Observer Synthesis using Linear Matrix Inequalities

4.7. Observer Synthesis using Linear Matrix Inequalities#

Note: This is an advanced topic

4.7.1. References#

Charkabarty, Ankush. “Guest Lecture: Exploiting Linear Matrix Inequalities In Control Systems Design”. (2015). Retrieved 29 March 2021.
Boyd, Stephen, et al. Linear matrix inequalities in system and control theory. Society for industrial and applied mathematics, 1994.
Caverly, Ryan James, and James Richard Forbes. “Lmi properties and applications in systems, stability, and control theory.” arXiv preprint arXiv:1903.08599 (2019).

4.7.2. Model Parameters#

4.7.3. Lyapunov Design#

Assuming no modeling error and ignoring disturbance inputs, the observer dynamics are described by

\frac{d e}{d t} = (A - L C) e

where $e = \hat{x} - x$ is the difference between the estimated and process states. Given a symmetric positive definite matrix $P$ , define the Lyapunov frunction $V (e)$ as $V (e) = e^{T} P e$ .

\begin{array}{r} \begin{aligned} \frac{d V}{d t} & = {\dot{e}}^{T} P e + e^{T} P \dot{e} \\ = e^{T} (A - L C)^{T} P e + e^{T} P (A - L C) e \\ = e^{T} (A^{T} P + P A - C^{T} L^{T} P - P L C) e \end{aligned} \end{array}

A sufficient condition for the global asympototic stability of observer is the left-hand side of this equation be negative for all $e \neq 0$ . This will be true if and only if $A^{T} P + P A - C^{T} L^{T} P - P L C$ is negative definite, i.e;,

e^{T} (A^{T} P + P A - C^{T} L^{T} P - P L C) e < 0 \forall e \neq 0

To provide some margin for robustness relative to model error, we will specify

\begin{array}{r} \frac{d V}{d t} \leq - γ V \end{array}

for some $γ > 0$ . When recast as a linear matrix inequality, we obtain

A^{T} P + P A - C^{T} L^{T} P - P L C ≺ - γ P

where the notation $Q ≺ 0$ implies $Q$ is negative definite.

As stated, given parameters $A$ , $C$ and $γ$ , the task is to find a symmetric positive definite $P ≻ 0$ and a matrix of observer gains $L$ which satistify the condition above. As stated, this is a bilinear relationship due to the terms $P L$ and $C^{T} L^{T}$ appearing in the expression. A standard ‘trick’ is to introduce a new decision variable $Y = P L$ to yield the linear matrix inequality

A^{T} P + P A - C^{T} Y^{T} - Y C + γ P ≺ 0

After finding a satisfactory solution $P ≻ 0$ and $Y$ , the desired observer gains are given by

L = P^{- 1} Y

The next challenge is to perform the necessary calculations.

The first challenge is that the above relationship is homogeneous. In other words, for any scale factor $α > 0$ , if $P$ and $Y$ satisfy the relationship then so do $α P$ and $α Y$ resulting in the same solution for $L$ . Without loss of generality, we can specify a specific solution by adding the constraints

P ≻ I

4.7.4. CVXPY#

!pip install cvxpy

# parameter estimates.
alpha = 0.00016       # watts / (units P * percent U1)
P1 = 200              # P units
P2 = 100              # P units
CpH = 4.46            # heat capacity of the heater (J/deg C)
CpS = 0.819           # heat capacity of the sensor (J/deg C)
Ua = 0.050            # heat transfer coefficient from heater to environment
Ub = 0.021            # heat transfer coefficient from heater to sensor
Uc = 0.0335           # heat transfer coefficient between heaters
Tamb = 21             # ambient room temperature

# state space model
A = np.array([[-(Ua + Ub + Uc)/CpH, Ub/CpH, Uc/CpH, 0], 
              [Ub/CpS, -Ub/CpS, 0, 0],
              [Uc/CpH, 0, -(Ua + Ub + Uc)/CpH, Ub/CpH],
              [0, 0, Ub/CpS, -Ub/CpS]])

Bu = np.array([[alpha*P1/CpH, 0], [0, 0], [0, alpha*P2/CpH], [0, 0]])

Bd = np.array([[Ua/CpH], [0], [Ua/CpH], [0]])

C = np.array([[0, 1, 0, 0], [0, 0, 0, 1]])

n = A.shape[0]
p = C.shape[0]

import numpy as np
import cvxpy as cp

P = cp.Variable((n, n), PSD=True)
Y = cp.Variable((n, p))

gamma = 1/10
constraints = [P >> np.eye(n)]
constraints += [A.T@P + P@A - C.T@Y.T - Y@C + gamma*P << 0]

prob = cp.Problem(cp.Minimize(0), constraints)
prob.solve()
L = np.linalg.inv(P.value)@Y.value
print(L)

[[0.56726483 0.34537519]
 [0.22509201 0.09832281]
 [0.34537519 0.56726483]
 [0.09832281 0.22509201]]

4.7.5. $H_{2}$ Optimal State Estimation#

Let’s now consider performance metrics for the observer. In particular, we assume our system is of the form

\begin{array}{r} \begin{aligned} \frac{d x}{d t} & = A x + B_{d} d \\ y & = C_{y} x \\ z & = C_{z} x \end{aligned} \end{array}

where $d$ are unmeasured disturbances, $y$ are process measurements, and $z$ are process variables we are attempting to estimate. Constructing an estimator

\begin{array}{r} \begin{aligned} \frac{d \hat{x}}{d t} & = A \hat{x} - L (\hat{y} - y) + B_{d} \hat{d} \\ \hat{y} & = C_{y} \hat{x} \end{aligned} \end{array}

and defining the state error in the usual way as $e_{x} = \hat{x} - x$ yields error dynamics given by

\begin{array}{r} \begin{aligned} \frac{d e}{d t} & = (A - L C_{y}) e + B_{d} (\hat{d} - d) \\ e_{z} & = C_{z} e \end{aligned} \end{array}

where $e_{z} = C_{z} e$ denotes the estimator error of interest. The design objective is to find values for $L$ that minimize the impact of changes in disturbance $\hat{d} - d$ on $e_{z}$ .

\begin{array}{r} \begin{aligned} [\begin{array}{c} P A + A^{T} P - Y C_{y} - C_{y}^{T} Y^{T} & P B_{d} \\ B_{d}^{T} P & - I \end{array}] & ≺ 0 \\ [\begin{array}{c} P & C_{z}^{T} \\ C_{z} & Z \end{array}] & ≻ 0 \\ T r (Z) & < ν \end{aligned} \end{array}

import numpy as np
import cvxpy as cp

P = cp.Variable((n, n), PSD=True)
Z = cp.Variable((p, p), PSD=True)
Y = cp.Variable((n, p))
nu = cp.Variable()

Cz = np.array([[1, 0, 0, 0], [0, 0, 1, 0]])

constraints = [cp.bmat([[A.T@P + P@A - Y@C - C.T@Y.T + np.eye(n), P@Bd], 
                       [Bd.T@P, -np.ones((1,1))]]) << 0]
constraints += [cp.bmat([[P, Cz.T], [Cz, Z]]) >> 0]
constraints += [cp.trace(Z) <= nu]
constraints += [nu >= 0]

prob = cp.Problem(cp.Minimize(nu), constraints)
prob.solve()
print(nu.value)
L = np.linalg.inv(P.value) @ Y.value
print(L)

0.01510843285193801
[[ 0.02450293 -0.00062794]
 [ 0.00531293  0.00279802]
 [-0.00062794  0.02450293]
 [ 0.00279802  0.00531293]]

import numpy as np
import cvxpy as cp

P = cp.Variable((5, 5), PSD=True)
Z = cp.Variable((1, 1), PSD=True)
Y = cp.Variable((5, 2))
nu = cp.Variable()


A_aug = np.vstack([np.hstack([A, Bd]), np.zeros([1, 5])])
Bu_aug = np.vstack([Bu, [[0, 0]]])
Bd_aug = np.vstack([np.zeros([4, 1]), [[1]]])
C_aug = np.hstack([C, np.zeros([2, 1])])

Cz_aug = np.array([[0, 0, 0, 0, 1]])


P = cp.Variable((5, 5), PSD=True)
Y = cp.Variable((5, 2))

gamma = 1/20
constraints = [P >> np.eye(5)]
constraints += [A_aug.T@P + P@A_aug - C_aug.T@Y.T - Y@C_aug + gamma*P << 0]

prob = cp.Problem(cp.Minimize(0), constraints)
prob.solve()
L = np.linalg.inv(P.value)@Y.value
print(L)

constraints = [cp.bmat([[A_aug.T@P + P@A_aug - Y@C_aug - C_aug.T@Y.T + np.eye(5), P@Bd_aug], 
                       [Bd_aug.T@P, -np.ones((1,1))]]) << 0]
constraints += [cp.bmat([[P, Cz_aug.T], [Cz_aug, Z]]) >> 0]
constraints += [cp.trace(Z) <= nu]
constraints += [nu >= 0]

prob = cp.Problem(cp.Minimize(0), constraints)
prob.solve()
print(P.value)
print(nu.value)
L = np.linalg.inv(P.value) @ Y.value
print(L)

---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-1-baf7d3ce8340> in <module>
      8 
      9 
---> 10 A_aug = np.vstack([np.hstack([A, Bd]), np.zeros([1, 5])])
     11 Bu_aug = np.vstack([Bu, [[0, 0]]])
     12 Bd_aug = np.vstack([np.zeros([4, 1]), [[1]]])

NameError: name 'A' is not defined

4.7.6. $H_{\infty}$ Optimal State Estimation#

import numpy as np
import cvxpy as cp

P = cp.Variable((n, n), PSD=True)
Y = cp.Variable((4, 2))

constraint = cp.bmat([[A.T@P + P@A - Y@C - C.T@Y.T + np.eye(n), P@Bd], 
                       [Bd.T@P, -2*np.ones((1,1))]]) << 0
prob = cp.Problem(cp.Minimize(0), [constraint])
prob.solve()
L = np.linalg.inv(P.value) @ Y.value
print(L)

[[ 0.13413964  0.02750461]
 [ 0.21936083 -0.01039676]
 [ 0.02750461  0.13413964]
 [-0.01039676  0.21936083]]

ev, _ = np.linalg.eig(A - L@C)
-1/ev

array([21.53881154+20.40481958j, 21.53881154-20.40481958j,
       24.71082258+21.80171218j, 24.71082258-21.80171218j])

(repr(L))

'array([[ 0.0247742 , -0.00252384],\n       [-0.00184805,  0.00579824],\n       [-0.00252384,  0.0247742 ],\n       [ 0.00579824, -0.00184805]])'