Initial Value Theorem

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Initial Value Theorem
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Classical Control Final Value Theorem
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1 Introduction to Initial Value Theorem

In mathematics, the initial value theorem is a theorem used to relate frequency domain expressions to the time domain behavior as time approaches zero.[1]

The Laplace Transform of LaTeX: x\left(t\right) is

LaTeX: \mathcal{L} \left[ x\left( t \right) \right]=X\left(s\right) Definition


where

LaTeX: X\left( s \right) = \int_{0}^\infty x \left( t \right) e^{-st}\,dt Definition


be the (one-sided) Laplace transform of LaTeX: x\left(t\right). The initial value theorem then states[2]

LaTeX: \lim_{t\to 0}x\left( t \right)=\lim_{s\to\infty}{sX \left( s \right)}



2 Continuous Time form of the Initial Value Theorem

LaTeX: \lim_{t\to 0}x\left( t \right)=\lim_{s\to\infty}{sX \left( s \right)} Continuous Time



2.1 Franklin et all's version[3]

The Initial Value Theorem states that it is always possible to determine the initial vlaue of the time function LaTeX: f \left( t \right) from its Laplace Transform. Mathematically this can be stated as:

LaTeX: \lim_{s \to \infty} sF \left( s \right) = f \left( 0^{+} \right) Eqn. 3.28



2.1.1 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 Eqn. 3.29


Consider when LaTeX: s \to \infty and rewrite as

LaTeX: \int_{0^{-}}^\infty e^{-st} \frac{df \left( t \right)}{dt} dt=\int_{0^{+}}^\infty e^{-st} \frac{df \left( t \right)}{dt} dt + \int_{0^{-}}^{0^{+}} e^{-st} \frac{df \left( t \right)}{dt} dt


Taking the limit of Eqn. 3.29 as LaTeX: s \to \infty, we get

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


The 2nd term on the right side approaches 0 because LaTeX: e^{-st} \to 0. So

LaTeX: \lim_{s \to \infty} \left[ sF \left(s \right) - f \left( 0^{-} \right) = \lim_{s \to \infty} \left[ f \left(0^{+} \right) - f \left( 0^{-} \right) \right] = f \left(0^{+} \right) - f \left( 0^{-} \right)


or

LaTeX: \lim_{s \to \infty} sF \left( s \right) = f \left( 0^{+} \right)



2.1.2 Example

Find the initial value of the signal

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


Answer:

LaTeX: y \left( 0^{+} \right) = \lim_{s \to \infty}sY \left( s \right)= \lim_{s \to \infty} s\frac{3}{s \left( s-2 \right)} = s \frac{1}{s^{2}} = \frac{\infty}{\infty^{2}}= 0



3 Discrete Time form of the Initial Value Theorem

LaTeX: \lim_{t\to 0}x\left( t \right)=\lim_{z\to\infty}{X \left( z \right)}=\lim_{z\to\infty}\frac{z-1}{z}{X \left( z \right)} Discrete Time



4 References

4.1 Notes

  1. http://fourier.eng.hmc.edu/e102/lectures/Laplace_Transform/node17.html, 4/3/09
  2. Cannon, pg 567
  3. Franklin et all, pg 105