Exercise 4

Problem 1: Controller parameterizations

This problem studies the equivalence between the below 2-dof controller parameterizations

1.A:

Show that the observer-based feedback control law $u=-L\hat x + l_r y_r$, where $\hat x$ is given by a Kalman filter, kan be written as

$U(s) = -C_{\text{fb}}Y(s) + C_\text{ff}Y_r(s)$,

with

$\begin{aligned} C_\text{fb}(s) & = L(sI - A + BL + KC)^{-1} K \\ C_\text{ff} & = (I - L(sI - A + BL + KC)^{-1}B)l_r \\ & = (I + L(sI - A + KC)^{-1}B)^{-1}l_r \end{aligned}$

1.B

Show that the controller above can be written as

$R(s)U = -S(s)Y + T(s)Y_r$,

with

$\begin{aligned} R(s) & = \det(sI - A + BL + KC) \\ T(s) & = l_r\cdot\det(sI - A + KC) \end{aligned}$

Hence, the observer polynomial equals the zero polynomial of the feedforward controller $C_\text{ff}$

Problem 2: Euclid, Bezout and Diophantine

This exercise takes you through Euclid's algorithm, the Bezout identity and Diophantine equations. Going through this takes a bit of work, but will give you the tools to understand and implement pole-placing algorithms.

Euclid's algorithm

Let $l$, $m$ and $n$ be natural numbers. If $m = kn$ for some natural number $k$, then $n$ is a divisor of $m$. If also $l = k'n$ for some natural number $k'$, then $m$ is a common divisor of $l$ and $m$.

Euclid's algorithm finds the greatest common divisor of two natural numbers.

2a)

 

•  Let $m > n$ and let $l$ be the greatest common divisor of $m$ and $n$ ($l = \textrm{gcd}(m, n)$). Prove that $l = \textrm{gcd}(m - n, n)$

Now, let $m_0 = m$, $n_0 = n$ and consider the recursion

$\begin{aligned} m_{i+1} & = \max(m_i-n_i, n_i) \\ n_{i+1} & = \min(m_i-n_i, n_i) \end{aligned}$.

2b)

• Show that if $n_i = 0$ for some $i$, then $m_i = gcd(m, n)$.
• Show that the algorithm converges in finite time.
• Compute $gcd(210, 84)$ using Euclid's algorithm

An improvement due to Gabriel Lamé (1844)

Let $m$ and $n$ be integers. Define $r_{-2} = m$ and $r_{-1} = n$. Given $r_{k-2}$ and $r_{k-1}$ perform division with remainder by finding an integer quotient $q_k$ and remainder $r_k$:

$r_{k-2} = q_k r_{k-1} + r_k$,

until $r_n = 0$. This gives a sequence $r_{-2} = m, r_{-1} = n,r_0, \ldots, r_{n} = 0$.

2c)

• Prove that the above recursion converges.
• Prove that $r_{n-1}$is a common divisor of $r_{-2}, r_{-1}, \ldots, r_{n-2}$
• Let $l$ be another common divisor of $m$ and $n$, show that $l$ also divides $r_{n-1}$
• Compute $gcd(210, 84)$ using the improved version.

Divisors of integers are defined analogously to natural numbers.

Bézout's identity

Bezout's identity: Assume that $m$ and $n$ are integers and that $d = \textrm{gcd}(m, n)$. Then there exist integers $x$ and $y$ such that $mx + ny = d$.

2d)

• Show that if $(x, y)$ solves $mx + ny = d$, then all solutions can be obtained from $\left( x - k\frac{n}{d}, y + k \frac{m}{d}\right)$.
• Use Euclid's algorithm (based on division with remainder) to prove Bezout's identity.
• Show that the extended euclidean algorithm (below) computes x and y with $r_{n-1} = d$, $x = s_{n-1}$ and $y = t_{n-1}$.

The extended euclidean algorithm adds two extra quantities $s_{-2} = 1, s_{-1} = 0, \ldots, s_{n-1}$ and $t_{-2} = 0, t_{-1}= 1, \ldots, t_{n-1}$ (in addition ro $r_k$ and $q_k$ that follow the recursive relationship

$\begin{aligned} r_{k-2} & = q_k r_{k-1} + r_k \\ s_{k-2} & = q_ks_{k-1} + s_k \\ t_{k-2} & = q_kt_{k-1} + t_k \end{aligned}$

Diophantine equations

Consider the linear Diophantine equation $ax + by = c$ where $a, b, c$ are fixed integers and $x, y$ are unknown integers to be solved for.

2e)

• Prove that the linear Diophantine equation has a solution if, and only if $c = k \cdot \textrm{gcd}(a, b)$.

All of the development until now concerned integers. So how does this relate to pole placement? In L7:Pole Placement, we reduced pole-placement to solving a polynomial equation known as a polynomial Diophantine Equation:

$AX + BY = C$

where $A, B,\ C$ are given polynomials and $X,\ Y$ are unknowns to be solved for.

Polynomials (over a field) share many of the algebraic properties of integers:

Property	Polynomials	Integers
Irreducability	$(s - a)$ for any $a$	primes: n = -1,1,2,3,5,7,11..
Fundamental theorem	Algebra: $X(S) = (s-a_1)^{n_1}(s-a_2)^{n_2}\cdots(s-a_n)^{n_n}$	Arithmetic: $m = a_1^{n_1}a_2^{n_2}\cdots a_n^{n_n}$, $a_i$ prime
greatest common divisor	product of shared irreducible factors (with multiplicities)	product of shared primes (with multiplicities)
euclidean division	Given $A(s), B(s) \neq 0$ there exist two unique polynomials $Q(s), R(s)$ with $\textrm{deg}R(s) < \textrm{deg}B(s)$ so that $A(s) = Q(s)B(s) + R(s)$	Given integers: $a, b \neq 0$ there exist two unique integers $q,r$ with $|r| < |b|$ so that $a = bq + r$

2f) Translate 2c-2e to polynomials.

Problem 3

Use Euclid's algorithm to find all solutions to the equation

$s^2x(s) + (0.5s + 1)y(s) = 1$

where $x$ and $y$ are polynomials. Use the results to find a solution to the equation

$s^2f(s) + (0.5s + 1)g(s) = (s^2 + 2\zeta_c\omega_c s + \omega_c^2)(s^2 + 2\zeta_o\omega_os + \omega_o^2)$

such that the polynomials $f$ and $g$ have degrees 2 and 1 respectively.

Problem 4

For polynomials $A, B, C$ with $\textrm{deg}(A) = \textrm{deg}(B) = n$ and unknown polynomials $X, Y$ with degrees $n - 1$.
The Diophantine equation can be posed as a set of linear equations:

$\begin{bmatrix} a_0 & 0 & \ldots & 0 & b_0 & 0 & \ldots & 0 \\ a_1 & a_0 & \ldots & 0 & b_1 & b_0 & \ldots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\ a_{n-1} & a_{n-2} & \ldots & a_0 & b_{n-1} & b_{n-2} & \ldots & b_0 \\ a_n & a_{n-1} & \ldots & a_1 & b_n & b_{n-1} & \ldots & b_1 \\ 0 & a_n & \ldots &a_2 & 0 & b_n & \ldots & b_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & \ldots & 0 & a_n & 0 & \ldots & 0 & b_n \end{bmatrix} \begin{bmatrix} x_0\\ \vdots \\x_{n-1}\\ y_0 \\ \vdots \\ y_{n-1} \end{bmatrix} = \begin{bmatrix} c_0 \\ c_1 \\c_2 \\ \vdots \\c_n \\c_{n+1} \\ c_{n+2} \\ \vdots \\ c_{2n - 1} \end{bmatrix}$

If $A$ and $B$ have common factors, what are the left and right null spaces?

Problem 5

Prove that if $\lim_{s \to 0} P(s) \neq 0$, then zero stationary error implies  $\lim_{s \to \textcolor{red}{0}} C(s) = \infty$. (Discuss what assumptions are needed for this statement...)

Problem 6

This exercise compares two ways of introducing integral action into observer-based feedback controllers. Apply your results to

$P_2(s) = 50 \frac{(1/5 - s)(s + 5)}{ (1/5 + s)(5 - s) (s + 10)}$

from exercise 3.1.

Design controllers for $P_2$ and construct code to find the controller polynomials R,S,T for state-feedback controller with observer that includes integral action:

a)  with u = -K\hat x + k_i x_i + k_r r, as in the basic course Ch 8. See Explicit integral state

b) with "input offset estimation", see Disturbance observer

Explicit integral state

The following procedure can be used to introuce integral action. Consider

$\begin{aligned} \dot x & = Ax + Bu \\ y & = Cx \end{aligned}$

Augment the state vector with an extra component

$x_i = \int_0^t e(\tau)d\tau= \int_0^t \left(y_r(\tau) - y(\tau)\right)d\tau$.

Taking $x_a = \begin{bmatrix} x \\ x_i \end{bmatrix}$, we get

$\begin{aligned} \dot x_a & = \begin{bmatrix} A & 0 \\ -C & 0 \end{bmatrix} x_a + \begin{bmatrix}B \\ 0 \end{bmatrix} u + \begin{bmatrix}0 \\ 1\end{bmatrix}y_r \\ y & = \begin{bmatrix}C & 0 \end{bmatrix} \end{aligned}$.

A state-feedback control law $u = -K_ax_a + k_r y_r$ then has a natural decomposition into $u = -Kx - k_ix_i + k_r y_r$.

The state $x_i$ is is unobservable from $y$, but it's known to the controller, as we construct it causally from the known quantities $y$ and $y_r$. Typically one uses the controller $u = -K\hat x - k_i x_i + k_r y_r$ where $\hat x$ is generated by a Kalman filter. See the figure below.

Disturbance Observer

The standard observer-based feedback controller is based on modeling disturbances as perturbations to the initial conditions. This model is too simplistic to be useful in practice. The method of introducing an explicit integrator mediates this somewhat, but is limited to precisely integrators. The following allows the engineer to include a large variety of disturbance models into observer-based design methods.

Consider a process that is affected by a disturbance $d$, modeled by

$\begin{aligned} \dot x & = Ax + B_u u + B_d d\\ y & = Cx \end{aligned}$.

The disturbance $d$, is modeled as

$\begin{aligned} \dot w & = A_w w \\ d & = C_w w \end{aligned}$

The matrix $A_w$ typically has eigenvalues close to the origin (models constant load disturbances), or on the imaginary axis (models sinusoidal disturbances). The augmented system becomes

$\begin{aligned} \dot x_a & = \begin{bmatrix} A & B_dC_w \\ 0 & A_w \end{bmatrix} x_a + \begin{bmatrix} B_u\\ 0\end{bmatrix} u \\ y & = \begin{bmatrix} C & 0 \end{bmatrix}x_a \end{aligned}$,

where the augmented state is given by $x_a = \begin{bmatrix} x \\ w \end{bmatrix}$. The state $w$ is not controllable, but it is observable. We construct the controller as$u = -K\hat x - K_w \hat w + k_r y_r$, where $\hat x_a = \begin{bmatrix} \hat x \\ \hat w \end{bmatrix}$ is computed by a Kalman filter. The closed-loop system is described by

$\begin{aligned} \dot x & = (A - B_u K)x + (B_d C_w - B_u K_w)w - B_uK \tilde x - B_u K_w \tilde w\\ \dot w & = A_w w \\ \dot{\tilde x} & = (A - LC)\tilde x +B_d C_w \tilde w \\ \dot{\tilde w} & = A_w \tilde w - L_w C \tilde x \end{aligned}$

This method is particularly effective if one can pick $K_w$ such that $B_d C_w = B_u K_w$. Input offset estimation builds on the model $y = P(u + d)$, i.e. $B_u = B_d$. This is sometimes called matched disturbance.