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PAGE 1

7.6 (continued)

\( u_c(t) \)

Figure: Graph of the unit step function u_c(t) starting at t=c.

\( \delta(t-c) \)

It turns out \[ \frac{d}{dt} u_c(t) = \delta(t-c) \]

Why?

Ramp function

As \( h \to 0 \) we get \( u_c(t) \)

Figure: Graph showing the ramp function becoming a step function as h approaches 0.

Derivative

Become \( \delta(t-c) \)

Figure: Graph showing the pulse narrowing and growing taller, approaching a delta function as h goes to 0.

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What \( f(t) \) has Laplace transform of 1?

\[ \mathcal{L} \{ \delta(t-c) \} = e^{-cs} \]

If \( c=0 \), \( \mathcal{L} \{ \delta(t) \} = 1 \) impulse at \( t=0 \)

Mass-spring-damper

\( ay'' + by' + cy = \delta(t) \) with \( y(0) = y'(0) = 0 \)

\( (as^2 + bs + c)Y = 1 \)

\( Y = \frac{1}{as^2 + bs + c} \) impulse response of the system

\[ y = \int_{0}^{t} g(t-\tau) \delta(\tau) d\tau \]

\( ay'' + by' + cy = f(t) \) with \( y(0) = y'(0) = 0 \)

\( (as^2 + bs + c)Y = F \)

\( Y = \underbrace{\frac{1}{as^2 + bs + c}}_{\text{impulse response}} F \)

Back to \( t \):

\[ y(t) = \int_{0}^{t} g(t-\tau) f(\tau) d\tau \]

Convolution

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Convolution and Impulse Responses

Convolution is treating \( f(t) \) as a bunch of impulses at different times. Collect impulse responses at various \( t \) and stack them.

Input Signal

The input signal \( f(t) \) is represented as a continuous curve sampled by discrete impulses.

Figure: Graph of input function f(t) with vertical red bars representing discrete impulses along the t-axis.

System Impulse Response

The system impulse response is denoted as \( g(t) \).

Impulse Responses at Different Times

Impulse at \( t = 0 \)

Figure: Impulse response starting at t=0.

\( t = t_1 \)

Figure: Impulse response shifted to start at t1.

\( t = t_2 \), etc.

Figure: Impulse response shifted to start at t2.

Stacking Responses

Stack them \( \rightarrow \) actual \( y(t) \)

Figure: Graph showing the summation of shifted impulse responses to form the total output y(t).

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Duhamel's Principle

\[ y = \int_{0}^{t} g(t - \tau) f(\tau) d\tau \]

\( g(t - \tau) \): impulse response
\( f(\tau) \): force input

\[ Y = G F \]

\[ Y = \frac{G}{s} \cdot Fs \]

Using some algebra and Laplace properties:

\[ y = h(t) f(0) + \int_{0}^{t} h(t - \tau) f'(\tau) d\tau \]

\( h(t) \): step response
\( f'(\tau) \): derivative of force

interchangeable \( \rightarrow \) Duhamel's principle

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2.4 Euler's Method

Numerical method to solve y' = f(t, y). Also applicable for higher order.

Basic idea: use tangent line approx. (like in Calculus I)

y(t) (unknown) but we have its slope at all (t, y):

\[ y' = f(t, y) \]

One point we know: (t_0, y_0)

Figure: Graph of unknown curve y(t) with a dashed tangent line at point (t0, y0) on t and y axes.

Tangent Line Equation

\[ y - y_0 = f(t_0, y_0)(t - t_0) \]

Slope from diff. eq.

\[ y = y_0 + f(t_0, y_0)(t - t_0) \]

If t - t_0 is "small" this should be a "good" approx.

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Figure: Diagram of iterative Euler steps showing tangent segments approximating a curve y(t).

y₁* is approx. y₁
y₂* is approx. y₂ by doing another tangent line approx.
Repeat until we reach target t

We can choose how far to move: t - t₀ = h "step size"

Example

\[ y' = 2y - 3t, \quad y(0) = 1 \]

Use step size of h = 0.25 to estimate y(0.5)

Two steps: t₀ = 0, target t = 0.5, step = 0.25

\[ t_0 = 0 \]

\[ y_0 = 1 \] (given)

\[ t_1 = t_0 + h = 0 + 0.25 = 0.25 \]

\[ y_1 = y_0 + f(t_0, y_0)(t_1 - t_0) = 1 + [2(1) - 3(0)](0.25) = 1.5 \]

Slope at previous point

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\[ t_2 = t_1 + h = 0.25 + 0.25 = 0.5 \]

Target \( t \), stop at end of this.

\[ y_2 = y_1 + f(t_1, y_1)(t_2 - t_1) = \dots = 2.0625 \text{ approx.} \]

Accuracy?

\[ y' = 2y - 3t, \quad y(0) = 1 \]

Solve \( \dots \)

\[ y = \frac{3}{4}(2t + 1) + \frac{1}{4}e^{2t} \]

\[ y(0.5) = 2.1796 \text{ (true)} \]

If \( h = 0.01 \) (50 steps)

\[ y \approx 2.1729 \]

Euler's method:

Error is proportional to \( h \).