Homework 9

Consider the polynomial $a(s) + Kb(s) = 0$ where $a$ and $b$ are polynomials. Show that the roots of this polynomial vary continuously as $K \in \mathbb{R}$ changes. That is, as $K$ approaches any fixed number, say 1, through a sequence, then the roots of the polynomial sequence converges to the roots of the polynomial with $K = 1$.

Hint: We can follow the three steps below:

a) First show that for any polynomial $p(s) = a_0 + a_1 s + \cdots + a_{n-1}s^{n-1} + s^n$, the matrix

$A = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -a_0 & -a_1 & -a_2 & \cdots & -a_{n-1} \end{bmatrix}$

has $p(s)$ as its characteristic polynomial.

b) From this, one can show through some algebraic analysis that eigenvalues are uniformly bounded where the bound continuously changes with the polynomial coefficients: Consider an eigenvalue $\lambda$ with eigenvector $v$ with $Av = \lambda v$ and $v = \begin{bmatrix} v_1 & \cdots & v_n \end{bmatrix}^T$. Then, for every $i$, we have that

$|\lambda| |v_i| = \left|\sum_j A(i,j) v(j)\right| \leq \left(\max_i \sum_j |A(i,j)|\right) \max_j |v_j|$

and hence

$|\lambda| \max_i |v_i| \leq \left(\max_i \sum_j |A(i,j)|\right) \max_j |v_j|$

and thus $|\lambda| \leq \max_i \sum_j |A(i,j)|$ for all eigenvalues. As a result $|\lambda_j| \leq \max(1, \sum_i |a_i|)$ for each $1 \leq j \leq n$. Alternatively, you can use a very useful result known as Gershgorin circle theorem. What matters is that the bound (on the eigenvalues) is uniformly bounded (as $K$ changes), since the obtained bound above changes continuously with $K$.

c) Now, as $K$ changes, the coefficients of the polynomial continuously change. Therefore, by part b), the roots of the polynomial are uniformly bounded for $K$ sufficiently close to 1. This implies that for every sequence $K_m \to 1$, the corresponding sequence of roots must contain a converging subsequence. Then, we can arrive at a contradiction for the following contrapositive argument: suppose that $K_m$ approaches 1 along some sequence but the roots do not converge to the roots of the polynomial with $K = 1$.

Complete the argument. As noted above, the argument applies for an arbitrary limit for the $K_m$ sequence, in place of 1.

Consider Figure 1 (a standard feedback control block diagram with $r \to (+) \to C(s) \to P(s) \to y$, with negative feedback from $y$ to the summing junction).

a) [Proportional Control] Let the plant and controller be given with $P(s) = \frac{1}{s^2}$ (double integrator dynamics) and $C(s) = k_p$ (such a controller is known as a proportional controller). Find the root locus as $k_p$ changes from 0 to $\infty$. Can this system be made stable by such a proportional controller?

b) [PD Control] Consider $P(s) = \frac{1}{s^2}$ (double integrator dynamics), and $C(s) = k_p + k_d s$ (the term PD-controller means proportional plus derivative control). Let $k_p = k_d = K$. Find the root locus as $K$ changes from 0 to $\infty$. Conclude, while comparing with part a above, that the addition of the derivative controller has pushed the poles to the left-half plane (thus, leading to stability!)

c) [Reference Tracking] For the system with the controller in part b), let $K > 0$: Let $r(t) = A1_{\{t \geq 0\}}$ for some $A \in \mathbb{R}$. Find $\lim_{t \to \infty} y(t)$. Hint: Apply the Final Value Theorem. We have that $R(s) = A\frac{1}{s}$ and with $Y(s) = A \frac{Ks+K}{s(s^2+Ks+K)}$ we have that $sY(s)$ has all poles on the left-half plane. By the final value theorem, the limit is $A$. Thus, the output asymptotically tracks the input signal.

Some engineering interpretation. $\frac{1}{s^2}$ can be viewed as a map from acceleration to position: $\frac{d^2p}{dt^2} = u$; part a) in the above suggests that if we only use position error we cannot have a stable tracking system; but if we use position and derivative (that is, velocity) information, then we can make the system stable. Furthermore, if we have a reference tracking problem, the output will indeed track the reference path.

Consider Figure 1 (same feedback block diagram as Problem 2).

a) Consider $C(s) = K$, $P(s) = \frac{1}{(s+1)^3}$. Is this system stable for a given $K > 0$. Explain through the Nyquist stability criterion.

b) Consider $P(s)C(s) = \frac{K}{(s+a)^3}$ with the controller in an error feedback form so that the closed loop transfer function is given by $\frac{P(s)C(s)}{1+P(s)C(s)}$. Is this system stable? Explain through the Nyquist stability criterion.

c) Let $P(s)C(s) = \frac{3}{(s+1)^3}$. Compute the gain stability margin. Draw the phase stability margin on the Nyquist curve.

Consider Figure 1 (same feedback block diagram). Let $C(s) = K$, $P(s) = \frac{s+1}{s(\frac{s}{10}-1)}$.

Study stability properties using either the root locus and Nyquist stability criteria.

a) Consider a linear system with feedback, which we assume to be stable: We generalize the observation above by viewing the input as one in $L_2(\mathbb{R}; \mathbb{C})$. Consider then the gain of a linear system with:

$\gamma := \sup_{u \in L_2(\mathbb{R};\mathbb{C}): \|u\|_2 \neq 0} \frac{\|y\|_2}{\|u\|_2}$

We know, by Parseval's theorem, that

$\gamma := \sup_{u \in L_2(\mathbb{R};\mathbb{C}): \|u\|_2 \neq 0} \frac{\|\mathcal{F}_{CC}(y)\|_2}{\|\mathcal{F}_{CC}(u)\|_2}$

By writing $i\omega$ instead of $i2\pi f$ in the following, we have

$\mathcal{F}_{CC}(y)(\omega) = \mathcal{F}_{CC}(u)(\omega) G(i\omega),$

where $G$ is the closed-loop transfer function with $s = i\omega$.

Show that

$\gamma = \sup_{u \in L_2(\mathbb{R};\mathbb{C}): \|u\|_2 \neq 0} \sqrt{\frac{\int_\omega |\mathcal{F}_{CC}(u)(i\omega) G(i\omega)|^2 d\omega}{\|\mathcal{F}_{CC}(u)\|^2}} = \sup_{\omega \in \mathbb{R}} |G(i\omega)| =: \|G\|_\infty$

b) Let us define a system to be $L_2$-stable if a bounded input, in the $L_2$-sense, leads to a bounded output in the $L_2$-sense. Show that BIBO stability of a linear system implies $L_2$-stability.

c) Prove the following statement: [Small Gain Theorem] Consider a feedback control system with closed-loop transfer function $G(s) = \frac{H_1(s)}{1 + H_1(s)H_2(s)}$, where $H_1$ and $H_2$ are stable. Suppose further that the gains of $H_1$ and $H_2$ are $\gamma_1$ and $\gamma_2$, respectively. Then, if $\gamma_1 \gamma_2 < 1$, the closed-loop system is stable.