Lecture 19 Practical Issues in PID Implementation Process Control - - PowerPoint PPT Presentation

Lecture 19 Practical Issues in PID Implementation

Process Control

• Prof. Kannan M. Moudgalya

IIT Bombay Wednesday, 4 September 2013

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Outline

• 1. Implementable derivative mode
• 2. Discretisation (minimal)
• 3. Handling communication mismatch
• 4. Setpoint kick

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PID controller

u(t) = u + Kc

• e(t) + 1

τi t e(w)dw + τd de(t) dt

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Process Control Practical Issues in PID Implementation

• 1. Noise handling in derivative mode

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Handling noise in error signal

PID control law is given by ∆u(t) = Kc

• e(t) + 1

τi t e(w)dw + τd de(t) dt

• ◮ Problem with derivative, thanks to noise

◮ Can be seen in Laplace transformed version

also: ∆U(s) = Kc(1 + 1 τis + τds)E(s)

◮ Can see that in Laplace transform also.

• function. Substitute s = jω.

To be explained in frequency response.

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Implementable PID Controller

◮ Recall ∆U(s) = Kc

• 1 + 1

τis + τds

• E(s)

◮ Problem happens at high s in the derivative

mode: magnitude is unbounded

◮ Constrain its magnitude at high frequency ◮ Filtered derivative mode: ◮ ∆u(t) = Kc

• 1 + 1

τis + τds 1 + τds

N

• e(t)

◮ N is a large number ◮ Typical values are 5, 10, etc.

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PID Control Implementation

◮ Discrete version is required for implementation ◮ Many ways to do this ◮ Digital Control by Moudgalya presents several

methods

◮ We present one approach now

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• 2. Introduction to Discretisation

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PID Controller Discretisation

Recall the PID control law: ∆u(t) = Kc

• e(t) + 1

τi t e(t)dt + τd de(t) dt

• Let us discretise, with ˜

u(n) = ∆u(t)|t=n ˜ u(n) = Kc [e(n)+ 1 τi

• Ts

e(0) + e(1) 2 + · · · + Ts e(n − 1) + e(n) 2

• + τD

e(n) − e(n − 1) Ts

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Process Control Practical Issues in PID Implementation

Calculate Change in PID Control

˜ u(n) = Kc [e(n)+ 1 τi

• Ts

e(0) + e(1) 2 + · · · + Ts e(n − 1) + e(n) 2

• + τD

e(n) − e(n − 1) Ts

• Calculate ˜

u(n) − ˜ u(n − 1) and simplify ˜ u(n) − ˜ u(n − 1) = s0e(n) + s1e(n − 1) + s2e(n − 2) s0 = Kc

• 1 + Ts

2τi + τd Ts

• , s2 = Kc

τd Ts s1 = Kc

• −1 + Ts

2τi − 2τd Ts

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Process Control Practical Issues in PID Implementation

• 3. Bumpless Control

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Preferable form of control law

◮ Recall the discrete time control law (blue):

˜ u(n) − ˜ u(n − 1) = s0e(n) + s1e(n − 1) + s2e(n − 2)

◮ Should we calculate this as

∆˜ u(n) = f(e(n), e(n − 1), e(n − 2))

◮ or as

˜ u(n) = g(˜ u(n − 1), e(n), e(n − 1), e(n − 2))

◮ What if ˜

u(n − 1) is not known exactly?

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Communication errors: An example

◮ Consider manual to automatic mode change ◮ Suppose that the controller thinks that the

valve position is 5%

◮ But the actual control valve position is at

+50% position and that it is working fine

◮ Suppose that the control law gives ∆˜

u = 0.1%

◮ First control law will output this as ∆˜

u = 0.1%

◮ So, 50.1% will get implemented ◮ The second control law will output this as

˜ u(n) = ˜ u(n − 1) + 0.1%

◮ So, the second control law will enforce 5.1%!

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Bumpless Control

Because control law is given in terms of ∆˜ u,

◮ Mismatch between the state of end control

element and what controller thinks it to be, may not matter much

◮ Hence known as bumpless transfer

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• 4. Setpoint Kick

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Recall PID Controller - Basic Design

u(t) = K

• e(t) + 1

τi t e(t)dt + τd de(t) dt

• U(s) = K(1 + 1

τis + τds)E(s) U(s)

△

= Sc(s) Rc(s)E(s) If integral mode is present, Rc(0) = 0. Filtered derivative mode: u(t) = K

• 1 + 1

τis + τds 1 + τds

N

• e(t)

N is a large number, of the order of 10 to 100.

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MCQ: Filtered derivative mode

We use filtered derivative mode, because

• 1. We want to reduce the adverse effect of

derivative mode

• 2. We want to make the controller stable
• 3. We want the offset to go to zero
• 4. We want the closed loop to be stable

Answer: 1

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Setpoint, Derivative, Proportional Kicks

◮ The standard PID control law has a

shortcoming.

◮ If there is a sudden change in the setpoint, ◮ both proportional and derivative modes will

introduce large jumps in the control effort, known as setpoint kick.

◮ The large change introduced by the derivative

mode is known as derivative kick.

◮ Proportional mode could also produce a large

control effort, known as proportional kick.

◮ Both derivative and proportional kicks are

generally not acceptable.

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MCQ: Setpoint Kick

Setpoint kick implies

• 1. The setpoint becomes a large value
• 2. The control effort becomes a large value
• 3. Offset becomes a large value
• 4. The plant becomes unstable

Answer: 2

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Solution to Setpoint Kick

Split the PID controller into two parts

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2-DOF Controller

y Tc Rc G = B A Sc Rc r u −

u = Tc Rc r − Sc Rc y Arrive at the following relation between r - y. y = Tc Rc B/A 1 + BSc/ARc r = BTc ARc + BSc r Error e, given by r − y is given by e =

• 1 −

BTc ARc + BSc

• r = ARc + BSc − BTc

ARc + BSc r

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Offset-Free Tracking of Steps with Integral

Use capitals to denote Laplace transform: E(s) = A(s)Rc(s) + B(s)Sc(s) − B(s)Tc(s) A(s)Rc(s) + B(s)Sc(s) R(s) lim

t→∞ e(t) = lim s→0 sA(s)Rc(s) + B(s)Sc(s) − B(s)Tc(s)

(A(s)Rc(s) + B(s)Sc(s))s Because the controller has integral action, Rc(0) = 0: e(∞) = Sc(s) − Tc(s) Sc(s)

• s=0

= Sc(0) − Tc(0) Sc(0)

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Previous slide - continued

This can be satisfied if one of the following is met: Tc = Sc Tc = Sc(0) Tc(0) = Sc(0)

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Solution to Setpoint Kick

◮ Split the PID controller ◮ Many solutions are possible

• see Digital Control by Moudgalya

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MCQ: Filtered derivative mode

We use filtered derivative mode, because

• 1. We want to make the controller stable
• 2. We want the offset to go to zero
• 3. We want an implementable derivative control

mode

• 4. We want the closed loop to be stable

Answer: 3

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What we learnt today

◮ Implementable derivative mode ◮ Discretisation ◮ Handling communication mismatch ◮ Setpoint kick

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Thank you

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