14. Observer Design

1. State Observer (Luenburger Observer (LO) )

1.1 Nominal LO Design

Given a linear system,

\[ \begin{aligned} \dot x(t) &= Ax(t) + Bu(t) \\ y(t) &= Cx(t) + Du(t) \end{aligned} \]

Since states cannot be measured directly, we can design a state observer to get the unknown states,

\[ \dot {\hat x} = A\hat x + Bu + L\underbrace{(y-C\hat x -Du)}_{\text{output estimation error}} \]

• \(\hat x\) is the state estimation

Given the state estimation error \(e = x - \hat x\), there have:

\[ \begin{aligned} \dot e &= \dot x - \dot{\hat x} \\ &= Ax + Bu - A \hat x - Bu - L(Cx+Du-C\hat x -Du) \\ &= (A-LC)e \end{aligned} \]

The eigenvalue for \((A-LC)\) have:

\[ \text{eig}(A-LC) = \text{eig}(A-LC)^T = \text{eig}(A^T-C^TL^T) \]

we can let:

• \(\tilde A = A^T\)
• \(\tilde B = C^T\)
• \(\tilde K = L^T\)

thus we have: \(A^T-C^TL^T = \tilde A - \tilde B \tilde K\), Given the observable matrix:

\[ \begin{aligned} &M_O = \begin{bmatrix} C \\ CA \\ \vdots \\ CA^{n-1}\end{bmatrix} \Rightarrow M_O^T = \begin{bmatrix} \tilde B & \tilde A \tilde B & \cdots & \tilde A^{n-1} \tilde B\end{bmatrix} \end{aligned} \]

A N&S condition for the design of an asymptotic observer with arbitrarily specified eigenvalue of the estimation error dynamics is that \((A,C)\) is observable.

Given a system \(p = 1\), we can apply Ackermann's formula,

\[ L^T = \begin{bmatrix} 0&\cdots&0&1\end{bmatrix}(M_O^T)^{-1} p^*(A^T) \]

When the system with \(p > 1\), \(C = \begin{bmatrix} C_1 \\ \vdots \\ C_p \end{bmatrix}\)

If \((A,C)\) is observable, \(\Rightarrow\) \(\exists i\), such that \((A, C_i)\) that is observable. We can apply Ackermann's formula on \((A,C_i)\):

\[ \begin{aligned} L &= \begin{bmatrix} L_1&\cdots&L_p\end{bmatrix} \\ &= \begin{bmatrix} 0&\cdots&0&L_i&0&\cdots&0\end{bmatrix} \end{aligned} \]

1.2 LO for the System With Disturbances

Given the linear system with disturbances,

\[ \begin{aligned} &\dot x = Ax + Bu + Md, &x\in\Re^n, u\in\Re^m\\ &y = Cx + Du + Nd, &y\in\Re^p, d\in\Re^r \end{aligned} \]

• If \(d\) is measurable, we can calculate the estimation error directly:

  \[ \dot{\hat x} = A \dot{\hat x} + Bu + Md + L(y - C\hat x - Du - Nd) \]

  We let \(e=x - \hat x\), thus we have:\(\dot e=(A-LC)e\)

• If \(d\) is not measurable:

  \[ \begin{aligned} \dot{\hat x} &= A \dot{\hat x} + Bu + L(y - C\hat x - Du) \\ e &= x - \hat x \\ \dot e &= (A-LC) e + (M-LN) d \end{aligned} \]

  If \(d(t) = \bar d\), \(\forall t \geq 0\),

  \[ e_\infty = -(A-LC)^{-1} (M-LN) \bar d \]

We can draw the system schematic:

1.3 Estimation of Constant Disturbances via Observer Design

Motivations:

1. You can estimate the state correctly
2. You can use the disturbances estimation for the disturbance compensation

Consider a system with disturbance:

To eliminate the disturbance, we can design a disturbance estimator:

If the disturbance is constant, \(d(t) = \bar d\), \(\forall t\),

\[ \begin{aligned} &\dot x = Ax + Bu + Md, &x\in\Re^n, u\in\Re^m\\ &y = Cx + Du + Nd, &y\in\Re^p, d\in\Re^r \end{aligned} \]

Since \(d\) is constant, \(\dot d = 0\), and we have:

\[ \begin{aligned} \begin{bmatrix} \dot x \\ \dot d \end{bmatrix} &= \underbrace{\begin{bmatrix} A&M \\ 0&0 \end{bmatrix}}_{\tilde A} \begin{bmatrix} x \\ d \end{bmatrix} + \underbrace{\begin{bmatrix} B \\ 0 \end{bmatrix}}_{\tilde B} u \\ y &= \underbrace{\begin{bmatrix} C&N \end{bmatrix}}_{\tilde C} \begin{bmatrix} x \\ d \end{bmatrix} + \underbrace{D}_{\tilde D} u \end{aligned} \]

The condition should be satisfied for designing the observer:

• \((\tilde A, \tilde C)\) must be observable \(\Leftrightarrow\) \((A,C)\) is observable

• \(\text{rank}\begin{bmatrix} A&M \\ 0&0 \end{bmatrix} = n+r \left\{\begin{aligned} &r \leq p \quad \text{more output than disturbances} \\ &\text{No invariant zero in 0 in the transfer matrix for } d \text{ to } y \end{aligned}\right.\)

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What are the close loop eigenvalues?

We assume the order of the system is \(n\),

\[ \begin{aligned} \text{system:}&\quad \left\{\begin{aligned} &\dot x = Ax + Bu \\ &y = Cx + Du \end{aligned}\right. \\ \text{control law:}&\quad u = -K\hat x + \gamma \\ \text{observer:}&\quad \dot{\hat x} = A \hat x + Bu + L(y-C\hat x -Du) \end{aligned} \]

Apply the controller and observer to the system:

\[ \begin{aligned} \dot x &= Ax - BK \hat x + B\gamma \\ \dot {\hat x} &= A\hat x -BK\hat x + B \gamma + L(y-C\hat x) \end{aligned} \\ \]

Since we have: \(e = x - \hat x\),

\[ \left\{\begin{aligned} \dot x &= (A-BK)x + BKe + B\gamma \\ \dot e &= (A-LC)e \end{aligned}\right. \]

1.4 Separation Principle

We can let:

\[ \tilde A = \begin{bmatrix} A-BK&BK \\ 0&A-LC \end{bmatrix} \]

The eigenvalue of \(\tilde A\) are the union of the eigenvalue of \(A-BK\) and \(A-LC\), it means that we can design controller and observer independently.

2. Stabilizing Regulator Transfer Matrix

We replaced observer with the transfer matrix \(R(s)\),

to find the expression of \(R(s)\),

\[ \begin{aligned} \text{control law: }& \quad u = -K\hat x \\ \text{observer: }& \quad \dot{\hat x} = A \hat x + Bu + L(y-C\hat x) \end{aligned} \]

We have:

\[ \begin{aligned} \dot{\hat x} &= \overbrace{(A-BK-LC)}^{\tilde A} \hat x + \overbrace{L}^{\tilde B}y \\ u &= \underbrace{-K}_{\tilde C}\hat x \end{aligned} \]

The transfer matrix is:

\[ R(s) = K(sI - (A-BK-LC))^{-1}L \]

3. Reduced Order Observer (RO)

For the linear system with the state space equation below:

\[ \left\{\begin{aligned} \dot x &= Ax + Bu \\ y &= Cx \end{aligned}\right. \]

We can transfer the system into canonical form,

\[ \begin{aligned} \tilde x &= Tx \\ \tilde A &= TAT^{-1} \\ \tilde B &= TB \\ \tilde C &= CT^{-1} = \begin{bmatrix} \mathbf I&\mathbf 0 \end{bmatrix} \end{aligned} \]

• \(T = \begin{bmatrix} C \\ T_1 \end{bmatrix}\), \(\det T \neq 0\)
• \((A,C)\) should be observable

We have the output part \(y\) and other unknown states \(x_v\), \(\tilde x = \begin{bmatrix} y \\ \tilde x_v \end{bmatrix}\), thus:

\[ \begin{aligned} \dot y &= \tilde A_{11}y + \tilde A_{12} \tilde x_v + \tilde B_1 u \\ \dot {\tilde x}_v &= \tilde A_{21}y + \tilde A_{22}\tilde x_v + \tilde B_2 u \end{aligned} \]

We can let:

• \(z = \dot y - \tilde A_{11}y - \tilde B_1 u = \tilde A_{12} \tilde x_v\)
• \(\eta = \tilde A_{21}y + \tilde B_2 u\)

and we can simplify the system:

\[ \left\{\begin{aligned} &\dot {\tilde x}_v = \tilde A_{22}\tilde x_v + \eta \\ &z = \tilde A_{12} \tilde x_v \end{aligned}\right. \]

Now this new system have \(n-p\) states,

\[ \begin{aligned} \dot{\hat{\tilde x}}_v &= \tilde A_{22} \hat{\tilde x}_v + \eta + L(z-\tilde A_{12}\hat{\tilde x}_v) \\ \dot{\tilde e}_v &= (\tilde A_{22} - L\tilde A_{12})\tilde e \end{aligned} \]

• \((\tilde A_{22}, \tilde A_{12})\) must be observable

For this system, we can apply the observer design:

\[ \begin{aligned} \dot{\hat{\tilde x}}_v &= \tilde A_{22} \hat{\tilde x}_v + \tilde A_{21}y + \tilde B_2 u + L(\dot y - \tilde A_{11}y - \tilde B_1 u-\tilde A_{12}\hat{\tilde x}_v) \\ \end{aligned} \]

Let \(w = \tilde x_v - Ly\),

\[ \begin{aligned} \dot w &= (\tilde A_{22} - L\tilde A_{12})w + (\tilde A_{21} - L\tilde A_{11})Ly + \tilde A_{21}y + \tilde B_2 u - L\tilde A_{11}y - L\tilde B_1 u \\ \end{aligned} \]

And we have: \(\hat{\tilde x}_v = w + Ly\),

\[ \hat x = T^{-1} \hat{\tilde x}_v = T^{-1} \begin{bmatrix} y\\ \hat{\tilde x}_v \end{bmatrix} \]

Example

\[ \left\{\begin{aligned} &\dot x_1 = x_2 \\ &\dot x_2 = u \\ &y = x_1 \end{aligned}\right. \]

• \(A = \begin{bmatrix} 0&1\\0&0 \end{bmatrix}\), \(B = \begin{bmatrix} 0\\1 \end{bmatrix}\), \(C = \begin{bmatrix} 1&0 \end{bmatrix}\)
• \((A,C)\) is observable

let \(z = \dot y\), We have \(\left\{\begin{aligned} &\dot y = x_2 \\ &\dot x_2 = u \\ &z = x_2 \end{aligned}\right.\),

from these equations:

\[ \begin{aligned} \dot {\hat x}_2 &= u + L(z - \hat x_2) \\ &= u + L(\dot y - \hat x_2) \end{aligned} \]

let \(w = \hat x_2 - Ly\),

\[ \begin{aligned} \dot w &= \dot{\hat x}_2 - L\dot y \\ &= u - L\hat x_2 \\ &= -L(\hat x_2 - Ly) - L^2y + u \\ &= -Lw - L^2y + u \end{aligned} \]

And because there have: \(\hat x_2 = w + Ly\), we can draw the block diagram:

Thus all the states could be obtained: \(\hat x = \begin{bmatrix} y \\ \hat x_2 \end{bmatrix}\)

4. Pole placement and State Observer in Discrete Time Systems

Given the discrete time system with state space equation:

\[ \begin{aligned} x(k+1) &= Ax(k) + Bu(k) \\ y(k) &= Cx(k) + Du(k) \end{aligned} \]

Assume all the states can be observed directly, thus:

• \(y(k) = x(k)\)

And we can design the state feedback controller for the system,

\[ u = -Kx(k) + \gamma(k) \]

Apply the controller to the system, we can get the closed loop function:

\[ x(k+1) = \underbrace{(A-BK)}_{\bar A_{cl}}x(k) + B\gamma(k) \]

To make sure the controller is feasible, it should meet the condition:

• \((A, B)\) is reachable

When designing the eigenvalues, we need to consider the stability regions of the system

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Since not all the state could be obtained directly, for estimating the state, there have following methods:

• Predictor: use the info up to \(K\) to estimate \(x(k+1)\)
• Filter: use the info up to \(K\) to estimate \(x(k)\)

4.1 Predictor

We can design a state estimator with predicted states:

\[ \hat x(k+1|k) = A\hat x(k|k-1) + Bu(k) + L(y(k) - C\hat x(k|k-1) - Du(k)) \]

And the estimation error is:

\[ \begin{aligned} e(k) &= x(k) - \hat x(k|k-1) \\ e(k+1) &= (A-LC)e(k) \end{aligned} \]

To make the observer design feasible,

• \((A, C)\) must be observable

If all the eigenvalues of \(A-LC\) are set to 0, then you get a deadbeat observer (for the finite number of \(x\), the estimation error goes to \(0\))

Example

\[ \bar A = \mathbf 0_{3\times 3} \]

\[ \begin{aligned} e(k+1) &= \bar Ae(k) = 0 \end{aligned} \]

4.2 Filter

We can design a state estimator with current states:

\[ \hat x(k+1|k+1) = A\hat x(k|k) + Bu(k) + L(y(k+1) - C(A\hat x(k|k) + Bu(k)) - Du(k+1)) \]

And the estimation error is:

\[ \begin{aligned} e(k+1|k+1) &= x(k+1) - \hat x(k+1|k+1) \\ &= (A-L\underbrace{CA}_{\tilde C})e(k|k) \end{aligned} \]

To make the observer design feasible,

• \((A, \tilde C)\) must be observable

Info

Proof:

\[ \tilde M_o = \begin{bmatrix} \tilde C \\ \tilde CA \\ \vdots \\ \tilde CA^{n-1} \end{bmatrix} = \begin{bmatrix} \tilde CA \\ \tilde CA^2 \\ \vdots \\ \tilde CA^{n} \end{bmatrix} = M_o A \]

This is observable when \(A\) is non singular and \((A, C)\) is observable

If \(A\) is singular, \((A, \tilde C)\) is not observable even if \((A,C)\) is observable and \(\lambda = 0\) is in the unobservable part.

Example

For a system:

\[ \begin{aligned} x(k+1) &= Ax(k) \\ z(k) &= CAx(k) \end{aligned} \]

• \(x(0) = v\), \(v\) is the eigenvector assume to \(\lambda = 0\)
• \(x(1) = Av = \lambda v = 0\)
• \(x(k) = 0\), \(k\geq 1\)
• \(z(0) = CAV = 0\)
• \(z(k) = 0\), \(k \geq 0\)

4.3 Regulator Transfer Matrix

For the discrete time system:

\[ \begin{aligned} x(k+1) &= Ax(k) + Bu(k) \\ y(k) &= Cx(k) \end{aligned} \]

Within the predictor:

\[ \hat x(k+1|k) = A\hat x(k|k-1) + Bu(k) + L(y(k) - C\hat x(k|k-1) - Du(k)) \]

And control law:

\[ u(k) = -Kx(k|k-1) \]

We can draw the system schematics:

And we can ge the expression for \(R(z)\):

\[ U(z) = -\underbrace{K(zI - A + BK + LC)^{-1}L}_{R(z)} Y(z) \]

• \(R(z)\) is strictly proper

Within the Filter Design:

\[ \hat x(k+1|k+1) = A\hat x(k|k) + Bu(k) + L(y(k+1) - C(A\hat x(k|k) + Bu(k)) - Du(k+1)) \]

The expression of \(R(z)\) becomes:

\[ U(z) = - \underbrace{K(zI - A+BK+LCA-LCBK)^{-1}Lz}_{R(z)} Y(z) \]

• \(R(z)\) is proper

Example

Consider a strict proper case for the SISO system:

\[ \begin{aligned} R(z) &= \frac{b_{n-1}z^{n-1} + b_{n-2}z^{n-2} + \dots + b_{1}z + b_0}{z^n + a_{n-1}z^{n-1} + \dots + a_{1}z + a_0} \\ &= \frac{(b_{n-1}z^{n-1} + b_{n-2}z^{n-2} + \dots + b_{1}z + b_0)z^{-n}}{(z^n + a_{n-1}z^{n-1} + \dots + a_{1}z + a_0)z^{-n}} \\ &= \frac{b_{n-1}z^{-1} + b_{-2}z^{n-2} + \dots + b_0z^{-n}}{1 + a_{n-1}z^{-1} + \dots + a_0z^{-n}} \\ \end{aligned} \]

We can let \(V(z) = R(z) Y(z)\),

\[ (1 + a_{n-1}z^{-1} + \dots + a_0z^{-n})V(z) = (b_{n-1}z^{-1} + b_{-2}z^{n-2} + \dots + b_0z^{-n}) Y(z) \]

Do the inverse z-transform, we can get:

\[ v(k) = -a_{n-1}v(k-1) - \dots - a_0 v(k-n) + b_{n-1}y(k-1) + \dots + b_0 y(k-n) \]

Example

And we consider the proper case for the SISO system:

\[ \begin{aligned} R(z) &= \frac{b_{n}z^{n} + b_{n-1}z^{n-1} + b_{n-2}z^{n-2} + \dots + b_{1}z + b_0}{z^n + a_{n-1}z^{n-1} + \dots + a_{1}z + a_0} \\ &= \frac{(b_{n}z^{n} + b_{n-1}z^{n-1} + b_{n-2}z^{n-2} + \dots + b_{1}z + b_0)z^{-n}}{(z^n + a_{n-1}z^{n-1} + \dots + a_{1}z + a_0)z^{-n}} \\ &= \frac{b_{n} + b_{n-1}z^{-1} + b_{-2}z^{n-2} + \dots + b_0z^{-n}}{1 + a_{n-1}z^{-1} + \dots + a_0z^{-n}} \\ \end{aligned} \]

We can let \(V(z) = R(z) Y(z)\),

\[ (1 + a_{n-1}z^{-1} + \dots + a_0z^{-n})V(z) = (b_{n}z^{n} + b_{n-1}z^{-1} + b_{-2}z^{n-2} + \dots + b_0z^{-n}) Y(z) \]

Do the inverse z-transform, we can get:

\[ v(k) = -a_{n-1}v(k-1) - \dots - a_0 v(k-n) + b_{n}y(k) + b_{n-1}y(k-1) + \dots + b_0 y(k-n) \]

This case use the instant data from \(y(k)\), which does not provide time for calculation

4.4 Estimate of Constant Disturbance via Observer

Consider the discrete time system with constant disturbance:

\[ \begin{aligned} x(k+1) &= Ax(k) + Bu(k) + Md(k)\\ y(k) &= Cx(k) + Du(k) + Nd(k) \end{aligned} \]

We can write the equation with extended form:

\[ \begin{aligned} \begin{bmatrix} x(k+1) \\ d(k+1) \end{bmatrix} &= \underbrace{\begin{bmatrix} A&M \\ 0&I \end{bmatrix}}_{\bar A} \begin{bmatrix} x(k) \\ d(k) \end{bmatrix} + \underbrace{\begin{bmatrix} B \\ 0 \end{bmatrix} u(k)}_{\bar B} \\ y(t) &= \underbrace{\begin{bmatrix} C&N \end{bmatrix}}_{\bar C} \begin{bmatrix} x(k) \\ d(k) \end{bmatrix} + Du(k) \end{aligned} \]

It should satisfy that \((\bar A, \bar C)\) is observable, which is equivalent to:

• \((A, C)\) is observable
• \(r \leq p\)
• No invariant zeros in \(z = 1\) in the transfer matrix of \(d \to y\)

5. Discrete Time RO

Given the discrete time system:

\[ \begin{aligned} x(k+1) &= Ax(k) + Bu(k) \\ y(k) &= Cx(k) \end{aligned} \]

We can transform it into canonical form:

\[ \tilde x = \underbrace{\begin{bmatrix} C \\ T_1 \end{bmatrix}}_T x = \begin{bmatrix} y \\ \tilde x_v \end{bmatrix} \]

And there have \(z(k) = Cx(k) = y(k) - Du(k)\)

• \((A, C)\) should be observable

\[ \begin{aligned} y(k+1) &= \tilde A_{11}y(k) + \tilde A_{12} \tilde x_v(k) + \tilde B_1 u(k) \\ \tilde x_v(k+1) &= \tilde A_{21}y(k) + \tilde A_{22} \tilde x_v(k) + \tilde B_2 u(k) \end{aligned} \]

• \(\tilde A = TAT^{-1} = \begin{bmatrix} \tilde A_{11}&\tilde A_{12} \\ \tilde A_{21}&\tilde A_{22} \end{bmatrix}\), \(\tilde B = TB = \begin{bmatrix} \tilde B_1 \\ \tilde B_2 \end{bmatrix}\)

And,

• \(z(k+1) = y(k+1) - \tilde A_{11}y(k) - \tilde B_1 u = \tilde A_{12} \tilde x_v(k)\)
• \(\eta(k) = \tilde A_{21}y(k) + \tilde B_2 u\)
• \(\tilde x_v(k) = \tilde A_{22} \tilde x_v(k) + \eta(k)\)

For the filter design, there have:

\[ \hat{\tilde x}_v(k+1|k+1) = \tilde A_{22} \tilde x_v(k|k) + \eta(k) + L(z(k+1) - \tilde A_{12} \tilde x_v(k|k)) \]

We know that \(\tilde x = Tx\), The full state could be expressed as:

\[ \hat x(k+1|k+1) = T^{-1}\begin{bmatrix} y(k+1) \\ \tilde x_v(k+1|k+1) \end{bmatrix} \]

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