The properties associated with the Laplace transform are similar to those of the Fourier transform. First, let's set define some notation, we will use the notation L{}L{} to denote the Laplace transform operation. Therefore we can write X(s)=Lx(t)X(s)=Lx(t) and x(t)=L-1X(s)x(t)=L-1X(s) for the forward and inverse Laplace transforms, respectively. We can also use the transform pair notation used earlier:
x
(
t
)
↔
X
(
s
)
x
(
t
)
↔
X
(
s
)
(1)
With this notation defined, lets now look at some properties.
Given that x1(t)↔X1(s)x1(t)↔X1(s) and x2(t)↔X2(s)x2(t)↔X2(s) then for any constants αα and ββ, we have
α
x
1
(
t
)
+
β
x
2
(
t
)
↔
α
X
1
(
s
)
+
β
X
2
(
s
)
α
x
1
(
t
)
+
β
x
2
(
t
)
↔
α
X
1
(
s
)
+
β
X
2
(
s
)
(2)
The linearity property follows easily using the definition of the Laplace transform.
The reason we call this the time delay property rather than the time shift property is that the time shift must be positive, i.e. if τ>0τ>0, then x(t-τ)x(t-τ) corresponds to a delay. If τ<0τ<0 then we would not be able to use the single-sided Laplace transform because we would have a lower integration limit of ττ, which is less than zero. To derive the property, lets evaluate the Laplace transform of the time-delayed signal
L
x
(
t
-
τ
)
=
∫
0
∞
x
(
t
-
τ
)
e
-
s
t
d
t
L
x
(
t
-
τ
)
=
∫
0
∞
x
(
t
-
τ
)
e
-
s
t
d
t
(3)
Letting γ=t-τγ=t-τ leads to t=γ+τt=γ+τ and dt=dγdt=dγ. Substituting these quantities into Equation 3 gives
L
x
(
t
-
τ
)
=
∫
-
τ
∞
x
(
γ
)
e
-
s
(
γ
+
τ
)
d
γ
=
e
-
s
τ
∫
-
τ
∞
x
(
γ
)
e
-
s
γ
d
γ
=
e
-
s
τ
∫
-
τ
0
x
(
γ
)
e
-
s
γ
d
γ
+
e
-
s
τ
∫
0
∞
x
(
γ
)
e
-
s
γ
d
γ
L
x
(
t
-
τ
)
=
∫
-
τ
∞
x
(
γ
)
e
-
s
(
γ
+
τ
)
d
γ
=
e
-
s
τ
∫
-
τ
∞
x
(
γ
)
e
-
s
γ
d
γ
=
e
-
s
τ
∫
-
τ
0
x
(
γ
)
e
-
s
γ
d
γ
+
e
-
s
τ
∫
0
∞
x
(
γ
)
e
-
s
γ
d
γ
(4)
where we note that the first integral in the last line is zero since x(t)=0,t<0x(t)=0,t<0. Therefore the time delay property is given by
L
x
(
t
-
τ
)
=
e
-
s
τ
X
(
s
)
L
x
(
t
-
τ
)
=
e
-
s
τ
X
(
s
)
(5)
This property is the Laplace transform corresponds to the frequency shift property of the Fourier transform. In fact, the derivation of the ss-shift property is virtually identical to that of the frequency shift property.
L
e
-
a
t
x
(
t
)
=
∫
0
∞
e
-
a
t
x
(
t
)
e
-
s
t
d
t
=
∫
0
∞
x
(
t
)
e
-
(
a
+
s
)
t
d
t
=
∫
0
∞
x
(
t
)
e
-
(
a
+
σ
+
j
Ω
)
t
d
t
=
X
(
s
+
a
)
L
e
-
a
t
x
(
t
)
=
∫
0
∞
e
-
a
t
x
(
t
)
e
-
s
t
d
t
=
∫
0
∞
x
(
t
)
e
-
(
a
+
s
)
t
d
t
=
∫
0
∞
x
(
t
)
e
-
(
a
+
σ
+
j
Ω
)
t
d
t
=
X
(
s
+
a
)
(6)
The ss-shift property also alters the region of convergence of the Laplace transform. If the region of convergence for X(s)X(s) is σ>σminσ>σmin, then the region of convergence for Le-atx(t)Le-atx(t) is σ>σmin-Re(a)σ>σmin-Re(a).
Let's begin by taking the derivative of the Laplace transform:
d
X
(
s
)
d
s
=
d
d
s
∫
0
∞
x
(
t
)
e
-
s
t
d
t
=
∫
0
∞
x
(
t
)
d
d
s
e
-
s
t
d
t
=
-
∫
0
∞
t
x
(
t
)
e
-
s
t
d
t
d
X
(
s
)
d
s
=
d
d
s
∫
0
∞
x
(
t
)
e
-
s
t
d
t
=
∫
0
∞
x
(
t
)
d
d
s
e
-
s
t
d
t
=
-
∫
0
∞
t
x
(
t
)
e
-
s
t
d
t
(7)
So we can write
L
t
x
(
t
)
=
-
d
X
(
s
)
d
s
L
t
x
(
t
)
=
-
d
X
(
s
)
d
s
(8)
This idea can be extended to multiplication by tntn. Letting y(t)=tx(t)y(t)=tx(t), if follows that
t
y
(
t
)
↔
-
d
Y
(
s
)
d
s
↔
d
2
X
(
s
)
d
s
2
t
y
(
t
)
↔
-
d
Y
(
s
)
d
s
↔
d
2
X
(
s
)
d
s
2
(9)
Proceeding in this manner, we find that
t
n
x
(
t
)
↔
(
-
1
)
n
d
n
X
(
s
)
d
s
n
t
n
x
(
t
)
↔
(
-
1
)
n
d
n
X
(
s
)
d
s
n
(10)
The time scaling property for the Laplace transform is similar to that of the Fourier transform:
L
x
(
α
t
)
=
∫
0
∞
x
(
α
t
)
e
-
s
t
d
t
=
1
α
∫
0
∞
x
(
γ
)
e
-
s
α
γ
d
γ
=
1
α
X
s
α
L
x
(
α
t
)
=
∫
0
∞
x
(
α
t
)
e
-
s
t
d
t
=
1
α
∫
0
∞
x
(
γ
)
e
-
s
α
γ
d
γ
=
1
α
X
s
α
(11)
where in the second equality, we made the substitution t=γαt=γα and dt=dγαdt=dγα.
The derivation of the convolution property for the Laplace transform is virtually identical to that of the Fourier transform. We begin with
L
∫
-
∞
∞
x
(
τ
)
h
(
t
-
τ
)
d
τ
=
∫
-
∞
∞
x
(
τ
)
L
h
(
t
-
τ
)
d
τ
L
∫
-
∞
∞
x
(
τ
)
h
(
t
-
τ
)
d
τ
=
∫
-
∞
∞
x
(
τ
)
L
h
(
t
-
τ
)
d
τ
(12)
Applying the time-delay property of the Laplace transform gives
∫
-
∞
∞
x
(
τ
)
L
h
(
t
-
τ
)
d
τ
=
H
(
s
)
∫
-
∞
∞
x
(
τ
)
e
-
s
τ
d
τ
=
H
(
s
)
X
(
s
)
∫
-
∞
∞
x
(
τ
)
L
h
(
t
-
τ
)
d
τ
=
H
(
s
)
∫
-
∞
∞
x
(
τ
)
e
-
s
τ
d
τ
=
H
(
s
)
X
(
s
)
(13)
If h(t)h(t) is the the impulse response of a linear time-invariant system, then we call H(s)H(s) the system function of the system. The frequency response results by setting s=jΩs=jΩ in H(s)H(s). The system function provides us with a very powerful means of determining the output of a linear time-invariant filter given the input signal. It will also enable us to determine a means of establishing the stability of a linear-time invariant filter, something which was not possible with the frequency response.
The Laplace transform of the derivative of a signal will be used widely. Consider
L
d
d
t
x
(
t
)
=
∫
0
-
∞
x
'
(
t
)
e
-
s
t
d
t
L
d
d
t
x
(
t
)
=
∫
0
-
∞
x
'
(
t
)
e
-
s
t
d
t
(14)
this can be integrated by parts:
u
=
e
-
s
t
v
'
=
x
'
(
t
)
u
'
=
-
s
e
-
s
t
v
=
x
(
t
)
u
=
e
-
s
t
v
'
=
x
'
(
t
)
u
'
=
-
s
e
-
s
t
v
=
x
(
t
)
(15)
which gives
L
d
d
t
x
(
t
)
=
u
v
0
-
∞
-
∫
0
-
∞
u
'
v
d
t
=
e
-
s
t
x
(
t
)
0
-
∞
+
∫
0
-
∞
s
x
(
t
)
e
-
s
t
d
t
=
-
x
(
0
-
)
+
s
X
(
s
)
L
d
d
t
x
(
t
)
=
u
v
0
-
∞
-
∫
0
-
∞
u
'
v
d
t
=
e
-
s
t
x
(
t
)
0
-
∞
+
∫
0
-
∞
s
x
(
t
)
e
-
s
t
d
t
=
-
x
(
0
-
)
+
s
X
(
s
)
(16)
therefore we have,
d
d
t
x
(
t
)
↔
s
X
(
s
)
-
x
(
0
-
)
d
d
t
x
(
t
)
↔
s
X
(
s
)
-
x
(
0
-
)
(17)
The previous derivation can be extended to higher order derivatives. Consider
y
(
t
)
=
d
x
(
t
)
d
t
↔
s
X
(
s
)
-
x
(
0
-
)
y
(
t
)
=
d
x
(
t
)
d
t
↔
s
X
(
s
)
-
x
(
0
-
)
(18)
it follows that
d
y
(
t
)
d
t
↔
s
Y
(
s
)
-
y
(
0
-
)
d
y
(
t
)
d
t
↔
s
Y
(
s
)
-
y
(
0
-
)
(19)
which leads to
d
2
d
t
2
x
(
t
)
↔
s
2
X
(
s
)
-
s
x
(
0
-
)
-
d
x
(
0
-
)
d
t
d
2
d
t
2
x
(
t
)
↔
s
2
X
(
s
)
-
s
x
(
0
-
)
-
d
x
(
0
-
)
d
t
(20)
This process can be iterated to get the Laplace transform of arbitrary higher order derivatives, giving
d
n
x
(
t
)
d
t
n
↔
s
n
X
(
s
)
-
s
n
-
1
x
(
0
-
)
-
∑
k
=
2
n
s
n
-
k
d
k
-
1
x
(
0
-
)
d
t
k
-
1
d
n
x
(
t
)
d
t
n
↔
s
n
X
(
s
)
-
s
n
-
1
x
(
0
-
)
-
∑
k
=
2
n
s
n
-
k
d
k
-
1
x
(
0
-
)
d
t
k
-
1
(21)
where it should be understood that
d
m
x
(
0
-
)
d
t
m
≡
d
m
x
(
t
)
d
t
m
t
=
0
-
,
m
=
1
,
...
,
n
-
1
d
m
x
(
0
-
)
d
t
m
≡
d
m
x
(
t
)
d
t
m
t
=
0
-
,
m
=
1
,
...
,
n
-
1
(22)
Let
g
(
t
)
=
∫
0
-
t
x
(
τ
)
d
τ
g
(
t
)
=
∫
0
-
t
x
(
τ
)
d
τ
(23)
it follows that
d
g
(
t
)
d
t
=
x
(
t
)
d
g
(
t
)
d
t
=
x
(
t
)
(24)
and g0-=0g0-=0. Moreover, we have
X
(
s
)
=
L
d
g
(
t
)
d
t
=
s
G
(
s
)
-
g
0
-
=
s
G
(
s
)
X
(
s
)
=
L
d
g
(
t
)
d
t
=
s
G
(
s
)
-
g
0
-
=
s
G
(
s
)
(25)
therefore
G
(
s
)
=
X
(
s
)
s
G
(
s
)
=
X
(
s
)
s
(26)
but since
G
(
s
)
=
L
∫
0
-
t
x
(
τ
)
d
τ
G
(
s
)
=
L
∫
0
-
t
x
(
τ
)
d
τ
(27)
we have
∫
0
-
t
x
(
τ
)
d
τ
↔
X
(
s
)
s
∫
0
-
t
x
(
τ
)
d
τ
↔
X
(
s
)
s
(28)
Now suppose x(t)x(t) has a non-zero integral over negative values of tt. We have
∫
∞
t
x
(
τ
)
d
τ
=
∫
-
∞
0
-
x
(
τ
)
d
τ
+
∫
0
-
t
x
(
τ
)
d
τ
∫
∞
t
x
(
τ
)
d
τ
=
∫
-
∞
0
-
x
(
τ
)
d
τ
+
∫
0
-
t
x
(
τ
)
d
τ
(29)
The quantity ∫-∞0-x(τ)dτ∫-∞0-x(τ)dτ is a constant for positive values of tt, and can be expressed as
u
(
t
)
∫
-
∞
0
-
x
(
τ
)
d
τ
u
(
t
)
∫
-
∞
0
-
x
(
τ
)
d
τ
(30)
it follows that
∫
∞
t
x
(
τ
)
d
τ
↔
∫
-
∞
0
-
x
(
τ
)
d
τ
s
+
X
(
s
)
s
∫
∞
t
x
(
τ
)
d
τ
↔
∫
-
∞
0
-
x
(
τ
)
d
τ
s
+
X
(
s
)
s
(31)
where we have used the fact that u(t)↔1s.u(t)↔1s.
The initial value theorem makes it possible to determine x(t)x(t) at t=0+t=0+ from X(s)X(s). From the derivative property of the Laplace transform, we can write
L
d
x
(
t
)
d
t
=
s
X
(
s
)
-
x
0
-
L
d
x
(
t
)
d
t
=
s
X
(
s
)
-
x
0
-
(32)
Taking the limit s→∞s→∞
lim
s
→
∞
∫
0
-
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
∫
0
-
∞
lim
s
→
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
lim
s
→
∞
∫
0
-
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
∫
0
-
∞
lim
s
→
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
(33)
There are two cases, the first is when x(t)x(t) is continuous at t=0t=0. In this case it is clear that dx(t)dte-st→0dx(t)dte-st→0 as s→∞s→∞, so Equation 33 can be written as
0
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
0
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
(34)
Since x(t)x(t) is continuous at t=0t=0, x0-=x0+x0-=x0+, the Initial Value Theorem follows,
x
0
+
=
lim
s
→
∞
s
X
(
s
)
x
0
+
=
lim
s
→
∞
s
X
(
s
)
(35)
The second case is when x(t)x(t) is discontinuous at t=0t=0. In this case, we use the fact that
d
x
(
t
)
d
t
t
=
0
=
x
0
-
-
x
0
+
δ
(
t
)
d
x
(
t
)
d
t
t
=
0
=
x
0
-
-
x
0
+
δ
(
t
)
(36)
For example, if we integrate the right-hand side of Equation 36 with x0-=0x0-=0 and x0+=1x0+=1, we get the unit step function, u(t)u(t). Proceeding as before, we have
lim
s
→
∞
∫
0
-
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
lim
s
→
∞
∫
0
-
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
∞
s
X
(
s
)
-
x
0
-
(37)
The left-hand side of Equation 37 can be written as
lim
s
→
∞
∫
0
-
0
+
x
0
-
-
x
0
+
δ
(
t
)
e
-
s
t
d
t
+
lim
s
→
∞
∫
0
+
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
lim
s
→
∞
∫
0
-
0
+
x
0
-
-
x
0
+
δ
(
t
)
e
-
s
t
d
t
+
lim
s
→
∞
∫
0
+
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
(38)
From the sifting property of the unit impulse, the first term in Equation 38 is
x
0
-
-
x
0
+
x
0
-
-
x
0
+
(39)
while the second term is zero since in the limit, the real part of ss goes to infinity. Substituting these results into the left-hand side of Equation 37 again leads to the initial value theorem, in Equation 35.
The Final Value Theorem allows us to determine
lim
t
→
∞
x
(
t
)
lim
t
→
∞
x
(
t
)
(40)
from X(s)X(s). Taking the limit as ss approaches zero in the derivative property gives
lim
s
→
0
∫
0
-
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
0
s
X
(
s
)
-
x
0
-
lim
s
→
0
∫
0
-
∞
d
x
(
t
)
d
t
e
-
s
t
d
t
=
lim
s
→
0
s
X
(
s
)
-
x
0
-
(41)
The left-hand-side of Equation 41 can be written as
∫
0
-
∞
lim
s
→
0
d
x
(
t
)
d
t
e
-
s
t
d
t
=
∫
0
-
∞
d
x
(
t
)
d
t
d
t
=
x
(
∞
)
-
x
0
-
∫
0
-
∞
lim
s
→
0
d
x
(
t
)
d
t
e
-
s
t
d
t
=
∫
0
-
∞
d
x
(
t
)
d
t
d
t
=
x
(
∞
)
-
x
0
-
(42)
Substituting this result back into Equation 41 leads to the Final Value Theorem
x
(
∞
)
=
lim
s
→
0
s
X
(
s
)
x
(
∞
)
=
lim
s
→
0
s
X
(
s
)
(43)
which is only valid as long as the limit x(∞)x(∞) exists.
Table 1: Laplace Transform properties.
| Property |
y
(
t
)
y
(
t
)
|
Y
(
s
)
Y
(
s
)
|
| Linearity |
α
x
1
(
t
)
+
β
x
2
(
t
)
α
x
1
(
t
)
+
β
x
2
(
t
)
|
α
X
1
(
s
)
+
β
X
2
(
s
)
α
X
1
(
s
)
+
β
X
2
(
s
)
|
| Time Delay |
x
(
t
-
τ
)
x
(
t
-
τ
)
|
X
(
s
)
e
-
s
τ
X
(
s
)
e
-
s
τ
|
| s-Shift |
x
(
t
)
e
-
a
t
x
(
t
)
e
-
a
t
|
X
(
s
+
a
)
)
X
(
s
+
a
)
)
|
| Multiplication by tt |
t
x
(
t
)
t
x
(
t
)
|
-
d
X
(
s
)
d
s
-
d
X
(
s
)
d
s
|
| Multiplication by tntn |
t
n
x
(
t
)
t
n
x
(
t
)
|
(
-
1
)
n
d
n
X
(
s
)
d
s
n
(
-
1
)
n
d
n
X
(
s
)
d
s
n
|
| Convolution |
x
(
t
)
*
h
(
t
)
x
(
t
)
*
h
(
t
)
|
X
(
j
Ω
)
H
(
j
Ω
)
X
(
j
Ω
)
H
(
j
Ω
)
|
| Differentiation |
d
x
(
t
)
d
t
d
x
(
t
)
d
t
|
s
X
(
s
)
-
x
0
-
s
X
(
s
)
-
x
0
-
|
| |
d
2
x
(
t
)
d
t
2
d
2
x
(
t
)
d
t
2
|
s
2
X
(
s
)
-
s
x
0
-
-
d
x
0
-
d
t
s
2
X
(
s
)
-
s
x
0
-
-
d
x
0
-
d
t
|
| |
d
n
x
(
t
)
d
t
n
d
n
x
(
t
)
d
t
n
|
s
n
X
(
s
)
-
s
n
-
1
x
0
-
-
∑
k
=
2
n
s
n
-
k
d
k
-
1
x
0
-
d
t
k
-
1
s
n
X
(
s
)
-
s
n
-
1
x
0
-
-
∑
k
=
2
n
s
n
-
k
d
k
-
1
x
0
-
d
t
k
-
1
|
| Integration |
∫
∞
t
x
(
τ
)
d
τ
∫
∞
t
x
(
τ
)
d
τ
|
∫
-
∞
0
-
x
(
τ
)
d
τ
s
+
X
(
s
)
s
∫
-
∞
0
-
x
(
τ
)
d
τ
s
+
X
(
s
)
s
|
| Initial Value Theorem |
x
0
+
=
lim
s
→
∞
s
X
(
s
)
x
0
+
=
lim
s
→
∞
s
X
(
s
)
|
| Final Value Theorem |
x
(
∞
)
=
lim
s
→
0
s
X
(
s
)
x
(
∞
)
=
lim
s
→
0
s
X
(
s
)
|
