Introduction to Computer Control Systems Lecture 4: Linearization + Basis of PID controller Dave Zachariah

Introduction to Computer Control Systems

Lecture 4: Linearization + Basis of PID controller

Dave Zachariah

Div. Systems and Control, Dept. Information Technology, Uppsala University

December 9, 2014

Today’s lecture: What and why?

Linearization

Why: Approximating real nonlinear systems with linear models.

PID controller

Why: Widely used, simple yet sufficiently powerful in many systems.

Today’s lecture: What and why?

Linearization

Why: Approximating real nonlinear systems with linear models.

PID controller

Why: Widely used, simple yet sufficiently powerful in many systems.

Linearization: Steady-state and deviation variables

y(t)

u(t) G

Most real systems arenonlinear and often modeled as:

˙x(t) = f (x(t), u(t)) y(t) = h(x(t), u(t)).

Linearization: Pick a point (x0, u0) of f (x, u). Use linearmodel to describe system behaviour around (x₀, u₀):

(˙x(t) = f (x(t), u(t)) y(t) = h(x(t), u(t)) ≈

(˙˜x(t) =A˜x(t) + B˜u(t)

˜

y(t) =C˜x(t) + D˜u(t) When system is at rest ˙x = f (x0, u0) =0, we call (x0, u0) a stationary point. Note: could be more than one!

Linearization: Steady-state and deviation variables

y(t)

u(t) G

Most real systems arenonlinear and often modeled as:

˙x(t) = f (x(t), u(t)) y(t) = h(x(t), u(t)).

Linearization: Pick a point (x0, u0) of f (x, u). Uselinear model to describe system behaviour around (x₀, u₀):

(˙x(t) = f (x(t), u(t)) y(t) = h(x(t), u(t)) ≈

(˙˜x(t) =A˜x(t) + B˜u(t)

˜

y(t) =C˜x(t) + D˜u(t)

When system is at rest ˙x = f (x0, u0) =0, we call (x0, u0) a stationary point. Note: could be more than one!

Linearization: Steady-state and deviation variables

y(t)

u(t) G

Most real systems arenonlinear and often modeled as:

˙x(t) = f (x(t), u(t)) y(t) = h(x(t), u(t)).

Linearization: Pick a point (x0, u0) of f (x, u). Uselinear model to describe system behaviour around (x₀, u₀):

(˙x(t) = f (x(t), u(t)) y(t) = h(x(t), u(t)) ≈

(˙˜x(t) =A˜x(t) + B˜u(t)

˜

y(t) =C˜x(t) + D˜u(t) When system is at rest ˙x = f (x0, u0) =0, we call (x0, u0) a stationary point. Note: could be more than one!

Linearization: Steady-state and deviation variables

Define deviation variables

x(t) = x(t) − x˜ 0

˜

u(t) = u(t) − u₀

Then linearize byfirst-order Taylor expansion around x0, u0:

˙˜x= ˙x − 0

= f (x0+ ˜x, u0+ ˜u)

≈ f (x₀, u₀)

| {z }

=0

+∇_xf(x₀, u₀)˜x+ ∇_uf(x₀, u₀)˜u

=A˜x+ B˜u,

where A and B given by simple derivatives,

a_ij = ∂f_i(x, u)/∂x_j b_i = ∂f_i(x, u)/∂u, evaluated at x = x₀ and u = u₀.

Linearization: Steady-state and deviation variables

Similarly, y0 = h(x0, u0) for the stationary point. Define

˜

y(t) = y(t) − y0

Then linearize byfirst-order Taylor expansion around x₀, u₀:

˜

y= h(x0+ ˜x, u0+ ˜u) − y0

≈ h(x₀, u₀) + ∇_xh(x₀, u₀)˜x+ ∇_uh(x₀, u₀)˜u − y₀

= ∇_xh(x₀, u₀)˜x+ ∇_uh(x₀, u₀)˜u + 0

=C˜x+ D˜u,

where C and D given by derivatives,

cj = ∂h(x, u)/∂xj D= ∂h(x, u)/∂u, evaluated at x = x0 and u = u0.

Linearization: Example

Nonlinearsystem:

 ˙x1

˙x2



= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



Find equilibrium point: ˙x = 0

0 0



= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



With constant input u ≡ u₀ we get solvex₁ and x₂: x1,0= 1

Kcu0 x2,0= 3

Kcu0 ⇒ x₀ =

 ₁

Kcu0 3 Kcu0



Linearization: Example

Nonlinearsystem:

 ˙x1

˙x2



= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



Find equilibrium point: ˙x = 0

0 0



= −K√ x1

x1−¹₃x2

 +

₁

cu 0



With constant input u ≡ u₀ we get solvex₁ and x₂: x1,0= 1

Kcu0 x2,0= 3

Kcu0 ⇒ x₀ =

 ₁

Kcu0 3 Kcu0



Linearization: Example

Nonlinearsystem:

 ˙x1

˙x2



= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



Find equilibrium point: ˙x = 0

0 0



= −K√ x1

x1−¹₃x2

 +

₁

cu 0



With constant input u ≡ u₀ we getsolve x₁ and x₂: x1,0= 1

Kcu0 x2,0 = 3

Kcu0 ⇒ x₀ =

 ₁

Kcu0 3 Kcu0



Linearization: Example, cont’d

Nonlinear system:

 ˙x1

˙x2



| {z }

˙ x

= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



| {z }

f (x,u)

Tolinearize around (u₀, x₀): Define

˜

u = u − u0 x˜= x − x0

weevaluatederivatives

a₁₁= ∂f₁/∂x₁ a₁₂= ∂f₁/∂x₂ b₁ = ∂f₁/∂u a₂₁= ∂f₂/∂x₁ a₂₂= ∂f₂/∂x₂ b₂ = ∂f₂/∂u at (u0, x0) so that

˙˜x1

˙˜x2



=

"

−₂^√K_x

1,0 0

1 −¹₃

#

| {z }

A

 ˜x1

˜ x2

 +

₁

c

0



|{z}

B

u

Linearization: Example, cont’d

Nonlinear system:

 ˙x1

˙x2



| {z }

˙ x

= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



| {z }

f (x,u)

Tolinearize around (u₀, x₀): Define

˜

u = u − u0 x˜= x − x0

weevaluatederivatives

a₁₁= ∂f₁/∂x₁ a₁₂= ∂f₁/∂x₂ b₁ = ∂f₁/∂u a₂₁= ∂f₂/∂x₁ a₂₂= ∂f₂/∂x₂ b₂ = ∂f₂/∂u at (u₀, x₀)

so that

˙˜x1

˙˜x2



=

"

−₂^√K_x

1,0 0

1 −¹₃

#

| {z }

A

 ˜x1

˜ x2

 +

₁

c

0



|{z}

B

u

Linearization: Example, cont’d

Nonlinear system:

 ˙x1

˙x2



| {z }

˙ x

= −K√ x₁ x1−¹₃x2

 +

₁

cu 0



| {z }

f (x,u)

Tolinearize around (u₀, x₀): Define

˜

u = u − u0 x˜= x − x0

weevaluatederivatives

a₁₁= ∂f₁/∂x₁ a₁₂= ∂f₁/∂x₂ b₁ = ∂f₁/∂u a₂₁= ∂f₂/∂x₁ a₂₂= ∂f₂/∂x₂ b₂ = ∂f₂/∂u at (u₀, x₀) so that

˙˜x1

˙˜x2



=

"

−₂^√K_x

1,0 0

1 −¹₃

#

| {z }

A

 ˜x1

˜ x2

 +

₁

c

0



|{z}

B

u

Feedback approach: Use control error e(t)

y u

r

+

− ^e Controller G

Control error:

e(t) =r(t)−y(t)

Q: How should controller determine input u(t)based on error e(t)?

Feedback approach: Use control error e(t)

y u

r

+

− ^e Controller G

Control error:

e(t) =r(t)−y(t)

Q: How should controller determine input u(t)based on error e(t)?

Feedback approach: PID controller

y u

r

+

− ^e Controller G

A popular approach, set input u(t)based on:

present value of e(t): u(t) = Ke(t) (Proportional)

historyof e(t): u(t) = KRt

τ =0e(τ ) (Integral) changeof e(t): u(t) = K ˙e(t) (Derivative) where K are tuning constants.

Feedback approach: PID controller

y u

r

+

− ^e Controller G

A popular approach, set input u(t)based on:

present value of e(t): u(t) = Ke(t) (Proportional) historyof e(t): u(t) = KRt

τ =0e(τ ) (Integral)

changeof e(t): u(t) = K ˙e(t) (Derivative) where K are tuning constants.

Feedback approach: PID controller

y u

r

+

− ^e Controller G

A popular approach, set input u(t)based on:

present value of e(t): u(t) = Ke(t) (Proportional) historyof e(t): u(t) = KRt

τ =0e(τ ) (Integral) changeof e(t): u(t) = K ˙e(t)(Derivative) where K are tuning constants.

PID controller: time-domain

y u

r

+

− ^e

H

A popular approach: the PIDcontroller u(t) = Kpe(t)

| {z }

P

+ Ki

Z t τ =0

e(τ )dτ

| {z }

I

+ Kd˙e(t)

| {z }

D

PID controller: Laplace-domain

y u

r

+

− ^e

H

A popular approach: the PIDcontroller U (s) = KpE(s) + Ki

1

sE(s) + K_dsE(s)

=



Kp+Ki

s + K_ds



| {z }

transfer functionH(s)

E(s),

where E(s) = R(s) − Y (s).

PID controller: Laplace-domain, cont’d

y u

r

+

− ^e

H

Final value theorem: If e(t) has a final value, then it can be computed by

t→∞lim e(t) = lim

s→0sE(s)

Static error analysis: Study limit of e(t) when r(t) is a constant. On the board: Let r(t) = C for t ≥ 0. Then e(t) goes to 0 when H(s)G(s) contains an integrator 1/s.

PID controller: Laplace-domain, cont’d

y u

r

+

− ^e

H

Final value theorem: If e(t) has a final value, then it can be computed by

t→∞lim e(t) = lim

s→0sE(s)

Static error analysis: Study limit of e(t) when r(t) is a constant.

On the board: Let r(t) = C for t ≥ 0. Then e(t) goes to 0 when H(s)G(s) contains an integrator 1/s.

PID controller: Laplace-domain, cont’d

y u

r

+

− ^e

H

Final value theorem: If e(t) has a final value, then it can be computed by

t→∞lim e(t) = lim

s→0sE(s)

Static error analysis: Study limit of e(t) when r(t) is a constant.

On the board: Let r(t) = C for t ≥ 0. Then e(t) goes to 0 when H(s)G(s) contains an integrator 1/s.

PID controller: Laplace-domain, cont’d

y u

r

+

− ^e

H

G

Total system from reference r(t) to output y(t):

Y (s) = Gtotal(s)R(s)

On the board: Derive Gtotal(s) = _1+H(s)G(s)^H(s)G(s)

PID controller: Laplace-domain, cont’d

y u

r

+

− ^e

H

G

Total system from reference r(t) to output y(t):

Y (s) = Gtotal(s)R(s) On the board: Derive G_total(s) = _1+H(s)G(s)^H(s)G(s)

PID controller: practical concerns

PID controllers

1 have typically bad performancewhen there is a significant time-delay in the loop. They work nicely for relativelysmall time-delays.

2 with derivative-term amplify noise. Add a low-pass filterto term by Kds ≈ _1+εK^Kds

ds, for a small ε.

3 candifferdepending on manufactures. It is important to know the configuration of the PID algorithm before tuning.

4 are often implemented in discrete timebut tuned using a continuous formulation

Today’s lecture: What and why?

Linearization

Why: Approximating real nonlinear systems with linear models.

PID controller

Why: Widely used, simple yet sufficiently powerful in many systems.