Setpoint Tracking in SS Systems 1. Setpoint Tracking in SS Systems - - PowerPoint PPT Presentation

1.

Setpoint Tracking in SS Systems

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k)

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action:

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k)

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error,

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k),

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) =

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) = xI(k) + r − Cx(k)

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) = xI(k) + r − Cx(k) Combine the two and form an augmented state equation:

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) = xI(k) + r − Cx(k) Combine the two and form an augmented state equation: x(k + 1) xI(k + 1)

• =

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) = xI(k) + r − Cx(k) Combine the two and form an augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) = xI(k) + r − Cx(k) Combine the two and form an augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1.

Setpoint Tracking in SS Systems In addition to the standard state space model, x(k + 1) = Ax(k) + Bu(k) y(k) = Cx(k) consider including an integral action: xI(k + 1) = xI(k) + e(k) e is the error, given by r − y(k), resulting in, xI(k + 1) = xI(k) + r − y(k) = xI(k) + r − Cx(k) Combine the two and form an augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Digital Control

1

Kannan M. Moudgalya, Autumn 2007

2.

Setpoint Tracking with Integral Mode

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

B Kp KI x(k) xI(k)

• +

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

= A −C 1

• −

B Kp KI x(k) xI(k)

• +

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

= A −C 1

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

= A −C 1

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

A − BK form.

2.

Setpoint Tracking with Integral Mode Consider the augmented state equation: x(k + 1) xI(k + 1)

• =

A −C 1 x(k) xI(k)

• +

B

• u(k) +

1

• r(k)

Introduce control law: u(k) = −

• Kp KI

x(k) xI(k)

• Closing the loop, the state equation becomes

= A −C 1 x(k) xI(k)

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

= A −C 1

• −

B Kp KI x(k) xI(k)

• +

1

• r(k)

A − BK form. Recall condition for pole placement.

Digital Control

2

Kannan M. Moudgalya, Autumn 2007

3.

Linear Quadratic Regulator - Formulation

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

Controllability matrix consists of B

• ,

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

Controllability matrix consists of B

• ,

AB −cB

• ,

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

Controllability matrix consists of B

• ,

AB −cB

• ,

A2B −cAB

• ,

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

Controllability matrix consists of B

• ,

AB −cB

• ,

A2B −cAB

• ,

A3B −cA2B

• ,

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

Controllability matrix consists of B

• ,

AB −cB

• ,

A2B −cAB

• ,

A3B −cA2B

• , · · · ,

An−1B −cAn−2B

3.

Linear Quadratic Regulator - Formulation When can we place the poles of A −C 1

• −

B Kp KI

• ?

When A −C 1

• ,

B

• is controllable!

Controllability matrix consists of B

• ,

AB −cB

• ,

A2B −cAB

• ,

A3B −cA2B

• , · · · ,

An−1B −cAn−2B

• This can be written as,

I 0 −C B AB A2B A3B · · · An−1B B AB A2B · · · An−2B

• Controllable if (A, B) is controllable.

Digital Control

3

Kannan M. Moudgalya, Autumn 2007

4.

Tandem Queue System

4.

Tandem Queue System

• Computing systems can be modelled as queueing

systesms.

4.

Tandem Queue System

• Computing systems can be modelled as queueing

systesms.

• Let us consider two such queues.

4.

Tandem Queue System

• Computing systems can be modelled as queueing

systesms.

• Let us consider two such queues.

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

Digital Control

4

Kannan M. Moudgalya, Autumn 2007

5.

Tandem Queue System

5.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Sequence of two queueing systems.

5.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Sequence of two queueing systems.
• Total time to complete a request (R) to be made small. R

is a sum of the two responses (R1, R2) in two queues.

5.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Sequence of two queueing systems.
• Total time to complete a request (R) to be made small. R

is a sum of the two responses (R1, R2) in two queues.

• Possible to regulate the response time by manipulating the

buffer size (K).

5.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Sequence of two queueing systems.
• Total time to complete a request (R) to be made small. R

is a sum of the two responses (R1, R2) in two queues.

• Possible to regulate the response time by manipulating the

buffer size (K).

• If buffer size is made small, response time will become small;

but some requests may not be honoured, owing to overflow.

Digital Control

5

Kannan M. Moudgalya, Autumn 2007

6.

Tandem Queue System

6.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Define states:

6.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Define states: x1(k)

x2(k)

• =

R1(k) − R1 R2(k) − R2

• Linearise. Nominal operating point: R1 = 2.5, R2 = 6.5

6.

Tandem Queue System

Total response time = R1 + R2 = R Queueing System 1 Queueing System 2 Buffer size (K) Response time (R1) Response time (R2)

• Define states: x1(k)

x2(k)

• =

R1(k) − R1 R2(k) − R2

• Linearise. Nominal operating point: R1 = 2.5, R2 = 6.5

A = 0.13 0.46 0.63

• ,

B = 0.069

• ,

C =

• 1 1
• Digital Control

6

Kannan M. Moudgalya, Autumn 2007

7.

Tandem Queue System

7.

Tandem Queue System A = 0.13 0.46 0.63

• ,

B = 0.069

• ,

C =

• 1 1

7.

Tandem Queue System A = 0.13 0.46 0.63

• ,

B = 0.069

• ,

C =

• 1 1
• Place poles of

A −C 1

• −

B Kp KI

• at desired loca-

tions

Digital Control

7

Kannan M. Moudgalya, Autumn 2007

8.

Steady State Solution to LQR Problem

8.

Steady State Solution to LQR Problem Recall the Discrete Riccati equation: S(k) = AT[S(k + 1) − S(k + 1)BR−1BTS(k + 1)]A + Q1

8.

Steady State Solution to LQR Problem Recall the Discrete Riccati equation: S(k) = AT[S(k + 1) − S(k + 1)BR−1BTS(k + 1)]A + Q1 Steady state solution: S(k) = S(k + 1) = S∞

8.

Steady State Solution to LQR Problem Recall the Discrete Riccati equation: S(k) = AT[S(k + 1) − S(k + 1)BR−1BTS(k + 1)]A + Q1 Steady state solution: S(k) = S(k + 1) = S∞ Substituting S∞ = AT[S∞ − S∞BR−1BTS∞]A + Q1

8.

Steady State Solution to LQR Problem Recall the Discrete Riccati equation: S(k) = AT[S(k + 1) − S(k + 1)BR−1BTS(k + 1)]A + Q1 Steady state solution: S(k) = S(k + 1) = S∞ Substituting S∞ = AT[S∞ − S∞BR−1BTS∞]A + Q1 Although we can solve this by iteration, it is not an easy problem.

Digital Control

8

Kannan M. Moudgalya, Autumn 2007

9.

Steady State Solution to LQR Problem

9.

Steady State Solution to LQR Problem Construct the control Hamiltonian matrix, Hc, given by Hc =

• A + BQ−1

2 BTA−TQ1 −BQ−1 2 BTA−T

−A−TQ1 A−T

9.

Steady State Solution to LQR Problem Construct the control Hamiltonian matrix, Hc, given by Hc =

• A + BQ−1

2 BTA−TQ1 −BQ−1 2 BTA−T

−A−TQ1 A−T

• Steady state solution to the Riccati equation:

9.

Steady State Solution to LQR Problem Construct the control Hamiltonian matrix, Hc, given by Hc =

• A + BQ−1

2 BTA−TQ1 −BQ−1 2 BTA−T

−A−TQ1 A−T

• Steady state solution to the Riccati equation:

S∞ = ΛIX−1

I

where

• XI ΛI

T is the eigenvector of Hc corresponding to the stable eigenvalues.

9.

Steady State Solution to LQR Problem Construct the control Hamiltonian matrix, Hc, given by Hc =

• A + BQ−1

2 BTA−TQ1 −BQ−1 2 BTA−T

−A−TQ1 A−T

• Steady state solution to the Riccati equation:

S∞ = ΛIX−1

I

where

• XI ΛI

T is the eigenvector of Hc corresponding to the stable eigenvalues. The steady state control law is given by u(k) = −K∞x(k)

9.

Steady State Solution to LQR Problem Construct the control Hamiltonian matrix, Hc, given by Hc =

• A + BQ−1

2 BTA−TQ1 −BQ−1 2 BTA−T

−A−TQ1 A−T

• Steady state solution to the Riccati equation:

S∞ = ΛIX−1

I

where

• XI ΛI

T is the eigenvector of Hc corresponding to the stable eigenvalues. The steady state control law is given by u(k) = −K∞x(k) where K∞ = (Q2 + BTS∞B)−1BTS∞A

Digital Control

9

Kannan M. Moudgalya, Autumn 2007

10.

Performance Index of Steady State LQR Prob- lem

10.

Performance Index of Steady State LQR Prob- lem

The performance index for the steady state control problem is J∞ = 1 2xT (0)S∞x(0)

10.

Performance Index of Steady State LQR Prob- lem

The performance index for the steady state control problem is J∞ = 1 2xT (0)S∞x(0) Notice that this control law K∞ can be calculated

• nce and for all at the very beginning.

Digital Control

10

Kannan M. Moudgalya, Autumn 2007