Final value theorem

In the following statements, the notation '${\displaystyle s\to 0}$' means that ${\displaystyle s}$ approaches 0, whereas '${\displaystyle s\downarrow 0}$' means that ${\displaystyle s}$ approaches 0 through the positive numbers.

Suppose that every pole of ${\displaystyle F(s)}$ is either in the open left half plane or at the origin, and that ${\displaystyle F(s)}$ has at most a single pole at the origin. Then ${\displaystyle sF(s)\to L\in \mathbb {R} }$ as ${\displaystyle s\to 0}$, and ${\displaystyle \lim _{t\to \infty }f(t)=L}$.[5]

Suppose that ${\displaystyle f(t)}$ and ${\displaystyle f'(t)}$ both have Laplace transforms that exist for all ${\displaystyle s>0}$. If ${\displaystyle \lim _{t\to \infty }f(t)}$ exists and ${\displaystyle \lim _{s\,\to \,0}{sF(s)}}$ exists then ${\displaystyle \lim _{t\to \infty }f(t)=\lim _{s\,\to \,0}{sF(s)}}$.[3]^{: Theorem 2.36}[4]^: 20[6]

Remark

Both limits must exist for the theorem to hold. For example, if ${\displaystyle f(t)=\sin(t)}$ then ${\displaystyle \lim _{t\to \infty }f(t)}$ does not exist, but ${\displaystyle \lim _{s\,\to \,0}{sF(s)}=\lim _{s\,\to \,0}{\frac {s}{s^{2}+1}}=0}$.[3]^{: Example 2.37}[4]^: 20

Suppose that ${\displaystyle f:(0,\infty )\to \mathbb {C} }$ is bounded and differentiable, and that ${\displaystyle tf'(t)}$ is also bounded on ${\displaystyle (0,\infty )}$. If ${\displaystyle sF(s)\to L\in \mathbb {C} }$ as ${\displaystyle s\to 0}$ then ${\displaystyle \lim _{t\to \infty }f(t)=L}$.[7]

Suppose that every pole of ${\displaystyle F(s)}$ is either in the open left half-plane or at the origin. Then one of the following occurs:

1. ${\displaystyle sF(s)\to L\in \mathbb {R} }$ as ${\displaystyle s\downarrow 0}$, and ${\displaystyle \lim _{t\to \infty }f(t)=L}$.
2. ${\displaystyle sF(s)\to +\infty \in \mathbb {R} }$ as ${\displaystyle s\downarrow 0}$, and ${\displaystyle f(t)\to +\infty }$ as ${\displaystyle t\to \infty }$.
3. ${\displaystyle sF(s)\to -\infty \in \mathbb {R} }$ as ${\displaystyle s\downarrow 0}$, and ${\displaystyle f(t)\to -\infty }$ as ${\displaystyle t\to \infty }$.

In particular, if ${\displaystyle s=0}$ is a multiple pole of ${\displaystyle F(s)}$ then case 2 or 3 applies (${\displaystyle f(t)\to +\infty }$ or ${\displaystyle f(t)\to -\infty }$).[5]

Suppose that ${\displaystyle f(t)}$ is Laplace transformable. Let ${\displaystyle \lambda >-1}$. If ${\displaystyle \lim _{t\to \infty }{\frac {f(t)}{t^{\lambda }}}}$ exists and ${\displaystyle \lim _{s\downarrow 0}{s^{\lambda +1}F(s)}}$ exists then

${\displaystyle \lim _{t\to \infty }{\frac {f(t)}{t^{\lambda }}}={\frac {1}{\Gamma (\lambda +1)}}\lim _{s\downarrow 0}{s^{\lambda +1}F(s)}}$

where ${\displaystyle \Gamma (x)}$ denotes the Gamma function.[5]

Final value theorems for obtaining ${\displaystyle \lim _{t\to \infty }f(t)}$ have applications in establishing the long-term stability of a system.

Suppose that ${\displaystyle f:(0,\infty )\to \mathbb {C} }$ is bounded and measurable and ${\displaystyle \lim _{t\to \infty }f(t)=\alpha \in \mathbb {C} }$. Then ${\displaystyle F(s)}$ exists for all ${\displaystyle s>0}$ and ${\displaystyle \lim _{s\,\to \,0^{+}}{sF(s)}=\alpha }$.[7]

Elementary proof[7]

Suppose for convenience that ${\displaystyle |f(t)|\leq 1}$ on ${\displaystyle (0,\infty )}$, and let ${\displaystyle \alpha =\lim _{t\to \infty }f(t)}$. Let ${\displaystyle \epsilon >0}$, and choose ${\displaystyle A}$ so that ${\displaystyle |f(t)-\alpha |<\epsilon }$ for all ${\displaystyle t>A}$. Since ${\displaystyle s\int _{0}^{\infty }e^{-st}\,dt=1}$, for every ${\displaystyle s>0}$ we have

${\displaystyle sF(s)-\alpha =s\int _{0}^{\infty }(f(t)-\alpha )e^{-st}\,dt;}$

hence

${\displaystyle |sF(s)-\alpha |\leq s\int _{0}^{A}|f(t)-\alpha |e^{-st}\,dt+s\int _{A}^{\infty }|f(t)-\alpha |e^{-st}\,dt\leq 2s\int _{0}^{A}e^{-st}\,dt+\epsilon s\int _{A}^{\infty }e^{-st}\,dt=I+II.}$

Now for every ${\displaystyle s>0}$ we have

${\displaystyle II<\epsilon s\int _{0}^{\infty }e^{-st}\,dt=\epsilon }$.

On the other hand, since ${\displaystyle A<\infty }$ is fixed it is clear that ${\displaystyle \lim _{s\to 0}I=0}$, and so ${\displaystyle |sF(s)-\alpha |<\epsilon }$ if ${\displaystyle s>0}$ is small enough.

Suppose that all of the following conditions are satisfied:

1. ${\displaystyle f:(0,\infty )\to \mathbb {C} }$ is continuously differentiable and both ${\displaystyle f}$ and ${\displaystyle f'}$ have a Laplace transform
2. ${\displaystyle f'}$ is absolutely integrable - that is, ${\displaystyle \int _{0}^{\infty }|f'(\tau )|\,d\tau }$ is finite
3. ${\displaystyle \lim _{t\to \infty }f(t)}$ exists and is finite

Then

${\displaystyle \lim _{s\to 0^{+}}sF(s)=\lim _{t\to \infty }f(t)}$.[8]

Remark

The proof uses the dominated convergence theorem.[8]

Let ${\displaystyle f:(0,\infty )\to \mathbb {C} }$ be a continuous and bounded function such that such that the following limit exists

${\displaystyle \lim _{T\to \infty }{\frac {1}{T}}\int _{0}^{T}f(t)\,dt=\alpha \in \mathbb {C} }$

Then ${\displaystyle \lim _{s\,\to \,0,\,s>0}{sF(s)}=\alpha }$.[9]

Suppose that ${\displaystyle f:[0,\infty )\to \mathbb {R} }$ is continuous and absolutely integrable in ${\displaystyle [0,\infty )}$. Suppose further that ${\displaystyle f}$ is asymptotically equal to a finite sum of periodic functions ${\displaystyle f_{\mathrm {as} }}$, that is

${\displaystyle |f(t)-f_{\mathrm {as} }(t)|<\phi (t)}$

where ${\displaystyle \phi (t)}$ is absolutely integrable in ${\displaystyle [0,\infty )}$ and vanishes at infinity. Then

${\displaystyle \lim _{s\to 0}sF(s)=\lim _{t\to \infty }{\frac {1}{t}}\int _{0}^{t}f(x)\,dx}$.[10]

Let ${\displaystyle f(t):[0,\infty )\to \mathbb {R} }$ and ${\displaystyle F(s)}$ be the Laplace transform of ${\displaystyle f(t)}$. Suppose that ${\displaystyle f(t)}$ satisfies all of the following conditions:

1. ${\displaystyle f(t)}$ is infinitely differentiable at zero
2. ${\displaystyle f^{(k)}(t)}$ has a Laplace transform for all non-negative integers ${\displaystyle k}$
3. ${\displaystyle f(t)}$ diverges to infinity as ${\displaystyle t\to \infty }$

Then ${\displaystyle sF(s)}$ diverges to infinity as ${\displaystyle s\to 0^{+}}$.[11]

Let ${\displaystyle h:[0,\infty )\to \mathbb {R} }$ be measurable and such that the (possibly improper) integral ${\displaystyle f(x):=\int _{0}^{x}h(t)\,dt}$ converges for ${\displaystyle x\to \infty }$. Then

${\displaystyle \int _{0}^{\infty }h(t)\,dt:=\lim _{x\to \infty }f(x)=\lim _{s\downarrow 0}\int _{0}^{\infty }e^{-st}h(t)\,dt.}$

This is a version of Abel's theorem.

To see this, notice that ${\displaystyle f'(t)=h(t)}$ and apply the final value theorem to ${\displaystyle f}$ after an integration by parts: For ${\displaystyle s>0}$,

${\displaystyle s\int _{0}^{\infty }e^{-st}f(t)\,dt={\Big [}-e^{-st}f(t){\Big ]}_{t=o}^{\infty }+\int _{0}^{\infty }e^{-st}f'(t)\,dt=\int _{0}^{\infty }e^{-st}h(t)\,dt.}$

By the final value theorem, the left-hand side converges to ${\displaystyle \lim _{x\to \infty }f(x)}$ for ${\displaystyle s\to 0}$.

To establish the convergence of the improper integral ${\displaystyle \lim _{x\to \infty }f(x)}$ in practice, Dirichlet's test for improper integrals is often helpful. An example is the Dirichlet integral.

Final value theorems for obtaining ${\displaystyle \lim _{s\,\to \,0}{sF(s)}}$ have applications in probability and statistics to calculate the moments of a random variable. Let ${\displaystyle R(x)}$ be cumulative distribution function of a continuous random variable ${\displaystyle X}$ and let ${\displaystyle \rho (s)}$ be the Laplace–Stieltjes transform of ${\displaystyle R(x)}$. Then the ${\displaystyle n}$-th moment of ${\displaystyle X}$ can be calculated as

${\displaystyle E[X^{n}]=(-1)^{n}\left.{\frac {d^{n}\rho (s)}{ds^{n}}}\right|_{s=0}}$

The strategy is to write

${\displaystyle {\frac {d^{n}\rho (s)}{ds^{n}}}={\mathcal {F}}{\bigl (}G_{1}(s),G_{2}(s),\dots ,G_{k}(s),\dots {\bigr )}}$

where ${\displaystyle {\mathcal {F}}(\dots )}$ is continuous and for each ${\displaystyle k}$, ${\displaystyle G_{k}(s)=sF_{k}(s)}$ for a function ${\displaystyle F_{k}(s)}$. For each ${\displaystyle k}$, put ${\displaystyle f_{k}(t)}$ as the inverse Laplace transform of ${\displaystyle F_{k}(s)}$, obtain ${\displaystyle \lim _{t\to \infty }f_{k}(t)}$, and apply a final value theorem to deduce ${\displaystyle \lim _{s\,\to \,0}{G_{k}(s)}=\lim _{s\,\to \,0}{sF_{k}(s)}=\lim _{t\to \infty }f_{k}(t)}$. Then

${\displaystyle \left.{\frac {d^{n}\rho (s)}{ds^{n}}}\right|_{s=0}={\mathcal {F}}{\Bigl (}\lim _{s\,\to \,0}G_{1}(s),\lim _{s\,\to \,0}G_{2}(s),\dots ,\lim _{s\,\to \,0}G_{k}(s),\dots {\Bigr )}}$

and hence ${\displaystyle E[X^{n}]}$ is obtained.

For example, for a system described by transfer function

${\displaystyle H(s)={\frac {6}{s+2}},}$

the impulse response converges to

${\displaystyle \lim _{t\to \infty }h(t)=\lim _{s\to 0}{\frac {6s}{s+2}}=0.}$

That is, the system returns to zero after being disturbed by a short impulse. However, the Laplace transform of the unit step response is

${\displaystyle G(s)={\frac {1}{s}}{\frac {6}{s+2}}}$

and so the step response converges to

${\displaystyle \lim _{t\to \infty }g(t)=\lim _{s\to 0}{\frac {s}{s}}{\frac {6}{s+2}}={\frac {6}{2}}=3}$

So a zero-state system will follow an exponential rise to a final value of 3.

For a system described by the transfer function

${\displaystyle H(s)={\frac {9}{s^{2}+9}},}$

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:

1. All non-zero roots of the denominator of ${\displaystyle H(s)}$ must have negative real parts.
2. ${\displaystyle H(s)}$ 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 ${\displaystyle 0+j3}$ and ${\displaystyle 0-j3}$.

If ${\displaystyle \lim _{k\to \infty }f[k]}$ exists and ${\displaystyle \lim _{z\,\to \,1}{(z-1)F(z)}}$ exists then ${\displaystyle \lim _{k\to \infty }f[k]=\lim _{z\,\to \,1}{(z-1)F(z)}}$.[4]^: 101

Final value of the system

${\displaystyle {\dot {\mathbf {x} }}(t)=\mathbf {A} \mathbf {x} (t)+\mathbf {B} \mathbf {u} (t)}$

${\displaystyle \mathbf {y} (t)=\mathbf {C} \mathbf {x} (t)}$

in response to a step input ${\displaystyle \mathbf {u} (t)}$ with amplitude ${\displaystyle R}$ is:

${\displaystyle \lim _{t\to \infty }\mathbf {y} (t)=-\mathbf {CA} ^{-1}\mathbf {B} R}$

The sampled-data system of the above continuous-time LTI system at the aperiodic sampling times ${\displaystyle t_{i},i=1,2,...}$ is the discrete-time system

${\displaystyle {\mathbf {x} }(t_{i+1})=\mathbf {\Phi } (h_{i})\mathbf {x} (t_{i})+\mathbf {\Gamma } (h_{i})\mathbf {u} (t_{i})}$

${\displaystyle \mathbf {y} (t_{i})=\mathbf {C} \mathbf {x} (t_{i})}$

where ${\displaystyle h_{i}=t_{i+1}-t_{i}}$ and

${\displaystyle \mathbf {\Phi } (h_{i})=e^{\mathbf {A} h_{i}}}$, ${\displaystyle \mathbf {\Gamma } (h_{i})=\int _{0}^{h_{i}}e^{\mathbf {A} s}\,ds}$

The final value of this system in response to a step input ${\displaystyle \mathbf {u} (t)}$ with amplitude ${\displaystyle R}$ is the same as the final value of its original continuous-time system. [12]