysis
We are dealing with linear ccts
Our Laplace Transforms will consist of rational functions
(ratios of polynomials in s) and exponentials like e-st
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-st is an entire function and has no poles anywhere
To understand linear cct respons
