We got a joke! Heaviside (British electrician) was told during a restaurant that mathematicians are finally using his formula, he replied: I don't need the chef to tell me the food was good. (a lil bit cynical!)
(Heaviside) Unit step function
The Heaviside step function is defined as:
$u(t-a)=\begin{cases}0,\quad t<a \\\\1,\quad a\leq t\end{cases}$
graph it, it just "switches on" when $t=a$
If we use laplace transforms we will see that solving equations with this Heaviside function is very natural. Using mouc wont work, voparam might work but it's messy.
Let's derive some fundamental LT properties involving the unit step function:
$$\mathcal{L}\{u(t-a)f(t-a)\}$$
$\int _{0} ^\infty e^{-st}u(t-a)f(t-a)\, dt$
$=\int _{0}^a 0 \, dt+\int _{a}^\infty e^{-st}\cdot 1f(t-a)\, dt$
let $x=t-a$
$=\int _{0} ^\infty e^{-s(x+a)}f(x)\, dx$
$=e^{-as}\int _{0} ^\infty e^{-sx}f(x)\, dx$
the right side is nothing but the LT of f. How nice!
$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)\quad\leftarrow\text{Very useful formula!}$$
Same process can be done to show the following:
$$\mathcal{L}\{u(t-a)f(t)\}=e^{-as}\mathcal{L}\{f(t+a)\} \leftarrow\text{Also useful}$$
We can express a case-function in terms of unit step functions.
Consider this function where multiple functions are being switched:
$f(t)=\begin{cases} 1, \quad t<1 \\\\ t, \quad 1\leq t\leq \pi \\\\ \sin(t), \quad \pi\leq t\end{cases}$
We express it using heaviside functions:
$f(t)=1-u(t-1)+tu(t-1)-tu(t-\pi)+\sin(t)u(t-\pi)$
think about it switching on and off certain parts of the equation at certain times t.
current in electric circuit $I(t)$ is defined by:
$$I''+I=g(t),\quad I(0)=I'(0)=0$$
where:
$$g(t)=\begin{cases}1,\quad 0\leq t<\pi \\\\ -1,\quad \pi\leq t\end{cases}$$
find $I(t)$
The capacitance here is huge, 1F! (I believe it comes from the coefficient of $y$ term)
and we are switching it instantaneously, lets imagine we don't blow anything up.
We can express $g$ using just one heaviside unit step function:
$I''+I=1-2u(t-\pi)$
$\mathcal{L}\{I(t)\}=J(s)$
$s\cdot I(0)=0 \qquad I'(0)=0$
Hit it with the LT!
$s^2J(s)+J=\frac{1}{s}-\mathcal{L}\{2u(t-\pi)\}$
using formula we derived earlier, ie: $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$
$s^2J(s)+J=\frac{1}{s}-e^{-\pi s}\frac{2}{s}$
$J=\underbrace{ \frac{1}{s(s^2+1)} }_{ =F(s) }- \frac{2}{s(s^2+1)}e^{-\pi s}$
$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^2+1}\}$ using partial fractions
$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=1-\cos t$
$J(s)=F(s)-2e^{-\pi s}F(s)$
Take the inverse LT of both sides:
$I(s)=1-\cos t-2\mathcal{L}^{-1}\{e^{-\pi s}F(s)\}$
How do we take the inverse of $e^{-\pi s}F(s)$?
Recall that we derived: $\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$
So: $u(t-a)f(t-a)=\mathcal{L}^{-1}\{e^{-as}F(s)\}$
Applying this formula gives:
$I(t)=1-\cos(t)-2u(t-\pi)(1-\cos(t-\pi))$
$$I(t)=1-\cos(t)-2u(t-\pi)(1+\cos(t))$$
end of lec 18