Download original notes
PAGE 1

7.4 (continued)

multiplication: if \( \mathcal{L} \{ h(t) \} = F(s) G(s) \), what is \( h(t) \)?

it is NOT \( f(t) g(t) \) in general

\[ \mathcal{L} \{ f(t) g(t) \} \neq F(s) G(s) \]

it is actually, the convolution integral

\[ f(t) * g(t) = \int_{0}^{t} f(\tau) g(t-\tau) d\tau = \int_{0}^{t} f(t-\tau) g(\tau) d\tau \]
\[ \mathcal{L} \{ f(t) * g(t) \} = F(s) G(s) \]

this integral has some important meanings we will see in 7.6

for now, it's an alternative to perform inverse Laplace transform

PAGE 2

for example, \( \mathcal{L}^{-1} \left\{ \frac{1}{s(s^2+1)} \right\} \)

option 1: partial fraction

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

option 2: integral property

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

option 3: convolution

\[ \mathcal{L}^{-1} \left\{ \underbrace{\frac{1}{s}}_{F(s)} \cdot \underbrace{\frac{1}{s^2+1}}_{G(s)} \right\} \]

\( f(t) = 1 \)

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

\[ = \int_{0}^{t} \underbrace{1}_{f(\tau)} \cdot \underbrace{\sin(t-\tau)}_{g(t-\tau)} d\tau = \int_{0}^{t} \underbrace{1}_{f(t-\tau)} \cdot \underbrace{\sin(\tau)}_{g(\tau)} d\tau \]\[ = -\cos(\tau) \Big|_0^t = -\cos(t) + 1 \]
PAGE 3

7.5 Discontinuous Input Functions

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

discontinuous is ok as long as it's piecewise continuous

one tool: 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 1 for t >= c.
= c." class="w-48 h-auto">

we can write many functions in terms of \( u_c(t) \)

\[ f(t) = \begin{cases} 0 & 0 \le t < 1 \\ \pi & 1 \le t < 2 \\ -10 & 2 \le t < \infty \end{cases} \]
Graph of a piecewise function with values 0, pi, and -10 over different intervals of t.
\[ = 0 + u_1(t)(\pi) + u_2(t)(-\pi - 10) \]

Switch on at \( t=1 \)

this happens at \( t=1 \)

reset to 0

what we want

PAGE 4

try this one:

\[ f(t) = \begin{cases} t & 0 \le t < 1 \\ 1 & t \ge 1 \end{cases} \]
Graph of a function that is linear (f(t)=t) for 0 <= t < 1 and constant (f(t)=1) for t >= 1.
= 1." class="w-48 h-auto">
\[ = t + u_1(t)(-t + 1) \]

reset to 0

desired level

\[ \mathcal{L} \{ u_c(t) \} = \int_{0}^{\infty} u_c(t) e^{-st} dt \quad \quad u_c = \begin{cases} 0 & t < c \\ 1 & t \ge c \end{cases} \]
\[ = \int_{0}^{c} 0 \cdot e^{-st} dt + \int_{c}^{\infty} 1 \cdot e^{-st} dt \]
\[ = -\frac{1}{s} e^{-st} \Big|_{t=c}^{t=\infty} = \frac{1}{s} e^{-sc} \]
\[ \mathcal{L} \{ u_c(t) \} = e^{-cs} \frac{1}{s} \]
PAGE 5

Laplace Transform of Delayed Functions

More general: \(\mathcal{L} \{ u_c(t) f(t-c) \}\)

Delayed activation of \(f(t)\) by \(c\)

Consider the function \(f(t) = t\). When we apply a unit step function delay, we see the same shape of \(t\) at a later time.

Coordinate graph showing f(t) equals t as a linear line passing through the origin.
Coordinate graph showing the function shifted right to start at point c on the horizontal t-axis.

Shift RIGHT by \(c\)

Derivation of the Transform

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

Where the unit step function is defined as:

\[ u_c = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases} \]

Applying the definition of \(u_c\), the integral becomes:

\[ = \int_{c}^{\infty} f(t-c) e^{-st} dt \]

Let \(\tau = t - c\), then \(t = \tau + c\) and \(d\tau = dt\). Substituting these into the integral:

\[ = \int_{0}^{\infty} f(\tau) e^{-s(\tau + c)} d\tau \]\[ = e^{-sc} \int_{0}^{\infty} f(\tau) e^{-s\tau} d\tau \]

Since \(\tau\) is a dummy variable, we can rename it to \(t\):

\[ \mathcal{L} \{ f(\tau) \} = \mathcal{L} \{ f(t) \} = F(s) \]
PAGE 6

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

Transform \(f(t-c)\) shifted back to origin: change \(t\) to \(t+c\) (LEFT shift by \(c\))

Example Calculation

Calculate the transform for a specific delayed function:

\[ \mathcal{L} \{ u_1(t) (1-t) \} = e^{-s} \mathcal{L} \{ 1 - (t+1) \} \]

Here, \(f(t-1) = 1-t\), so \(f(t) = 1-(t+1) = -t\)

\[ = e^{-s} \mathcal{L} \{ -t \} = \frac{-e^{-s}}{s^2} \]