\chapter{Describing Function Analysis}
The describing function method allows us to apply familiar frequency domain techniques, used in the linear system analysis, to the analysis of a class of nonlinear dynamic systems. The describing function method can be viewed as a ``harmonic linearization'' of a nonlinear element. The method provides a ``linear approximation'' to the nonlinear element based on the assumption that the input to the nonlinear element is a sinusoid of known, constant amplitude. The fundamental harmonic of the element's output is compared with the input sinusoid to determine the steady-state amplitude and phase relation. This relation is the describing function for the nonlinear element. The describing function method was developed during World War II. The first to introduce the method in the open literature seems to be be A.~Tustin of England.

The describing function method is based on the Fourier series. In our review of Fourier series we use the notion and properties of general scalar product that we discuss in the following section.

\section{General Scalar Product}
Let $V$ be the set of of continuous functions on some interval, say the interval $[\alpha,\beta]\subset \mb{\Bb R}$, that is, $V=C[\alpha,\beta]$.
\begin{defn}
The {\em scalar product} of two real-valued functions $u, v \in V$ is the number
\[
\langle u,v \rangle = \int_{\alpha}^{\beta}u(t)v(t)dt.
\]
\end{defn}
Using simple properties of the integral we can verify that the above scalar product satisfies the following conditions:
\begin{enumerate}
\item for all $u, v \in V$ we have $\langle u,v \rangle = \langle v,u \rangle$,
\item if $u, v, w$ are elements of $V$, then $\langle u,v+w \rangle = \langle u,v \rangle + \langle u,w \rangle$,
\item if $x$ is a number, then $\langle xu,v \rangle = x\langle u,v \rangle = \langle u,xv \rangle$,
\item for all $v \in V$ we have $\langle v,v \rangle \ge 0$, and $\langle v,v \rangle > 0$ if $v \ne 0$.
\end{enumerate}
The functions $u, v \in V$ are said to be {\em orthogonal\/} if $\langle u,v \rangle = 0$. A set of functions from $V$ is said to be {\em mutally orthogonal\/} if each distinct pair of functions in the set is orthogonal.
\begin{example}\rm
The functions $\psi_n(t)=\sin\,n\omega t$ and $\phi_n(t)=\cos\,n\omega t$, $n=1,2,\ldots$, form a mutually orthogonal set of functions on the interval $[t+0,t_0+{2\pi}/\omega]$. This can be verified by direct integration. For example, for positive $m$ and $n$ such that $m\ne n$ we have
\begin{eqnarray*}
\langle \psi_m,\psi_n \rangle&=&\int_{t_0}^{t_0+{2\pi}/\omega}\sin\,m\omega t\,\sin\,n\omega t\, dt\\
\qquad &{}&=\frac{1}{2}\int_{t_0}^{t_0+{2\pi}/\omega}\left( \cos\,(m-n)\omega t-\cos\,(m+n)\omega
t\right)dt\\
\qquad &{}&=\frac{1}{2\omega} \left.\left(\frac{ \sin\,(m-n)\omega t}{m-n}-\frac{\sin\,(m+n)\omega t}{m+n}\right)\right\arrowvert_{t_0}^{t_0+{2\pi}\omega}\\
\qquad &{}&=0 
\end{eqnarray*}
Note that $m+n\ne 0$ because $m$ and $n$ are positive. Using similar computations we can show that for $m\ne n$
\[
\langle \phi_m,\phi_n \rangle = 0,
\]
and for all positive $m$, $n$
\[
\langle \phi_m,\psi_n \rangle = 0.
\]
\end{example}\markex
We define the {\em norm\/} of $v\in V$ to be
\[
\|v\|=\sqrt{\langle v,v \rangle}.
\]
\begin{example}\rm
Let $\psi_n(t)=\sin\,n\omega t$ be defined on the interval $[t_0,t_0+{2\pi}/\omega]$. Then,
\begin{eqnarray*}
\|\psi_n\|&=&\int_{t_0}^{t_0+{2\pi}/\omega}\sin\,n\omega t\,\sin\,n\omega t\, dt\\
&=& \int_{t_0}^{t_0+{2\pi}/\omega}\sin^2n\omega t\, dt\\
&=& \frac{1}{2}\int_{t_0}^{t_0+{2\pi}/\omega}\left(1-\cos\,2n\omega t\right)\, dt\\
&=& \frac{1}{2}\left.\left( t-\frac{\sin\,2n\omega t}{2n\omega}\right)\right\arrowvert_{t_0}^{t_0+{2\pi}/\omega}\\
&=&\frac{\pi}{\omega}
\end{eqnarray*}
\end{example}\markex

So far we discussed only continuous functions. In many applications we have to work with more general functions, specifically with piecewise continuous functions.
\begin{defn}
A function $f$ is said to be {\em piecewise continuous\/} on an interval $[a, b]$ if the interval can be partitioned by a finite number of points say $t_i$ so that
\begin{enumerate}
\item $a=t_0<t_1<\ldots<t_n=b$;
\item the function $f$ is continuous on each open subinterval $(t_{i-1},t_i)$,
\item the function $f$ approaches a finite limit as the end points of each subinterval are approached from within the subinterval, that is, the limits
\[
\lim_{{h \to 0}\atop {h>0}}f(t_i-h),\quad \mb{and}\quad \lim_{{h \to 0}\atop {h>0}}f(t_i+h)
\]
both exist.
\end{enumerate}
\end{defn}
The graph of a piecewise continuous function is shown in Figure~\ref{fig_piececon}.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/piececon.bmp x=4in y=3in}
\caption{A graph of a piecewise continuous function}
\label{fig_piececon}
\end{figure}
In further discussion, we assume that two picewise functions $f$ and $g$ are equal if they have the same values at the points where they are continuous. Thus, the functions shown in Figure~\ref{fig_disfuns} are considered to be equal, though they differ at the points of discontinuity.
\begin{figure}
\vspace{3in}
\hskip1in\special{bmp:c:/mybook/disfuns.bmp x=4in y=3in}
\caption{Two piecewise continuous functions that we consider to be equal.}
\label{fig_disfuns}
\end{figure}
\begin{thm}
Let $C[a,b]$ be the space of functions that are piecewise continuous on the interval $[a,b]$. Let $f \in C[a,b]$. Then, $\|f\|=0$ if, and only if $f(x)=0$ for all but finite number of points $x$ in the interval $[a,b]$.
\end{thm}
{\bf Proof}\quad $(\Leftarrow)$ It is clear that if $f(x)=0$ except for a finite number of $x \in [a,b]$, then
\[
\|f\|^2=\int_a^b f^2(x)dx =0.
\]


$(\Rightarrow)$ Suppose $f$ is piecewise continuous on $[a,b]$, and let $a=t_0<t_1<\ldots<t_n=b$ be a partition of the interval $[a,b]$ such that $f$ is continuous on each subinterval $[t_{i-1},t_i]$ except possibly at the end points. Suppose that $\| f \|=0$. Then, also $\| f \|^2=0= \langle f,f \rangle$. The above means that
\[
\int_a^b f^2(x)dx=0,
\]
where the integral is the sum of the integrals over the intervals $[t_{i-1},t_i]$, that is,
\[
\int_a^b f^2(x)dx=\sum_{i=1}^n \int_{t_{i-1}}^{t_i}f^2(x)dx=0.
\]
Each integral satisfies
\[
\int_{t_{i-1}}^{t_i}f^2(x)dx \ge 0.
\]
Hence each such integral must be equal to $0$. The function $f$ is continuous on $(t_{i-1},t_i)$. Therefore,
\[
f^2(x)=0 \quad \mb{for} \quad t_{i-1} < x < t_i,
\]
which means that
\[
f(x)=0 \quad \mb{for} \quad t_{i-1} < x < t_i.
\]
Hence, $f(x)=0$ except at a finite number of points. The proof is complete.\\
\hspace*{5.5in} $\Box$

The above theorem implies that the scalar product on the space of piecewise continuos functions $C[a,b]$ is positive definite. In other words, for all piecewise continuos functions $f \in C[a,b]$ we have $\langle f,f \rangle \ge 0$ and $\langle f,f \rangle >0$ if $f \ne 0$ for all but a finite number of points $x$ in the interval $[a,b]$. 

The space of piecewise continuos functions $C[a,b]$ is not finite dimensional. 



\section{Fourier Series}
We begin with the definition of a periodic function.
\begin{defn}
A function $f:\mb{\Bb R} \to \mb{\Bb R}$ is said to be periodic with the period $T$ if the domain of $f$ contains $t+T$ whenever it contains $t$, and if
\[
f(t+T)=f(t)
\]
for every value of $t$. The smallest positive value of $T$ for which $f(t+T)=f(t)$ is called the fundamental period of $f$.
\end{defn}
It follows from the above definition that if $T$ is a period of $f$, Then $2T$ is also a period, and so is any integral multiple of $T$.
\begin{figure}
\vspace{1in}
\setlength{\unitlength}{.05in}
\begin{center}
\begin{picture}(40,20)
\thicklines
\put(10,5){\framebox(20,10){}}
\put(11,6){\framebox(18,8){}}
\put(4,10){\vector(1,0){6}}
\put(30,10){\vector(1,0){6}}
\end{picture}
\end{center}
\caption{A symbol for a nonlinear element.}
\label{fig_non_el}
\end{figure}