Saturday, July 23, 2011

Impulse function notes definition


Definition

First consider a function $\delta_\Delta(t)$ defined as

\begin{displaymath}\delta_\Delta(t)\stackrel{\triangle}{=}\left\{ \begin{array}{... ... \leq t < \Delta/2$ } \\ 0 & \mbox{else} \end{array} \right. \end{displaymath}


Note that the area A underneath this function is always 1, independent of the parameter $\Delta$:

\begin{displaymath}A=\int_{-\infty}^{\infty} \delta_\Delta(t) dt =\int_{-\Delta/2}^{\Delta/2} \delta_\Delta(t) dt =\frac{1}{\Delta} \Delta=1 \end{displaymath}


When $\Delta$ approaches zero, this function $\delta_\Delta(t)$ becomes the impulse function $\delta(t)$ with 0 duration and infinite intensity:

\begin{displaymath}\delta(t)\stackrel{\triangle}{=}\lim_{\Delta \rightarrow 0} \... ...fty & \mbox{if $t=0$ } \\ 0 & \mbox{else} \end{array} \right. \end{displaymath}


which still covers a unit area or ``energy'':

\begin{displaymath}A=\int_{-\infty}^{\infty} \delta(t) dt = \int_{0^-}^{0^+} \delta(t) dt =1 \end{displaymath}


The range of integral does not have to be the entire axis ( $-\infty < t < \infty$). The integral of this function is 1 so long as its range covers the moment of the impulse, such as the infinitesimal interval 0- <t< 0+ covering the moment t=0.The impulse function $\delta(t)$ can be shifted by an arbitrary amount $\tau$ along the time axis to become $\delta(t-\tau)$ so that the impulse occurs when $t-\tau=0$ (i.e. $t=\tau$) instead of t=0, we have

\begin{displaymath}A=\int_{\tau^-}^{\tau^+} \delta(t-\tau) dt= \int_{\tau^-}^{\tau^+} \delta(t-\tau) dt =1 \end{displaymath}


where the integral range $\tau^- <t< \tau^+$ is an infinitesimal interval covering the moment $t=\tau$.The impulse function $\delta(x-x_0)$ can be physically interpreted as a one dimensional density of a unit point mass or charge located at x=x0. The infinite value of the impulse represents the density distribution, rather than the absolute amount, of the unit mass or charge. The actual amount (unity) is represented by the integration of the impulse over the range containing the point x=x0. Consequently, the dimension of the impulse function is

\begin{displaymath}[\delta(x)]=1/[x] \end{displaymath}

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