17.3: External Stability

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The inputs in Figure 17.5 are related to the signals \(y_{1}\), and \(y_{2}\) as follows:

\[\begin{array}{l}
y_{1}=H_{1}\left(y_{2}+r_{1}\right) \\
y_{2}=H_{2}\left(y_{1}+r_{2}\right)
\end{array}\nonumber\]

which can be written as

\[\left[\begin{array}{cc}
I & -H_{1} \\
-H_{2} & I
\end{array}\right]\left[\begin{array}{l}
y_{1} \\
y_{2}
\end{array}\right]=\left[\begin{array}{cc}
H_{1} & 0 \\
0 & H_{2}
\end{array}\right]\left[\begin{array}{l}
r_{1} \\
r_{2}
\end{array}\right] \ \tag{17.7}\]

We assume that the interconnection in Figure 17.5 is well-posed. Let the map \(\mathcal{T}\left(H_{1}, H_{2}\right)\) be defined as follows:

\[\left(\begin{array}{l}
y_{1} \\
y_{2}
\end{array}\right)=\mathcal{T}\left(H_{1}, H_{2}\right)\left(\begin{array}{c}
r_{1} \\
r_{2}
\end{array}\right)\nonumber\]

From the relations 17.7 the form of the map \(\mathcal{T}\left(H_{1}, H_{2}\right)\) is given by

\[\mathcal{T}\left(H_{1}, H_{2}\right)=\left[\begin{array}{cc}
\left(I-H_{1} H_{2}\right)^{-1} H_{1} & \left(I-H_{1} H_{2}\right)^{-1} H_{1} H_{2} \\
\left(I-H_{2} H_{1}\right)^{-1} H_{2} H_{1} & \left(I-H_{2} H_{1}\right)^{-1} H_{2}
\end{array}\right]\nonumber\]

We term the interconnected system externally \(p\)-stable if the map \(\mathcal{T}\left(H_{1}, H_{2}\right)\) is \(p\)- stable. In our finite-order LTI case, what this requires is precisely that the poles of all the entries of the rational matrix

be in the open left half of the complex plane.

External stability guarantees that bounded inputs \(r_{1}\), and \(r_{2}\) will produce bounded responses \(y_{1}\), \(y_{2}\), \(u_{1}\), and \(u_{2}\). External stability is guaranteed by asymptotic stability (or internal stability) of the state-space description obtained through the process described in our discussion of well-posedness. However, as noted in earlier chapters, it is possible to have external stability of the interconnection without asymptotic stability of the state-space description (because of hidden unstable modes in the system - an issue that will be discussed much more in later chapters). On the other hand, external stability is stronger than input/output stability of the mapping \(\left(I-H_{1} H_{2}\right)^{-1} H_{1}\) between \(r_{1}\) and \(y_{1}\), because this mapping only involves a subset of the exposed or external variables of the interconnection.

Example 17.3

Assume we have the configuration in Figure 17.5, with \(H_{1}=\frac{s-1}{s+1}\) and \(H_{2}=-\frac{1}{s-1}\). The transfer function relating \(r_{1}\) to \(y_{1}\) is

\[\begin{aligned}
\frac{H_{1}}{1-H_{1} H_{2}} &=\frac{s-1}{s+1}\left(1+\frac{1}{s+1}\right)^{-1} \\
&=\left(\frac{s-1}{s+1}\right)\left(\frac{s+1}{s+2}\right) \\
&=\frac{s-1}{s+2}
\end{aligned}\nonumber\]

Since the only pole of this transfer function is at \(s = -2\), the input/output relation between \(r_{1}\) and \(y_{1}\) is stable. However, consider the transfer function from \(r_{2}\) to \(u_{1}\), which is

\[\begin{aligned}
\frac{H_{2}}{1-H_{1} H_{2}} &=\frac{1}{s-1}\left(\frac{1}{1+\frac{1}{s+1}}\right) \\
&=\frac{s+1}{(s-1)(s+2)}
\end{aligned}\nonumber\]

This transfer function is unstable, which implies that the closed-loop system is externally unstable.