\chapter{Operational Calculus}
In this chapter, we present a tool, named operational calculus, that we will be using in the analysis and design of control system. Operational calculus were developed by the Polish mathematician Jan Mikusi\'{n}ski in the fifties and his results were published in~\cite{mikusinski83}.

The starting point of operational calculus is the notion of {\em convolution} of two functions. We consider the class of functions that are defined and continuous in the interval $0\le t< \infty$. We denote the class of such function by $\cal C$. Functions of class $\cal C$ that we consider may take real or complex values.
\begin{defn}
The convolution of the functions $a(t)$ and $b(t)$ is the function
\[
c(t)=\int_0^ta(t-\tau)b(\tau)d\tau.
\]
\end{defn}
We denote the convolution of the functions $a$ and $b$ by
\[
c(t)=(a\star b)(t).
\]
When there is no ambiguity we drop the argument and write
\[
c=a\star b.
\]
\begin{example}\rm
Given $a(t)=t$ and $b(t)=\sin t$. Using the defintion of convolution we obtain
\begin{eqnarray*}
c(t) &=& (a\star b)(t)\\
&=&\int_0^ta(t-\tau)b(\tau)d\tau\\
&=&\int_0^t(t-\tau)\sin \tau d\tau\\
&=&-\left. t\cos \tau\right\arrowvert_{\tau=0}^{\tau=t}-\int_0^t\tau\sin \tau d\tau\\
&=& -t\cos t +t -\left.\left(-\tau\cos \tau +\sin \tau  \right)\right\arrowvert_{\tau=0}^{\tau=t}\\
&=& t-\sin t.  
\end{eqnarray*}
\end{example}
We can verify the above calculations using MATLAB by typing the following commands
\begin{verbatim}
f='(t-s)*sin(s)';
int(f,'s',0,'t')
\end{verbatim}

The convoultion operation has many properties of ordinary multiplication operation. In particular, one can show the following properties:
\begin{eqnarray*}
a\star b &=& b \star a \quad \mb{commutative law}\\
a\star (b_1+b_2) &=& a\star b_1 + a\star b_2 \quad \mb{distributive law}\\
(a\star b)\star c &=& a\star (b \star c) \quad \mb{associative law}\\
a\star 0 &=& 0\star a =0.
\end{eqnarray*}
However, there are other properties of ordinary multiplication that convolution operation does not have. For example, it is not true in general that $f\star 1=f$. To see this note that
\begin{eqnarray*}
(f\star 1)(t) &=& \int_0^tf(t-\tau)d \tau\\
&=& \int_0^tf(t-\tau)d\tau.
\end{eqnarray*}
If for example $f(t)=\cos t$, then
\begin{eqnarray*}
(f\star 1)(t) &=& \int_0^t\cos (t-\tau)d \tau\\
&=& -\left.\sin(t-\tau)\right\arrowvert_{\tau=0}^{\tau=t}\\
&=& \sin t.
\end{eqnarray*}
Clearly $(f\star 1) \ne f$. Similarly, it may not be true that $f \star f$ is nonnegative. For example, let $f=\sin t$. Then,
\begin{eqnarray*}
(f\star f)(t) &=& \int_0^t\sin(t-\tau) \sin \tau d\tau\\
&=& \int_0^t(\sin t \cos \tau -\cos t \sin \tau)\sin \tau d\tau\\
&=& \sin t \int_0^t\cos \tau \sin \tau d\tau - \cos t \int_0^t \sin ^2\tau d\tau\\
&=& \frac{1}{2}\sin t \sin ^2 t-\cos t(\frac{1}{2}t-\frac{1}{2}\sin t \cos t)\\
&=& \frac{1}{2}(\sin t -t \cos t).
\end{eqnarray*}
Note that $(f\star f)(2 \pi)=-\pi <0$.

Observe that if $a,b \in {\cal C}$, then their convolution $c=a\star b \in {\cal C}$. Furthermore, if $a, b \in {\cal C}$ are not identically zero, then neither is $a\star b$ identically equal to zero. This important property of the convolution was proved by E.~Titchmarsh in 1924---see the book by Mikusi\'{n}ski for the proof of the result.

An inverse operation to convolution leads to the notion of an {\em operator}. Specifically, consider the fraction $a/b=c$, of continuous functions $a$ and $b$. This fraction can be represented equivalently as $a=b\star c$. If for two given functions $a$ and $b\ne 0$ of class ${\cal C}$ there is no continuous function $c$ such that $a=b\star c$, then the fraction $a/b=c$ is called an operator.