Initial value theorem

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

It is also known under the abbreviation IVT.

Let

 F(s) = \int_0^\infty f(t) e^{-st}\,dt

be the (one-sided) Laplace transform of ƒ(t). The initial value theorem then says[2]

\lim_{t\to 0}f(t)=\lim_{s\to\infty}{sF(s)}. \,

Proof[edit]

Based on the definition of Laplace transform of derivative we have:

sF(s)=f(0^-)+\int_{t=0^-}^{\infty}e^{-st}f^{'}(t)dt

thus:

\lim_{s \to \infty} sF(s)=\lim_{s \to \infty} [f(0^-)+\int_{t=0^-}^{\infty}e^{-st}f^{'}(t)dt]

But \lim_{s \to \infty}e^{-st} is indeterminate between t=0- to t=0+; to avoid this, the integration can be performed in two intervals:

\lim_{s \to \infty} [\int_{t=0^-}^{\infty}e^{-st}f^{'}(t)dt]
=\lim_{s \to \infty}\{\lim_{\epsilon \to 0^+}[\int_{t=0^-}^{\epsilon}e^{-st}f^{'}(t)dt] + \lim_{\epsilon \to 0^+}[\int_{t=\epsilon}^{\infty}e^{-st}f^{'}(t)dt]\}

In the first expression where 0-<t<0+, e-st=1. In the second expression, the order of integration and limit-taking can be changed. Also \lim_{s \to \infty}e^{-st}(t) where 0+<t<∞ is zero. Therefore:[3]

\begin{align}
\lim_{s \to \infty} [\int_{t=0^-}^{\infty}e^{-st}f^{'}(t)dt] &=\lim_{s \to \infty}\{\lim_{\epsilon \to 0^+}[\int_{t=0^-}^{\epsilon}f^{'}(t)dt]\} + \lim_{\epsilon \to 0^+}\{\int_{t=\epsilon}^{\infty}\lim_{s \to \infty}[e^{-st}f^{'}(t)dt]\}\\
&=f(t)|_{t=0^-}^{t=0^+} + 0\\
&= f(0^+)-f(0^-)+0\\
\end{align}

By substitution of this result in the main equation we get:

\lim_{s \to \infty} sF(s)=f(0^-)+f(0^+)-f(0^-)=f(0^+)

See also[edit]

Notes[edit]

  1. ^ http://fourier.eng.hmc.edu/e102/lectures/Laplace_Transform/node17.html
  2. ^ Robert H. Cannon, Dynamics of Physical Systems, Courier Dover Publications, 2003, page 567.
  3. ^ Robert H., Jr. Cannon (4 May 2012). Dynamics of Physical Systems. Courier Dover Publications. p. 569. ISBN 978-0-486-13969-2.