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6.4 (continued)

Non-sinusoid periodic force

\[ \mathcal{L}\{u_c(t)\} = e^{-cs} \frac{1}{s} \]
\[ y'' + y = f(t) \quad y(0) = y'(0) = 0 \]
\[ f(t) = u_0(t) + 2 \sum_{k=1}^{\infty} (-1)^k u_{k\pi}(t) \] \[ = u_0(t) - 2u_{\pi}(t) + 2u_{2\pi}(t) - 2u_{3\pi}(t) + 2u_{4\pi}(t) - \dots \]
Graph of a square wave function f(t) alternating between 1 and -1 with period 2 pi.
Diagram of a mass-spring system with mass m=1 and spring constant k=1, y positive downwards.

Laplace transform of \( y'' + y = u_0(t) - 2u_{\pi}(t) + 2u_{2\pi}(t) - 2u_{3\pi}(t) + \dots \)

\[ s^2 Y - s y(0) - y'(0) + Y = \frac{1}{s} - 2e^{-\pi s} \frac{1}{s} + 2e^{-2\pi s} \frac{1}{s} - 2e^{-3\pi s} \frac{1}{s} + \dots \]

Note: \( y(0) = 0 \) and \( y'(0) = 0 \) (given).

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

Partial Fraction Decomposition

\[ \frac{1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1} \] \[ 1 = A(s^2+1) + (Bs+C)(s) \] \[ 1 = (A+B)s^2 + Cs + A \]
\[ A = 1, \quad B = -1, \quad C = 0 \]
\[ \frac{1}{s(s^2+1)} = \frac{1}{s} - \frac{s}{s^2+1} \]

\( \downarrow \) transforms to \( 1 \) and \( \cos(t) \)

Back to \( t \) domain

\[ y = 1 - \cos(t) - 2u_{\pi}(t)(1 - \cos(t-\pi)) + 2u_{2\pi}(t)(1 - \cos(t-2\pi)) \] \[ - 2u_{3\pi}(t)(1 - \cos(t-3\pi)) + 2u_{4\pi}(t)(1 - \cos(t-4\pi)) - \dots \]

Using trigonometric identities: \( \cos(t-\pi) = -\cos(t) \) and \( \cos(t-2\pi) = \cos(t) \)

\[ y = 1 - \cos(t) - 2u_{\pi}(t)(1 + \cos(t)) + 2u_{2\pi}(t)(1 - \cos(t)) - 2u_{3\pi}(t)(1 + \cos(t)) \] \[ + 2u_{4\pi}(t)(1 - \cos(t)) - \dots \]
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Piecewise Definition of the Solution

\[ y = \begin{cases} 1 - \cos(t) & 0 \le t < \pi \\ -1 - 3\cos(t) & \pi \le t < 2\pi \\ 1 - 5\cos(t) & 2\pi \le t < 3\pi \\ -1 - 7\cos(t) & 3\pi \le t < 4\pi \\ 1 - 9\cos(t) & 4\pi \le t < 5\pi \\ \vdots \end{cases} = (-1)^n - (2n+1)\cos(t) \quad \begin{matrix} n\pi \le t < (n+1)\pi \\ n = 0, 1, 2, 3, \dots \end{matrix} \]

As \(t\) increases, the magnitude of \(\cos(t)\) increases.

Hand-drawn sketch of a wave with increasing amplitude over time on a y vs t coordinate system.

Resonance Analysis

Resonance occurs because the:

  • Fundamental frequency \(\sqrt{\frac{k}{m}} = 1\)
  • Input force frequency is:
\[ \text{period} = \frac{2\pi}{\text{freq}} \] \[ 2\pi = \frac{2\pi}{\text{freq}} \rightarrow \text{freq} = 1 \]
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Numerical Plot of the Resonant System

The following graph illustrates the numerical solution for the system, showing the characteristic linear growth in amplitude over time due to resonance.

A computer-generated plot showing an oscillating wave with amplitude linearly increasing from 0 to 30 over 50 units of time.
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Oscillatory Response Visualization

The following graph illustrates a system response characterized by an oscillating signal with increasing amplitude over time, superimposed on a square wave reference signal.

A graph showing a blue oscillating curve with increasing amplitude and an orange square wave on a coordinate system.
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6.5 Impulse Functions

very short acting input

A simple coordinate graph showing a vertical arrow at t_0 representing an impulse function.

\(\delta(t - t_0)\)

Dirac Delta Function

or

impulse function

(not a true function \(\rightarrow\) generalized function)

Building the Impulse Function

build this with step up and step down

A rectangular pulse function graph centered at t_0 with width 2\tau and height 1/(2\tau), area equals 1.

shrink \(\tau\)

A series of narrowing and taller rectangular pulses converging to a vertical arrow at t_0.

\(\rightarrow \delta(t - t_0)\)

\[ \int_{-\infty}^{\infty} \delta(t - t_0) dt = 1 \]
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Laplace Transform of the Dirac Delta Function

\[ \mathcal{L} \{ \delta(t-t_0) \} = ? \]

We approximate the delta function using a rectangular pulse and take the limit as the pulse width goes to zero:

\[ \begin{aligned} &\mathcal{L} \left\{ \frac{1}{2\tau} u_{t_0-\tau} - \frac{1}{2\tau} u_{t_0+\tau} \right\} \\ &= \frac{1}{2\tau} \left[ \mathcal{L} \{ u_{t_0-\tau} \} - \mathcal{L} \{ u_{t_0+\tau} \} \right] = \frac{1}{2\tau} \left( \frac{e^{-(t_0-\tau)s}}{s} - \frac{e^{-(t_0+\tau)s}}{s} \right) \\ \\ &= \frac{1}{2\tau} \frac{e^{-t_0s}}{s} \left( e^{\tau s} - e^{-\tau s} \right) = \frac{1}{s} e^{-t_0s} \left( \frac{e^{\tau s} - e^{-\tau s}}{2\tau} \right) \end{aligned} \]

Now, we take the limit as \(\tau \to 0\):

\[ \lim_{\tau \to 0} \frac{e^{-t_0s}}{s} \left( \frac{e^{\tau s} - e^{-\tau s}}{2\tau} \right) \to \frac{0}{0} \text{ l'Hospital's Rule!} \]
\[ \begin{aligned} &= \frac{e^{-t_0s}}{s} \lim_{\tau \to 0} \left( \frac{s e^{\tau s} + s e^{-\tau s}}{2} \right) = e^{-t_0s} \lim_{\tau \to 0} \left( \frac{s e^{\tau s} + s e^{-\tau s}}{2s} \right) \\ &= e^{-t_0s} \left( \frac{2s}{2s} \right) = e^{-t_0s} \end{aligned} \]
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\[ \mathcal{L} \{ \delta(t-t_0) \} = e^{-t_0s} \]
\[ \mathcal{L} \{ u_c(t) \} = e^{-cs} \frac{1}{s} \]

Example: Mass-Spring System with Impulse

Let's try solving the differential equation:

\[ y'' + y = \delta(t-\pi), \quad y(0)=1, \, y'(0)=0 \]
A mass-spring system diagram showing a spring with constant k=1 attached to a mass m=1.
\[ s^2 Y - s y(0) - y'(0) + Y = e^{-\pi s} \]
\[ (s^2 + 1) Y = s + e^{-\pi s} \]
\[ Y = \frac{s}{s^2+1} + e^{-\pi s} \frac{1}{s^2+1} \]

Note: \(\frac{s}{s^2+1}\) corresponds to \(\cos(t)\) and \(\frac{1}{s^2+1}\) corresponds to \(\sin(t)\).

Back to \(t\):

\[ y = \cos(t) + u_{\pi}(t) (\sin(t-\pi)) \]

NOT a \(\delta(t-\pi)\)!

Input is short but effect is not (think of hitting a baseball).