Basics of Automation and Control I

Three-term controller Ziegler-Nichols tuning rules Summary

Lecture 13: PID controllers

Paweł Malczyk

Division of Theory of Machines and Robots Institute of Aeronautics and Applied Mechanics Faculty of Power and Aeronautical Engineering Warsaw University of Technology

1 Three-term controller

2 Ziegler-Nichols tuning rules

3 Summary

1 Three-term controller Introduction Basic control functions P controller PI controller PD controller PID controller

2 Ziegler-Nichols tuning rules

3 Summary

Fig. 1: Block diagram of a process with feedback controller

1 The three-term controller, i.e. proportional-integral-derivative (PID) controller, is a control loop feedback mechanism widely used in many industrial control systems. 2 PID controllers appear in many different forms: as stand-alone controllers, as part of hierarchical, distributed control systems and built into embedded components. 3 By tuning the three parameters in the PID controller algorithm, the controller can provide control action designed for specific process requirements (benign process dynamics, moderate performance).

Fig. 2: Block diagram of feedback control system with ideal PID controller The input/output relation for an ideal PID controller*: [ ∫ ] 1 t de(t) u(t) = k e(t) + e(τ)dτ + T = p T d dt ∫i 0 (1) t de(t) = kpe(t) + ki e(τ)dτ + kd 0 dt e(t) = r(t) − y(t) – error signal, u(t) – control signal,

kp – proportional gain, Ti – integral time, Td – derivative time, k = kp – integral gain, k = k T – derivative gain. i Ti d p d * Modified versions of the PID controller are used in practical applications.

© Paweł Malczyk. Basics of Automation and Control I Lecture 13: PID controllers 5 / 35 Three-term controller Ziegler-Nichols tuning rules Summary Basic control functions Ideal PID controller: [ ∫ ] t 1 de(t) u(t) = kp e(t) + e(τ)dτ + T T d dt i 0 ( ) (2) ( ) → U s 1 C(s) = = kp 1 + + Tds E(s) Tis

Fig. 3: Block diagram of ideal PID controller • P controller (Ti = ∞, Td = 0) C(s) = kp (3)

• PI controller (Td = 0)( ) 1 C(s) = kp 1 + (4) Tis

• PD controller (Ti = ∞) C(s) = kp(1 + Tds) (5) Fig. 4: PID interpretation → Gentle introduction to PID controllers: https://www.youtube.com/watch?v=4Y7zG48uHRo. © Paweł Malczyk. Basics of Automation and Control I Lecture 13: PID controllers 6 / 35 Three-term controller Ziegler-Nichols tuning rules Summary Mechanical system

Ns N Fig. 5: Układ mechaniczny. Stałe fizyczne: m = 1 kg, b = 2 m , k = 5 m Goal: choose a control force f(t) (in N) such that the mass will stop at the desired position (say 1 m from equilibrium). Y(s) 1 Plant tranfer function G(s) = F(s) = ms2+bs+k . Reference signal (desired position): r(t) = yd · 1(t) (yd = 1 m). In steady-state the force f(t) need to balance the spring force, i.e.: f = k · yd · 1(t), then 1 yss = lim y(t) = lim sY(s) = lim sG(s)F(s) = lim sG(s)kyd = G(0)kyd = yd t→∞ s→0 s→0 s→0 s

Fig. 6: Mechanical system. Physical constants: m = 1 kg, b = 2 Ns , k = 5 N , and g = 10 m m m s2 Y(s) 1 Plant transfer function G(s) = F(s) = ms2+bs+k . Disturbance – gravity force: d(t) = −mg · 1(t).

Reference signal (desired position): r(t) = yd · 1(t) (yd = 1 m). In steady-state the force f(t) need to balance the sum of spring force and the gravity force, i.e.: u = (k · yd + mg) · 1(t). Assumption: perfect knowledge of parameters and disturbance...

True values: m, b, k. Estimated params: m, b, k. Real parameters and disturbance signals are

usually unknown. Fig. 7: Mechanical system with parametric uncertainty

Control signal: u = (k · yd + mg) · 1(t). Input signal: f = (kyd + (m − m)g) · 1(t). Steady-state response: − 1 1 − yss = lim y(t) = lim sY(s) = lim sG(s)(kyd+(m m)g) = (kyd+(m m)g) t→∞ s→0 s→0 s k Error signal: 1 1 e = y − y = y − (ky + (m − m)g) = ((k − k)y + (m − m)g) ss ss ss d k d k d The greater the uncertainty is, the greater the steady-state error is. Transient-response is not controlled. Open-loop control is sensitive to parametric uncertainty.

Fig. 8: Feedback control system

Assumption: u(t) = kp(r(t) − y(t)) – force proportional to the error. Open-loop transfer function: L(s) = C(s)G(s).

Y(s) L(s) kp Closed-loop transfer function: T(s) = = = 2 . R(s) 1+L(s) ms +bs+k+kp

kp – virtual spring coefficient. Let us calculate the response y(t) and steady-state error e(t).

Fig. 9: Feedback control system The response: Y(s) = T(s)R(s) + G(s)S(s)D(s). Sensitivity function: 1 ms2 + bs + k ( ) = = S s 2 1 + L(s) ms + bs + k + kp Load disturbance sensitivity function: 1 ( ) = ( ) ( ) = W s G s S s 2 ms + bs + k + kp · → yd Reference signal: r(t) = yd 1(t) R(s) = s − · → − mg Disturbance: d(t) = mg 1(t) D(s) = s .

Fig. 10: Feedback control system Steady-state response:

yd mg yss = lim y(t) = lim sY(s) = lim s(T(s) + W(s)(− )) = t→∞ s→0 s→0 s s − kp − 1 = ydT(0) mgW(0) = yd mg k + kp k + kp

If k → ∞, then kp → 1 and 1 → 0. In consequence: y = y . p k+kp k+kp ss d Conclusion: → disturbance rejection (even unknown) and trajectory tracking.

Fig. 11: P controller, C(s) = kp Effects of P controller Proportional term, P, causes a corrective control actuation proportional to the error. The system with P controller will usually have nonzero steady-state errors. As kp increases, then the static position error decreases.

As kp increases, the stability margins decrease and the system may become unstable. As kp increases, the BW (as well as overshoot and settling time) increases. It is difficult to ensure both good transient response accuracy and steady- state performance.

kp=30 40 1.4 30 1.2 20 10 1 r(t) 0 0.8 -10 -20 y(t) 0.6 kp=30 kp=25 -30 0.4 kp=20

Magnitude (dB) Magnitude -40 0.2 kp=15 -50 -60 0 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t kp=30 0 30 25 -30 20 -60 15 10 -90 G C u(t) 5 kp=15 -120 L 0 kp=20 Phase (deg) Phase S -5 kp=25 -150 T -10 kp=30 -180 -15 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t

Fig. 12: P controller characteristics  Comment 1 If kp ↗, then Mp ↗, tp ↘, ts ↗, ess ↘, stability issues.

kp=30 40 1.4 30 1.2 20 10 1 r(t) 0 0.8

-10 y(t) 0.6 -20 kp=30 kp=25 Magnitude (dB) Magnitude G -30 0.4 kp=20 C -40 L kp=15 S 0.2 -50 T -60 0 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t Fig. 13: P controller characteristics

( ) 1 Fig. 14: PI controller, C(s) = kp 1 + Tis Effects of PI controller The integral term, I, gives a correction proportional to the integral of the error. PI largely reduces the steady-state errors (compared to P controller). It may cause the closed loop system less stable (or even unstable). PI may slow down the transient response. Filtering out high-frequency noise. Windup - interaction of integral action and saturations.

kp=30, Ti=1 kp=30 80 1.6 60 1.4 1.2 40 1 20 0.8 0.6 0 y(t) 0.4 Ti=1 -20 0.2 Ti=5 0 -40 Ti=10 Magnitude (dB) Magnitude -0.2 G C L S T -60 -0.4 Ti=100 -80 -0.6 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t kp=30, Ti=1 kp=30 0 30 25 -30 20 -60 15 10 -90 G 5 C u(t) 0 Ti=1 -120 L -5 Ti=5 Phase (deg) Phase S Ti=10 -150 -10 T -15 Ti=100 -180 -20 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t

Fig. 15: PI controller characteristics  Comment 2 If Ti ↘, then Mp ↗, ts ↗, ↘, ess ↘.

kp=30, Ti=1 kp=30 80 1.6 G 1.4 60 C L 1.2 40 S 1 T 20 0.8 0.6 0 y(t) 0.4

-20 0.2 Magnitude (dB) Magnitude -40 0 Ti=1 -0.2 Ti=5 -60 Ti=10 -0.4 Ti=100 -80 -0.6 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t Fig. 16: PI controller characteristics

Fig. 17: PD controller, C(s) = kp(1 + Tds) Effects of PD controller The derivative term, D, gives a predictive capability yielding a control action proportional to the rate of change of the error. Derivative action tends to have a stabilizing effect. PD controller adds damping and reduces maximum overshoot. It improves the transient response (decreases rise and settling time). PD controller improves relative stability (increases gain and phase margins). Possibly accentuating noise at higher frequencies. Filtered version of the PD (or PID) controller is used due to derivative kick, ( ) = ( + 1 ) then, C s kp 1 Tds Td . Typical values of N are 8 to 20. N s+1

kp=30, Td=0.1 kp=30 80 1.4 1.2 60 G C L S T 1 40 0.8 20 0.6 0.4 0 y(t) 0.2 Td=0 -20 0 Td=0.01 Td=0.05 Magnitude (dB) Magnitude -0.2 -40 -0.4 Td=0.1 -60 -0.6 10-2 10-1 100 101 102 103 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t kp=30, Td=0.1 kp=30 120 30 90 25 60 20 30 15 0 10 -30 G u(t) 5 -60 C Td=0

-90 L 0 Td=0.01 Phase (deg) Phase -120 S -5 Td=0.05 -150 T -10 Td=0.1 -180 -15 10-2 10-1 100 101 102 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t

Fig. 18: PD controller characteristics  Comment 3 If Td ↗, then Mp ↘, tp ∼↗, ts ↘.

kp=30, Td=0.1 kp=30 80 1.4 G 1.2 60 C L 1 S 40 T 0.8 0.6 20

0.4 y(t) 0

0.2 Magnitude (dB) Magnitude -20 0 Td=0 -0.2 Td=0.01 -40 Td=0.05 -0.4 Td=0.1 -60 -0.6 10-2 10-1 100 101 102 103 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t Fig. 19: PD controller characteristics

( ) 1 Fig. 20: PID controller, C(s) = kp 1 + + Tds Tis Effects of PID controller The features of each of the PI and PD controllers are utilized. Improves both steady-state errors as well as transient response specifications. The PID controller algorithm involves three separate constant parameters that need to be tuned.

kp=30, Td=0.1, Ti=1 kp=30, Td=0.1, Ti=1 80 1.4 60 1.2 40 1 20 0.8 0 y(t) 0.6 -20 0.4

-40 y(t) Magnitude (dB) Magnitude -60 G C L S T 0.2 r(t) -80 0 10-3 10-2 10-1 100 101 102 103 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t kp=30, Td=0.1, Ti=1 kp=30, Td=0.1, Ti=1 120 50 90 60 40 30 30 0 -30 20 -60 u(t)

-90 10 Phase (deg) Phase -120 0 -150 G C L S T u(t) -180 -10 10-3 10-2 10-1 100 101 102 103 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t

Fig. 21: PID controller characteristics

kp=30, Td=0.1, Ti=1 kp=30, Td=0.1, Ti=1 80 1.4

60 1.2

40 1 20 0.8

0 y(t) 0.6

-20 Magnitude (dB) Magnitude 0.4 -40

-60 0.2 y(t) G C L S T r(t) -80 0 10-3 10-2 10-1 100 101 102 103 0 1 2 3 4 5 6 7 8 9 10 Frequency (rad/sec) t Fig. 22: PID controller characteristics

Fig. 23: Ideal P, PI, PD and PID controllers 1 PID controllers are applicable to many control problems, and often perform satisfactorily. 2 Ideal PID are linear controllers with constant coefficients. 3 There are many variations of PID controllers associated with practical control applications (e.g. preventing from integrator windup).

1 Three-term controller

2 Ziegler-Nichols tuning rules Introduction First method Second method Comments Example

3 Summary

Fig. 24: Ideal PID controller and plant If a mathematical model of the plant is known, then it is possible to design the PID controller that will meet the transient and steady-state specifications of the closed-loop system. If the mathematical model cannot be easily obtained, one should select the controller parameters experimentally, which known as a process of controller tuning. Experimental Ziegler and Nichols tuning rules for PID controllers as a starting point for fine tuning.

Experimental response of the plant due to Y(s) a unit-step input. The transfer function U(s) may be approximated by Y(s) Ke−τs G(s) = = (6) U(s) Ts + 1 T – time constant, τ – delay, K – gain.  Comment 4 The controller parameters are designed to result in 25% amplitude decay ratio in one Fig. 25: S-shaped response of a closed-loop system due to a unit-step input. The plant period. involves neither integrators nor dominant complex conjugate closed-loop poles Ziegler-Nichols tuning rule based on a step response of a plant

Type kp Ti Td T ∞ P Kτ 0 T τ PI 0.9 Kτ 0.3 0 T PID 1.2 Kτ 2τ 0.5τ

Fig. 26: Closed-loop system with P controller and sustained oscillation with period Tcr Ziegler-Nichols tuning rule based 1. Set Ti = ∞, Td = 0 and increase kp on a critical period Tcr from 0 to a critical value kcr, where the output exhibits sustained oscillations. Type kp Ti Td

2. Find the critical period Tcr. P 0.5kcr ∞ 0 1 3. Select the parameters kp, Ti and Td PI 0.45kcr 1.2 Tcr 0 according to the table. PID 0.6kcr 0.5Tcr 0.125Tcr  Comment 5 The controller parameters are designed to result in 25% amplitude decay ratio in one period.  Comment 6 One can use Nyquist plot to find kcr, and Tcr.

1 Ziegler-Nichols tuning rules (and other tuning rules) have been widely used in process control systems. 2 The plant dynamics may be known or may be not precisely known. 3 Experience has shown that the Z-N rules provide acceptable closed-loop response for many systems. 4 The empirical PID tuning strategies are only starting points in the process to obtain the right controller. 5 Tuning method (by Astrom & Hagglund) based on relay feedback is available.

 Example 1 Consider the control system shown in the Fig. 27. The process Y(s) 1 has transfer function G(s) = U(s) = (Ts+1)2 . Apply the Ziegler-Nichols tuning rule to determine the constants of the P, PI, and PID controller.

Fig. 27: Liquid-level control system

− Ke τs Process model G(s) = Ts+1 . K = 1 τ = t = 12.72sec 1 T = t3 − t1 = 122.737sec Type k T T p i d T ∞ P Kτ 0 PI 0.9 T τ 0 Kτ 0.3 T PID 1.2 Kτ 2τ 0.5τ Type k T T p i d − . ∞ − P 9 6491 0 PI 8.6842 42.3576 0 PID 11.579 25.44 6.36 Fig. 28: Process reaction curve ( ) 1 CP(s) = 9.6491, CPI(s) = 8.6842 1 + , ( ) 42.3576s 1 CPID(s) = 11.579 1 + + 6.36s 25.44s

− − −

−

Fig. 29: Closed loop system responses for various controllers

1 What is the main objective of introducing proportional control, integral control or derivative control? 2 What is a P, PI, PD, PID controller? Write the input-output relation, unit-step response and Bode plots. 3 Give the effects of P,PI, PD and PID controllers on the feedback control system performance. 4 Formulate and comment on the two Ziegler-Nichols tuning rules for a PID controller. Discuss the range of applicability of both methods.

1 Three-term controller

2 Ziegler-Nichols tuning rules

3 Summary