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
