From cbb11563f84a9296f607ad48ba55ce5bb86bbd0d Mon Sep 17 00:00:00 2001
From: Sasserisop <mail@sasserisop.com>
Date: Mon, 6 Nov 2023 10:28:14 -0700
Subject: [PATCH] revised heaviside

---
 ...(Heaviside) Unit step function (lec 18).md | 66 +++++++++++--------
 content/_index.md                             |  2 +-
 2 files changed, 39 insertions(+), 29 deletions(-)

diff --git a/content/(Heaviside) Unit step function (lec 18).md b/content/(Heaviside) Unit step function (lec 18).md
index 153006a..ca58afc 100644
--- a/content/(Heaviside) Unit step function (lec 18).md	
+++ b/content/(Heaviside) Unit step function (lec 18).md	
@@ -1,46 +1,56 @@
-
-# (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
\ No newline at end of file
diff --git a/content/_index.md b/content/_index.md
index ac04513..8df141b 100644
--- a/content/_index.md
+++ b/content/_index.md
@@ -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.)