Relation between frequency- and time-domain behavior at large time
In mathematical analysis , the final value theorem (FVT) is one of several similar theorems used to relate frequency domain expressions to the time domain behavior as time approaches infinity.[ 1] [ 2] [ 3] [ 4]
Mathematically, if
f
(
t
)
{\displaystyle f(t)}
in continuous time has (unilateral) Laplace transform
F
(
s
)
{\displaystyle F(s)}
, then a final value theorem establishes conditions under which
lim
t
→
∞
f
(
t
)
=
lim
s
→
0
s
F
(
s
)
{\displaystyle \lim _{t\to \infty }f(t)=\lim _{s\,\to \,0}{sF(s)}}
Likewise, if
f
[
k
]
{\displaystyle f[k]}
in discrete time has (unilateral) Z-transform
F
(
z
)
{\displaystyle F(z)}
, then a final value theorem establishes conditions under which
lim
k
→
∞
f
[
k
]
=
lim
z
→
1
(
z
−
1
)
F
(
z
)
{\displaystyle \lim _{k\to \infty }f[k]=\lim _{z\to 1}{(z-1)F(z)}}
An Abelian final value theorem makes assumptions about the time-domain behavior of
f
(
t
)
{\displaystyle f(t)}
(or
f
[
k
]
{\displaystyle f[k]}
) to calculate
lim
s
→
0
s
F
(
s
)
{\displaystyle \lim _{s\,\to \,0}{sF(s)}}
.
Conversely, a Tauberian final value theorem makes assumptions about the frequency-domain behaviour of
F
(
s
)
{\displaystyle F(s)}
to calculate
lim
t
→
∞
f
(
t
)
{\displaystyle \lim _{t\to \infty }f(t)}
(or
lim
k
→
∞
f
[
k
]
{\displaystyle \lim _{k\to \infty }f[k]}
)
(see Abelian and Tauberian theorems for integral transforms ).
Deducing limt → ∞ f (t ) [ edit ]
In the following statements, the notation '
s
→
0
{\displaystyle s\to 0}
' means that
s
{\displaystyle s}
approaches 0, whereas '
s
↓
0
{\displaystyle s\downarrow 0}
' means that
s
{\displaystyle s}
approaches 0 through the positive numbers.
Standard Final Value Theorem [ edit ]
Suppose that every pole of
F
(
s
)
{\displaystyle F(s)}
is either in the open left half plane or at the origin, and that
F
(
s
)
{\displaystyle F(s)}
has at most a single pole at the origin. Then
s
F
(
s
)
→
L
∈
R
{\displaystyle sF(s)\to L\in \mathbb {R} }
as
s
→
0
{\displaystyle s\to 0}
, and
lim
t
→
∞
f
(
t
)
=
L
{\displaystyle \lim _{t\to \infty }f(t)=L}
.[ 5]
Suppose that
f
(
t
)
{\displaystyle f(t)}
and
f
′
(
t
)
{\displaystyle f'(t)}
both have Laplace transforms that exist for all
s
>
0
{\displaystyle s>0}
. If
lim
t
→
∞
f
(
t
)
{\displaystyle \lim _{t\to \infty }f(t)}
exists and
lim
s
→
0
s
F
(
s
)
{\displaystyle \lim _{s\,\to \,0}{sF(s)}}
exists then
lim
t
→
∞
f
(
t
)
=
lim
s
→
0
s
F
(
s
)
{\displaystyle \lim _{t\to \infty }f(t)=\lim _{s\,\to \,0}{sF(s)}}
.[ 3] : Theorem 2.36 [ 4] : 20 [ 6]
Remark
Both limits must exist for the theorem to hold. For example, if
f
(
t
)
=
sin
(
t
)
{\displaystyle f(t)=\sin(t)}
then
lim
t
→
∞
f
(
t
)
{\displaystyle \lim _{t\to \infty }f(t)}
does not exist, but
lim
s
→
0
s
F
(
s
)
=
lim
s
→
0
s
s
2
+
1
=
0
{\displaystyle \lim _{s\,\to \,0}{sF(s)}=\lim _{s\,\to \,0}{\frac {s}{s^{2}+1}}=0}
.[ 3] : Example 2.37 [ 4] : 20
Improved Tauberian converse Final Value Theorem [ edit ]
Suppose that
f
:
(
0
,
∞
)
→
C
{\displaystyle f:(0,\infty )\to \mathbb {C} }
is bounded and differentiable, and that
t
f
′
(
t
)
{\displaystyle tf'(t)}
is also bounded on
(
0
,
∞
)
{\displaystyle (0,\infty )}
. If
s
F
(
s
)
→
L
∈
C
{\displaystyle sF(s)\to L\in \mathbb {C} }
as
s
→
0
{\displaystyle s\to 0}
then
lim
t
→
∞
f
(
t
)
=
L
{\displaystyle \lim _{t\to \infty }f(t)=L}
.[ 7]
Extended Final Value Theorem [ edit ]
Suppose that every pole of
F
(
s
)
{\displaystyle F(s)}
is either in the open left half-plane or at the origin. Then one of the following occurs:
s
F
(
s
)
→
L
∈
R
{\displaystyle sF(s)\to L\in \mathbb {R} }
as
s
↓
0
{\displaystyle s\downarrow 0}
, and
lim
t
→
∞
f
(
t
)
=
L
{\displaystyle \lim _{t\to \infty }f(t)=L}
.
s
F
(
s
)
→
+
∞
∈
R
{\displaystyle sF(s)\to +\infty \in \mathbb {R} }
as
s
↓
0
{\displaystyle s\downarrow 0}
, and
f
(
t
)
→
+
∞
{\displaystyle f(t)\to +\infty }
as
t
→
∞
{\displaystyle t\to \infty }
.
s
F
(
s
)
→
−
∞
∈
R
{\displaystyle sF(s)\to -\infty \in \mathbb {R} }
as
s
↓
0
{\displaystyle s\downarrow 0}
, and
f
(
t
)
→
−
∞
{\displaystyle f(t)\to -\infty }
as
t
→
∞
{\displaystyle t\to \infty }
.
In particular, if
s
=
0
{\displaystyle s=0}
is a multiple pole of
F
(
s
)
{\displaystyle F(s)}
then case 2 or 3 applies (
f
(
t
)
→
+
∞
{\displaystyle f(t)\to +\infty }
or
f
(
t
)
→
−
∞
{\displaystyle f(t)\to -\infty }
).[ 5]
Generalized Final Value Theorem [ edit ]
Suppose that
f
(
t
)
{\displaystyle f(t)}
is Laplace transformable. Let
λ
>
−
1
{\displaystyle \lambda >-1}
. If
lim
t
→
∞
f
(
t
)
t
λ
{\displaystyle \lim _{t\to \infty }{\frac {f(t)}{t^{\lambda }}}}
exists and
lim
s
↓
0
s
λ
+
1
F
(
s
)
{\displaystyle \lim _{s\downarrow 0}{s^{\lambda +1}F(s)}}
exists then
lim
t
→
∞
f
(
t
)
t
λ
=
1
Γ
(
λ
+
1
)
lim
s
↓
0
s
λ
+
1
F
(
s
)
{\displaystyle \lim _{t\to \infty }{\frac {f(t)}{t^{\lambda }}}={\frac {1}{\Gamma (\lambda +1)}}\lim _{s\downarrow 0}{s^{\lambda +1}F(s)}}
where
Γ
(
x
)
{\displaystyle \Gamma (x)}
denotes the Gamma function .[ 5]
Final value theorems for obtaining
lim
t
→
∞
f
(
t
)
{\displaystyle \lim _{t\to \infty }f(t)}
have applications in establishing the long-term stability of a system .
Deducing lims → 0 s F (s ) [ edit ]
Abelian Final Value Theorem [ edit ]
Suppose that
f
:
(
0
,
∞
)
→
C
{\displaystyle f:(0,\infty )\to \mathbb {C} }
is bounded and measurable and
lim
t
→
∞
f
(
t
)
=
α
∈
C
{\displaystyle \lim _{t\to \infty }f(t)=\alpha \in \mathbb {C} }
. Then
F
(
s
)
{\displaystyle F(s)}
exists for all
s
>
0
{\displaystyle s>0}
and
lim
s
→
0
+
s
F
(
s
)
=
α
{\displaystyle \lim _{s\,\to \,0^{+}}{sF(s)}=\alpha }
.[ 7]
Elementary proof [ 7]
Suppose for convenience that
|
f
(
t
)
|
≤
1
{\displaystyle |f(t)|\leq 1}
on
(
0
,
∞
)
{\displaystyle (0,\infty )}
, and let
α
=
lim
t
→
∞
f
(
t
)
{\displaystyle \alpha =\lim _{t\to \infty }f(t)}
. Let
ϵ
>
0
{\displaystyle \epsilon >0}
, and choose
A
{\displaystyle A}
so that
|
f
(
t
)
−
α
|
<
ϵ
{\displaystyle |f(t)-\alpha |<\epsilon }
for all
t
>
A
{\displaystyle t>A}
. Since
s
∫
0
∞
e
−
s
t
d
t
=
1
{\displaystyle s\int _{0}^{\infty }e^{-st}\,dt=1}
, for every
s
>
0
{\displaystyle s>0}
we have
s
F
(
s
)
−
α
=
s
∫
0
∞
(
f
(
t
)
−
α
)
e
−
s
t
d
t
;
{\displaystyle sF(s)-\alpha =s\int _{0}^{\infty }(f(t)-\alpha )e^{-st}\,dt;}
hence
|
s
F
(
s
)
−
α
|
≤
s
∫
0
A
|
f
(
t
)
−
α
|
e
−
s
t
d
t
+
s
∫
A
∞
|
f
(
t
)
−
α
|
e
−
s
t
d
t
≤
2
s
∫
0
A
e
−
s
t
d
t
+
ϵ
s
∫
A
∞
e
−
s
t
d
t
=
I
+
I
I
.
{\displaystyle |sF(s)-\alpha |\leq s\int _{0}^{A}|f(t)-\alpha |e^{-st}\,dt+s\int _{A}^{\infty }|f(t)-\alpha |e^{-st}\,dt\leq 2s\int _{0}^{A}e^{-st}\,dt+\epsilon s\int _{A}^{\infty }e^{-st}\,dt=I+II.}
Now for every
s
>
0
{\displaystyle s>0}
we have
I
I
<
ϵ
s
∫
0
∞
e
−
s
t
d
t
=
ϵ
{\displaystyle II<\epsilon s\int _{0}^{\infty }e^{-st}\,dt=\epsilon }
.
On the other hand, since
A
<
∞
{\displaystyle A<\infty }
is fixed it is clear that
lim
s
→
0
I
=
0
{\displaystyle \lim _{s\to 0}I=0}
, and so
|
s
F
(
s
)
−
α
|
<
ϵ
{\displaystyle |sF(s)-\alpha |<\epsilon }
if
s
>
0
{\displaystyle s>0}
is small enough.
Suppose that all of the following conditions are satisfied:
f
:
(
0
,
∞
)
→
C
{\displaystyle f:(0,\infty )\to \mathbb {C} }
is continuously differentiable and both
f
{\displaystyle f}
and
f
′
{\displaystyle f'}
have a Laplace transform
f
′
{\displaystyle f'}
is absolutely integrable - that is,
∫
0
∞
|
f
′
(
τ
)
|
d
τ
{\displaystyle \int _{0}^{\infty }|f'(\tau )|\,d\tau }
is finite
lim
t
→
∞
f
(
t
)
{\displaystyle \lim _{t\to \infty }f(t)}
exists and is finite
Then
lim
s
→
0
+
s
F
(
s
)
=
lim
t
→
∞
f
(
t
)
{\displaystyle \lim _{s\to 0^{+}}sF(s)=\lim _{t\to \infty }f(t)}
.[ 8]
Remark
The proof uses the dominated convergence theorem .[ 8]
Final Value Theorem for the mean of a function [ edit ]
Let
f
:
(
0
,
∞
)
→
C
{\displaystyle f:(0,\infty )\to \mathbb {C} }
be a continuous and bounded function such that such that the following limit exists
lim
T
→
∞
1
T
∫
0
T
f
(
t
)
d
t
=
α
∈
C
{\displaystyle \lim _{T\to \infty }{\frac {1}{T}}\int _{0}^{T}f(t)\,dt=\alpha \in \mathbb {C} }
Then
lim
s
→
0
,
s
>
0
s
F
(
s
)
=
α
{\displaystyle \lim _{s\,\to \,0,\,s>0}{sF(s)}=\alpha }
.[ 9]
Final Value Theorem for asymptotic sums of periodic functions [ edit ]
Suppose that
f
:
[
0
,
∞
)
→
R
{\displaystyle f:[0,\infty )\to \mathbb {R} }
is continuous and absolutely integrable in
[
0
,
∞
)
{\displaystyle [0,\infty )}
. Suppose further that
f
{\displaystyle f}
is asymptotically equal to a finite sum of periodic functions
f
a
s
{\displaystyle f_{\mathrm {as} }}
, that is
|
f
(
t
)
−
f
a
s
(
t
)
|
<
ϕ
(
t
)
{\displaystyle |f(t)-f_{\mathrm {as} }(t)|<\phi (t)}
where
ϕ
(
t
)
{\displaystyle \phi (t)}
is absolutely integrable in
[
0
,
∞
)
{\displaystyle [0,\infty )}
and vanishes at infinity. Then
lim
s
→
0
s
F
(
s
)
=
lim
t
→
∞
1
t
∫
0
t
f
(
x
)
d
x
{\displaystyle \lim _{s\to 0}sF(s)=\lim _{t\to \infty }{\frac {1}{t}}\int _{0}^{t}f(x)\,dx}
.[ 10]
Final Value Theorem for a function that diverges to infinity [ edit ]
Let
f
(
t
)
:
[
0
,
∞
)
→
R
{\displaystyle f(t):[0,\infty )\to \mathbb {R} }
and
F
(
s
)
{\displaystyle F(s)}
be the Laplace transform of
f
(
t
)
{\displaystyle f(t)}
. Suppose that
f
(
t
)
{\displaystyle f(t)}
satisfies all of the following conditions:
f
(
t
)
{\displaystyle f(t)}
is infinitely differentiable at zero
f
(
k
)
(
t
)
{\displaystyle f^{(k)}(t)}
has a Laplace transform for all non-negative integers
k
{\displaystyle k}
f
(
t
)
{\displaystyle f(t)}
diverges to infinity as
t
→
∞
{\displaystyle t\to \infty }
Then
s
F
(
s
)
{\displaystyle sF(s)}
diverges to infinity as
s
→
0
+
{\displaystyle s\to 0^{+}}
.[ 11]
Final Value Theorem for improperly integrable functions (Abel's theorem for integrals)[ edit ]
Let
h
:
[
0
,
∞
)
→
R
{\displaystyle h:[0,\infty )\to \mathbb {R} }
be measurable and such that the (possibly improper) integral
f
(
x
)
:=
∫
0
x
h
(
t
)
d
t
{\displaystyle f(x):=\int _{0}^{x}h(t)\,dt}
converges for
x
→
∞
{\displaystyle x\to \infty }
. Then
∫
0
∞
h
(
t
)
d
t
:=
lim
x
→
∞
f
(
x
)
=
lim
s
↓
0
∫
0
∞
e
−
s
t
h
(
t
)
d
t
.
{\displaystyle \int _{0}^{\infty }h(t)\,dt:=\lim _{x\to \infty }f(x)=\lim _{s\downarrow 0}\int _{0}^{\infty }e^{-st}h(t)\,dt.}
This is a version of Abel's theorem .
To see this, notice that
f
′
(
t
)
=
h
(
t
)
{\displaystyle f'(t)=h(t)}
and apply the final value theorem to
f
{\displaystyle f}
after an integration by parts : For
s
>
0
{\displaystyle s>0}
,
s
∫
0
∞
e
−
s
t
f
(
t
)
d
t
=
[
−
e
−
s
t
f
(
t
)
]
t
=
o
∞
+
∫
0
∞
e
−
s
t
f
′
(
t
)
d
t
=
∫
0
∞
e
−
s
t
h
(
t
)
d
t
.
{\displaystyle s\int _{0}^{\infty }e^{-st}f(t)\,dt={\Big [}-e^{-st}f(t){\Big ]}_{t=o}^{\infty }+\int _{0}^{\infty }e^{-st}f'(t)\,dt=\int _{0}^{\infty }e^{-st}h(t)\,dt.}
By the final value theorem, the left-hand side converges to
lim
x
→
∞
f
(
x
)
{\displaystyle \lim _{x\to \infty }f(x)}
for
s
→
0
{\displaystyle s\to 0}
.
To establish the convergence of the improper integral
lim
x
→
∞
f
(
x
)
{\displaystyle \lim _{x\to \infty }f(x)}
in practice, Dirichlet's test for improper integrals is often helpful. An example is the Dirichlet integral .
Final value theorems for obtaining
lim
s
→
0
s
F
(
s
)
{\displaystyle \lim _{s\,\to \,0}{sF(s)}}
have applications in probability and statistics to calculate the moments of a random variable . Let
R
(
x
)
{\displaystyle R(x)}
be cumulative distribution function of a continuous random variable
X
{\displaystyle X}
and let
ρ
(
s
)
{\displaystyle \rho (s)}
be the Laplace–Stieltjes transform of
R
(
x
)
{\displaystyle R(x)}
. Then the
n
{\displaystyle n}
-th moment of
X
{\displaystyle X}
can be calculated as
E
[
X
n
]
=
(
−
1
)
n
d
n
ρ
(
s
)
d
s
n
|
s
=
0
{\displaystyle E[X^{n}]=(-1)^{n}\left.{\frac {d^{n}\rho (s)}{ds^{n}}}\right|_{s=0}}
The strategy is to write
d
n
ρ
(
s
)
d
s
n
=
F
(
G
1
(
s
)
,
G
2
(
s
)
,
…
,
G
k
(
s
)
,
…
)
{\displaystyle {\frac {d^{n}\rho (s)}{ds^{n}}}={\mathcal {F}}{\bigl (}G_{1}(s),G_{2}(s),\dots ,G_{k}(s),\dots {\bigr )}}
where
F
(
…
)
{\displaystyle {\mathcal {F}}(\dots )}
is continuous and
for each
k
{\displaystyle k}
,
G
k
(
s
)
=
s
F
k
(
s
)
{\displaystyle G_{k}(s)=sF_{k}(s)}
for a function
F
k
(
s
)
{\displaystyle F_{k}(s)}
. For each
k
{\displaystyle k}
, put
f
k
(
t
)
{\displaystyle f_{k}(t)}
as the inverse Laplace transform of
F
k
(
s
)
{\displaystyle F_{k}(s)}
, obtain
lim
t
→
∞
f
k
(
t
)
{\displaystyle \lim _{t\to \infty }f_{k}(t)}
, and apply a final value theorem to deduce
lim
s
→
0
G
k
(
s
)
=
lim
s
→
0
s
F
k
(
s
)
=
lim
t
→
∞
f
k
(
t
)
{\displaystyle \lim _{s\,\to \,0}{G_{k}(s)}=\lim _{s\,\to \,0}{sF_{k}(s)}=\lim _{t\to \infty }f_{k}(t)}
. Then
d
n
ρ
(
s
)
d
s
n
|
s
=
0
=
F
(
lim
s
→
0
G
1
(
s
)
,
lim
s
→
0
G
2
(
s
)
,
…
,
lim
s
→
0
G
k
(
s
)
,
…
)
{\displaystyle \left.{\frac {d^{n}\rho (s)}{ds^{n}}}\right|_{s=0}={\mathcal {F}}{\Bigl (}\lim _{s\,\to \,0}G_{1}(s),\lim _{s\,\to \,0}G_{2}(s),\dots ,\lim _{s\,\to \,0}G_{k}(s),\dots {\Bigr )}}
and hence
E
[
X
n
]
{\displaystyle E[X^{n}]}
is obtained.
Example where FVT holds [ edit ]
For example, for a system described by transfer function
H
(
s
)
=
6
s
+
2
,
{\displaystyle H(s)={\frac {6}{s+2}},}
the impulse response converges to
lim
t
→
∞
h
(
t
)
=
lim
s
→
0
6
s
s
+
2
=
0.
{\displaystyle \lim _{t\to \infty }h(t)=\lim _{s\to 0}{\frac {6s}{s+2}}=0.}
That is, the system returns to zero after being disturbed by a short impulse. However, the Laplace transform of the unit step response is
G
(
s
)
=
1
s
6
s
+
2
{\displaystyle G(s)={\frac {1}{s}}{\frac {6}{s+2}}}
and so the step response converges to
lim
t
→
∞
g
(
t
)
=
lim
s
→
0
s
s
6
s
+
2
=
6
2
=
3
{\displaystyle \lim _{t\to \infty }g(t)=\lim _{s\to 0}{\frac {s}{s}}{\frac {6}{s+2}}={\frac {6}{2}}=3}
So a zero-state system will follow an exponential rise to a final value of 3.
Example where FVT does not hold [ edit ]
For a system described by the transfer function
H
(
s
)
=
9
s
2
+
9
,
{\displaystyle H(s)={\frac {9}{s^{2}+9}},}
the final value theorem appears to predict the final value of the impulse response to be 0 and the final value of the step response to be 1. However, neither time-domain limit exists, and so the final value theorem predictions are not valid. In fact, both the impulse response and step response oscillate, and (in this special case) the final value theorem describes the average values around which the responses oscillate.
There are two checks performed in Control theory which confirm valid results for the Final Value Theorem:
All non-zero roots of the denominator of
H
(
s
)
{\displaystyle H(s)}
must have negative real parts.
H
(
s
)
{\displaystyle H(s)}
must not have more than one pole at the origin.
Rule 1 was not satisfied in this example, in that the roots of the denominator are
0
+
j
3
{\displaystyle 0+j3}
and
0
−
j
3
{\displaystyle 0-j3}
.
Deducing limk → ∞ f [k ] [ edit ]
Final Value Theorem [ edit ]
If
lim
k
→
∞
f
[
k
]
{\displaystyle \lim _{k\to \infty }f[k]}
exists and
lim
z
→
1
(
z
−
1
)
F
(
z
)
{\displaystyle \lim _{z\,\to \,1}{(z-1)F(z)}}
exists then
lim
k
→
∞
f
[
k
]
=
lim
z
→
1
(
z
−
1
)
F
(
z
)
{\displaystyle \lim _{k\to \infty }f[k]=\lim _{z\,\to \,1}{(z-1)F(z)}}
.[ 4] : 101
Final value of linear systems [ edit ]
Continuous-time LTI systems [ edit ]
Final value of the system
x
˙
(
t
)
=
A
x
(
t
)
+
B
u
(
t
)
{\displaystyle {\dot {\mathbf {x} }}(t)=\mathbf {A} \mathbf {x} (t)+\mathbf {B} \mathbf {u} (t)}
y
(
t
)
=
C
x
(
t
)
{\displaystyle \mathbf {y} (t)=\mathbf {C} \mathbf {x} (t)}
in response to a step input
u
(
t
)
{\displaystyle \mathbf {u} (t)}
with amplitude
R
{\displaystyle R}
is:
lim
t
→
∞
y
(
t
)
=
−
C
A
−
1
B
R
{\displaystyle \lim _{t\to \infty }\mathbf {y} (t)=-\mathbf {CA} ^{-1}\mathbf {B} R}
Sampled-data systems [ edit ]
The sampled-data system of the above continuous-time LTI system at the aperiodic sampling times
t
i
,
i
=
1
,
2
,
.
.
.
{\displaystyle t_{i},i=1,2,...}
is the discrete-time system
x
(
t
i
+
1
)
=
Φ
(
h
i
)
x
(
t
i
)
+
Γ
(
h
i
)
u
(
t
i
)
{\displaystyle {\mathbf {x} }(t_{i+1})=\mathbf {\Phi } (h_{i})\mathbf {x} (t_{i})+\mathbf {\Gamma } (h_{i})\mathbf {u} (t_{i})}
y
(
t
i
)
=
C
x
(
t
i
)
{\displaystyle \mathbf {y} (t_{i})=\mathbf {C} \mathbf {x} (t_{i})}
where
h
i
=
t
i
+
1
−
t
i
{\displaystyle h_{i}=t_{i+1}-t_{i}}
and
Φ
(
h
i
)
=
e
A
h
i
{\displaystyle \mathbf {\Phi } (h_{i})=e^{\mathbf {A} h_{i}}}
,
Γ
(
h
i
)
=
∫
0
h
i
e
A
s
d
s
{\displaystyle \mathbf {\Gamma } (h_{i})=\int _{0}^{h_{i}}e^{\mathbf {A} s}\,ds}
The final value of this system in response to a step input
u
(
t
)
{\displaystyle \mathbf {u} (t)}
with amplitude
R
{\displaystyle R}
is the same as the final value of its original continuous-time system.[ 12]
^ Wang, Ruye (2010-02-17). "Initial and Final Value Theorems" . Archived from the original on 2017-12-26. Retrieved 2011-10-21 .
^ Alan V. Oppenheim; Alan S. Willsky; S. Hamid Nawab (1997). Signals & Systems . New Jersey, USA: Prentice Hall. ISBN 0-13-814757-4 .
^ a b c Schiff, Joel L. (1999). The Laplace Transform: Theory and Applications . New York: Springer. ISBN 978-1-4757-7262-3 .
^ a b c d Graf, Urs (2004). Applied Laplace Transforms and z-Transforms for Scientists and Engineers . Basel: Birkhäuser Verlag. ISBN 3-7643-2427-9 .
^ a b c Chen, Jie; Lundberg, Kent H.; Davison, Daniel E.; Bernstein, Dennis S. (June 2007). "The Final Value Theorem Revisited - Infinite Limits and Irrational Function". IEEE Control Systems Magazine . 27 (3): 97–99. doi :10.1109/MCS.2007.365008 .
^ "Final Value Theorem of Laplace Transform" . ProofWiki . Retrieved 12 April 2020 .
^ a b c Ullrich, David C. (2018-05-26). "The tauberian final value Theorem" . Math Stack Exchange .
^ a b Sopasakis, Pantelis (2019-05-18). "A proof for the Final Value theorem using Dominated convergence theorem" . Math Stack Exchange .
^ Murthy, Kavi Rama (2019-05-07). "Alternative version of the Final Value theorem for Laplace Transform" . Math Stack Exchange .
^ Gluskin, Emanuel (1 November 2003). "Let us teach this generalization of the final-value theorem". European Journal of Physics . 24 (6): 591–597. doi :10.1088/0143-0807/24/6/005 .
^ Hew, Patrick (2020-04-22). "Final Value Theorem for function that diverges to infinity?" . Math Stack Exchange . [permanent dead link ]
^ Mohajeri, Kamran; Madadi, Ali; Tavassoli, Babak (2021). "Tracking Control with Aperiodic Sampling over Networks with Delay and Dropout". International Journal of Systems Science . 52 (10): 1987–2002. doi :10.1080/00207721.2021.1874074 .