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# (Heaviside) Unit step function
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!)
Anyways, 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.
#ex
compute:
$\mathcal{L}\{u(t-a)f(t-a)\}$
$\int _{0} ^\infty e^{-st}u(t-a)f(t-a)\, dt$
$=\int _{a}^\infty e^{-st}\cdot 1f(t-a)\, dt$
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!
$$\mathcal{L}\{u(t-a)f(t-a)\}=e^{-as}F(s)$$
Very useful formula! ^
$\mathcal{L}\{u(t-a)f(t)\}=\mathcal{L}\{u(t-a)f(t+a-a)\}=e^{-as}\mathcal{L}\{f(t+a)\}$
$\mathcal{L}^{-1}\{e^{-as}F(s)\}=u(t-a)f(t-a)$
#ex
multiple functions being switched:
$f(t)=\begin{cases} 1, \quad t<1 \\ t, \quad 1\leq t\leq \pi \\ \sin(t), \quad \pi\leq t\end{cases}$
then we express it using heaviside functions:
$f(t)=1-u(t-1)+tu(t-1)-tu(t-\pi)+\sin(t)u(t-\pi)$
think it about switching on certain parts of the equation at certain times t.
#ex
current in electric circuit I(t) is defined by:
$I''+I=g(t),\quad I(0)=I'(0)=0$
$g(t)=\begin{cases}1,\quad 0\leq t<\pi \\ -1,\quad \pi\leq t\end{cases}$
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.
$I''+I=1-2u(t-\pi)$
$\mathcal{L}\{I(t)\}=J(s)$
$s\cdot I(0)=s\cdot I'(0)=0$
using formula we derived in earlier ex:
$s^2J(s)+J=\frac{1}{s}-\frac{2e^{-\pi s}}{s}$
$J=\underbrace{ \frac{1}{s(s^2+1)} }_{ =F(s) }-2 \frac{1}{s(s^2+1)}e^{-\pi s}$
$\mathcal{L}^{-1}\left\{ \frac{1}{s(s^2+1)} \right\}=\mathcal{L}^{-1}\{\frac{1}{s}-\frac{s}{s^2+1}\}=1-\cos t$
$=1-\cos(t)-2u(t-\pi)(1-\cos(t+\pi))$
$$=1-\cos(t)-2u(t-\pi)(1+\cos(t))$$
#end of lec 18