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7.4 (continued)

Multiplication of Laplace Transforms

\( \mathcal{L} \{ f(t) + g(t) \} = \mathcal{L} \{ f(t) \} + \mathcal{L} \{ g(t) \} \)

but \( \mathcal{L} \{ f(t) \} \mathcal{L} \{ g(t) \} \neq \mathcal{L} \{ f(t) g(t) \} \) in general,

it turns out

\[ \begin{aligned} \mathcal{L} \left\{ \int_{0}^{t} f(\tau) g(t-\tau) d\tau \right\} &= \mathcal{L} \left\{ \int_{0}^{t} f(t-\tau) g(\tau) d\tau \right\} \\ &= F(s) G(s) \end{aligned} \]

can show this is true by using definition and swapping order of integration

Convolution Integral

\[ \begin{aligned} \int_{0}^{t} f(\tau) g(t-\tau) d\tau &= \int_{0}^{t} f(t-\tau) g(\tau) d\tau \\ &= f(t) * g(t) \end{aligned} \]
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more to say about convolution in 7.5, 7.6

for now, it can be used as another way to do inverse transforms

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

Option 1: Partial Fraction

\[ \frac{1}{s(s^2+1)} = \frac{A}{s} + \frac{Bs+C}{s^2+1} \]

Option 2: Integral Property

\[ \mathcal{L} \left\{ \int_{0}^{t} f(\tau) d\tau \right\} = \frac{F(s)}{s} \]
\[ \mathcal{L}^{-1} \left\{ \frac{\left(\frac{1}{s^2+1}\right)}{s} \right\} = \int_{0}^{t} \text{inverse } d\tau \]

Option 3: Convolution

\[ \mathcal{L}^{-1} \left\{ \underbrace{\frac{1}{s}}_{F(s)} \underbrace{\frac{1}{s^2+1}}_{G(s)} \right\} = \int_{0}^{t} f(\tau) g(t-\tau) d\tau = \int_{0}^{t} f(t-\tau) g(\tau) d\tau \]

\( f(t) = 1 \)

\( g(t) = \sin(t) \)

\[ \begin{aligned} &= \int_{0}^{t} (1) \sin(t-\tau) d\tau = \int_{0}^{t} 1 \cdot \sin(\tau) d\tau \\ &= -\cos(\tau) \Big|_0^t \\ &= -\cos(t) + 1 \end{aligned} \]
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7.5 Discontinuous Input Functions

\[ay'' + by' + cy = f(t)\]

Note: \(f(t)\) is discontinuous but at least piecewise continuous.

To model discontinuities, the function we can use is the unit step function.

\[u_c(t) = u(t-c) = \begin{cases} 1 & \text{if } t \ge c \\ 0 & \text{else} \end{cases}\]
Graph of the unit step function u_c(t) which is 0 for t < c and jumps to 1 at t = c.

Let's use it to model:

\[f(t) = \begin{cases} 0 & 0 \le t < 1 \\ \pi & 1 \le t < 2 \\ -10 & 2 \le t < \infty \end{cases}\]
Graph of a piecewise function f(t) with steps at 0, pi, and -10 over different intervals of t.
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In terms of unit steps

\[f(t) = 0 + u_1(t)(\pi) + u_2(t)(-\pi - 10) = \pi u_1(t) - \pi u_2(t) - 10 u_2(t)\]\[= \pi(u_1 - u_2) - 10 u_2\]
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\[f(t) = \begin{cases} t & 0 \le t < 1 \\ 1 & t \ge 1 \end{cases}\]
\[= t + u_1(t)(-t + 1)\]
Graph of f(t) showing a linear increase from 0 to 1 for t between 0 and 1, then constant at 1 for t > 1.
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Laplace transform of unit step functions

\[\mathcal{L}\{u_c(t)\} = \int_0^\infty u_c(t) e^{-st} dt = \int_0^c u_c(t) e^{-st} dt + \int_c^\infty u_c(t) e^{-st} dt\]

Note: \(u_c(t) = 0\) if \(t < c\) and \(u_c(t) = 1\) if \(t \ge c\).

\[= \int_c^\infty e^{-st} dt = -\frac{1}{s} e^{-st} \Big|_c^\infty = \frac{1}{s} e^{-cs}\]
\[\mathcal{L}\{u_c(t)\} = e^{-cs} \frac{1}{s}\]
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Laplace Transform of Delayed Functions

Next, we consider the Laplace transform of a delayed function:

\[ \mathcal{L} \{ u_c(t) f(t-c) \} \]

This represents the delayed activation of \( f(t) \) by \( c \).

Two coordinate graphs: the top shows f(t)=t starting at origin; the bottom shows f(t-c) shifted right to start at c.
Visualizing the delay by \( c \)

The same picture is shifted RIGHT by \( c \) \( \rightarrow \) change \( t \) to \( t + c \).

Derivation of the Transform

\[ \begin{aligned} \mathcal{L} \{ u_c(t) f(t-c) \} &= \int_{0}^{\infty} u_c(t) f(t-c) e^{-st} dt \\ &= \int_{c}^{\infty} f(t-c) e^{-st} dt \end{aligned} \]

Let \( \tau = t - c \), then \( t = \tau + c \) and \( d\tau = dt \).

\[ \begin{aligned} &= \int_{0}^{\infty} f(\tau) e^{-s(\tau+c)} d\tau \\ &= e^{-sc} \int_{0}^{\infty} f(\tau) e^{-s\tau} d\tau \end{aligned} \]
\[ \mathcal{L} \{ u_c(t) f(t-c) \} = e^{-sc} \mathcal{L} \{ f(t) \} \]

Do NOT transform \( f(t-c) \) but transform \( f(t) \).

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Example: Applying the Shift Property

Example:

\[ \mathcal{L} \{ u_1(t) (1-t) \} \]

Shift LEFT by 1 \( \rightarrow \) \( t \) changes to \( t + 1 \)

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