20.8: Appendix

Necessity of the small gain condition for robust stability can be proved by showing that if $$\sigma_{\max }\left[M\left(j \omega_{0}\right)\right]>1$$ for some $$\omega_{0}$$, we can construct a $$\Delta$$ of norm less than one, such that the resulting closed-loop map $$G_{zv}$$ is unstable. This is done as follows. Take the singular value decomposition of $$M\left(j \omega_{0}\right)$$,

$M\left(j \omega_{0}\right)=U \Sigma V^{\prime}=U\left[\begin{array}{ccc} \sigma_{1} & & \\ & \ddots & \\ & & \sigma_{n} \end{array}\right] V^{\prime} \label{20.17}$

Since $$\sigma_{\max }\left[M\left(j \omega_{0}\right)\right]>1, \sigma_{1}>1$$. Then $$\Delta\left(j \omega_{0}\right)$$ can be constructed as:

$\Delta\left(j \omega_{0}\right)=V\left[\begin{array}{cccc} 1 / \sigma_{1} & & \\ & 0 & \\ & & \ddots \\ & & & 0 \end{array}\right] U^{\prime} \label{20.18}$

Clearly, $$\sigma_{\max } \Delta\left(j \omega_{0}\right)<1$$. We then have

$(I-M \Delta)^{-1}\left(j \omega_{0}\right)= I-U\left[\begin{array}{cccc} \sigma_{1} & & \\ & \sigma_{2} & \\ & & \ddots \\ & & & \sigma_{n} \end{array}\right] \quad V^{\prime} V\left[\begin{array}{cccc} 1 / \sigma_{1} & & & \\ & 0 & & \\ & & \ddots & \\ & & & 0 \end{array}\right] U^{\prime} \ =U\left[I-\left[\begin{array}{cccc} 1 & & \\ & 0 & \\ & & \ddots & \\ & & & 0 \end{array}\right]\right] U^{\prime} =U\left[\begin{array}{cccc} 0 & & & \\ & 1 & & \\ & & \ddots & \\ & & & 1 \end{array}\right] U^{\prime} \label{20.19}$

which is singular. Only one problem remains, which is that $$\Delta\left(s\right)$$ must be legitimate as the transfer function of a stable system, evaluating to the proper value at $$s=j \omega_{0}$$, and having its maximum singular value over all $$\omega$$ bounded below 1. The value of the destabilizing perturbation at $$\omega_{0}$$ is given by

$\Delta_{0}\left(j \omega_{0}\right)=\frac{1}{\sigma_{\max }\left(M\left(j \omega_{0}\right)\right)} v_{1} u_{1}^{\prime} \nonumber$

Write the vectors $$v_{1}$$ and $$u_{1}^{\prime}$$ as

$v_{1}=\left[\begin{array}{c} \pm\left|a_{1}\right| e^{j \theta_{1}} \\ \pm\left|a_{2}\right| e^{j \theta_{2}} \\ \vdots \\ \pm\left|a_{n}\right| e^{j \theta_{n}} \end{array}\right], \quad u_{1}^{\prime}=\left[\begin{array}{ccc} \pm\left|b_{1}\right| e^{j \phi_{1}} & \pm\left|b_{2}\right| e^{j \phi_{2}} & \cdots & \pm\left|b_{n}\right| e^{j \phi_{n}} \end{array}\right] \label{20.20}$

where $$\theta_{i}$$ and $$\phi_{i}$$ belong to the interval [0, $$pi$$). Note that we used $$\pm$$ in the representation of the vectors $$v_{1}$$ and $$u_{1}^{\prime}$$ so that we can restrict the angles $$\theta_{i}$$ and $$\phi_{i}$$ to the interval [0, $$pi$$). Now we can choose the nonnegative constants $$\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} \text { and } \beta_{1}, \beta_{2}, \cdots, \beta_{n}$$ such that the phase of the function $$\frac{s-\alpha_{i}}{s+\alpha_{i}}$$ at $$s=j \omega_{0}$$ is $$\theta_{i}$$, and the phase of the function $$\frac{s-\beta_{i}}{s+\beta_{i}}$$ at $$s=j \omega_{0}$$ is $$\phi_{i}$$. Now the destabilizing $$\Delta\left(s\right)$$ is given by

$\Delta(s)=\frac{1}{\sigma_{\max }\left(M\left(j \omega_{0}\right)\right)} g(s) h^{T}(s)$

where

$g(s)=\left[\begin{array}{c} \pm\left|a_{1}\right| \frac{s-\alpha_{1}}{s+\alpha_{n}} \\ \pm\left|a_{2}\right| \frac{s-\alpha_{2}}{s+\alpha_{2}} \\ \vdots \\ \pm\left|a_{n}\right| \frac{s-\alpha_{n}}{s+\alpha_{n}} \end{array}\right], \quad h(s)=\left[\begin{array}{c} \pm\left|b_{1}\right| \frac{s-\beta_{1}}{s+\beta_{1}} \\ \pm\left|b_{2}\right| \frac{s-\beta_{2}}{s+\beta_{2}} \\ \vdots \\ \pm\left|b_{n}\right| \frac{s-\beta_{n}}{s+\beta_{n}} \end{array}\right]$