Control Systems - Chibum Lee · 2015. 6. 1. · Chibum Lee -Seoultech Advanced Control Systems...

Control Systems

Compensator

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Advanced Control SystemsChibum Lee -Seoultech

Outline

Combined control law and estimator

Reference tracking input with the estimator and

controller

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Regulator with estimated states

Plant:

𝑥 = 𝐴𝑥 + 𝐵𝑢 + 𝐵𝑤𝑤

𝑦 = 𝐶𝑥

Control law:

𝑢 = −𝐾 𝑥

Estimator:

𝑥 = 𝐴 𝑥 + 𝐵𝑢 + 𝐿(𝑦 − 𝐶 𝑥)

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Compensator: Combined Control and Estimator

Plant: 𝑥 = 𝐴𝑥 + 𝐵𝑢 + 𝐵𝑤𝑤, 𝑦 = 𝐶𝑥

Estimator: 𝑥 = 𝐴 𝑥 + 𝐵𝑢 + 𝐿(𝑦 − 𝐶 𝑥)

Control law: 𝑢 = −𝐾 𝑥

Combining these equations:

𝑥 = 𝐴𝑥 − 𝐵𝐾 𝑥 + 𝐵𝑤𝑤

𝑥 = 𝐴 𝑥 − 𝐵𝐾 𝑥 + 𝐿𝐶(𝑥 − 𝑥)

• Thus the closed-loop state equations are:

𝑥 𝑥=

𝐴 −𝐵𝐾𝐿𝐶 𝐴 − 𝐵𝐾 − 𝐿𝐶

𝑥 𝑥+

𝐵𝑤0

𝑤

𝑦 = 𝐶 0𝑥 𝑥

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These equations may also be written in terms of the estimation

error 𝑥 = 𝑥 − 𝑥

𝑥 = 𝐴𝑥 − 𝐵𝐾 𝑥 + 𝐵𝑤𝑤

𝑥 = 𝐴 𝑥 − 𝐵𝐾 𝑥 + 𝐿𝐶(𝑥 − 𝑥)

• Subtracting: 𝑥 = 𝐴 − 𝐿𝐶 𝑥 + 𝐵𝑤𝑤

• Substituting: 𝑥 = 𝑥 − 𝑥 in the first equation yields:

𝑥 = 𝐴 − 𝐵𝐾 𝑥 + 𝐵𝐾 𝑥 + 𝐵𝑤𝑤

With this alternative set of state variables the closed-loop system

equations are:

𝑥 𝑥=

𝐴 − 𝐵𝐾 𝐵𝐾0 𝐴 − 𝐿𝐶

𝑥 𝑥+

𝐵𝑤𝐵𝑤

𝑤

𝑦 = 𝐶 0𝑥 𝑥

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The closed-loop characteristic polynomial is thus:

𝛼(𝑠) = det(𝑠𝐼 − 𝐴𝑐𝑙) = det𝑠𝐼 − 𝐴 + 𝐵𝐾 −𝐵𝐾

0 𝑠𝐼 − 𝐴 + 𝐿𝐶

which is block triangular. Hence

𝛼 𝑠 = det 𝑠𝐼 − 𝐴 + 𝐵𝐾 ⋅ det 𝑠𝐼 − 𝐴 + 𝐿𝐶

= 𝛼𝑘 𝑠 ⋅ 𝛼𝐿(𝑠)

• Thus, the poles of the CL system consist of the poles obtained by

full state feedback through 𝐾, together with the estimator poles

determined by the selection of 𝐿

• That is, the control law and the estimator can be designed

independently of each other

The controlled plant poles are not changed by the

design of estimator

Separation Principle

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By including the feedback law 𝑢 = −𝐾 𝑥 , the estimator

equation 𝑥 = 𝐴 𝑥 + 𝐵𝑢 + 𝐿(𝑦 − 𝐶 𝑥) becomes

𝑥 = 𝐴 − 𝐵𝐾 − 𝐿𝐶 𝑥 + 𝐿𝑦

𝑢 = −𝐾 𝑥

Transfer function for compensator:

𝐻 𝑠 =𝑈(𝑠)

𝑌(𝑠)= −𝐾 𝑠𝐼 − 𝐴 + 𝐵𝐾 + 𝐿𝐶 −1𝐿

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Selection of estimator poles

CL char. poly: 𝛼 𝑠 = 𝛼𝑘 𝑠 ⋅ 𝛼𝐿(𝑠)

The estimator poles (roots of 𝛼𝐿 𝑠 = 0) are usually chosen to be

between 2 to 6 times faster than the controller poles (roots of

𝛼𝑘 𝑠 = 0)

Clearly there is no direct effect of choosing faster estimator poles

on control effort. However, consider the effect of sensor noise.

Assume that the sensed output is 𝑦 = 𝐶𝑥 + 𝑣. The estimation error

dynamics then become:

𝑥 = 𝐴 − 𝐿𝐶 𝑥 + 𝐵𝑤𝑤 − 𝐿𝑣

Thus, the higher estimator gains required to produce faster

estimator dynamics will further amplify the sensor noise, thereby

corrupting the state estimates

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Symmetric root locus for SISO system estimator

Estimator dynamics with 'process noise' 𝑤 and 'sensor noise' 𝑣 :

The process noise 𝑤 can represent unknown disturbances and errors

in the plant model parameters

Recall the estimator equations: 𝑥 = 𝐴 𝑥 + 𝐵𝑢 + 𝐿(𝑦 − 𝐶 𝑥)

If 𝑤 is large:

• our plant model is uncertain, and hence a poor predictor

• we would put greater emphasis on the sensor data to correct the model

predictions larger 𝐿, faster estimator poles

If 𝑣 is large:

• our measurements are uncertain, and hence provide poor corrections

• we would put greater emphasis on the model predictions smaller 𝐿,

slower estimator poles

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𝑥 = 𝐴 − 𝐿𝐶 𝑥 + 𝐵𝑤𝑤 − 𝐿𝑣


Symmetric root locus for SISO system estimator

In optimal estimation theory

• the process noise is modeled as white noise with variance 𝜎𝑤2

• the sensor noise is modeled as white noise with variance 𝜎𝑣2

The estimator pole locations which will minimize the variance of the

state estimation error can be found from the solution of the

symmetric root locus (SRL) equation: 1 + 𝜌𝑁𝑒 𝑠 𝑁𝑒(−𝑠)

𝐷 𝑠 𝐷(−𝑠)= 0

where 𝜌 = 𝜎𝑤2/𝜎𝑣2 is a measure of the plant model uncertainty

relative to the measurement uncertainty, and

𝑍(𝑠)

𝑊(𝑠)= 𝐶 𝑠𝐼 − 𝐴 −1𝐵𝑤 =

𝑁𝑒(𝑠)

𝐷(𝑠)

is the plant transfer function from the process noise input to the

measured output

If the process noise 𝑤 and control input 𝑢 are additive (𝐵𝑤 = 𝐵),

the same SRL can be used for controller and estimator pole

selection 10


Ex: regulation of water level in two-tank system

The system can be described

Required closed-loop time

constants are 1 = 0.2 s, 2 = 0.05 s.

Assume

For

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Controller: For closed-loop time constants 1 = 0.2 s, 2 = 0.05 s; the

desired CL char. poly is

• Demonstrate Ackermann's formula:

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Estimator: Try estimator poles = 2 x controller poles;

• i.e., desired CL char. poly is α = (s +10)(s + 40) = s2 + 50s + 400

• Demonstrate Ackermann's formula:

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Complete system

• Equivalent feedback compensator:

• Transfer function for compensator:

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Compensator for reduced order estimator

The similar approach can be carried out for reduced order estimator

Control law is

Substituting it into the equation for reduced order estimator

where

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MATLAB design of regulator with state estimator

Select desired closed-loop poles dclp

• e.g. by use of symmetric root locus

Use place or acker to get state feedback gains

• K = place(A, B, dclp)

Select desired estimator poles dep

• e.g. by use of symmetric root locus

Use place or acker to get estimator gains

• L = place(A’, C’, dep)’

Form feedback compensator (regulator)

• H = reg(G, K, L) % or

• H = reg(G,K,L,sensors,known,controls)

Connect regulator to plant

• Gcl = feedback(G, H, +1) % or

• Gcl = feedback(G,H,feedin,feedout,+1)16


Ex. DC Servo-full order Compensator

Use the state-space pole placement method to design a compensator for

the DC servo system with the transfer function

𝐺(𝑠) =10

𝑠(𝑠 + 2)(𝑠 + 8)

Using a state description in OCF, place the control poles at 𝑝𝑐 = [−1.42 −

1.04 ± 2.14𝑗] locations and the full-order estimator poles at 𝑝𝑒 = [−4.25 −

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𝐴 =−10 1 0−16 0 10 0 0

, 𝐵 =0010

𝐶 = 1 0 0 , 𝐷 = 0


Ex. DC Servo-full order Compensator

The estimator gain Lt=place(A’,C’, pe), L=Lt’

𝐿 =0.561.4216

The compensator transfer function


𝑌(𝑠)= −𝐾 𝑠𝐼 − 𝐴 + 𝐵𝐾 + 𝐿𝐶 −1𝐿

= −190(𝑠 + 0.432)(𝑠 + 2.10)

(𝑠 − 1.88)(𝑠 + 2.94 ± 8.32𝑗)

Compensator has an unstable pole at 𝑠 = 1.88

but CL poles are stable.

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1 + 𝐾10

𝑠(𝑠 + 2)(𝑠 + 8)

(𝑠 + 0.432)(𝑠 + 2.10)

(𝑠 − 1.88)(𝑠 + 2.94 ± 8.32𝑗)= 0


Ex. DC Servo-reduced order compensator

Design a compensator for the DC servo system using the same

control poles at 𝑝𝑐 = [−1.42 − 1.04 ± 2.14𝑗] but the reduced-order

estimator poles at 𝑝𝑒 = −4.24 ± 4.24𝑗 positions (𝜔𝑛 = 6 rad/sec,

𝜁 = 0.707)

Sol)

𝐴𝑎𝑎 𝐴𝑎𝑏𝐴𝑏𝑎 𝐴𝑏𝑏

=−10 1 0−16 0 10 0 0

,𝐵𝑎𝐵𝑏

=0010

The estimator characteristic polynomials det 𝑠𝐼 − 𝐴𝑏𝑏 − 𝐿𝐴𝑎𝑏 = 0

The desired polynomials 𝛼𝑒 𝑠 = 𝑠2 + 2 ⋅ 0.707 ⋅ 6𝑠 + 36

Solving for 𝐿 using place 𝐿 =8.536

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Ex. DC Servo-reduced order compensator

The compensator transfer function


𝑌(𝑠)= 𝐶𝑟 𝑠𝐼 − 𝐴𝑟

−1𝐵𝑟 + 𝐷𝑟

= 20.92(𝑠 − 0.735)(𝑠 + 1.871)

𝑠 + 0.990 ± 6.120𝑗

Compensator has an unstable zero

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1 + 𝐾10

𝑠(𝑠 + 2)(𝑠 + 8)

(𝑠 − 0.735)(𝑠 + 1.871)

𝑠 + 0.990 ± 6.120𝑗= 0

Control Systems - Chibum Lee · 2015. 6. 1. · Chibum Lee -Seoultech Advanced Control Systems...

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Transcript of Control Systems - Chibum Lee · 2015. 6. 1. · Chibum Lee -Seoultech Advanced Control Systems...