Lecture #14:
THE LAPLACE TRANSFORM - METHOD OF SOLUTION
Motivation:
- Continue to describe methods for representing signals as superpositions of complex exponential functions
- Develop efficient methods for analyzing LTI systems
Outline:
- Laplace transform of the family of singularity functions
- More on the region of convergence
- Analysis of networks with the Laplace transform — the impedance method
- Finding inverse transforms — partial fraction expansion
- Historical perspective — Oliver Heaviside
Review
- The Laplace transform represents a time function as a superposition of complex exponentials.
- A time function is related uniquely to a Laplace transform if the ROC is specified.
- If the Laplace transform of a sum of causal and anti-causal exponential time functions exists, its ROC is a strip in the s-plane parallel to the jω-axis.
I. LAPLACE TRANSFORMS OF SINGULARITY FUNCTIONS
1/ Unit impulse function
L
{
δ
(
t
)
}
=
∫
−
∞
∞
δ
(
t
)
e
−
st
dt
L
{
δ
(
t
)
}
=
∫
−
∞
∞
δ
(
t
)
e
−
st
dt
size 12{L lbrace δ \( t \) rbrace = Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {δ \( t \) e rSup { size 8{ - ital "st"} } ital "dt"} } {}
Recall the definition of the unit impulse
∫
−
∞
∞
δ
(
t
)
f
(
t
)
dt
=
f
(
0
)
∫
−
∞
∞
δ
(
t
)
f
(
t
)
dt
=
f
(
0
)
size 12{ Int cSub { size 8{ - infinity } } cSup { size 8{ infinity } } {δ \( t \) f \( t \) ital "dt"} =f \( 0 \) } {}
Hence,
L
{
δ
(
t
)
}
=
1
L
{
δ
(
t
)
}
=
1
size 12{L lbrace δ \( t \) rbrace =1} {}
for all values of s. The region of convergence is the entire s plane.
2/ Unit impulse function delayed — use of properties
The Laplace transform of an impulse located at t = 0 is
L
{
δ
(
t
)
}
=
1
L
{
δ
(
t
)
}
=
1
size 12{L lbrace δ \( t \) rbrace =1} {}
Using the delay property,
x
(
t
)
⇔
L
X
(
s
)
x
(
t
−
T
)
⇔
L
X
(
s
)
e
−
sT
x
(
t
)
⇔
L
X
(
s
)
x
(
t
−
T
)
⇔
L
X
(
s
)
e
−
sT
alignl { stack {
size 12{x \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} X \( s \) } {} #
x \( t - T \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} X \( s \) e rSup { size 8{ - ital "sT"} } {}
} } {}
the Laplace transform of the delayed impulse is
L
{
δ
(
t
−
T
)
}
=
e
−
sT
L
{
δ
(
t
−
T
)
}
=
e
−
sT
size 12{L lbrace δ \( t - T \) rbrace =e rSup { size 8{ - ital "sT"} } } {}
and the region of convergence is the whole s plane.
Two-minute miniquiz problem
Problem 5-1
Find the Laplace transform including the ROC for
x
(
t
)
=
e
−
2
(
t
−
4
)
u
(
t
−
4
)
x
(
t
)
=
e
−
2
(
t
−
4
)
u
(
t
−
4
)
size 12{x \( t \) =e rSup { size 8{ - 2 \( t - 4 \) } } u \( t - 4 \) } {}
Solution
We use the Laplace transform of the causal exponential time function and time delay property to solve this problem.
e
−
2t
u
(
t
)
⇔
L
1
s
+
2
for
σ
>
−
2
e
−
2
(
t
−
4
)
u
(
t
−
4
)
⇔
L
1
s
+
2
e
−
4s
for
σ
>
−
2
e
−
2t
u
(
t
)
⇔
L
1
s
+
2
for
σ
>
−
2
e
−
2
(
t
−
4
)
u
(
t
−
4
)
⇔
L
1
s
+
2
e
−
4s
for
σ
>
−
2
alignl { stack {
size 12{e rSup { size 8{ - 2t} } u \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} { {1} over {s+2} } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ> - 2} {} #
e rSup { size 8{ - 2 \( t - 4 \) } } u \( t - 4 \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} { {1} over {s+2} } e rSup { size 8{ - 4s} } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ> - 2 {}
} } {}
3/ Singularity functions and their relatives
The Laplace transform of a unit impulse is
δ
(
t
)
⇔
L
1
for
all
s
δ
(
t
)
⇔
L
1
for
all
s
size 12{δ \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} 1 matrix {
{} # {}
} ital "for" matrix {
{} # {}
} ital "all" matrix {
{} # {}
} s} {}
and from the Laplace transform of a causal exponential with α = 0 we have the Laplace transform of a causal step function
u
(
t
)
⇔
L
1
s
for
σ
>
0
u
(
t
)
⇔
L
1
s
for
σ
>
0
size 12{u \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} { {1} over {s} } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ>0} {}
Note this fits together with the time differentiation property
dx
(
t
)
dt
⇔
L
sX
(
s
)
dx
(
t
)
dt
⇔
L
sX
(
s
)
size 12{ { { ital "dx" \( t \) } over { ital "dt"} } { matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } } cSup {} matrix {
{} # {}
} ital "sX" \( s \) } {}
since in a generalized function sense
δ
(
t
)
=
du
(
t
)
dt
⇔
L
L
{
δ
(
t
)
}
=
s
1
s
=
1
δ
(
t
)
=
du
(
t
)
dt
⇔
L
L
{
δ
(
t
)
}
=
s
1
s
=
1
size 12{δ \( t \) = { { ital "du" \( t \) } over { ital "dt"} } matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} L lbrace δ \( t \) rbrace =s left [ { {1} over {s} } right ]=1} {}
We use the multiplication by t property
tx
(
t
)
⇔
L
−
dX
(
s
)
ds
tx
(
t
)
⇔
L
−
dX
(
s
)
ds
size 12{ ital "tx" \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} - { { ital "dX" \( s \) } over { ital "ds"} } } {}
to obtain
tu
(
t
)
⇔
L
−
d
ds
1
s
=
1
s
2
for
σ
>
0
tu
(
t
)
⇔
L
−
d
ds
1
s
=
1
s
2
for
σ
>
0
size 12{ ital "tu" \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} - { {d} over { ital "ds"} } left [ { {1} over {s} } right ]= { {1} over {s rSup { size 8{2} } } } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ>0} {}
and use it again to obtain
t
2
u
(
t
)
⇔
L
−
d
ds
1
s
2
=
1
s
3
for
σ
>
0
t
2
u
(
t
)
⇔
L
−
d
ds
1
s
2
=
1
s
3
for
σ
>
0
size 12{t rSup { size 8{2} } u \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} - { {d} over { ital "ds"} } left [ { {1} over {s rSup { size 8{2} } } } right ]= { {1} over {s rSup { size 8{3} } } } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ>0} {}
which implies that by induction
t
n
u
(
t
)
⇔
L
n
!
s
n
+
1
for
σ
>
0
t
n
u
(
t
)
⇔
L
n
!
s
n
+
1
for
σ
>
0
size 12{t rSup { size 8{n} } u \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} { {n!} over {s rSup { size 8{n+1} } } } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ>0} {}
or
t
n
−
1
(
n
−
1
)
!
u
(
t
)
⇔
L
1
s
n
for
σ
>
0
t
n
−
1
(
n
−
1
)
!
u
(
t
)
⇔
L
1
s
n
for
σ
>
0
size 12{ { {t rSup { size 8{n - 1} } } over { \( n - 1 \) !} } u \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} { {1} over {s rSup { size 8{n} } } } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ>0} {}
4/ Summary of singularity functions and their relatives
5/ Wild and crazy singularity functions
Since taking the derivative of a time function corresponds to multiplying the Laplace transform by s we can contemplate the derivative of the unit impulse called the unit doublet.
dδ
(
t
)
dt
=
δ
.
(
t
)
⇔
L
s
dδ
(
t
)
dt
=
δ
.
(
t
)
⇔
L
s
size 12{ { {dδ \( t \) } over { ital "dt"} } = {δ} cSup { size 8{ "." } } \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} s} {}
This process can be continued by taking successive derivatives of the impulse to form the unit triplet which has Laplace transform
s2s2 size 12{s rSup { size 8{2} } } {}, unit quadruplet, etc. In general, the nth derivative of the unit impulse has a Laplace transform
snsn size 12{s rSup { size 8{n} } } {}. We shall consider the usefulness of these higher order singularity functions later!
6/ General comments on the ROC
- Unit impulse
⇒⇒ size 12{ drarrow } {} ROC is the whole s plane.
- Finite duration, absolutely integrable time function
⇒⇒ size 12{ drarrow } {} ROC is the whole s plane.
- Time shifting a time function does not change its ROC.
- Right-sided time function
⇒⇒ size 12{ drarrow } {} ROC is to the right of the rightmost pole.
- Left-sided time function
⇒⇒ size 12{ drarrow } {} ROC is to left of the left-most pole.
- A sum of causal and anti-causal exponential time functions that has a Laplace transform
⇒⇒ size 12{ drarrow } {} ROC is a strip in the s plane.
- There are no poles in the ROC.
- Some time functions do not have Laplace transforms, e.g.,
x(t)=e−tx(t)=e−t size 12{x \( t \) =e rSup { size 8{ - t} } } {} for all t.
II. ANALYSIS OF NETWORKS WITH THE LAPLACE TRANSFORM — THE IMPEDANCE METHOD KIRCHHOFF’S LAWS
Kirchhoff’s current and voltage laws are algebraic equations that link the branch variables in a network,
∑
node
i
k
(
t
)
=
0
and
∑
loop
v
k
(
t
)
=
0
∑
node
i
k
(
t
)
=
0
and
∑
loop
v
k
(
t
)
=
0
size 12{ Sum cSub { size 8{ ital "node"} } {i rSub { size 8{k} } \( t \) } =0 matrix {
{} # {}
} ital "and" matrix {
{} # {}
} Sum cSub { size 8{ ital "loop"} } {v rSub { size 8{k} } \( t \) } =0} {}
If we take the Laplace transform of these equations then we obtain
∑
node
I
k
(
s
)
=
0
and
∑
loop
V
k
(
s
)
=
0
∑
node
I
k
(
s
)
=
0
and
∑
loop
V
k
(
s
)
=
0
size 12{ Sum cSub { size 8{ ital "node"} } {I rSub { size 8{k} } \( s \) } =0 matrix {
{} # {}
} ital "and" matrix {
{} # {}
} Sum cSub { size 8{ ital "loop"} } {V rSub { size 8{k} } \( s \) } =0} {}
Hence, the Laplace transforms of the branch variables satisfy KCL and KVL.
1/ Constitutive relations
a/ Resistance, capacitance, and inductance
b/ Impedance and admittance
The ratios of voltage to current and current to voltage are system functions with special names. The impedance is defined as
Z
(
s
)
=
V
(
s
)
I
(
s
)
Z
(
s
)
=
V
(
s
)
I
(
s
)
size 12{Z \( s \) = { {V \( s \) } over {I \( s \) } } } {}
and the admittance is defined as
Y
(
s
)
=
I
(
s
)
V
(
s
)
Y
(
s
)
=
I
(
s
)
V
(
s
)
size 12{Y \( s \) = { {I \( s \) } over {V \( s \) } } } {}
The impedance and admittance of the resistance, capacitance, and inductance are
2/ Significance
- The equilibrium equations of a network involve KCL, KVL, and the constitutive relations.
- KCL and KVL are algebraic equations for both time functions and their Laplace transforms.
- In terms of time functions, the constitutive relations involve derivatives.
- In terms of Laplace transforms, the constitutive relations are algebraic.
3/ Conclusions
- Analysis of a network in the time domain leads to differential equations.
- Analysis of a network in the Laplace transform domain leads to algebraic equations.
Thus, R, L, and C networks can be analyzed using methods developed for resistive networks. These include: use of series and parallel combinations, voltage and current dividers, as well as Thévenin’s and Norton’s equivalents. Methods work just as well for any system (e.g., mechanical, acoustic, chemical, etc.) that is analogous to an electric circuit.
4/ Example — impulse response of an RLC network
We wish to find the output voltage
vo(t)vo(t) size 12{v rSub { size 8{o} } \( t \) } {} for the RLC network to an impulse of input voltage,
v
i
(
t
)
=
δ
(
t
)
v
i
(
t
)
=
δ
(
t
)
size 12{v rSub { size 8{i} } \( t \) =δ \( t \) } {}
The numbers show the values of the capacitance, resistance, and inductance in farads, ohms, and henries, respectively
The response of an LTI system to an impulse is important and is called the impulse response and is usually designated by h(t).
a/ Solution to network by impedance method
The first step is to redraw the network in terms of Laplace transforms of variables and the impedances of the elements.
The impedance is shown next to each network element.
Vi(s) is divided between the voltage on the capacitance and that on the parallel resistance and inductance combination. The impedance of the parallel resistance and inductance is
Z
RL
=
1
2
s
+
3
=
s
3s
+
2
Z
RL
=
1
2
s
+
3
=
s
3s
+
2
size 12{Z rSub { size 8{ ital "RL"} } = { {1} over { { {2} over {s} } +3} } = { {s} over {3s+2} } } {}
and therefore the output voltage is
V
o
(
s
)
=
s
3s
+
2
1
s
+
s
3s
+
2
V
i
(
s
)
=
s
2
s
2
+
3s
+
2
V
i
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
V
i
(
s
)
V
o
(
s
)
=
s
3s
+
2
1
s
+
s
3s
+
2
V
i
(
s
)
=
s
2
s
2
+
3s
+
2
V
i
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
V
i
(
s
)
size 12{V rSub { size 8{o} } \( s \) = { { { {s} over {3s+2} } } over { { {1} over {s} } + { {s} over {3s+2} } } } V rSub { size 8{i} } \( s \) = { {s rSup { size 8{2} } } over {s rSup { size 8{2} } +3s+2} } V rSub { size 8{i} } \( s \) = { {s rSup { size 8{2} } } over { \( s+1 \) \( s+2 \) } } V rSub { size 8{i} } \( s \) } {}
b/ Laplace transform of output voltage
The next step is to find the Laplace transform of the input voltage. Since
v
i
(
t
)
=
δ
(
t
)
V
i
(
s
)
=
1
for
all
s
v
i
(
t
)
=
δ
(
t
)
V
i
(
s
)
=
1
for
all
s
alignl { stack {
size 12{v rSub { size 8{i} } \( t \) =δ \( t \) } {} #
V rSub { size 8{i} } \( s \) =1 matrix {
{} # {}
} ital "for" matrix {
{} # {}
} ital "all" matrix {
{} # {}
} s {}
} } {}
Therefore, since
V
o
(
s
)
=
H
(
s
)
V
i
(
s
)
V
o
(
s
)
=
H
(
s
)
V
i
(
s
)
size 12{V rSub { size 8{o} } \( s \) =H \( s \) V rSub { size 8{i} } \( s \) } {}
where
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
size 12{V rSub { size 8{o} } \( s \) =H \( s \) = { {s rSup { size 8{2} } } over { \( s+1 \) \( s+2 \) } } } {}
This shows that the Laplace transform of the impulse response of a system equals the system function,
h
(
t
)
⇔
L
H
(
s
)
h
(
t
)
⇔
L
H
(
s
)
size 12{h \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} H \( s \) } {}
Since H(s) characterizes the system, so does h(t).
c/ Region of convergence of system function
What is the ROC of this system function? Because the network is a passive RLC network, the system is causal, i.e., the impulse response cannot precede the occurrence of the impulse. Thus, the ROC is to the right of the rightmost pole, i.e., σ > −1. So we have the following pole-zero diagram and ROC for
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
for
σ
>
1
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
for
σ
>
1
size 12{V rSub { size 8{o} } \( s \) =H \( s \) = { {s rSup { size 8{2} } } over { \( s+1 \) \( s+2 \) } } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ>1} {}
d/ Partial fraction expansion of the Laplace transform of the output voltage
The Laplace transform of the output voltage is
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
for
σ
>
−
1
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
for
σ
>
−
1
size 12{V rSub { size 8{o} } \( s \) =H \( s \) = { {s rSup { size 8{2} } } over { \( s+1 \) \( s+2 \) } } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ> - 1} {}
Note that H(s) is an improper rational function. A rational function is a ratio of polynomials. A proper rational function has a denominator polynomial whose order exceeds that of the numerator. The first step in finding the voltage as a function of time is to expand H(s) into a polynomial and a proper rational function.
H
(
s
)
=
P
(
s
)
+
H
P
(
s
)
H
(
s
)
=
P
(
s
)
+
H
P
(
s
)
size 12{H \( s \) =" P" \( s \) +" H" rSub { size 8{P} } \( s \) } {}
where P(s) is a polynomial and
Hp(s)Hp(s) size 12{H rSub { size 8{p} } \( s \) } {} is a proper rational function.
e/ Synthetic division
We can synthetically divide the denominator into the numerator of
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
=
s
2
s
2
+
3s
+
2
V
o
(
s
)
=
H
(
s
)
=
s
2
(
s
+
1
)
(
s
+
2
)
=
s
2
s
2
+
3s
+
2
size 12{V rSub { size 8{o} } \( s \) =H \( s \) = { {s rSup { size 8{2} } } over { \( s+1 \) \( s+2 \) } } = { {s rSup { size 8{2} } } over {s rSup { size 8{2} } +3s+2} } } {}
as follows
s
2
s
2
+
3s
+
2
=
(
s
2
+
3s
+
2
)
−
(
3s
+
2
)
s
2
+
3s
+
2
s
2
s
2
+
3s
+
2
=
(
s
2
+
3s
+
2
)
−
(
3s
+
2
)
s
2
+
3s
+
2
size 12{ { {s rSup { size 8{2} } } over {s rSup { size 8{2} } +3s+2} } = { { \( s rSup { size 8{2} } +3s+2 \) - \( 3s+2 \) } over {s rSup { size 8{2} } +3s+2} } } {}
to obtain
V
o
(
s
)
=
H
(
s
)
=
1
−
3s
+
2
(
s
+
1
)
(
s
+
2
)
V
o
(
s
)
=
H
(
s
)
=
1
−
3s
+
2
(
s
+
1
)
(
s
+
2
)
size 12{V rSub { size 8{o} } \( s \) =H \( s \) =1 - { {3s+2} over { \( s+1 \) \( s+2 \) } } } {}
f/ Partial fraction expansion
We can expand the proper rational function in a partial fraction expansion of the form
H
P
(
s
)
=
−
3s
+
2
(
s
+
1
)
(
s
+
2
)
=
A
s
+
1
+
B
s
+
2
H
P
(
s
)
=
−
3s
+
2
(
s
+
1
)
(
s
+
2
)
=
A
s
+
1
+
B
s
+
2
size 12{H rSub { size 8{P} } \( s \) = - { {3s+2} over { \( s+1 \) \( s+2 \) } } = { {A} over {s+1} } + { {B} over {s+2} } } {}
The coefficient A is found as follows
[
(
s
+
1
)
H
P
(
s
)
]
s
=
−
1
=
[
(
s
+
1
)
A
s
+
1
+
(
s
+
1
)
B
s
+
2
]
s
=
−
1
=
A
[
(
s
+
1
)
H
P
(
s
)
]
s
=
−
1
=
[
(
s
+
1
)
A
s
+
1
+
(
s
+
1
)
B
s
+
2
]
s
=
−
1
=
A
size 12{ \[ \( s+1 \) H rSub { size 8{P} } \( s \) \] rSub { size 8{s= - 1} } = \[ \( s+1 \) { {A} over {s+1} } + \( s+1 \) { {B} over {s+2} } \] rSub { size 8{s= - 1} } =A} {}
Therefore,
A
=
3
−
2
−
1
+
2
=
1
A
=
3
−
2
−
1
+
2
=
1
size 12{A= { {3 - 2} over { - 1+2} } =1} {}
By a similar argument
B
=
6
−
2
−
2
+
1
=
−
4
B
=
6
−
2
−
2
+
1
=
−
4
size 12{B= { {6 - 2} over { - 2+1} } = - 4} {}
so that
V
o
(
s
)
=
H
(
s
)
=
1
+
1
s
+
1
−
4
s
+
2
V
o
(
s
)
=
H
(
s
)
=
1
+
1
s
+
1
−
4
s
+
2
size 12{V rSub { size 8{o} } \( s \) =H \( s \) =1+ { {1} over {s+1} } - { {4} over {s+2} } } {}
g/ Inverse Laplace transform of output voltage
The partial fraction expansion shows that
V
o
(
s
)
=
H
(
s
)
=
1
+
1
s
+
1
−
4
s
+
2
for
σ
>
−
1
V
o
(
s
)
=
H
(
s
)
=
1
+
1
s
+
1
−
4
s
+
2
for
σ
>
−
1
size 12{V rSub { size 8{o} } \( s \) =H \( s \) =1+ { {1} over {s+1} } - { {4} over {s+2} } matrix {
{} # {}
} ital "for" matrix {
{} # {}
} σ> - 1} {}
Therefore,
v
o
(
t
)
=
h
(
t
)
=
δ
(
t
)
+
e
−
t
u
(
t
)
−
4
e
−
2t
u
(
t
)
v
o
(
t
)
=
h
(
t
)
=
δ
(
t
)
+
e
−
t
u
(
t
)
−
4
e
−
2t
u
(
t
)
size 12{v rSub { size 8{o} } \( t \) =h \( t \) =δ \( t \) +e rSup { size 8{ - t} } u \( t \) - 4e rSup { size 8{ - 2t} } u \( t \) } {}
h/ Physical interpretation of result
v
o
(
t
)
=
h
(
t
)
=
δ
(
t
)
+
e
−
t
u
(
t
)
−
4
e
−
2t
u
(
t
)
v
o
(
t
)
=
h
(
t
)
=
δ
(
t
)
+
e
−
t
u
(
t
)
−
4
e
−
2t
u
(
t
)
size 12{v rSub { size 8{o} } \( t \) =h \( t \) =δ \( t \) +e rSup { size 8{ - t} } u \( t \) - 4e rSup { size 8{ - 2t} } u \( t \) } {}
How can we explain the impulse response of this circuit in physical terms. There are three critical times: (1) at t = 0,
vo(t)vo(t) size 12{v rSub { size 8{o} } \( t \) } {} has a unit impulse and a discontinuity of value −3; (2) for t > 0,
vo(t)vo(t) size 12{v rSub { size 8{o} } \( t \) } {} consists of complex exponentials at the frequencies −1 and −2; (3) as t→∞,
vo(t)→0vo(t)→0 size 12{v rSub { size 8{o} } \( t \) rightarrow 0} {}.
The voltages and currents in the network must satisfy KVL and KCL plus the constitutive relations of the elements.
- The reasoning at t = 0 is tricky. If the impulse in
vi(t)vi(t) size 12{v rSub { size 8{i} } \( t \) } {} appeared in
vc(t)vc(t) size 12{v rSub { size 8{c} } \( t \) } {} that would cause a doublet in current that cannot be matched to satisfy KCL. Therefore, the impulse appears in vo(t) which causes an impulse in
iR(t)=3δ(t)iR(t)=3δ(t) size 12{i rSub { size 8{R} } \( t \) =3δ \( t \) } {} which flows through the capacitance to cause a step
vC(t)=3u(t)vC(t)=3u(t) size 12{v rSub { size 8{C} } \( t \) =3u \( t \) } {} which appears as an initial step in
vo(t)vo(t) size 12{v rSub { size 8{o} } \( t \) } {}.
- After the impulse occurs, the capacitance has an initial voltage and the inductance has an initial current, i.e., the network is energized. All voltages and currents now relax exponentially at the natural frequencies of −1 and −2.
- Since the network is lossy, the natural frequencies are in the left-half of the s plane all voltages and current decay to zero.
Two-minute miniquiz problem
Problem 5-2
Consider the network shown below.
The input voltage vi(t) is
v
i
(
t
)
=
e
−
t
u
(
t
)
v
i
(
t
)
=
e
−
t
u
(
t
)
size 12{v rSub { size 8{i} } \( t \) =e rSup { size 8{ - t} } u \( t \) } {}
Determine vo(t).
Solution
The system function is
H
(
s
)
=
V
o
(
s
)
V
i
(
s
)
=
1
s
1
2
+
1
s
=
2
s
+
2
H
(
s
)
=
V
o
(
s
)
V
i
(
s
)
=
1
s
1
2
+
1
s
=
2
s
+
2
size 12{H \( s \) = { {V rSub { size 8{o} } \( s \) } over {V rSub { size 8{i} } \( s \) } } = { { { {1} over {s} } } over { { {1} over {2} } + { {1} over {s} } } } = { {2} over {s+2} } } {}
The Laplace transform of the input voltage is
V
i
(
s
)
=
1
s
+
1
V
i
(
s
)
=
1
s
+
1
size 12{V rSub { size 8{i} } \( s \) = { {1} over {s+1} } } {}
Therefore,
V
o
(
s
)
=
V
i
(
s
)
H
(
s
)
=
2
(
s
+
1
)
(
s
+
2
)
=
2
s
+
1
−
2
s
+
2
V
o
(
s
)
=
V
i
(
s
)
H
(
s
)
=
2
(
s
+
1
)
(
s
+
2
)
=
2
s
+
1
−
2
s
+
2
size 12{V rSub { size 8{o} } \( s \) =V rSub { size 8{i} } \( s \) H \( s \) = { {2} over { \( s+1 \) \( s+2 \) } } = { {2} over {s+1} } - { {2} over {s+2} } } {}
and
v
o
(
t
)
=
2
e
−
t
u
(
t
)
−
2
e
−
2t
u
(
t
)
v
o
(
t
)
=
2
e
−
t
u
(
t
)
−
2
e
−
2t
u
(
t
)
size 12{v rSub { size 8{o} } \( t \) =2e rSup { size 8{ - t} } u \( t \) - 2e rSup { size 8{ - 2t} } u \( t \) } {}
III. CONCLUSION — LAPLACE TRANSFORM METHOD FOR FINDING THE RESPONSE OF AN LTI SYSTEM
- Find Laplace transform of input
x(t)⇔LX(s)x(t)⇔LX(s) size 12{x \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} X \( s \) } {}.
- Determine system function H(s) from
- impulse response of system
h(t)⇔LH(s)h(t)⇔LH(s) size 12{h \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} H \( s \) } {};
- structural model of system using impedance method PLUS knowledge about causality, stability. etc.;
- differential equation PLUS knowledge about causality, stability, etc.
- Determine Laplace transform of output Y (s) = H(s)X(s).
- Determine output time function
y(t)⇔LY(s)y(t)⇔LY(s) size 12{y \( t \) matrix {
{} # {}
} { dlrarrow } cSup { size 8{L} } matrix {
{} # {}
} Y \( s \) } {}.
This method can be summarized as follows
IV. HISTORICAL PERSPECTIVE
Oliver Heaviside (1850-1925)
James Clerk Maxwell (1831-1879) died of cancer at age 48 before his ideas on electromagnetic theory could be completely worked out and disseminated. That job was left to three younger men known as the Maxwellians — Oliver Lodge, George Francis FitzGerald, and Oliver Heaviside (shown on the left).
- Born in London on May 18, 1850.
- Nephew of Charles Wheatstone a pioneer in telegraphy who sparked Oliver’s interest in electrical science.
- He had a serious hearing defect and difficulties in school which he quit at age 16. He was largely self-taught.
- Worked as a telegrapher from age 18 to 24 at which time he retired.
- He was supported by his parents first and then his brother. His needs were modest and his family regarded him as a genius.
- He had no academic appointment, attended scientific meetings very rarely, and published largely in an electrical trade journal The Electrician.
- He was a recluse, worked in a small room that he kept extremely hot and filled with pipe smoke. He was combative with a caustic wit — “a first-rate oddity”. He was devoid of social skills and avoided social contacts.
- He made many important contributions to science, mathematics, and especially to electrical engineering, including:
- He introduced the concepts of inductance, capacitance, and impedance (labelled it Z).
- He was first to write Maxwell’s equations in the modern (vector) form.
- He solved problems of signal propagation in the atmosphere and in cables.
- He used operational calculus to solve differential equations and electric networks. He defined his resistance operator p = d/dt to calculate impedances directly from circuits.
- He was a contemporary of James Clerk Maxwell, Charles Darwin, Michael Faraday, George Stokes, William Thomson (Lord Kelvin). He corresponded with many of these and other scientists and was highly respected by the leading scientists of his day.
- He died February 3, 1925.
Exercises .
Solutions of Exercises.
