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\begin{document}
\section*{Quiz 11}


\rule{\textwidth}{0.5ex}
{\Large\bf Background}. {\bf Switches and Impulses}

\begin{quote}\Large\sf
Laplace's method  solves differential equations. It is the premier
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-Steiltjes  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
The fundamental equation is
$\int_{-\infty}^\infty F(x)\delta(x-a)dx = F(a)$. Symbol $\delta(t-a)$ is not manipulated as an ordinary function, but regarded as $du(t-a)/dt$ in a Riemann-Stieltjes integral.
\end{itemize}

{\bf THEOREM} ({\sf Second Shifting Theorem}). Let $f(t)$ and $g(t)$ be piecewise
continuous and of exponential order. Then for $a\ge 0$,
$$\begin{array}{l@{\hspace{3pc}}l}
\mbox{\bf Forward table} & \mbox{\bf Backward table} \\[2ex]
\dd\L\left( f(t-a)u(t-a)\right) = \dd e^{-as}\L(f(t)) &
\dd  e^{-as}\L(f(t)) = \dd\L\left( f(t-a)u(t-a)\right) \\[2ex]
\dd\L(g(t)u(t-a)) =\dd e^{-as}\L\left( \left. g(t)\right|_{t:=t+a}\right) &
\dd e^{-as}\L(f(t))=\dd \L\left( \left.f(t)u(t)\right|_{t:=t-a}\right). \\
\end{array}
$$

\vfill\newpage
\rule{\textwidth}{0.5ex}
{\Large\bf Problem 1}. Solve the following by Laplace methods.


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


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

$$ (1)~ \L((t-1)u(t-1)) \quad (2)~ \L(e^t\ramp(t-2)),
\quad (3)~ \L(5\pulse(t,2,4)).
$$

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

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

\end{itemize}

\vfill\newpage
\rule{\textwidth}{0.5ex}
{\Large\bf Problem 2}. Solve the following Dirac impulse problem.


\begin{itemize}

\item[{\bf (c)}] Dirac Impulse and the Second Shifting theorem. Solve the following forward table problems.

$$ (1)~ \L(2\delta(t-5)),
\quad (2)~ \L(2\delta(t-1)+5\delta(t-3)), \quad (3)~ \L(e^t \delta(t-2)).
$$

The sum of Dirac impulses in (2) is called an {\bf impulse train}. The numbers 2 and 5 represent the applied {\bf impulse} at times 1 and 3, respectively.
\end{itemize}

\rule{\textwidth}{0.5ex}
\ifsolutions
\subsection*{Solutions}




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

 (1)~ $\L((t-1)u(t-1))=\L\left(f(t-a)u(t-a)\right)$ where $f(t)=t$ and $a=1$. Then the second shifting theorem implies
  $\L((t-1)u(t-1))=e^{-as}\L(f(t))=e^{-s}\L(t)=\frac{1}{s^2}e^{-s}$. An observation is that this problem is identical to the problem $\L(\ramp(t-1))$.

  (2)~ $\L(\ramp(t-2))=\L((t-2)u(t-2))=e^{-as}\L(f(t))=e^{-2s}\L(t)=e^{-2s}\frac{1}{s^2}$, as in part (a). The first shifting theorem (canceled $e^{at}$ is corrected by shift $s=s-a$) then applies: $\L(e^t\ramp(t-2))=\left.\L(\ramp(t-2))\right|_{s=s-1} = \left.\frac{e^{-2s}}{s^2}\right|_{s=s-1} = \frac{e^{2-2s}}{(s-1)^2}$.

 (3)~ $\L(5\pulse(t,2,4)) = 5(\L(u(t-2))-\L(u(t-4)) ) = \frac{5e^{-2s}}{s} - \frac{5e^{-4s}}{s}$.

\bigskip

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

(1)~  $\L(f)=\frac{e^{-2s}}{s} = e^{-2s}\L(1)=e^{-as}\L(g)$ where $a=2$ and $g(t)=1$. Then the second shifting theorem implies $\L(f)=\L(\left. g(t)u(t)\right|_{t=t-a}) = \L(\left. 1\cdot u(t)\right|_{t=t-a}) = \L(u(t-2))$.
Lerch implies $f=u(t-2)$.

(2)~  $\L(f)=\frac{e^{-s}}{(s+1)^2}=e^{-as}\L(g(t))$ where $a=1$ and $\L(g(t))=\frac{1}{(s+1)^2} = \L(te^{-t})$ (first shifting theorem). Then $\L(f)=e^{-as}\L(g(t))=\L(\left. g(t)u(t)\right|_{t=t-a}) = \L(\left. te^{-t}u(t)\right|_{t=t-1}) =\L((t-1)e^{1-t}u(t-1))$.
Lerch implies $f=e^{1-t}(t-1)u(t-1)=e^{1-t}\ramp(t-1)$.

(3)~  $\L(f)=e^{-s}\frac{3}{s} - e^{-2s}\frac{3}{s}=\L(3u(t-1))-\L(3u(t-2))$, because $\L(u(t-a))=\frac{e^{-as}}{s}$.  Lerch implies $f=3u(t-1)-3u(t-2)=3\pulse(t,1,2)$.

\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)~ $\L(2\delta(t-5)) = 2\L(\delta(t-a))$ where $a=5$. Then $\L(\delta(t-a))=e^{-as}$ implies $\L(2\delta(t-5))=2e^{-5s}$.

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

(2)~ $\L(e^t \delta(t-2))= \left.\L(\delta(t-2))\right|_{s=s-1}$ by the first shifting theorem. Then $\L(\delta(t-2))=e^{-2s}$ implies $\L(e^t \delta(t-2))= \left.e^{-2s}\right|_{s=s-1} = e^{2-2s}$.


\rule{\textwidth}{0.5ex}
\else
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\fi
\section*{Reference: The Riemann-Stieltjes Integral}

{\bf Definition}

The Riemann-Stieltjes integral of a real-valued function f of a real variable with respect to a real monotone non-decreasing function g is denoted by

   $$ \int_a^b f(x) \, dg(x)$$

and defined to be the limit, as the mesh of the partition

   $$ P=\{ a = x_0 < x_1 < \cdots < x_n = b\}$$

of the interval $[a, b]$ approaches zero, of the approximating Riemann–Stieltjes sum

    $$S(P,f,g) = \sum_{i=0}^{n-1} f(c_i)(g(x_{i+1})-g(x_i))$$

where $c_i$ is in the $i$-th subinterval $[x_i, x_{i+1}]$. The two functions $f$ and $g$ are respectively called the {\bf integrand} and the {\bf integrator}.

The {\bf limit} is a number $A$, the value of the Riemann-Stieltjes integral. The meaning of the limit: Given $\varepsilon > 0$, then there exists $\delta > 0$ such that for every partition $ P=\{ a = x_0 < x_1 < \cdots < x_n = b\}$ with $\mbox{\bf mesh}(P)=\max_{~0\le i< n~}(x_{i+1}-x_i) < \delta$, and for every choice of points $c_i$ in $[x_i, x_{i+1}]$,

    $$|S(P,f,g)-A| < \varepsilon. \, $$


%\vfill\newpage
%\rule{\textwidth}{0.5ex}
%{\Large\bf Problem 3}. {\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}
%
%{\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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