PlanetPhysics/Laplace Transform of Dirac's Delta

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A Dirac $\delta$ symbol can be interpreted as a linear functional, i.e. a linear mapping from a [[../Bijective/|function]] space, consisting e.g. of certain real functions, to $ℝ$ (or $ℂ$), having the property

$\delta [f]=f(0).$ One may think this as the [[../NormInducedByInnerProduct/|inner product]] $⟨f,\delta ⟩=\underset{0}{\overset{\mathrm{\infty }}{\int }}f(t)\delta (t)dt$ of a function $f$ and another "function" $\delta$, when the well-known formula $\underset{0}{\overset{\mathrm{\infty }}{\int }}f(t)\delta (t)dt=f(0)$ is true.\, Applying this to\, $f(t):=e^{-st}$,\, one gets $\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}\delta (t)dt=e^{-0},$

i.e. the [[../2DLT/|Laplace transform]]

$\begin{array}{c}ℒ\{\delta (t)\}=1.\end{array}$

By the delay [[../Formula/|theorem]], this result may be generalised to $ℒ\{\delta (t-a))\}=e^{-as}.$\\

When introducing a so-called "Dirac delta function", for example

$\begin{array}{c}{\eta }_{\epsilon }(t):=\{\begin{array}{c}\frac{1}{\epsilon }\text{for}0\le t\le \epsilon ,\\ 0\text{for}t>\epsilon ,\end{array}\end{array}$

as an "approximation" of Dirac delta, we obtain the Laplace transform $ℒ\{{\eta }_{\epsilon }(t)\}=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}{\eta }_{\epsilon }(t)dt=\underset{0}{\overset{\epsilon }{\int }}\frac{e^{-st}}{\epsilon }dt+\underset{\epsilon }{\overset{\mathrm{\infty }}{\int }}{e}^{-st}\cdot 0dt=\frac{1}{\epsilon }\underset{0}{\overset{\epsilon }{\int }}{e}^{-st}dt=\frac{1-e^{-\epsilon s}}{\epsilon s}.$ As the Taylor expansion shows, we then have $\underset{\epsilon \to 0+}{\lim }ℒ\{{\eta }_{\epsilon }(t)\}=1,$ according to ref.(2).

Laplace transform of Dirac delta

The Dirac delta , $\delta$, can be correctly defined as a linear functional, i.e. a linear mapping from a function space, consisting e.g. of certain real functions, to $ℝ$ (or $ℂ$), having the property $\delta [f]=f(0).$ One may think of this as an inner product $⟨f,\delta ⟩=\underset{0}{\overset{\mathrm{\infty }}{\int }}f(t)\delta (t)dt$ of a function $f$ and another "function" $\delta$, when the well-known formula $\underset{0}{\overset{\mathrm{\infty }}{\int }}f(t)\delta (t)dt=f(0)$ holds.\, By applying this to \, $f(t):=e^{-st}$,\, one gets $\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}\delta (t)dt=e^{-0},$ i.e. the Laplace transform

$\begin{array}{c}ℒ\{\delta (t)\}=1.\end{array}$

By the delay theorem, this result may be generalised to: $ℒ\{\delta (t-a)\}=e^{-as}.$