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7.5 (continued)

Last time:

\[ \mathcal{L} \{ u_c(t) \cdot f(t-c) \} = e^{-cs} \mathcal{L} \{ f(t) \} \]
transform of \( f(t-c) \) shifted LEFT by \( c \): \( t \to t+c \)

for example,

\[ \mathcal{L} \{ u_{10}(t) e^{-2t} \} \]
shift LEFT by 10: \( t \to t+10 \)
\[ = e^{-10s} \mathcal{L} \{ e^{-2(t+10)} \} \]\[ = e^{-10s} e^{-20} \mathcal{L} \{ e^{-2t} \} \]\[ = e^{-10s} e^{-20} \frac{1}{s+2} \]

\( t \) to \( s \): shift LEFT (\( t \to t+c \)), transform, \( u_c \to e^{-cs} \)

back to \( s \) is everything opposite way

\( s \) to \( t \): \( e^{-cs} \to u_c \), inverse transform, shift RIGHT (\( t \to t-c \))

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for example,

\[ \mathcal{L}^{-1} \left\{ e^{-\pi s} \frac{2}{s^3} \right\} \]
\[ \mathcal{L}^{-1} \left\{ \frac{2}{s^3} \right\} = t^2 \] shifted RIGHT by \( \pi \)
\[ = u_{\pi}(t) \cdot (t-\pi)^2 \]

let's solve this:

\( y'' + y = f(t) \)     \( y(0) = y'(0) = 0 \)

\[ f(t) = \begin{cases} 0 & 0 \le t < 3 \\ t-3 & t \ge 3 \end{cases} \]\[ = u_3(t) \cdot (t-3) \]
Graph of f(t) which is zero for t less than 3 and a line with slope 1 starting at t equals 3.

transform both sides

shift LEFT by 3: \( t \to t+3 \)
\[ s^2 Y - s y(0) - y'(0) + Y = e^{-3s} \mathcal{L} \{ (t+3)-3 \} \]\[ s^2 Y + Y = e^{-3s} \mathcal{L} \{ t \} \]\[ (s^2 + 1) Y = e^{-3s} \frac{1}{s^2} \]
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\[ Y = e^{-3s} \left( \frac{1}{s^2(s^2+1)} \right) \rightarrow \frac{As+B}{s^2} + \frac{Cs+D}{s^2+1} = \dots = \frac{1}{s^2} - \frac{1}{s^2+1} \]

Preliminary Inverse Transform

\[ \mathcal{L}^{-1} \left\{ \frac{1}{s^2} - \frac{1}{s^2+1} \right\} = t - \sin(t) \]

back to t: \( e^{-3s} \rightarrow u_3 \)

inv. transform, then shift RIGHT by 3: \( t \rightarrow t-3 \)

\[ y(t) = u_3 \cdot \left[ (t-3) - \sin(t-3) \right] \]
\[ = \begin{cases} 0 & 0 < t < 3 \\ t-3 - \sin(t-3) & t \geq 3 \end{cases} \]
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7.6 Impulse Function

Used to model a short-acting input (e.g. hitting a baseball).

We can build this using a step up and then a step down.

Graph of a rectangular pulse function f(t) from -a to a with height 1/(2a) and area 1.
\[ f(t) = \frac{1}{2a} [u_{-a} - u_a] \]

Shrink a: \( \lim_{a \to 0} f(t) \)

Sequence of graphs showing the rectangular pulse narrowing and growing taller as a approaches 0.
Graph of the impulse function delta(t) represented as a vertical line at t=0.

impulse function \( \delta(t) \)

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The Dirac Delta Function

The Dirac delta function, denoted as \(\delta(t)\), is defined by its properties at the origin and its integral over the entire real line:

\[\delta(t) = \begin{cases} +\infty & t=0 \\ 0 & \text{else} \end{cases}\]
\[\int_{-\infty}^{\infty} \delta(t) dt = 1\]

Similarly, for a shifted delta function \(\delta(t-c)\):

\[\delta(t-c) = \begin{cases} +\infty & t=c \\ 0 & \text{else} \end{cases}\]
Graph of f(t) vs t showing a vertical spike at t=c on the horizontal axis.
Figure 1: Visual representation of the Dirac delta function spike at \(t=c\).

Laplace Transform of \(\delta(t-c)\)

Let's find \(\mathcal{L}\{\delta(t-c)\}\) by treating the delta function as a limit of rectangular pulses:

\[\begin{aligned} &= \mathcal{L} \left\{ \lim_{a \to 0} \frac{1}{2a} [u_{c-a} - u_{c+a}] \right\} \\ &= \lim_{a \to 0} \frac{1}{2a} [\mathcal{L}\{u_{c-a}\} - \mathcal{L}\{u_{c+a}\}] \\ &= \lim_{a \to 0} \frac{1}{2a} \left[ \frac{e^{-(c-a)s}}{s} - \frac{e^{-(c+a)s}}{s} \right] \\ &= \frac{1}{s} \lim_{a \to 0} \frac{1}{2a} [e^{-cs}e^{as} - e^{-cs}e^{-as}] \end{aligned}\]
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\[\begin{aligned} &= \frac{e^{-cs}}{s} \lim_{a \to 0} \frac{e^{as} - e^{-as}}{2a} \quad \text{l'Hospital's Rule} \\ &= \frac{e^{-cs}}{s} \lim_{a \to 0} \frac{se^{as} + se^{-as}}{2} \\ &= \frac{e^{-cs}}{s} \cdot \frac{2s}{2} = e^{-cs} \end{aligned}\]
\[\mathcal{L}\{\delta(t-c)\} = e^{-cs}\]

looks similar to \(\mathcal{L}\{u_c\} = \frac{e^{-cs}}{s}\)

Recall the definition of the shifted delta function:

\[\delta(t-c) = \begin{cases} \infty & t=c \\ 0 & \text{else} \end{cases}\]

When multiplied by a function \(f(t)\):

\[\delta(t-c)f(t) = \begin{cases} f(c) & t=c \\ 0 & \text{else} \end{cases}\]

\(\delta\) "samples" \(f(t)\) at \(t=c\)

Graph showing a curve f(t) and a vertical spike at t=c where the spike height is f(c).
Figure 2: Sampling property of the delta function at \(t=c\). Overlapped portion spike height is \(f(c)\).

Sampling Property Laplace Transform

\[\mathcal{L}\{\delta(t-c)f(t)\} = f(c)e^{-cs}\]
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Revisit Earlier Mass-Spring

Consider the differential equation for a mass-spring system:

\[y'' + y = f(t), \quad y(0) = y'(0) = 0\]

Where the forcing function is a Dirac delta impulse at \(t = 3\):

\[f(t) = \delta(t - 3)\]

Substituting the forcing function into the equation:

\[y'' + y = \delta(t - 3)\]

Solving using Laplace Transforms

Taking the Laplace transform of both sides:

\[s^2 Y + Y = e^{-3s}\]

Solving for \(Y\):

\[Y = e^{-3s} \frac{1}{s^2 + 1}\]

Inverse Laplace Transform

To go back to the time domain \(t\):

  • The term \(e^{-3s}\) corresponds to a unit step function shift: \(e^{-3s} \to u_3\)
  • The term \(\frac{1}{s^2 + 1}\) corresponds to a sine function: \(\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + 1} \right\} = \sin(t)\)
  • Shift RIGHT by 3: \(t \to t - 3\)
\[y(t) = u_3 \cdot \sin(t - 3)\]

Note: This is not \(\delta(t - 3)\) in the \(t\) domain.

Back to \(u_3\) because the effect is long-lasting even though the input is short-acting (a baseball continues to fly and doesn't drop to the ground right away).