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Handout Lesson 36, Impulses and Delta “Functions”

Textbook Section(s).

This lesson is based on Section 7.6 of your textbook by Edwards, Penney, and Calvis.

Impulses.

In our last lesson we used the unit step functions, \(u_a(t)\) to model on/off switches. Today, we will attempt to model sudden bursts such as lightening or hammer strikes. Consider an external force that is only applied momentarily (\(a \leq t \leq b\)) where \([a,b]\) is a very short interval. The effect of a force often depends only on the impulse \(p\) given by
\begin{gather*} p=\int_a^b f(t) \, dt \end{gather*}
Let’s construct a simple function with a unit impulse.
described in detail following the image
An axis system with a horizontal \(t\)-axis and a vertical \(y\)-axis
\(d_{a,\epsilon}(t)= \begin{cases} \\ \spc{2.5in} \\ \\ \spc{2.5in} \\ \end{cases}\)
\(\int_a^{a+\epsilon} d_{a,\epsilon}(t) \, dt = \spc{1in} \quad \text{ for all } a>0, \, \epsilon>0\)
Question What type of "function" would have an instantaneous unit impulse?
ANSWER:
\begin{gather*} \delta_a(t) = \lim_{\epsilon \rightarrow 0} d_{a,\epsilon}(t) = \begin{cases} \\ \spc{3in}\\ \\ \spc{3in}\\ \end{cases} \end{gather*}
Clearly, \(\delta_a\) is not a function in the real numbers. It is sometimes called a generalized function or a distribution.

Definition 248. Dirac delta function.

\begin{align*} \delta(t) = \delta_0(t) \defn \begin{cases} \infty \amp \text{if } t=0 \\ 0 \amp \text{if } t \neq 0 \end{cases}\\ \delta_a(t) = \delta(t-a) \defn \begin{cases} \infty \amp \text{if } t=a \\ 0 \amp \text{if } t \neq a \end{cases} \end{align*}
Justification:

\(\mathscr{L}\left\{\delta_c(t)\right\}\).

Number 17 on the table of Laplace transforms gives the following formula:
Justification:
\begin{align*} \mathscr{L}\{\delta_c(t)\} \amp = \mathscr{L}\{\delta(t-c)\}\\ \amp = \int_0^{\infty} e^{-st}\delta_c(t) \, dt\\ \amp = e^{-sc} \end{align*}

Differential equations containing \(\delta_c(t)\).

In this section, we will solve an IVP in which the differential equation contains a \(\delta\) function.

Example 251. An IVP with a \(\delta\) function.

Consider the IVP.
\begin{gather} x''+4x=\delta(t)+\delta(t-\pi) \qquad x(0)=x'(0)=0\tag{#} \end{gather}
  1. What is the meaning of the right-hand side of the differential equation in this IVP?
  2. What would a solution \(x(t)\) of (#) mean?
  3. Solve the IVP in (#).
More workspace on the next page
The graph of the solution is shown below
described in detail following the image
On \([0,\pi]\text{,}\) the graph is a sine function with period \(\pi\) and amplitude \(\frac{1}{2}\text{.}\) On \((\pi,\infty)\text{,}\) the graph is a sine function with period \(\pi\) and amplitude \(1\text{.}\)

Transfer Functions and Weight Functions.

Consider the initial value problem
\begin{gather} ax''+bx'+cx=f(t) \qquad x(0)=x'(0)=0\tag{#} \end{gather}
If we apply the Laplace transform to both sides of (#), we get
\begin{gather*} (as^2+bs+c)X(s)=F(s) \end{gather*}
so
\begin{gather} X(s) = \frac{F(s)}{as^2+bs+c} = \frac{1}{as^2+bs+c} \cdot F(s)\tag{#} \end{gather}

Definition 252.

The transfer function of the IVP (#), denoted \(W(s)\text{,}\) is given by
\begin{gather*} W(s) = \frac{1}{as^2+bs+c}. \end{gather*}
The weight function (or impulse response function), denoted \(w(t)\text{,}\) is given by
\begin{gather*} w(t)= \mathscr{L}^{-1}\{W(s)\} = \mathscr{L}^{-1}\left\{\frac{1}{as^2+bs+c}\right\}. \end{gather*}
NOTE: Your textbook uses \(\tau\) instead of \(u\text{,}\) but I think it is easier to distinguish between \(u\) and \(t\) than \(\tau\) and \(t\) when you are writing.

Example 254. Transfer and weight functions.

Find an integral formula for the solution of the IVP.
\begin{gather*} x''+6x'+9x=f(t) \qquad x(0)=x'(0)=0 \end{gather*}