Final Value Theorem

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Final Value Theorem
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Classical Control Initial 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)}


Answer:

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