Download presentation

Presentation is loading. Please wait.

Published byHayden Middlemiss Modified over 2 years ago

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

2
STABILITY OF LINEAR FEEDBACK

3
PROBLEM DEFINITION For people paralyzed from the neck down, the ability to drive themselves around in motorized wheelchairs is highly desirable. A proposed system uses velocity sensors mounted in the headgear at 900 intervals, so that forward, left, right, or reverse directions can be commanded. Output of the headgear sensor is proportional to the magnitude of the head movements. The block diagram for this system is shown in figure 1. Here, typical values for the time constants are 1 = 0.5 s, 3 = 1 s, and 4 = 1/4 s.

4
Block Diagram

5
Using MATLAB do the following 1) Determine the limiting gain K = K1K2K3 for a stable system. 2) When the gain K is set equal to 1/3 of the limiting value, determine if the settling time to within 2% of the final value of the system is less than 4 s. 3) Determine the value of gain that results in a system with a settling time of 4 s. Also, obtain the value of the roots of the characteristic equation when the settling time is equal to 4 s.

6
Part 1)

7
Routh-Hurwitz Table Routh-Hurwitz Table: s3 1 14 0 s2 7 8+8K 0 s1 A 0 0 s0 8+8K 0 0 A = -(8+8K-98) = 98-8-8K 7 7 For stability, A > 0, therefore 0 < 98 - 8 - 8K Therefore range for stability is given by: 0 < K < 11.25

8
Part 2) the closed-loop transfer function was found to be: T(s) = 8K s3 + 7s2 + 14s + 8 + 8K. In calculating the settling time, we assume the validity of a second order approximation, allowing the use of the dominant pole pair to find settling time as: Ts = 4 ζwn where ζwn = σd = -1 * real part of dominant poles..

9
Part 2) K = 3.7333 Open-loop system Transfer function: 3.733 --------------------------------------------------- 0.125 s^3 + 0.875 s^2 + 1.75 s + 1 Closed-loop system Transfer function: 3.733 ---------------------------------------------------------- 0.125 s^3 + 0.875 s^2 + 1.75 s + 4.733 P = -5.70955929441205 -0.645220352793972 + 2.4931592351599i -0.645220352793972 - 2.4931592351599i settling_time = 6.19943246160627

10
Part 2)

11
Part 3) For the system to be practical, a settling time of 4 seconds is required. Making use of the second order equation, settling time Ts = 4 / ζwn where ζwn = - real part of the dominant closed-loop pole pair, the settling time is calculated for each K value from 0.1 to the limiting gain or until one yielding a result of 4 seconds is found. The step response, transfer functions and the roots of the characteristic equation are displayed for this value of K.

12
Part 3) For K = 1.5 settling time is approximated as 4 seconds. Open-loop transfer function with K = 1.5 Transfer function: 1.5 ---------------------------------------------------- 0.125 s^3 + 0.875 s^2 + 1.75 s + 1 Closed-loop transfer function with K = 1.5 Transfer function: 1.5 ------------------------------------------------------ 0.125 s^3 + 0.875 s^2 + 1.75 s + 2.5 Roots of the characteristic equation for K = 1.5 P = -4.99999999999999 -1 + 1.73205080756888i -1 - 1.73205080756888i Settling time for K = 1.5 settling_time = 3.9999

13
Part 3)

14
Animation

15
Problem 2

16
PROBLEM DEFINITION The goal of vertical takeoff and landing (VTOL) aircraft is to achieve operation from relatively small airport and yet operate as normal aircraft in level flight. An aircraft taking off in a form similar to a missile (on end) is inherently unstable. A control system using adjustable jets can control the vehicle

17
Block Diagram

18
Use MATLAB a) Find and plot closed loop poles in s-plane and discuss their location for K=100. b) Determine the range of gain K for which the system is stable, marginally stable and unstable. c) Determine and plot the roots of the characteristic equation for gain K obtained in part "b", which makes the system to be marginally stable and for selected gain K that makes the system unstable including poles locations from part "a" giving full comment. d) Plot step responses of the system for K=100, selected system gain, which makes the system to be unstable and the obtained gain K in part "b" which makes the system to be marginally stable, giving comments on the obtained results.

19
Part a) MATLAB results for K=100: P1= -3.1269+7.9403i ; zeros: z1= -2 P2= -3.1269-7.9403i P3= -2.7463

20
Part a)

21
Part b) K<12.857143 - unstable K=12.875143 - marginally stable K>12.875143 - stable

22
Part C) The roots for selected K=2 (which make the system unstable) are: P1=-9.8531 P2=0.4266+0.4733i P3=0.4266-0.4733i

23
Part C)

24
The roots for K=12.857143 (from part b) are: P1=-9 P2=0+1.6903i P3=0-1.6903i zeros: z1=-2

25
Part C)

26
Part D) Output step responses for K=100, K=12.8571 & K=2 Output step response of the aircraft control system

27
Output step response for K=100

28
Output step response for K=12.85714

29
Output step response for K=2

31
Problem 3

32
Consider the potential for a robot steering a motorcycle. The block diagram of the system model is shown in Figure 1. Determine the range of K for stable operation of the motorcycle when α1=g/h=9, α2=V 2 /hc=2.7, and α3=V/hc=1.35 where g is the gravity, c is the distance between the wheels of the motorcycle and h is the height of the centre of gravity. We assume the motorcycle is moving with a constant velocity V=2m/s. The time constant of the controller is =0.2 s, and K>0.

33
Figure 1

34
Part a Using MATLAB do the following: 1) plot the Step Response for the Physical System Dynamics only 2) plot the Root Locus and Step Response for the Closed-Loop system with system gain (K) is 1

35
Figure 2

36
Part b For α3=V/hc=3, using MATLAB do the following: 1)plot the Root Locus and Step Response for the Closed-Loop system with system gain, K=0.1 and 57 degree (1 rad) setpoint 2) plot the Root Locus and Step Response for this system with angle equal to 0 radian (0 degree setpoint)

37
Part a 1) In this part, the step response of just the Physical System Dynamics is to be displayed. From figure 1 of the problem definition, the Physical system is given by the transfer function: 1 = 1 for α 1 = 9. s 2 - α 1 s 2 - 9

38
Transfer function: 1/s^2 - 9

39
Part a 2) The system defined by the block diagram in figure 1 of the problem, has non-unity feedback. Therefore the root locus is found using the product of G 1 (s) and H(s), and the Matlab function rlocus(GH). Here, G 1 (s) is the product of the controller and dynamics transfer function. The step response can then be found for the closed loop transfer function T(s) = feedback(G 1 (s),H(s)) with gain K=1.

40
Part a 2) RESULTS Controller Transfer function: 1.35 s + 2.7 0.2 s + 1 Dynamics Transfer function: 1 s^2 - 9 Closed-Loop System Transfer function: 1.35 s + 2.7 1.55 s^3 + 6.4 s^2 + 4.95 s - 6.3

43
Part b 1) The system defined by the block diagram in figure 2 of the problem, has non-unity feedback. Therefore the root locus is found using the product of G 1 (s) and H(s), and the Matlab function rlocus(GH). Here, G 1 (s) is the product of the controller and dynamics transfer function. The step response can then be found for the closed loop transfer function T(s) = feedback(G 1 (s),H(s)) with gain K=0.1.

44
Controller Transfer function: 3 s^2 - 6 s - 9 0.2 s + 1 Dynamics Transfer function: 1 s^2 – 9 Closed-Loop System Transfer function: 3 s^2 - 6 s - 9 0.3 s^4 + 2.6 s^3 - 5.6 s^2 - 11.4 s - 9.9

47
Part b 2) For the system described by figure 2 of the problem, the root locus and step response are found for an angle of 0 radians (0 degrees). This example corresponds to vertical travel of the motorbike. The system gain is again 0.1, but this time the input is 0 degrees.

48
Controller Transfer function: 3 s^2 - 6 s - 9 0.2 s + 1 Dynamics Transfer function: 1 s^2 – 9 Closed-Loop System Transfer function: 3 s^2 - 6 s - 9 0.3 s^4 + 2.6 s^3 - 5.6 s^2 - 11.4 s - 9.9

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google