1. Feedback Control Systems
Dr. Basil Hamed
Electrical & Computer Engineering
Islamic University of Gaza

2. Root Locus

3. PROBLEM DEFINITION
The supersonic passenger jet control system requires good quality handling and comfortable flying conditions.   An automatic flight control system can be designed for SST (Supersonic Transport) vehicles. The desired characteristics of the dominant roots of the control system shown in Figure have a ζ=  The characteristics of the aircraft are ωn =2.5, ζ=0.30, and τ= 0.1. The gain factor K1, however, will vary over the range 0.02 (at medium-weight cruise conditions) to 0.20 (at light-weight descent conditions).

4. Using MATLAB do the following
1) Sketch the pole-zero map as a function of the loop gain K1K2
2) Determine the gain K2 necessary to yield roots with ζ= when the aircraft is in the medium-cruise condition
3) With the gain K2 as found in 2), determine the ζ of the roots when the gain K1 results from the condition of light descent

5. Block Diagram

6. Part 1)  
From the pole-zero map, we can see the system remains stable for all values of loop gain K1.K2

7. Part 2)
Results shown here indicate that a gain of K2 = for the aircraft in the medium-cruise condition, will yield a damping ratio of approximately
In this part, we shall assume medium-weight cruise conditions, with K1=0.02.   By iterating over values from 1 to , we aim to find the value for K2 which will result in a system damping ratio of ζ=0.707.   For each gain value, ζ will be calculated using the location of the dominant closed-loop poles. Thus, ζ = Cos Θ = -real(pole)/ωn, where ωn is the natural frequency and is given by the radial distance from the origin to the pole. Here we assume a valid second order approximation with higher order poles much further away from the dominant pole pair.   From this, it is also assumed that the dominant pole pair will be stored last in the ordered poles vector, making location 5 the dominant pole used for the calculations.   Once a damping ratio of between and is found, the corresponding gain K2 is displayed and the program terminates

8. Part 3)
In this part ,the value of K1 changes to the light descent condition with K1=0.2, and we use the value of K2 (from part 2) equal to Again we use the equation (-real(pole))/ωn to calculate the damping ratio.

9. Actuator Transfer function: 10 ------------ s + 10
Controller Transfer function: s^ s s^ s
Actuator Transfer function: s + 10
Aircraft Dynamics Transfer function: s s^ s
Closed-loop system Transfer function: e004 s^ e005 s^ e006 s s^ s^ e004 s^ e005 s^ e006 s e006
poles = 1.0e+002 *
i , i, -0.10, i, i
required zeta =0.7902
Using the same value for K2 = , but for light-weight descent conditions with K1=0.2, the results show change in damping ratio from to

10. Animation

11. Problem 2

12. PROBLEM DEFINITION
The elevator in a modern office building travels at a top speed of 25 feet per second and is still able to stop within one eighth of an inch of the floor outside. The transfer function of the unity feedback elevator position control is shown next slide

13. Block Diagram

14. Using MATLAB do the following
a) Sketch the root locus for the unity feedback system above.
b) Plot the step response of the system for K1=1.
c) Determine the gain K when the complex roots have a damping ratio of
d) Find the percent overshoot OS%, and peak time for the gain K at point (c).
e) Plot the step response of the system with gain K obtained in part (b).

15. a) Root locus of unity feedback system

16. b) The output step response of the system

17. c) From MATLAB results it can be seen gain K for damping ratio
is equal to K= This gain is calculated by MATLAB code for interactively selected point where the root locus crosses
               

18. dominant pole that gives K=41.8962 is -0.6516+0.4057i

19. d) Obtained OS% and Tp for gain K=41.8962 calculated in part (c)
Tp=7.7927sec
Comparing the calculated and simulated (see step output characteristic for K= ) values for OS% and Tp we can notice very small difference

20. e) The resulting output step response of the elevator system

21. Animation

22. Problem 3

23. PROBLEM DEFINITION
Automatic control of helicopters is necessary because, unlike fixed- wing aircraft, which possess a fair degree of inherent stability, the helicopter is quite unstable. A helicopter control system that utilises an automatic control loop plus a pilot stick control is shown in the figure below. When the pilot is not using the control stick, the switch may be considered to be open. The dynamics of the helicopter are represented by the transfer function

24. Block Diagram

25. a) With the pilot control loop open (hands-off control), plot the root locus for the automatic stabilization loop. Determine the gain K2 that results in a damping for the complex roots equal to  ζ=
b) For the gain K2 obtained in part (a), determine the steady-state error due to a wind gust Td(s)=1/s.
c) With the pilot loop added, draw the root locus as K1 varies from zero to infinity when K2 is set at the value calculated in part (a).
d) Recalculate the steady-state error of part (b) when K1 is equal to a suitable  value based on the root locus.
e) Plot  closed loop system step responses when the pilot control loop is open and when the pilot loop is added (switch closed).  

26. a) Using MATLAB the plot of the Root Locus diagram for the pilot control loop

27. K1= 1.56, K2 =0.7475

28. b) the gain K2=1.6 and K2=0.74 obtained in part (a) the calculated steady-state errors due to wind gust Td(s)=1/s

29. from the MATLAB results we can see that the values of steady-state errors are: ess1= 3.8552                for interactively selected K2= ess1= 5.9386                for interactively selected K2=0.7475

30. c) The resulting root locus diagram for added pilot control loop when K1 varies from zero to infinity and K2 is set at the value calculated in part (a)

31. d) The steady-state error when K1 is equal to suitable value based on the pilot control loop

33. Using the final value theorem the steady-state error is

34. K1=2.0602

35. e) The closed loop system output step responses when the pilot control loop is open and when the pilot control loop is added for K2=1.5665,  K2=0.7475

37. The piloted output step response (switch closed) for K1=2
The piloted output step response (switch closed) for K1= and K2= is

38. Animation