Feedback Control Systems

Dr. Basil Hamed

Electrical & Computer Engineering

Islamic University of Gaza

STEADY-STATE ERROR

PROBLEM DEFINITION

The unity feedback system below with gain

K=100 represents an elevator in one of many

city's multistorey buildings

Each of the elevator's floor's buttons is the input of the system and is represented by unit step command. Assuming that the fourth (4) floor button is pressed on the ground floor in order to reach a floor 4

Using MATLAB do the following

1. Find:    a) The system type.    b) Position coefficient Kp, Velocity coefficient Kv and

acceleration coefficient Ka in terms of gain K.    c) Steady-state error for inputs u(t), tu(t) and t2u(t) in terms of

gain K2. Find the value of gain K to yield a 30% error in the steady-state.3. Plot the step response for the K value found in part 2, and give

comment on the result. Estimate rise time Tr, Peak time Tp, settling time Ts and percent overshoot from the plot.

4. Repeat the task from part 3 for K=100 and compare the obtained results with those from part 3.

Part 1) The system described in the problem definition has the forward

transfer function :           G(s) =            K(s+10)        

                      s(s+1)(s+20)(s+50)

If we consider the denominator of G(s), we see that the number of pure iterations in the forward path is 1.  Therefore this is a type 1 system.  Verification of this result can be found by calculating the static error coefficients :

 Positional constant Kp :   Kp=lims->0 G(s) = K/0 = Inf. Velocity constant Kv :      Kv=lims->0 sG(s) = Kx10/1x20x50=K/100  Acceleration constant Ka :     Ka=lims->0 s2G(s) = Kx0/100 = 0.

These results are consistent with the definition of a type 1 system.

Part 2)

For a 30% steady-state error, and using the expressions for Kv found in part 1.

    

Therefore a gain of K=333.33 will result in a 30% steady-state error for a ramp input.

The solution to this question is also shown using Matlab in part 3.

Part 3)

K = 333.3333

Forward transfer function: Transfer function:               333.3 s + 3333-------------------------------------s^4 + 71 s^3 + 1070 s^2 + 1000 s

Closed-loop system: Transfer function:                    333.3 s + 3333---------------------------------------------s^4 + 71 s^3 + 1070 s^2 + 1333 s + 3333

Part 3)

Part 4)

K =  100

Forward transfer function: Transfer function:               100 s + 1000-------------------------------------s^4 + 71 s^3 + 1070 s^2 + 1000 s

Closed-loop system: Transfer function:                    100 s + 1000---------------------------------------------s^4 + 71 s^3 + 1070 s^2 + 1100 s + 1000

Part 4)

Animation

Problem 2

PROBLEM DEFINITION

A space telescope is to be launched to carry out astronomical experiments .   The pointing control system is desired to achieve 0.01 minute of arc and track solar objects with apparent motion up to 0.21 minute per second.   The system is illustrated in Figure. The control system is shown in block diagram.   Assume that τ1= 1 second and τ2= 0 ( an approximation ).

Block Diagram

Using MATLAB do the following

1) Determine the gain K=K1K2 required so that the response to a step command is as rapid as reasonable with an overshoot of less than 5%

2) Determine the steady-state error of the system for a step and a ramp input

3) Determine the value of K1K2 for an ITAE optimal system for (1) a step input and (2) a ramp input

Part 1)

The value of gain K=K1K2=10 and the transfer function is equal to:

10 s + 10---------------s^2 + 10 s + 10

Part 1)

The value of gain K=K1K2=16 and the transfer function is equal to:

16 s + 16------------------------ s^2 + 16 s + 16

Part 1)

The value of corresponding gain K=K1K2=20 and the transfer function is equal to:

20 s + 20---------------s^2 + 20 s + 20

Part 2)

Part 2)

Part 2)

Part 2)

Part 3) K=K1.K2 = 1.96

Ramp response of original system (b) and for ITAE optimal system (g)

K=K1.K2=10.24

Animation