view 'Laplace transform of Dirac's delta distribution
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2010-10-06 22:12:45
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bci1
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A Dirac $\delta$ symbol can be interpreted as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to $\mathbb{R}$ (or $\mathbb{C}$), having the property
$$\delta[f] \;=\; f(0).$$
One may think this as the inner product
$$\langle f,\,\delta\rangle \;=\; \int_0^\infty\!f(t)\delta(t)\,dt$$
of a function $f$ and another ``function'' $\delta$, when the well-known \PMlinkescapetext{formula}
$$\int_0^\infty\!f(t)\delta(t)\,dt \;=\; f(0)$$
is true.\, Applying this to\, $f(t) := e^{-st}$,\, one gets
$$\int_0^\infty\!e^{-st}\delta(t)\,dt \;=\; e^{-0},$$
i.e. the Laplace transform
\begin{align}
\mathcal{L}\{\delta(t)\} \;=\; 1.
\end{align}
By the delay theorem, this result may be generalised to
$$\mathcal{L}\{\delta(t\!-\!a))\} \;=\; e^{-as}.$$\\
When introducing a so-called ``Dirac delta function'', for example
\begin{align*}
\eta_\varepsilon(t) \;:=\;
\begin{cases}
\frac{1}{\varepsilon} \quad \mbox{for}\;\; 0 \le t \le \varepsilon,\\
0 \quad \mbox{for} \qquad t > \varepsilon,
\end{cases}
\end{align*}
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2010-03-05 13:32:27
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bci1
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% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
taken out from preamble
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
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