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

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content/_index.md
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@ -18,7 +18,7 @@ Good luck on midterms! <3 -Oct 18 2023
[Resonance & AM (lec 13-14)](resonance-am-lec-13-14.html)
[Laplace transform (lec 14-16)](laplace-transform-lec-14-16.html)
[Solving IVP's using Laplace transform (lec 17-18)](solving-ivps-using-laplace-transform-lec-17-18.html)
[(Heaviside) Unit step function (lec 18)](heaviside-unit-step-function-lec-18.html) (raw notes, not reviewed or revised yet.)
[(Heaviside) Unit step function (lec 18)](heaviside-unit-step-function-lec-18.html)
[Periodic functions (lec 19)](periodic-functions-lec-19.html) (raw notes, not reviewed or revised yet.)
[Convolution (lec 19-20)](convolution-lec-19-20.html) (raw notes, not reviewed or revised yet.)
[Dirak δ-function (lec 21)](dirak-δ-function-lec-21.html) (raw notes, not reviewed or revised yet.)