Stable Feedback and State Estimation Techniques

Posted on Jan 4, 2025 in Electronics

Stable Feedback: General Case

Pole-placement design procedure is achieved by setting

det(sI – A + BK ) = s ⁿ+a_n–1s^n–1++a₁s +a₀

92

14.1 Stable Feedback: General Case      93

\Solve for n unknown K₁ ,…, K_n . The equations are linear!

K can be computed directly from this formula if and only if (A,B) is controllable.

Characteristic polynomialÞ det(sI–A+BK)=0

\s ⁿ+(a_n–1+ K_n)s ^n–1+  +(a₁+ K₂)s +(a₀+ K₁)=0

Since desired characteristic polynomial is

a_c (s) = sⁿ+a_n–1s^n–1+  +a₁s+a₀= 0

\K_i=a_i–1– a_i–1i =1,2,…n

Alternatively, K can be computed according to the Ackermann’s Formula as follows:

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94    Linear system: Lecture 14

Let the desired characteristic polynomial be

a_c (s) = s²+ (l₁+l₂ )s +l₁l₂= 0

Based on Ackermann’s formula, need to computeC_x

Example 2: Let the design specifications require a critically damped system with asettling time of 1 sec., i.e, 4t=1.

If we compute the closed-loop transfer function, we get the figure 14-2.

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14.2 State Estimation: Full-Order Observer  95

.5            1           1.5t

Figure 14-2.System plot for example 2.

14.2  State Estimation: Full-Order Observer

14.2.1 Theorem

The estimator equation is expressed as

We need to choose matrices F, G, and H such that xˆ(t) is accurate estimate of x(t). Then in the control system, the estimated states are used to generate the feedback control, i.e. u = –Kxˆ. To do this, the transfer function fromutoxˆ_imust be equal to the transferfunction from u to x_i , i =1,2,…,n, that is

ˆ

^X i ^(s)₌^X i^(s)_{i =1,2,…n}

U _i(s)U _i(s)

The Laplace Transform of state equations are

^ìsX (s)- x(0)= AX (s)+ BU (s), x( 0 ) is unknown

ï

í

ï_îY (s)= CX (s), A, B, C, and D known

By neglecting i.c. Þ X (s) = (sI – A)^–1BU (s)

The Laplace Transform of estimator is (neglecting i.c.)

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96    Linear system: Lecture 14

ˆ

_{We want}^X i^(s)₌ ^Xi^(s)

U (s)U (s)

\(sI – A)^–1B = (sI – F)^–1[H+ GC(sI – A)^–1B]Collecting (sI – A)^–1B terms

[I -(sI – F)^–1 GC](sI – A)^–1 B =(sI – F)^–1 H

Since (sI – F)^–1 (sI – F) = I \(sI – F)^–1[sI– F – GC](sI – A)^–1B = (sI – F)^–1H \[sI – F – GC](sI – A)^–1 B = H Þ(sI – A)^–1 B =(sI – F – GC)^–1 H

This equation is satisfied if we choose H=B and F+GC=A

\F = A – GC  and  H = B

G is chosen such that an acceptable transient response or frequency response for the state estimator is achieved.

\xˆ= Axˆ+ Bu + Gy – GCxˆ

ìïxˆ= Axˆ+ Bu + G( y – yˆ)

í

ïyˆ= Cxˆ

î

Plant

Since y=Cx , hencee = Ax -( A – GC) xˆ- GC x =( A – GC) (x – xˆ)

\e =( A – GC) e

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14.2 State Estimation: Full-Order Observer       97

The error in the estimation of the states is governed by the above dynamic equation with the characteristic polynomial: det(sI–A+GC)=0

The gain vector G is chosen to make the dynamics of the estimator faster than the open-loop system dynamics (~ 2 to 4 times faster).

14.2.2  Design of the State Estimator

1)xˆ=( A – GC)xˆ+ Bu + Gy , the characteristic polynomial isdet(sI–A+GC)=0

2)Let the desired char. poly. be a_e (s) , that reflects the desired transient behavior:

a_e (s) = sⁿ+a_n–1s^n–1+  +a₁s+a₀= 0

3)The gain matrix G is calculated to satisfy det(sI – A + GC) =a_e (s)

4) Using the transformationxo= To-1xtransforms the system into observable

̇

Canonical form

[][]

{

Now A – GC becomes

[]

\det(sI – A + GC) =a_e (s) = s ⁿ+a_n–1s^n–1+  +a₁s+a₀

\ g_i=a_i–1– a_i–1i =1,2,…n

Now using Ackermann’s formula we have [if and only if (A, C) is observable]

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98   Linear system: Lecture 14

Acontroller was designed for the characteristic polynomial a_c (s) = s ²+ 8s + 32 , which yielded a time constant t=.25sec,x=.707

We choose the estimator to be critically damped with a  t=.1 sec.

\a_e (s) = (s +10)²= s ²+ 20s +100

The transfer function of the controller-estimator is

G_ec(s)= K(sI – A + BK + GC)^–1 G

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14.2 State Estimation: Full-Order Observer  99

Figure 14-5.Block diagram for example 4.

The characteristic polynomial of the closed-loop system is

Figure 14-6.Signal flow graph.

Defining

̇

̇

̇

[̇][][][]

y

≠

.511.5t

Figure 14-7.Estimator response.

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100    Linear system: Lecture 14

14.2.3  Closed-Loop System Characteristics

The characteristic polynomial with full state feedback is and the characteristic polynomial of a state estimator is

We now derive the characteristic polynomial of the closed-loop system.

The 2n roots of the closed-loop characteristic polynomial are then the n roots of the pole-placement design plus the n roots of the estimator. This is fortunate, since otherwise the roots from the pole-placement would have been shifted by the addition of the estimator. This is called the separation principle.

Example 6: From the previous examples we have

We have

or equivalently from()

[]

The above characteristic polynomial is obtained.

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14.3 Reduced-Order Estimators101

14.3  Reduced-Order Estimators

Estimators developed so far are called full-order estimators since all states of the plant are estimated. However, usually from output measurement some states are directly available. Hence it is not logical to estimate states that are directly measured.

Assume without loss of generality that

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102    Linear system: Lecture 14

Thus the error dynamic is e=x_e–xˆ_eÞe=(A_ee–G_eA_1e)e

Thus, the characteristic polynomial of the reduced-order estimator is

a_e (s) = det(sI – A_ee+ G_e A_1e ) = 0

The above estimator requires measurement of y . To relax this requirement, define a new variable xˆ_e1=xˆ_e–G_ey or xˆ_e=xˆ_e1+G_ey

Substituting in observer dynamic yields

xˆ_e1+ G_e y =( A_ee– G_e A_1e)(xˆ_e1+ G_e y)+ A_e1 y + B_eu + G_e( y – a₁₁ y – b₁u)

\xˆ_e1=( A_ee– G_e A_1e)xˆ_e1+( A_e1– G_e a₁₁+ A_ee G_e– G_e A_1eG_e) y +(B_e– G_eb₁)u

The control u is

u = –K₁ y – K_e xˆ_e= –K₁ y – K_e(xˆ_e1+ G_e y)

Figure 14-8.Block diagram of the combined plant and reduced order observer.

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14.3 Reduced-Order Estimators103

Example 7: {^{̇[         ]        [ ]}the controller designed earlier yielded

As for the estimator of reduced order (1^st order), let

From Ackermann’s formula:

̇

̇

The estimated value ofis then

Example 8:If we combine plant equations with the estimator and the control we have

The signal flow graph becomes

1-8u1

-10

-100¹⁰

Figure 14-9.System signal flow graph.

The characteristic polynomial is

⏟⏟

Now opening the graph from u we calculate

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104    Linear system: Lecture 14

Controller  – estimatorplant

R=0₊

Figure 14-10.Closed-loop combined plant and controller estimator.

√

14.4  Reduced-Order State Estimator: General Case

ìx = Ax + Bu,x Î R ⁿ,  u Î R ^p

Consider ^ïí

ïy = Cx  , y Î R ^q

î

Assumption:  C has full rank,

^{0  ù}

^úé y ù

^ú ê ú

ê_ë z ú_û

Iú

n –q _û

which is an estimate of the original vector  x.

y

Figure 14-11.Block diagram for system with reduced order estimator.

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15 Lecture 15

Objectives

1)Tracking Problem

2)Closed-Loop Pole-Zero Assignment

3)General Case Using Phase Variable Canonical Form

15.1  Tracking Problem

ìx = Ax + Bu

ï

^ïíu = –Kx + K _r r

ï

îïy = x1

Figure 15-1.State feedback diagram.

It is assumed that K is designed to meet a desired closed-loop characteristics. Problem is to design to satisfy the tracking requirement.

⏟, where a choice foris to defined as

106

15.1 Tracking Problem107

Figure 15-2.State diagram for example 1.

For the general case, if≠, instead,

then we express

since we want the system to be driven by the difference between the input and output.

Comparing with

{

Hence one gain has to be determined from other design criteria.

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