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C. Results from Analytic Function Theory


C.8.1 Poisson's Integral for the Half-Plane

Theorem C.9  Consider a contour $C$ bounding a region $D$. $C$ is a clockwise contour composed by the imaginary axis and a semicircle to the right, centered at the origin and having radius $R\rightarrow \infty$. This contour is shown in Figure C.4. Consider some $z_0=x_0+jy_0$ with $x_0>0$.

Let $f(z)$ be a real function of $z$, analytic inside $D$ and of at least the order of $z^{-1}$; $f(z)$ satisfies


\begin{displaymath} \lim_{\vert z\vert\rightarrow \infty}\vert z\vert\vert f(z)\vert=\beta \hspace{10mm}0 \leq \beta <\infty \hspace{10mm}z\in D \end{displaymath} (C.8.1)

then


\begin{displaymath} f(z_0)=-\frac{1}{2\pi }\int_{-\infty}^{\infty}\frac{f(j\omega)}{j\omega-z_0}d\omega \end{displaymath} (C.8.2)

Moreover, if (C.8.1) is replaced by the weaker condition


\begin{displaymath} \lim_{\vert z\vert\rightarrow \infty}\frac{\vert f(z)\vert}{\vert z\vert }=0 \hspace{10mm}z\in D \end{displaymath} (C.8.3)

then


\begin{displaymath} f(z_0)=\frac{1}{\pi }\int_{-\infty}^{\infty}f(j\omega) \frac{x_0}{x_0^2+(y_0-\omega)^2}d\omega \end{displaymath} (C.8.4)


Figure C.4: RHP encircling contour
\begin{figure} \begin{center} \leavevmode \begin{picture} (0,0)% \epsfig{file=... ...t}{\mddefault}{\updefault}$C_\infty$ }}} \end{picture} \end{center} \end{figure}

Proof

Applying Theorem C.8, we have


\begin{displaymath} f(z_0)=-\frac{1}{2\pi j}\oint_C \frac{f(z)}{z-z_0}dz= -\frac... ...z-z_0}dz- \frac{1}{2\pi j}\int_{C_\infty} \frac{f(z)}{z-z_0}dz \end{displaymath} (C.8.5)

Now, if $f(z)$ satisfies (C.8.1), it behaves like $z^{-1}$ for large $\vert z\vert$, i.e., $ \frac{f(z)}{z-z_0}$ is like $z^{-2}$. The integral along $C_{\infty}$ then vanishes and the result (C.8.2) follows.

To prove (C.8.4) when $f(z)$ satisfies (C.8.3), we first consider $z_1$, the image of $z_0$through the imaginary axis, i.e., $z_1=-x_0+jy_0$. Then $ \frac{f(z)}{z-z_1}$ is analytic inside $D$, and, on applying Theorem C.7, we have that


\begin{displaymath} 0=-\frac{1}{2\pi j}\oint_C \frac{f(z)}{z-z_1}dz \end{displaymath} (C.8.6)

By combining equations (C.8.5) and (C.8.6), we obtain


\begin{displaymath}f(z_0)=-\frac{1}{2j\pi }\oint_C\left( \frac{f(z)}{z-z_0}- \f... ... -\frac{1}{2j\pi }\oint_C f(z)\frac{z_0-z_1}{(z-z_0)(z-z_1)}dz \end{displaymath} (C.8.7)

Because $C=C_i\cup C_\infty$, the integral over $C$ can be decomposed into the integral along the imaginary axis , $C_i$, and the integral along the semicircle of infinite radius, $C_{\infty}$. Because $f(z)$ satisfies (C.8.3), this second integral vanishes, because the factor $\frac{z_0-z_1}{(z-z_0)(z-z_1)}$ is of order $z^{-2}$ at $\infty$.

Then


\begin{displaymath} f(z_0)=-\frac{1}{2\pi }\int_{-\infty}^{\infty}f(j\omega) \frac{z_0-z_1}{(j\omega-z_0)(j\omega-z_1)}d\omega \end{displaymath} (C.8.8)

The result follows upon replacing $z_0$ and $z_1$ by their real; and imaginary-part decompositions.

$\Box \Box \Box $

Remark C.1  One of the functions that satisfies (C.8.3) but does not satisfy (C.8.1) is $f(z)=\ln g(z)$, where $g(z)$ is a rational function of relative degree $n_r\neq 0$. We notice that, in this case,


\begin{displaymath}\lim_{\vert z\vert\rightarrow \infty} \left[\frac{ \vert\ln g... ...ty} \frac{\vert K\vert \vert n_r \ln R+jn_r\theta \vert }{R}=0 \end{displaymath} (C.8.9)

where $K$ is a finite constant and $\theta$ is an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}] $.

Remark C.2   Equation (C.8.4) equates two complex quantities. Thus, it also applies independently to their real and imaginary parts. In particular,


\begin{displaymath} \Re\{f(z_0)\}=\frac{1}{\pi }\int_{-\infty}^{\infty}\Re\{f(j\omega)\} \frac{x_0}{x_0^2+(y_0-\omega)^2}d\omega \end{displaymath} (C.8.10)

This observation is relevant to many interesting cases. For instance, when $f(z)$ is as in remark C.1,


\begin{displaymath}\Re\{ f(z) \}=\ln \vert g(z)\vert \end{displaymath} (C.8.11)

For this particular case, and assuming that $g(z)$ is a real function of $z$, and that $y_0=0$, we have that (C.8.10) becomes


\begin{displaymath}\ln \vert g(z_0)\vert =\frac{1}{\pi }\int_{0}^{\infty}\ln \vert g(j\omega)\vert \frac{2x_0}{x_0^2+(y_0-\omega)^2}d\omega \end{displaymath} (C.8.12)

where we have used the conjugate symmetry of $g(z)$.