The Laplace Transform
Note: Some topics have been repeated in the slides. Just for better
understanding. Not for classroom teaching.
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The Laplace Transform
Laplace transform converts time domain problems into
functions of a complex variable, s, that is related to the
frequency response of the system
Laplace
Transform
Time domain
lesson10et438a.pptx Complex frequency 2
lesson10et438a.pptx 3
Basic Laplace Transform Pairs
5
5 u s (t) →
s
4
3 sin(4t ) → 3 2 =4
s + 16
10 1
10 t
s2 e −2 t s+2
− at 1 s
t e 3 cos( t ) 3 2
(s + a ) 2 (s + 1)
1
e5 t s−5
Laplace Theorems
Laplace of an unknown function
Capitalize unknown function name
L (f1 ( t ) ) = F1 (s) Replace t with s
Examples
Laplace
Operator L (i1 ( t ) ) = I1 (s)
Symbol
L (v1 ( t ) ) = V1 (s)
Linearity of transform - can multiply by constant
If L (f1 ( t ) ) = F1 (s) and L (f 2 ( t ) ) = F2 (s)
Then L (a f1 ( t ) + b f 2 ( t ) ) = a F1 (s) + b F2 (s)
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Laplace Transforms of Calculus Operators
Laplace Transform turns derivative into multiplication by s
If L (f1 ( t ) ) = F1 (s) Subtract any
non-zero
initial
d
Then L f1 ( t ) = s F1 (s) − f1 (0) conditions
dt
For higher order derivatives 0 initial conditions reduces
formula to
d2 d d2 2
L 2 f1 ( t ) = s (s F1 (s) − f1 (0)) − f1 (0) L 2 f1 ( t ) = s F1 (s)
dt dt dt
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Laplace Transforms of Calculus Operators
Laplace turns integration into division by s
If L (f1 ( t ) ) = F1 (s)
d 1
Then L f1 ( t ) dt = F1 (s)
dt s
Examples from circuit analysis:
1
Capacitor voltage L (v C ( t ) ) = L i C ( t ) dt
C
1
C
v C ( t ) = i C ( t ) dt 1 1
VC (s) = I C (s) =
1
I C (s)
C s Cs
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More Examples From Circuit Analysis
Find the Laplace relationship for inductor voltage
d
vL (t) = L i L (t)
dt
d
L (v L ( t ) ) = L L i L ( t )
dt
VL (s) = L s I L (s)
Laplace relationship for resistor voltage
v R (t) = R i R (t) L (v R ( t ) ) = L (R i R ( t ) )
VR (s) = R I R (s)
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Laplace Transforms and Impedance
Remember phasor analysis is only valid for sinusoidal steady-state. Turns
ac analysis into an analysis similar to the dc. (Ohm's law)
Resistance R
Inductive Reactance XL = j L = 2 f j = 90
1 1 1
Capacitive Reactance XC = = − j - j = = −90
j C C j
Since Laplace variable represents frequency, it's possible to replace j with s
and s with j. If s is replaced with j, analysis reverts to phasors We can find
the frequency response of a dynamic system by converting differential
equation into Laplace domain and replacing s with j. Sweeping frequency
produces Bode plot of system.
lesson10et438a.pptx 9
Laplace Transforms and Impedance
Laplace “Impedances” (Ohm’s Law) Impedance (Phasors)
Ls =
VL (s) VL ( j)
Inductors Inductors jL =
I L (s) I L ( j)
1 VC (s) 1 VC ( j)
Capacitors = Capacitors =
Cs I C (s) jC I C ( j)
V (s) VR ( j)
Resistors R= R Resistors R=
I R (s) I R ( j)
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Q3. In the circuit shown in Figure, switch 1 is closed at t = 0 and then, at t = t’=
4ms, switch 2 is opened. Find the current i(t) in the intervals 0 < t < t’ and t > t’
using Laplace Transform.
1 50
100 2
100V
i (t )
0.1H
Q3. In the circuit shown in Figure, switch is closed at t =0. Find the current i(t)
for t > 0 using Laplace Transform.
Representation of R, L, and C in the S-domain
Using differentiation and integration
Laplace property
Laplace Circuit Analysis
Time domain to complex frequency domain
v 1 (0)
R! _ L2 R2
+ _
+ C1 v 2 (0) i2 (t) _
C2
VA(t) + VB (t)
_ L1
i1 (t) +
v 1 (0)
_ L2i2(0)
R! + s sL 2 +
_
R2
1 1
sC 1 sC 2 _
v 2 (0)
+ +
VA(s) + s
VB(s)
_ _
sL 1
+
L1i1(0)
_
Laplace Circuit Analysis
Circuit Application:
Given the circuit below. Assume zero IC’s. Use Laplace to find vc(t).
The time domain circuit:
t=0 100
+
+
2u(t) V 0.001 F
v c (t)
_
_ Laplace circuit
t=0
2 1000 100
Vc ( s ) = s s
+
1000
100 + +
s 2 1000
I(s) s Vc(s)
s _
20
Vc ( s ) = _
s( s + 10)
Laplace Circuit Analysis
Circuit Application:
t=0
100
+
+
2 1000
I(s) s Vc(s)
s _
_
20 2 2
Vc ( s ) = = −
s( s + 10) s s + 10
vc ( t ) = 2 − 2e −10t u( t )
Laplace Circuit Analysis
Circuit Application:
Given the circuit below. Assume vc(0) = - 4 V. Use Laplace to find vc(t).
The time domain circuit:
t=0 100
+
+
2u(t) V 0.001 F
v c (t)
_
_
Laplace circuit:
t=0
100
2 4 1000
+ = I ( s )100 +
+
s
1000
s s 2
+ s
I(s) _ Vc(s)
s _ 4
6 s
100 I ( s ) = + _
s + 10
Laplace Circuit Analysis
Circuit Application:
t=0
100 1
2
1000
+ − 100 I ( s ) − Vc ( s ) = 0
+ s s
2
I(s) _ Vc(s)
s _ 4
s + 2 6
_
− = Vc ( s )
s s + 10
2 − 4 s + 20 A B
Vc ( s ) = = +
s( s + 10) s s + 10 Check the boundary conditions
3 vc(0) = - 4 V
2 6
Vc ( s ) = −
s s + 10
vc(oo) = 2 V
v ( t ) = 2 − 6e −10t u( t )
Laplace Circuit Analysis
Circuit Application: Find i0(t) using Laplace
1 2 2H
+ i0 (t)
Time Domain 4u(t) 1 e-tu(t)
_ 1F
1 2 2s
+
Laplace 4 1
s I 1 (s) 1 I 2 (s) 1 I 3 (s)
_ s s+1
Laplace Circuit Analysis
Circuit Application: Find i0(t) using Laplace
1 2 2s
+
4 1
s I 1 (s) 1 I 2 (s) 1 I 3 (s)
_ s s+1
Mesh 1
( s + 1) I 2 ( s) 4
I1 ( s ) − =
s s s
( s + 1) I1 ( s ) − I 2 ( s ) = 4
Laplace Circuit Analysis
Circuit Application: Find i0(t) using Laplace
1 2
+
4 1
s I 1 (s) 1 I 2 (s) 1 I 3 (s)
_ s s+1
Mesh 2
−1 3s + 1
I1 ( s ) + I 2 ( s) − I 3 ( s) = 0
s s
−1 3s + 1 1
I1 ( s ) + I 2 ( s) − =0
s s s +1
s
− I1 ( s ) + ( 3 s + 1) I 2 ( s ) =
s +1
− ( s + 1) I1 ( s ) + ( s + 1)(3 s + 1) I 2 ( s ) = s
Laplace Circuit Analysis
Circuit Application: Find i0(t) using Laplace
+
4 1
s I 1 (s) 1 I 2 (s) 1 I 3 (s)
_ s s+1
( s + 1) I1 ( s ) − I 2 ( s ) = 4 Add these 2
equations
− ( s + 1) I1 ( s ) + ( s + 1)( 3 s + 1) I 2 ( s ) = s
s ( 3 s + 4) I 2 ( s ) = s + 4
Laplace Circuit Analysis
Circuit Application: Find i0(t) using Laplace
+
4 1
s I 1 (s) 1 I 2 (s) 1 I 3 (s)
_ s s+1
s( 3 s + 4) I 2 ( s ) = s + 4
(1 )( s + 4)
I ( s) = 3
1
= − 3
2
2
s( s + 4 ) s s+ 4
3 3
2 − 43t
i2 ( t ) = [ 1 − e ] u( t )
3
