The Dirac Delta Function

The Dirac Delta FunctionSwapnil Sunil JainDecember 27, 2006

The Dirac Delta Function

Definition

The Dirac delta function can be defined as

$\delta (t)\triangleq \underset{ϵ\to 0}{lim}{\displaystyle \frac{1}{ϵ}}\mathrm{\Pi }\left({\displaystyle \frac{t}{ϵ}}\right)$

where $\mathrm{\Pi }(t)$ is the pulse function. The delta function can also be defined as

or it can also be defined as an operator s.t.

$\delta (t)[f(t)]\triangleq \underset{ϵ\to 0}{lim}{\displaystyle \frac{1}{ϵ}}{\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}}\mathrm{\Pi }\left({\displaystyle \frac{t}{ϵ}}\right)f(t)𝑑t$

Properties

$\text{1. }{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (t-a)𝑑t=f(a)$

Proof:

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (t-a)𝑑t$$=\underset{ϵ\to 0}{lim}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{a-ϵ}}f(t)\delta (t-a)𝑑t+{\displaystyle {\int }_{1-ϵ}^{a+ϵ}}f(t)\delta (t-a)𝑑t+{\displaystyle {\int }_{a+ϵ}^{\mathrm{\infty }}}f(t)\delta (t-a)𝑑t\right)$

$=\underset{ϵ\to 0}{lim}\left({\displaystyle {\int }_{a-ϵ}^{a+ϵ}}f(t)\delta (t-a)𝑑t\right)$

$=f(a)\underset{ϵ\to 0}{lim}\left({\displaystyle {\int }_{a-ϵ}^{a+ϵ}}\delta (t-a)𝑑t\right)=f(a)(1)=f(a)$

$\text{2. }{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (t)𝑑t=f(0)$

Proof: Readily seen if we set $a=0$ in Property #1.

$\text{3. }{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (at)𝑑t={\displaystyle \frac{1}{a}}f(0)$

Proof: Set $u=at$, then

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (at)𝑑t$$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f({\displaystyle \frac{u}{a}})\delta (u){\displaystyle \frac{du}{a}}={\displaystyle \frac{1}{a}}f(0)$

$\text{4. }{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (at-t_0)𝑑t={\displaystyle \frac{1}{a}}f({\displaystyle \frac{t_0}{a}})$

Proof: Set $u=at-t_0$, then

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (at-t_0)𝑑t$$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f({\displaystyle \frac{u-t_0}{a}})\delta (u){\displaystyle \frac{du}{a}}={\displaystyle \frac{1}{a}}f({\displaystyle \frac{t_0}{a}})$

$\text{5. }{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)\delta (g(t))𝑑t={\displaystyle \sum _{\{x_i|g(x_i)=\mathrm{\hspace{0.33em}0}\}}}{\displaystyle \frac{1}{|g^{\prime }(x_i)|}}f(x_i)$

$\text{6. }{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t){\delta }^{\prime }(t)𝑑t=-f^{\prime }(0)$

Proof: If we let $u=f(t)$ and $dv={\delta }^{\prime }(t)dt$, then, using integration by parts,

${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t){\delta }^{\prime }(t)𝑑t$$={f(t)\delta (t)|}_{-\mathrm{\infty }}^{\mathrm{\infty }}-{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{f}^{\prime }(t)\delta (t)𝑑t=-f^{\prime }(0)$