feedback control systems

ECM2105 - Control Engineering

Dr Mustafa M Aziz (2013)

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FEEDBACK CONTROL SYSTEMS 1. Control System Design 2. Open and Closed-Loop Control Systems 3. Why Closed-Loop Control? 4. Case Study --- Speed Control of a DC Motor 5. Steady-State Errors in Unity Feedback Control Systems

1. Control System Design 1. Establishing system goals (objectives), e.g. to control the velocity of a motor accurately − What do you want to control? e.g. speed, position, level, temperature, ... − What do you want to achieve? e.g. fast response, less vibration, ... − How do you translate them into control terms? e.g. overshoot, rising time, ... 2. Obtaining a model of the process, the actuator, and the sensor. E.g. G(s) = k/(Ts+1) − Differential equations − Transfer functions − Block diagrams − State space equations 3. Designing a controller (PID, root locus, frequency response, state space) − Proportional (P) control − Integral (I) control − Proportional-Derivative (PD) control − Proportional-Integral (PI) control − Proportional-Derivative-Integral (PID) control − State space based control 4. Simulation (MATLAB) or experimental test.

2. Open and Closed-Loop Control Systems

Design requirements and specification: − − − − −

An excellent transient response in terms of rise-time, size of overshoot and settling-time; An excellent steady-state response in terms of steady-state error; Acceptable stability margins; Robustness in terms of disturbance rejection; and Robustness in terms of sensitivity to parameter changes.

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Open-loop control system: utilises an actuating device to control the process (plant) directly without using feedback. Input

Actuating device

Process or plant

Output

Closed-loop control system: uses a measurement of the output and feedback of this signal to compare it with the desired input (reference or command).

R(s)

E(s)

+_

U(s)

Gc(s)

H(s)

Notation: R(s): Laplace transform of the input signal; U(s): Laplace transform of the control or actuating signal; Y(s): Laplace transform of the output signal; E(s): Laplace transform of the error signal; Gc(s): Transfer function of the controller; G(s): Transfer function of the process or plant; H(s): Transfer function of the sensor.

2

G(s)

Y(s)



________________________________________________________________________________ 3. Why Closed-Loop Control? An advantage of the closed-loop control system is the fact that the use of feedback makes the system response relatively insensitive to external disturbances (e.g. temperature and pressure) and internal variations in system parameters (e.g. component tolerances) which are not known or predicted. 3.1. Sensitivity of control systems to parameter variations Suppose that G(s) changes to G(s)+∆G(s) due to the environment, ageing, etc. For small ∆G(s) 1) For K = 500, two complex poles: s = -500+j1500 and s = -500-j1500 (ζ < 1) A compromise should be found between the need to increase K and the need to avoid the above problems. 4.10. Summary

Feedback (closed-loop) control can be used to stabilise systems, speed up the transient response, improve the steady-state characteristics, provide disturbance rejection, and decrease the sensitivity to parameter variations.

Proportional feedback control reduces errors and improves transient responses, but the high gain (large K) may lead to many problems.

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________________________________________________________________________________ 5. Steady-State Errors in Unity Feedback Control Systems

When a command input (desired output) is applied to a control system it is generally hoped that after any transient effects have died away the system output will settle down to the command value. The error with any system is the difference between the required (desired) output signal, i.e. the reference input signal which specifies what is required, and the actual output signal. For the unityfeedback control system shown in the figure, the error is:

R(s) E(s) = R (s) − Y(s) =

R (s) 1 + G (s)

E(s)

Y(s) G(s)

+ _

Using the final-value theorem (assuming the system is stable), the steady-state error is: sR (s) s → 0 1 + G (s )

e ss = e(∞) = lim e( t ) = lim sE(s) = lim t →∞

s →0

A general representation of G(s) is: G (s) =

K (s m + b m −1s m −1 + L + b1s + b 0 ) s N (s n + a n −1s n −1 + L + a 1s + a 0 )

where K is a constant and m and n are integers and neither a0 nor b0 is zero. N is an integer, the value of which is called the type of the system. Thus if N = 0 then the system is said to be type 0, if N = 1 then type 1, if N = 2 then type 2 and so on. The type number is thus the number of 1/s factors in the open-loop transfer function G(s). Since 1/s represents integration, the type number is the number of integrators in the open-loop transfer function. Exercise: What are the type numbers for the systems shown in the following figures?

(s + 1) (s − 1)(s + 6)

Y(s)

K

(s + 1) s(s − 1)(s + 6)

Y(s)

K

R(s) +

−

R(s) +

−

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________________________________________________________________________________ 5.1. Static error constants

Just as we defined the damping ratio, the natural frequency, settling time, percentage overshoot etc. as performance specifications for the transient response of a system, we can use static error constants to specify the steady-state error characteristics of control systems. 5.1.1. Static position error constant Kpo The steady-state error of the unity-feedback control system for a unit-step input is:

s 1 = s →0 s(1 + G (s )) 1 + G (0)


s→0

The static position error constant Kpo is defined by: K po = lim G (s) = G (0) s →0

Thus, the steady-state error in terms of the static position error constant Kpo is; 1 e ss = 1 + K po For a type 0 system, Kpo = C where C is a constant. For a type 1 system, Kpo = ∞. Hence, for a type 0 system, the static position error constant Kpo is finite, while for a type 1 or higher system, Kpo is infinite. This is, ess = 1/(1+C) for a type 0 system, and ess = 0 for a type 1 or higher system. Exercise: Find the static position error constants and steady-state errors, respectively, of the systems shown in the following figures for a unit-step input.

R(s) +

−

s +1 s+2

R(s)

Y(s)

+

16

−

s +1 s(s + 2)

Y(s)



________________________________________________________________________________ 5.1.2. Static velocity error constant Kv The steady-state error of a system with a unit-ramp (velocity) input is: s 1 = lim s →0 s (1 + G (s )) s →0 sG (s )


s →0

2

The static velocity error constant Kv is defined by: K v = lim sG (s) = s →0

1 e ss

For a type 0 system, K v = lim sG (s) = 0 . For a type 1 system, K v = lim sG (s) = C . For a type 2 or s →0

s →0

higher system, K v = lim sG (s) = ∞ . s →0

5.1.3. Static acceleration error constant Ka The steady-state error of a system with a unit-parabolic (acceleration) input, r(t) = t2/2, is: e ss = e(∞) = lim e( t ) = lim sE(s) = lim t →∞

s →0

s →0

s 1 = lim 2 s → 0 s (1 + G (s)) s G (s) 3

The static acceleration error constant Ka is defined by: K a = lim s 2 G (s) = s →0

1 e ss

For a type 0 system, K a = lim s 2 G (s) = 0 . For a type 1 system, K a = lim s 2 G (s) = 0 . For a type 2 s →0

s →0

system, K a = lim s G (s) = C . For a type 3 or higher system, K a = lim s 2 G (s) = ∞ . 2

s →0

s →0

Exercise: A unity feedback system has the following forward transfer function:

G (s) =

1000(s + 8) (s + 7)(s + 9)

(a) Evaluate the system type, Kpo, Kv, and Ka. (b) Use your answers to (a) to find the steady-state errors for the unit-step, unit-ramp, and unitparabolic inputs.

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Exercise: For the unity feedback system shown, determine the value of K to yield a 10% error in the steady-state when the input is r(t) = 1.

s + 12 (s + 14)(s + 18)

R(s) +

K −

18

Y(s)



________________________________________________________________________________ 5.2. Steady-State Error for Disturbances

D(s) R(s)

E(s) +

−

+

Gc(s)

+

Y(s)

G(s)

The error component when D(s) = 0 is: E(s) = R(s) - YR(s) where the output due to the reference input, YR(s), is given by: YR (s) =

G c (s)G (s) R (s) 1 + G c (s)G (s)

The steady-state error value due to R(s) is given by the final value theorem: sR (s) s →0 1 + G (s )G (s) c

e R (∞) = lim sE(s) = lim s →0

When R(s) = 0, the error in this case is given by:

E(s) = 0 - YD(s)

where the output due to the disturbance, YD(s), is: YD (s) =

G (s) D(s) 1 + G c (s)G (s)

The steady-state error value due to D(s) is found using final value theorem: − sG (s)D(s) s →0 1 + G (s )G (s ) c

e D (∞) = lim sE(s) = lim s →0

The steady-state error due to the reference and disturbance inputs is: e(∞) = eR(∞) + eD(∞)

Exercise: For the block diagram shown below, determine the stead-state error component due to a unit-step disturbance.

D(s) R(s) +

100

+

−

19

+

1 s(s + 25)

Y(s)



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TUTORIAL PROBLEM SHEET 5 1. List the major advantages and disadvantages of closed-loop control systems. 2. For the system shown in the figure, what are the steady-state errors when a unit-step input is applied to the following open-loop transfer functions G(s): 10 (s + 1)(s + 2) 6(s + 3) (b) G (s) = (s + 6)(s + 2) 10 (c) G (s) = s(s + 1)(s + 2) (a) G (s) =

R(s)

E(s) +

G(s)

Y(s)

−

3. What are the type numbers for the systems shown in the following figures? R(s) +

K −

(s + 1) (s − 1)(s + 6)

Y(s)

(s + 1) s(s − 1)(s + 6)

Y(s)

(a) R(s) +

K −

(b) 4. Determine the steady-state error for the system shown in question (1) above with: 2(s + 1) G (s) = 2 s (s + 2) 2 when subjected to the input r(t) = 1 + t + t . Plot the time response y(t) with MATLAB. 5. Find the static position error constants and steady-state errors, respectively, of the systems shown in the following figures for a unit-step input. R(s) +

−

s +1 s+2

R(s)

Y(s)

+

−

(a)

s +1 s(s + 2)

(b)

Plot the time response y(t) for a unit-step input using MATLAB.

20

Y(s)



________________________________________________________________________________ 6. For the system shown in the following figure: R(s) +

Y(s)

1 s(s + 5)(s + 10)

K −

(a) What value of K will yield a steady-state error in position of 0.01 for an input of r(t) = t/10? (b) What is the value of Kv for the value of K found in (a)? (c) Plot the time response using MATLAB. 7. Find the total steady-state error due to a unit-step input and a unit-step disturbance in the system shown below. D(s) R(s) +

1 s+5

−

+

+

100 s+2

Y(s)

8. Consider the system shown in the following figure with G(s) = 5/(s + 2). What are the values of the gain K which will achieve the following design specifications?

D(s) R(s) +

K

+

+

Y(s) G(s)

−

(a) Steady-state error due to a unit-step input R(s) to be less than 0.1. (b) Steady-state error due to a unit-step change in disturbance D(s) to be less than 0.1. (c) Validate the above results using MATLAB, that is to plot the time response for a unit-step input and a unit-step disturbance, respectively. 9. Redo question (8) for G (s) =

10 . s + 10s + 50 2

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