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PAGE 1

6.5 (Continued)

last time: \( \delta(t - t_0) \)

\[ \mathcal{L} \{ \delta(t - t_0) \} = e^{-t_0 s} \]
Graph of the Dirac delta function as a vertical spike at t_0 on a t-axis.
\[ \int_{-\infty}^{\infty} \delta(t - t_0) dt = 1 \]
\[ \int_{-\infty}^{\infty} \delta(t - t_0) f(t) dt = f(t_0) \]

extract overlap of \( \delta(t - t_0) \) and \( f(t) \)

Graph showing a curve f(t) and a delta spike at t_0, indicating the sampling property.

from integral

Example

\( y'' + \frac{1}{2}y' + y = \delta(t - 1) \quad y'(0) = y(0) = 0 \)

Diagram of a mass-spring-damper system with m=1, k=1, and gamma=1/2.

intuitively, a damped oscillation after impulse

as \( t \to \infty, y \to 0 \) (damped)

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Laplace transform both sides

\( s^2 Y - sy(0) - y'(0) + \frac{1}{2}(sY - y(0)) + Y = e^{-s} \)

zero initial conditions

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

complete square

\[ = e^{-s} \frac{1}{s^2 + \frac{1}{2}s + \frac{1}{16} + 1 - \frac{1}{16}} \]
\[ = e^{-s} \frac{1}{(s + \frac{1}{4})^2 + \frac{15}{16}} = e^{-s} \frac{\frac{\sqrt{15}}{4}}{(s + \frac{1}{4})^2 + (\frac{\sqrt{15}}{4})^2} \cdot \frac{4}{\sqrt{15}} \]

\( \mathcal{L}^{-1} \{ \frac{b}{(s-a)^2 + b^2} \} = e^{at} \sin bt \)

\( \mathcal{L}^{-1} \{ \frac{s-a}{(s-a)^2 + b^2} \} = e^{at} \cos bt \)

\( e^{-\frac{1}{4}t} \sin(\frac{\sqrt{15}}{4}t) \)

\( y = u_1(t) \cdot \frac{4}{\sqrt{15}} e^{-\frac{1}{4}(t-1)} \sin \left( \frac{\sqrt{15}}{4}(t-1) \right) \)

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Impulse Response Analysis

\[y = \begin{cases} 0 & 0 \le t < 1 \\ \frac{4}{\sqrt{15}} e^{-\frac{1}{4}(t-1)} \sin\left(\frac{\sqrt{15}}{4}(t-1)\right) & t \ge 1 \end{cases}\]
A coordinate graph showing an underdamped sine wave starting at t=1 on the horizontal axis.

underdamped

\(\max y \approx 0.7\) at \(t \approx 2.36\)

Variation with Damping Coefficient \(\gamma = \frac{1}{4}\)

if we had used \(\gamma = \frac{1}{4}\), \(y(0) = y'(0) = 0\)

\[y'' + \frac{1}{4}y' + y = \delta(t-1)\]

\(\vdots\) (solved same way)

\[y = u_1(t) \cdot \frac{8}{\sqrt{63}} e^{-\frac{1}{8}(t-1)} \sin\left(\frac{\sqrt{63}}{8}(t-1)\right)\]

\(\max y \approx 0.8\) at \(t \approx 2.46\)

weak \(\gamma \rightarrow\) larger displacement (longer time)

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Case: No Damping (\(\gamma = 0\))

if \(\gamma = 0\), \(y'' + y = \delta(t-1)\), \(y(0) = y'(0) = 0\)

Solved the same way,

\[y = u_1(t) \sin(t-1) \quad \max y = 1 \text{ at } t = 1 + \frac{\pi}{2} \approx 2.57\]

System Stabilization

now back \(y'' + \frac{1}{2}y' + y = \delta(t-1)\), \(y(0) = y'(0) = 0\)

A graph of a decaying sine wave with an arrow pointing to a specific time t1 where the motion ceases.

apply another impulse after one period (back to equilibrium) such that motion stops. (\(y=0, t \ge t_1\))

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\( t_1 = ? \) quasi-period is \( \frac{2\pi}{\sqrt{15}/4} = \frac{8\pi}{\sqrt{15}} \) (wait this long after \( t = 1 \))

\( t_1 = 1 + \frac{8\pi}{\sqrt{15}} \)

magnitude of 2nd impulse: \( K \)

so, new eq. is

\[ y'' + \frac{1}{2}y' + y = \delta(t-1) + K \delta(t-(1+\frac{8\pi}{\sqrt{15}})) \quad y(0)=y'(0)=0 \]

(solved same way)

\[ Y = e^{-s} \frac{1}{(s+\frac{1}{4})^2 + \frac{15}{16}} + K e^{-(1+\frac{8\pi}{\sqrt{15}})s} \frac{1}{(s+\frac{1}{4})^2 + \frac{15}{16}} \]
\[ y = u_1(t) \frac{4}{\sqrt{15}} e^{-\frac{1}{4}(t-1)} \sin\left(\frac{\sqrt{15}}{4}(t-1)\right) + K u_{1+\frac{8\pi}{\sqrt{15}}}(t) e^{-\frac{1}{4}(t-1-\frac{8\pi}{\sqrt{15}})} \sin\left(\frac{\sqrt{15}}{4}(t-1-\frac{8\pi}{\sqrt{15}})\right) \]
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at \( t = 1 + \frac{8\pi}{\sqrt{15}} \) and beyond, \( y = 0 \)

\[ \sin\left(\frac{\sqrt{15}}{4}(t-1)\right) = \sin\left(\frac{\sqrt{15}}{4}(t-1-\frac{8\pi}{\sqrt{15}})\right) \]

because the two times are one quasi-period apart

so,

\[ 0 = \frac{4}{\sqrt{15}} e^{-\frac{1}{4}(\frac{8\pi}{\sqrt{15}})} + K e^{-\frac{1}{4}(t-1)} e^{\frac{2\pi}{\sqrt{15}}} \rightarrow \begin{array}{|c|} \hline K = -e^{-\frac{2\pi}{\sqrt{15}}} \\ \hline \end{array} \]
\[ y'' + \frac{1}{2}y' + y = \delta(t-1) - e^{-2\pi/\sqrt{15}} \delta(t-1-\frac{8\pi}{\sqrt{15}}) \]

\( < 1 \rightarrow \) smaller impulse because the damper already removed some energy