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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.

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

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

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

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

• http://fourier.eng.hmc.edu/e102/lectures/Laplace_Transform/node17.html
• Robert H. Cannon, Dynamics of Physical Systems, Courier Dover Publications, 2003
• 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