2.1 KiB
(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