Final Value Theorem
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## 1 Introduction to Final Value Theorem[1]

The Final Value Theorem allows the evaluation of the steady-state value of a time function from its Laplace transform. The final value theorem is only valid if $LaTeX: X \left( s \right)$ is stable (all poles are in th left half plane). If all poles are in the left half plane then

 $LaTeX: \lim_{t \to \infty} y \left( t \right) = \lim_{s \to 0} sY \left( s \right)$ 3.26

## 2 Continuous Time Final Value Theorem[2]

 $LaTeX: \lim_{t \to \infty} y \left( t \right) = \lim_{s \to 0} sY \left( s \right)$ 3.26

### 2.1 Proof

 $LaTeX: \mathcal{L} \left \{ \frac{df}{dt} \right \}=sF \left(s \right) - f \left( 0^{-} \right)=\int_{0^{-}}^\infty e^{-st} \frac{df}{dt} dt$

Consider when $LaTeX: s \to 0$ and rewrite as

 $LaTeX: \lim_{s \to 0} \left[sF \left(s \right) - f \left( 0^{-} \right) \right] = lim_{s \to 0} \left(\int_{0}^\infty e^{-st} \frac{df}{dt} dt \right) = \lim_{t \to \infty} \left[ f \left( t \right) - f \left( 0 \right) \right]$

Then

 $LaTeX: \lim_{t \to \infty} f \left( t \right) = \lim_{s \to 0} sF \left( s \right)$ Result 1

Partial fractions can be used to see this another way

 $LaTeX: F \left( s \right) = \frac{C_{1}}{s-p_{1}}+\frac{C_{2}}{s-p_{2}}+ \cdots + \frac{C_{n}}{s-p_{n}}$

If $LaTeX: p_{1}=0$ and all other $LaTeX: p_{i}<0$ then $LaTeX: C_{1}$ becomes the steady-state value of $LaTeX: f \left( t \right)$ and

 $LaTeX: C_{1}=\lim_{t \to \infty} f \left( t \right) = sF \left(s \right) |_{s=0}$ Result 2

As you can see Result 1 is the same as Result 2.

### 2.2 Example

#### 2.2.1 The Right Way

Find the steady-state value of the system

 $LaTeX: Y \left( s \right) = \frac{3 \left( s+2 \right)}{s \left(s^2+2s+10 \right)}$

Applying the final value theorem:

 $LaTeX: y \left( \infty \right) = sY \left( s \right) |_{s=0}=\frac{3*2}{10}=0.6$

#### 2.2.2 The Wrong Way

Find the final value of the signal

 $LaTeX: Y \left( s \right) = \frac{3}{s \left( s-2 \right)}$

 $LaTeX: y \left( \infty \right) = sY\left( s \right) |_{s=0}=-\frac{3}{2}$

However,

 $LaTeX: y \left( t \right) = \left( -\frac{3}{2} + \frac{3}{2}e^{2t} \right) 1 \left( t \right)$

Remember that the final value theorem is only valid when all the poles are in the left half plane ($LaTeX: p \le 0$). Clearly this system has a pole at +2 and therefore the final value theorem does not apply.

## 3 DiscreteTime Final Value Theorem

 $LaTeX: \lim_{t \to \infty} x \left( t \right) = \lim_{s \to 0} sX \left( s \right) = \lim_{z \to 1} \left( z-1 \right) X \left( z \right)$

## 4 References

• Franklin, G. F., Emami-Naeini, A., and Powell, J. D. 1993 Feedback Control of Dynamic Systems. 3rd. Addison-Wesley Longman Publishing Co., Inc. ISBN 0201527472

### 4.1 Notes

1. Franklin et all, pp. 102-105
2. Franklin et all, pp. 102-105