On the special problem: up to 10 points of the course. Due same time as last HW. Today topic is related to special problem. can be done with computer simulation or theory.
Back to robust stability. Important to distinguish between dependent and independent uncertainty When we have intervals \([a_i^{-},a_i^{+}]\) then each coefficient is independent of any other coefficient. But we can also have coefficients that are dependents on each others, or correlated. When we do, we call them \(q's\) and write \(P(s,q) = s^n + \sum _{k=0}^{n-1} a_k(q) s^k \). For example \[ P(s,q) = s^3 + (6+q_1+2q_2)s^2 + (q_1+4)s+ (q_3+6+q_2) \] This generalized the interval polynomial because we can now write \[ P(s,q) = s^n + \sum _{k=0}^{n-1} q_k s^k \]
This framework handles non-linear dependence on \(q\). For example, assume we have problem with interval matrix. i.e. study the stability of interval matrix \[ A(q) = \left ( \begin{array}{cc} -1+q(1) & 2+q(2) \\ q(3) & -2+q(4) \end{array} \right ) \] Where \(|q(i)|\leq r\). We now find the determinant of \(A(q)\) as function of \(q_i\) which comes to \[ P(s)=s^2+s(3-q_4-q_1)+(2-q_4-2q_1+q_1q_2-2q_3-q_3q_2) \] reader: how large \(r\) can be with robust stability still guaranteed
We can over-bound by applying Kharitonov polynomials. Let \(r=\frac{1}{2}\) then \[ P(s) = s^2 + [2,4]s+ \dots \] And now check stability of the four polynomials. Note that, if the result is negative, i.e. Kharitonov polynomials say that \(P(s)\) is not stable, it might still be stable. So then we need to try other tests to check. An example is now given using \(P(s) = s^4+s^3+2qs^2+s+q\) where \(1.5\leq q leq 4\) and by over-bounding, we obtain the interval polynomial \[ P(s)=s^4+s^3+2[-1.5,4]s^2+s+[1.5,4] \]
reader: Show the above leads to 4 polynomials which one or more will be unstable, hence by over-bounding, we conclude it is not stable. But the it is robustly stable, which can be verified using Routh table.
Lyapunov Lemma:
This is important for non-linear systems. For linear systems it is not as important. It is also useful with dependent certainty. To start with, let us forget about uncertainty for now. Initial \(q=0\) and consider \(\dot{x}=Ax\). Lyapunov method will generalize. But using eigenvalues to check for stability will only work in linear systems.
What does Lyapunov says about stability of \(\dot{x}=Ax\)? Need energy like function \(V(x)=x^Tx=\sum _{i=1}^{n} x_i^2\).
We will study behavior of \(V(x(t))\) along trajectory \(x(t)\). If \(V(x)\) goes to zero, then \(x(t)=0\). It is easier to study scalar function \(V(x)\) for stability.
To determine if \(V(x)\) goes to zero, we look at \(\frac{dV(x)}{dt} < 0\) \begin{align*} \frac{dv}{dt} &= \frac{d}{dx} x^Tx \\ &= x^TAx + \dot{x}^Tx \\ &= x^TAx + (Ax)^Tx\\ &= x^TAx + x^TA^Tx\\ &= x^T(A+A^T)x \end{align*}
iff \((A+A^T)\) is negative definite, then stable. But this definition of \(V(x)\) is not satisfactory. Here is a counter example. Consider stable system \(A=\left ( \begin{array}{cc} -1&3\\ 0&-2\end{array} \right )\) but \(A+A^T= \left ( \begin{array}{cc} -2&3\\3&-4\end{array} \right )\) so not stable. (ps. I must have copied something wrong. Since is stable, need to check notes with someone else).
So this energy function \(V(x)=x^Tx\) is not good and need to try a better one.
lemma the \(n\times n\) matrix \(A\) is stable the following conditions is satisfied: Given any \(n\times n\) positive definite symmetric matrix \(Q\), the equation \(A^TP+PA=-Q\) has positive definite symmetric solution \(P\)
Back to the above example. Solving it, using \(Q=\left ( \begin{array}{cc} 1&0\\0&1\end{array} \right )\), gives
Back to energy function. Use \(V(x)=x^TPx\) instead of \(V(x)=x^Tx\). If stable, this energy function will satisfy \(\frac{dv}{dt} = -x^TQx <0\).
So for robust stability, use Lyapunov method above and not the eigenvalues method.
perturbation on \(A\):
Consider system \(\dot{x} = \left [ A_o+\Delta A(q) \right ] x\) and \(q\) is bounded as before, and we are interested in robust stability. \(q\) above can even be time changing and the method will still work. But eigenvalues method will not work here. Assuming \(A_o\) is stable without loss of generality, the system is stable if
\[ \Vert \Delta A(t) \Vert _2 \leq \frac{\lambda _{min} Q}{2 \lambda _{max} P} \]