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C. Results from Analytic Function Theory


C.10 Bode's Theorems

We will next review some fundamental results due to Bode.

Theorem C.12 (Bode integral in the half plane)  Let $l(z)$ be a proper real, rational function of relative degree $n_r$. Define


\begin{displaymath}g(z)\stackrel{\rm\triangle}{=}(1+l(z))^{-1} \end{displaymath} (C.10.1)

and assume that $g(z)$ has neither poles nor zeros in the closed RHP. Then


\begin{displaymath} \int_0^\infty\ln \vert g(j\omega)\vert d\omega=\begin{cases}... ...{\rm\triangle}{=}\lim_{z\rightarrow \infty}zl(z)$ } \end{cases}\end{displaymath} (C.10.2)

Proof

Because $\ln g(z)$ is analytic in the closed RHP,


\begin{displaymath}\oint_C \ln g(z)dz=0 \end{displaymath} (C.10.3)

where $C=C_i\cup C_\infty$ is the contour defined in Figure C.4.

Then


\begin{displaymath} \oint_C \ln g(z) dz=j\int_{-\infty}^\infty \ln g(j\omega)d\omega- \int_{C_\infty} \ln (1+l(z))dz \end{displaymath} (C.10.4)

For the first integral on the right-hand side of Equation (C.10.4), we use the conjugate symmetry of $g(z)$ to obtain


\begin{displaymath}\int_{-\infty}^\infty \ln g(j\omega)d\omega= 2\int_{0}^\infty \ln \vert g(j\omega)\vert d\omega \end{displaymath} (C.10.5)

For the second integral, we notice that, on $C_{\infty}$, $l(z)$ can be approximated by


\begin{displaymath}\frac{a}{z^{n_r}} \end{displaymath} (C.10.6)

The result follows upon using Example C.7 and noticing that $a=\kappa$ for $n_r=1$.

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Remark C.4  If $g(z)=\left(1+e^{-z\tau}l(z)\right)^{-1}$ for $\tau>0$, then result (C.10.9) becomes


\begin{displaymath} \int_0^\infty \ln \vert g(j\omega)\vert d\omega=0 \hspace{10mm}\forall n_r>0 \end{displaymath} (C.10.7)

The proof of (C.10.7) follows along the same lines as those of Theorem C.12 and by using the result in Example C.8.

Theorem C.13 (Modified Bode integral)  Let $l(z)$ be a proper real, rational function of relative degree $n_r$. Define


\begin{displaymath}g(z)\stackrel{\rm\triangle}{=}(1+l(z))^{-1}\hspace{10mm} \end{displaymath} (C.10.8)

Assume that $g(z)$ is analytic in the closed RHP and that it has $q$ zeros in the open RHP, located at $\zeta_1, \zeta_2,\ldots, \zeta_q$ with $\Re (\zeta_i)>0$. Then


\begin{displaymath} \int_0^\infty\ln \vert g(j\omega)\vert d\omega=\begin{cases}... ...{\rm\triangle}{=}\lim_{z\rightarrow \infty}zl(z)$ } \end{cases}\end{displaymath} (C.10.9)

Proof

We first notice that $\ln g(z)$ is no longer analytic on the RHP. We then define


\begin{displaymath}\tilde{g}(z)\stackrel{\rm\triangle}{=}g(z)\prod_{i=1}^{q}\frac{z+\zeta_i}{z-\zeta_i} \end{displaymath} (C.10.10)

Thus, $\ln \tilde{g}(z)$ is analytic in the closed RHP. We can then apply Cauchy's integral in the contour $C$ described in Figure C.4 to obtain


\begin{displaymath} \oint_C\ln \tilde{g}(z)dz=0=\oint_C\ln g(z)dz+ \sum_{i=1}^{q}\oint_C\ln \frac{z+\zeta_i}{z-\zeta_i}dz \end{displaymath} (C.10.11)

The first integral on the right-hand side can be expressed as


\begin{displaymath}\oint_C\ln g(z)dz=2j\int_0^\infty\ln \vert g(j\omega)\vert d\omega+\int_{C_\infty}\ln g(z)dz \end{displaymath} (C.10.12)

where, by using Example C.7.


\begin{displaymath}\int_{C_\infty}\ln g(z)dz=\begin{cases}0&\mbox{ for $n_r>1$ }... ...riangle}{=}\lim_{z\rightarrow \infty}zl(z)$\space } \end{cases}\end{displaymath} (C.10.13)

The second integral on the right-hand side of Equation (C.10.11) can be computed as follows:


\begin{displaymath}\oint_C\ln \frac{z+\zeta_i}{z-\zeta_i}dz= j\int_{-\infty}^\in... ...ta_i}d\omega+ \int_{C_\infty}\ln \frac{z+\zeta_i}{z-\zeta_i}dz \end{displaymath} (C.10.14)

We note that the first integral on the right-hand side is zero, and by using Example C.9, the second integral is equal to $-2j\pi\zeta_i$. Thus, the result follows.

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Remark C.5  Note that $g(z)$ is a real function of $z$, so


\begin{displaymath}\sum_{i=1}^{q}\zeta_i=\sum_{i=1}^{q}\Re\{\zeta_i\} \end{displaymath} (C.10.15)

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Remark C.6  If $g(z)=\left(1+e^{-z\tau}l(z)\right)^{-1}$ for $\tau>0$, then the result (C.10.9) becomes


\begin{displaymath} \int_0^\infty \ln \vert g(j\omega)\vert d\omega=\pi \sum_{i=1}^{q}\Re\{\zeta_i\} \hspace{10mm}\forall n_r>0 \end{displaymath} (C.10.16)

The proof of (C.10.16) follows along the same lines as those of Theorem C.13 and by using the result in Example C.8.

Remark C.7  The Poisson, Jensen, and Bode formulae assume that a key function is analytic, not only inside a domain $D$, but also on its border $C$. Sometimes, there may exist singularities on $C$. These can be dealt with by using an infinitesimal circular indentation in $C$, constructed so as to leave the singularity outside $D$. For the functions of interest to us, the integral along the indentation vanishes. This is illustrated in Example C.6 for a logarithmic function, when $D$ is the right-half plane and there is a singularity at the origin.

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