17.3: External Stability

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

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

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.