MATH201/content/(Heaviside) Unit step function (lec 18).md

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

#heaviside 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.

#ex #heaviside 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