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7.5 (Continued)

last time: \(\mathcal{L} \{u_c(t) f(t-c)\} = e^{-cs} \mathcal{L} \{f(t+c)\}\)

transform after shifting \(f(t-c)\) back to origin LEFT by \(c\)

\(t\) turns into \(t+c\)

for example, \(\mathcal{L} \{u_{10}(t) e^{-2t}\}\)

\[\begin{aligned} &= e^{-10s} \mathcal{L} \{ e^{-2(t+10)} \} \\ &= e^{-10s} e^{-20} \mathcal{L} \{ e^{-2t} \} \\ &= e^{-10s} e^{-20} \frac{1}{s+2} \end{aligned}\]

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

back to \(t\) is above in reverse / opposite

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

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for example, \(\mathcal{L}^{-1} \{e^{-\pi s} \frac{2}{s^3}\}\)

\[\mathcal{L}^{-1} \left\{ \frac{2}{s^3} \right\} = t^2\]

\(= u_{\pi}(t) \cdot (t-\pi)^2\) shifted RIGHT by \(\pi\)

full example of solving diff. eq.

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

\[f(t) = \begin{cases} 0 & 0 \le t < 3 \\ t-3 & t \ge 3 \end{cases}\]

transform:

Coordinate graph of f(t) vs t. The function is zero until t=3, where it begins a linear upward slope.

\(f(t) = u_3(t) \cdot (t-3)\)

\[\begin{aligned} s^2 Y - s y(0) - y'(0) + Y &= F(s) \\ (s^2 + 1) Y &= e^{-3s} \mathcal{L} \{ (t+3) - 3 \} \end{aligned}\]

shift LEFT by 3 : \(t \to t+3\)

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

Applying partial fraction decomposition to the term in parentheses:

\[ \frac{As+B}{s^2} + \frac{Cs+D}{s^2+1} = \frac{1}{s^2} - \frac{1}{s^2+1} \]

Preliminary Inverse Transform

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

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

Inverse transform, shift RIGHT by 3: \( t \rightarrow t-3 \)

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

Used to model a short-acting input (e.g., kicking a ball).

We can construct an impulse function from unit step functions.

Graph of a rectangular pulse function f(t) centered at the origin with width 2a and height 1/(2a).
\[ f(t) = \frac{1}{2a} [u_{-a}(t) - u_a(t)] \]

Area = 1

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

Sequence of narrowing and taller rectangular pulses, all with area 1, approaching a spike at t=0.
→
Graph of the Dirac delta function delta(t), represented as a vertical arrow at the origin.

Impulse function \( \delta(t) \)

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The Dirac Delta Function

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

Shifted Delta Function

\[\delta(t-c) = \begin{cases} +\infty & \text{if } t = c \\ 0 & \text{else} \end{cases}\]
Graph of f(t) vs t showing a vertical impulse at t=c.

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

Let's find \(\mathcal{L}\{\delta(t-c)\}\)

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

looks like unit step: \(\mathcal{L}\{u_c(t)\} = \frac{e^{-cs}}{s}\)

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

"Sampled impulse"

Graph showing a function f(t) and a vertical impulse at t=c, labeled 'overlapped portion'.
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\[ \mathcal{L} \{ \delta(t-c) f(t) \} = f(c) e^{-cs} \]

Revisit earlier example:

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

Let \( f(t) = \delta(t-3) \). The differential equation becomes:

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

Taking the Laplace transform of both sides:

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

To find the inverse Laplace transform, we note:

  • \( e^{-3s} \rightarrow u_3 \)
  • \( \mathcal{L}^{-1} \left\{ \frac{1}{s^2 + 1} \right\} = \sin(t) \)

Shift RIGHT by 3: \( t \rightarrow t-3 \)

\[ y(t) = u_3(t) \cdot \sin(t-3) \]

Note: does NOT come back to \( t \) as impulse

Because effect due to impulse doesn't go away.

(kicking a ball the ball flies forever until another input)