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\begin{document}
\section*{Chapter 7: Laplace Theory}

{\Large\bf Background}. {\bf Switches and Impulses}

\begin{quote}\Large\sf
Laplace's method  solves differential equations. It is the preferred
method for solving equations containing switches or impulses.
\end{quote}

\def\BB{}
\def\AA{{\bf Unit Step}~~}
\begin{topics}[\BB]{\AA}
\Writeup{\bf Unit Step}
{ Define $u(t-a) =
\left\{\begin{array}{ll} 1 & t \ge a, \\ 0 & t <
a.\end{array}\right.$. It is a {\large\bf switch}, turned on at $t=a$.}




\Writeup{\bf Ramp}{Define $\ramp(t-a)=(t-a)u(t-a)=\left\{\begin{array}{ll} t-a & t \ge a, \\ 0 & t <
a.\end{array}\right.$, whose graph shape is a continuous {\large\bf  ramp} at $45$-degree incline starting at $t=a$.}


\Writeup{\bf Unit Pulse}{Define
$\pulse(t,a,b)
=\left\{
\begin{array}{ll}
1 & a \le t < b, \\
0 & \mbox{otherwise}
\end{array}\right. = u(t-a)-u(t-b)$.
The switch is {\large\bf ON} at time $t=a$
and then {\large\bf OFF} at time $t=b$.}


\end{topics}


\subsection*{Impulse of a Force}

Define the {\large\bf impulse} of an applied force $F(t)$ on time interval $a\le t\le b$ by the equation
$$\mbox{Impulse of $F$} = \int_{a}^{b} F(t)dt=\left(\frac{\int_{a}^{b} F(t)dt}{b-a}\right)\,(b-a) = \mbox{Average Force $\times$ Duration Time}.$$

\subsection*{Dirac Unit Impulse} A Dirac impulse acts like a hammer hit, a brief injection of energy into a system. It is a special idealization of a real hammer hit, in which only the {\bf impulse} of the force is deemed important, and not its magnitude nor duration.

Define the {\bf Dirac Unit Impulse} by the equation $\delta(t-a)=\dd\frac{du}{dt}(t-a)$, where $u(t-a)$ is the unit step. Symbol $\delta$ makes sense {\sf only under an integral sign}, and the integral in question must be a generalized Riemann integral (definition pending), with new evaluation rules. Symbol $\delta$ is an abbreviation like {\bf etc} or {\bf e.g.}, because it abbreviates a paragraph of descriptive text.

\begin{itemize}\sf

\item
Symbol $M\delta(t-a)$ represents an ideal impulse of magnitude $M$ at time $t=a$. Value $M$ is the change in momentum, but $M\delta(t-a)$ contains no detail about the applied force or the duraction. A common force approximation for a hammer hit of very small duration $2h$ and impulse $M$ is Dirac's approximation $$F_h(t)=\frac{M}{2h}\pulse(t,a-h,a+h).$$

\item
Symbol $\delta$ is not manipulated as an ordinary function. It is a special modeling tool
with rules for application and rules for algebraic manipulation.
\end{itemize}

{\bf THEOREM} ({\sf Second Shifting Theorem}). Let $f(t)$ and $g(t)$ be piecewise
continuous and of exponential order. Then for $a\ge 0$,

$$\begin{array}{rcl}\dd
e^{-as}\L(f(t))&=& \dd\L\left( \left.f(t)u(t)\right|_{t:=t-a}\right), \\
\dd
\L(g(t)u(t-a)) &=&\dd e^{-as}\L\left( \left. g(t)\right|_{t:=t+a}\right).\end{array}
$$

\vfill\newpage
\rule{\textwidth}{0.5ex}
{\Large\bf Sample Problem 2}. Solve the following by Laplace methods.


\begin{itemize}
\item[{\bf (a)}]


Forward table. Compute the Laplace integral for the unit step, ramp and pulse, in these special cases:

$$ (1)~ \L(10u(t-\pi)) \quad (2)~ \L(\ramp(t-2\pi)),
\quad (3)~ \L(10\pulse(t,3,5)).
$$

\item[{\bf (b)}]  Backward table. Find $f(t)$ in the following special cases.

$$ (1)~ \L(f)=\frac{5e^{-3s}}{s} \quad (2)~ \L(f)=\frac{e^{-4s}}{s^2}
\quad (3)~ \L(f)=\frac{5}{s}\left(e^{-2s} - e^{-3s}\right).
$$

\end{itemize}

{\Large\bf Sample Problem 3}. Solve the following Dirac Impulse and Second Shifting theorem problems.


\begin{itemize}
\item[{\bf (c)}] Forward table problems.

$$ (1)~ \L(10\delta(t-\pi)),
\quad (2)~ \L(5\delta(t-1)+10\delta(t-2)+15\delta(t-3)), \quad (2)~ \L((t-\pi) \delta(t-\pi)).
$$

The sum of Dirac impulses in (2) is called an {\bf impulse train}.
\end{itemize}

\rule{\textwidth}{0.5ex}
\newpage
\subsection*{Solutions to sample problems 2 and 3}




{\bf Solution (a)}. The forward second shifting theorem applies.

 (1)~ $\L(10u(t-\pi))=\L\left(g(t)u(t-a)\right)$ where $g(t)=10$ and $a=\pi$. Then
  $\L(10u(t-\pi))=\L\left(g(t)u(t-a)\right)=e^{-as}\L\left(\left.g(t)\right|_{t=t+a}\right)=e^{-\pi s}\L\left(\left. 10\right|_{t=t+\pi}\right)=\frac{10}{s}e^{-\pi s}$.

  (2)~ $\L(\ramp(t-2\pi))=\L((t-2\pi)u(t-2\pi))=\L\left(\left.tu(t)\right|_{t=t-2\pi})\right)=e^{-2\pi s}\L(t)=\frac{1}{s^2}e^{-2\pi s}$.

 (3)~ $\L(10\pulse(t,3,5))=10\L(u(t-3)-u(t-5))=\frac{10}{s}(e^{3s}-e^{-5s})$.

\bigskip

{\bf Solution (b)}. Presence of an exponential $e^{-as}$ signals step $u(t-a)$ in the answer, the main tool bing the backward second shifting theorem.

(1)~ $\L(f)=\frac{5e^{-3s}}{s}=e^{-3s}\frac{5}{s} = e^{-3s}\L(5)=\L(\left.5u(t)\right|_{t=t+3})=\L(5u(t-3))$. Lerch implies $f=5u(t-3)$.

(2)~ $\L(f)=\frac{e^{-4s}}{s^2}=\frac{e-as}\L(t)$ where $a=4$. Then $\L(f)=\frac{e-as}\L(t)=\L(\left.tu(t)\right|_{t=t-a})=\L((t-4)u(t-4))=\L(\ramp(t-4))$. Lerch implies $f=\ramp(t-4)$.

(3)~ $\L(f)=e^{-2s}\frac{5}{s} - e^{-3s}\frac{5}{s}=\L(5u(t-2))-\L(5u(t-3))=\L(5\pulse(t,2,3))$. Lerch implies $f=5\pulse(t,2,3)$.

\bigskip

{\bf Solution (c)}. The main result for Dirac unit impulse $\delta$ is the equation
$$\int_)^\infty g(t)\delta(t-a)dt = g(a),$$
valid for $g(t)$ continuous on $0\le t < \infty$. When $g(t)=e^{-st}$, then the equation implies the Laplace formula $\L(\delta(t-a))=e^{-as}$.


(1)~ $\L(10\delta(t-\pi))= 10e^{-\pi s}$, by the displayed equation with $g(t)=10e^{-st}$,
or by using linearity and the formula $\L(\delta(t-a))=e^{-as}$.


(2)~ $\L(5\delta(t-1)+10\delta(t-2)+15\delta(t-3))=
5\L(\delta(t-1))+10\L(\delta(t-2))+15\L(\delta(t-3))=5e^{-s}+10e^{-2s}+15e^{-3s}$.

(3)~ $\L((t-\pi) \delta(t-2\pi)) = \int_0^\infty (t-\pi)e^{st}\delta(t-2\pi)dt = \left. (t-\pi)e^{-st}\right|_{t=2\pi} = \pi e^{-2\pi s}$, using $g(t)=(t-\pi)e^{-st}$ and $a=2\pi$ in the equation.


\vfill\newpage
\rule{\textwidth}{0.5ex}
{\Large\bf Sample Problem 4}. {\bf Experiment to Find the Transfer Function $h(t)$}


Consider a second order problem
$$ax''(t)+bx'(t)+cx(t) = f(t)$$
which by Laplace theory has a particular solution solution defined as the convolution of the transfer function $h(t)$ with the input $f(t)$,
$$x_p(t) = \int_0^t f(w)h(t-w)dw.$$
Examined in this problem is another way to find $h(t)$, which is the system response to a Dirac unit impulse with zero data. Then $h(t)$ is the solution of
$$ah''(t)+bh'(t)+ch(t) = \delta(t),\quad h(0)=h'(0)=0.$$


\rule{\textwidth}{0.5ex}
{\Large\bf The Problem}. Assume $a,b,c$ are constants and define $g(t)=\int_0^t h(w)dw$.

{\bf (a)}~ Show that $g(0)=g'(0)=0$, which means $g$ has zero data.

{\bf (b)}~ Let $u(t)$ be the unit step. Argue that $g$ is the solution of
$$ag''(t)+bg'(t)+cg(t) = u(t),\quad g(0)=g'(0)=0.$$

 \begin{quote}\sf
 The fundamental theorem of calculus says that $h(t)=g'(t)$. Therefore, to compute the transfer function $h(t)$, find the response $g(t)$ to the unit step with zero data, followed by computing the derivative $g'(t)$, which equals $h(t)$.

 The experimental impact is important. Turning on a switch creates a unit step,
generally easier than designing a hammer hit.
 \end{quote}

{\bf (c)}~ Illustrate the method for finding the transfer function $h(t)$ in the special case
$$x''(t) + 2x'(t)+5x(t) = f(t).$$


\rule{\textwidth}{0.5ex}
\subsection*{Solutions to sample problem 4}

{\bf (a)}~ $g(0)=\int_0^0 h(w)dw = 0$, $g'(0)=h(0)=0$.


{\bf (b)}~ Let $u(t)$ be the unit step. Initial data was decided in part (a). The Laplace applied to $ag''(t)+bg'(t)+cg(t) = u(t)$ gives $(as^2+bs+c)\L(g) = \L(u(t))$. Then $\L(g)=\L(h(t))\L(u(t))=\L(h(t))\frac{1}{s}\L\left(\int_0^t h(r)du\right)$ by the {\bf integral theorem}. Lerch's theorem then says $g(t)=\int_0^t h(r)dr$.

{\bf (c)}~ For equation $x''(t) + 2x'(t)+5x(t) = f(t)$ we replace $x(t)$ by $g(t)$ and $f(t)$ by the unit step $u(t)$, then solve $g''(t) + 2g'(t)+5g(t) = u(t)$, obtaining $\L(g)=\frac{1}{s}\frac{1}{s^2+2s+5}=\L(\frac15-\frac{1}{10}e^{-t}(2\cos(2t)+\sin(2t)))$. Then $g(t)=\frac15-\frac{1}{10}e^{-t}(2\cos(2t)+\sin(2t))$ and $h(t)=g'(t)=\frac12e^{-t}\sin(2t)$.

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