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7.5 The Dirac Delta Function: Class Notes

7.5a

Consider a function that represents a short burst of constant energy. Start with two unit step functions, with $0<a<b$:

Two unit step functions on the same axes. $𝒰(t-a)$ is 0 for $0\le t<a$ and jumps to 1 at $t=a$. $𝒰(t-b)$ is 0 for $0\le t<b$ and jumps to 1 at $t=b$.

$$𝒰(t-a)=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<a}\hfill \\ {\displaystyle 1,}\hfill & {\displaystyle t\ge a}\hfill \end{array}$$

$$𝒰(t-b)=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<b}\hfill \\ {\displaystyle 1,}\hfill & {\displaystyle t\ge b}\hfill \end{array}$$

Subtracting them turns the function on at $t=a$ and off again at $t=b$:

$$𝒰(t-a)-𝒰(t-b)=\{\begin{array}{cc}{\displaystyle 0-0,}\hfill & {\displaystyle 0\le t<a}\hfill \\ {\displaystyle 1-0,}\hfill & {\displaystyle a\le t<b}\hfill \\ {\displaystyle 1-1,}\hfill & {\displaystyle t\ge b}\hfill \end{array}=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<a}\hfill \\ {\displaystyle 1,}\hfill & {\displaystyle a\le t<b}\hfill \\ {\displaystyle 0,}\hfill & {\displaystyle t\ge b}\hfill \end{array}$$

Now center a pulse like this at $t_0$, give it width $2a$, and give it height ${\displaystyle \frac{1}{2a}}$, so that its area is $A=1$:

A rectangular pulse: the graph is 0, then has height ${\displaystyle \frac{1}{2a}}$ for $t_0-a\le t<t_0+a$, then is 0 again. The rectangle has width $2a$ and area $A=1$.

Definition

The unit impulse function is defined as

$${\displaystyle {\delta }_a(t-t_0)=\{\begin{array}{cc}{\displaystyle 0,}\hfill & {\displaystyle 0\le t<t_0-a}\hfill \\ {\displaystyle \frac{1}{2a},}\hfill & {\displaystyle t_0-a\le t<t_0+a}\hfill \\ {\displaystyle 0,}\hfill & {\displaystyle t\ge t_0+a}\hfill \end{array}}$$

If $a$ is small, this represents a large amplitude for a short time around $t_0$. Note that ${\displaystyle {\int }_0^{\infty }{\delta }_a(t-t_0)dt=1}$, since the area of the rectangle is ${\displaystyle 2a·\frac{1}{2a}=1}$ no matter what $a$ is.

Three rectangular pulses ${\delta }_a(t-t_0)$ side by side, shaded (in magenta). From left to right, the width $2a$ gets smaller and the height ${\displaystyle \frac{1}{2a}}$ gets larger, while the shaded area stays equal to 1.

Definition

The Dirac delta “function” is ${\displaystyle \delta (t-t_0)=\underset{a\to 0}{lim}{\delta }_a(t-t_0)}$.

It possesses the properties

$$\text{i) }\delta (t-t_0)=\{\begin{array}{cc}{\displaystyle \infty ,}\hfill & {\displaystyle t=t_0}\hfill \\ {\displaystyle 0,}\hfill & {\displaystyle t\ne t_0}\hfill \end{array}$$

$${\displaystyle \text{ii) }{\int }_0^{\infty }\delta (t-t_0)dt=1}$$

Theorem: Transform of the Dirac Delta Function

For $t_0>0$, $ℒ\{\delta (t-t_0)\}=e^{-s{t}_0}$.

Proof

Write the pulse as a difference of two unit step functions, one turning on at $t_0-a$ and one at $t_0+a$:

$${\displaystyle {\delta }_a(t-t_0)=\frac{1}{2a}\left[𝒰\right(t-(t_0-a))-𝒰(t-(t_0+a)\left)\right]}$$

Transform each step function:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{{\delta }_a(t-t_0)\}}& {\displaystyle =\frac{1}{2a}[\frac{e^{-s(t_0-a)}}{s}-\frac{e^{-s(t_0+a)}}{s}]}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-s{t}_0}\left(\frac{e^{as}-e^{-as}}{2as}\right)}\hfill \end{array}}$$

Now let $a\to 0$. The limit has the form $\frac{0}{0}$, so use L’Hôpital’s rule, differentiating with respect to $a$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle ℒ\{\delta (t-t_0)\}}& {\displaystyle =ℒ\{\underset{a\to 0}{lim}{\delta }_a(t-t_0)\}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-s{t}_0}\underset{a\to 0}{lim}\left(\frac{e^{as}-e^{-as}}{2as}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-s{t}_0}\underset{a\to 0}{lim}\frac{s{e}^{as}+s{e}^{-as}}{2s}}\hfill \\ \hfill {\displaystyle }& {\displaystyle =e^{-s{t}_0}}\hfill \end{array}}$$

Note that when $t_0=0$, we have $ℒ\{\delta (t)\}=1$.

This is $e^{-s(0)}=1$.

7.5b

Example

Use the Laplace transform to solve the given initial-value problem.

$$y″+y=\delta (t-2\pi )+\delta (t-4\pi ),y(0)=1,y\prime (0)=0$$

Solution

Transform each term of the equation, using $y(0)=1$ and $y\prime (0)=0$:

$$s^{2}Y(s)-s+Y(s)=e^{-2\pi s}+e^{-4\pi s}$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle (s^2+1)Y(s)}& {\displaystyle =s+e^{-2\pi s}+e^{-4\pi s}}\hfill \\ \hfill {\displaystyle Y(s)}& {\displaystyle =\frac{s}{s^2+1}+\frac{e^{-2\pi s}}{s^2+1}+\frac{e^{-4\pi s}}{s^2+1}}\hfill \end{array}}$$

Take the inverse transform. Each exponential shifts $\sin t$ and turns it on at that time:

$$y(t)=\cos t+𝒰(t-2\pi )\sin (t-2\pi )+𝒰(t-4\pi )\sin (t-4\pi )$$

Since sine has period $2\pi$, $\sin (t-2\pi )=\sin t=\sin (t-4\pi )$.

$$y″-7y\prime +6y=e^t+\delta (t-2)+\delta (t-4),y(0)=0,y\prime (0)=0$$

Solution

Transform each term; both initial values are 0:

$${\displaystyle s^{2}Y(s)-7sY(s)+6Y(s)=\frac{1}{s-1}+e^{-2s}+e^{-4s}}$$

Since $s^2-7s+6=(s-1)(s-6)$,

$${\displaystyle Y(s)=\frac{1}{(s-1{)}^2(s-6)}+\frac{e^{-2s}}{(s-1)(s-6)}+\frac{e^{-4s}}{(s-1)(s-6)}}$$

Two partial fraction decompositions are needed: one for the first term, and one that works for both of the last two terms.

$${\displaystyle \frac{1}{(s-1{)}^2(s-6)}=\frac{A}{s-1}+\frac{B}{(s-1{)}^2}+\frac{C}{s-6}\Rightarrow A=-\frac{1}{25},B=-\frac{1}{5},C=\frac{1}{25}}$$

$${\displaystyle \frac{1}{(s-1)(s-6)}=\frac{A}{s-1}+\frac{B}{s-6}\Rightarrow A=-\frac{1}{5},B=\frac{1}{5}}$$

So

$${\displaystyle Y(s)=-\frac{1/25}{s-1}-\frac{1/5}{(s-1{)}^2}+\frac{1/25}{s-6}+e^{-2s}(\frac{1/5}{s-6}-\frac{1/5}{s-1})+e^{-4s}(\frac{1/5}{s-6}-\frac{1/5}{s-1})}$$

Take the inverse transform. Each exponential becomes a unit step, and $t$ is replaced by $t-2$ or $t-4$ in the function it multiplies: