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PAGE 1

7.6 (continued)

\( u_c(t) \)

Figure: Graph of a unit step function starting at t=c with height 1.

\( \delta(t-c) \)

Figure: Graph of a Dirac delta function represented as a vertical arrow at t=c.

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

why?

approx. \( u_c(t) \) as a ramp

Figure: Graph of a ramp function starting at t=c and reaching 1 at t=c+h.

\( h \to 0 \)

\( \longrightarrow \)

Figure: Graph showing the ramp function approaching a step function as h goes to 0.

deriv.

\( h \to 0 \)

\( \longrightarrow \)

Figure: Graph showing the rectangular pulse approaching a delta function at t=c as h goes to 0.

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

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

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

mass-spring-damper

\[ ay'' + by' + cy = \delta(t) \quad y(0) = y'(0) = 0 \]\[ (as^2 + bs + c)Y = 1 \]\[ Y = \underbrace{\frac{1}{as^2 + bs + c}}_{\text{impulse response}} \quad y(t) = \mathcal{L}^{-1} \left\{ \frac{1}{as^2 + bs + c} \right\} = g(t) \]

\[ ay'' + by' + cy = f(t) \quad y(0) = y'(0) = 0 \]\[ (as^2 + bs + c)Y = F \]\[ Y = \underbrace{\frac{1}{as^2 + bs + c}}_{\text{imp. response}} \cdot F \]

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Convolution

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

Note: \(g(t-\tau)\) is the shifted impulse response.

What is this doing?

Sampling \(f(t)\) as a bunch of impulses at different times.
Find impulse responses then stack them.

\(f(t)\): infinitely many impulses under the curve.

Figure: Graph of force f(t) showing a continuous curve sampled by discrete vertical impulse lines at t1, t2, and t3.

The following graphs illustrate the individual responses at different time steps:

Figure: Response y(t) starting at t=0, showing a decaying oscillatory wave.

Figure: Response y(t) starting at t=t1, showing the same decaying wave shifted to the right.

Figure: Response y(t) starting at t=t2, showing the wave shifted further to the right.

Convolution superimposes all onto one function of \(t\).

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

Note: \(g\) is the impulse response.

Do some algebra in \(s\) domain and using properties of Laplace:

Duhamel's Principle

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

Where:

\(h(t)\) is the step response (response due to step input).
\(f'(\tau)\) is the derivative of force.

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2.4 Euler's Method

Numerical method to solve \( y' = f(t, y) \), also applicable to higher-order equations.

Basic idea: tangent line approx.

\( y(t) \) (unknown). We know its slope \( y' = f(t, y) \) at \( (t, y) \).

Suppose we know one point (initial condition) \( (t_0, y_0) \).

Figure: Coordinate graph with t and y(t) axes showing a curve y(t) and a dashed tangent line at point (t0, y0).

Tangent line at \( (t_0, y_0) \)

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

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

\[ \approx \text{true } y(t) \]

→ Move a little bit on the tangent line → \( y(t_1) = y_1 \)

Repeat as needed

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Figure: A sketch showing a true curve y and two segments of tangent lines approximating it at points t0, t1, and t2.

\( y_1 \) approx. of true \( y_1 \)

Typically we choose a fixed step size \( h \) : \( t_n = t_{n-1} + h \)

Example

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

Use \( h = 0.25 \) to estimate \( y(0.5) \)

\( t_0 = 0 \), \( y_0 = 1 \) (Given)

\( t_1 = t_0 + h = 0 + 0.25 = 0.25 \)

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

Note: slope at prev. point

\( t_2 = t_1 + h = 0.25 + 0.25 = 0.5 \)

\( y_2 = y_1 + f(t_1, y_1)(t_2 - t_1) = \dots = 2.0625 \)

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Accuracy of Numerical Solutions

Accuracy? We can actually solve the differential equation:

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

The exact solution is:

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

The true value at \(t = 0.5\) is:

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

Improving Accuracy

To improve, use a smaller step size \(h\) (more steps).

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

\[y(0.5) = 2.1729\]