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Feedback Control Systems

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Feedback Control Systems

Dr. Basil Hamed

Electrical & Computer Engineering

Islamic University of Gaza

Root Locus

The supersonic passenger jet control system requires good quality handling and comfortable flying conditions.Â Â An automatic flight control system can be designed for SST (Supersonic Transport) vehicles.Â The desired characteristics of the dominant roots of the control system shown in Figure have a Î¶= 0.707.Â The characteristics of the aircraft are Ï‰n =2.5, Î¶=0.30, and Ï„= 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).

1) Sketch the pole-zero map as a function of the loop gain K1K2

2) Determine the gain K2 necessary to yield roots with Î¶= 0.707 when the aircraft is in the medium-cruise condition

3) With the gain K2 as found in 2), determine the Î¶ of the roots when the gain K1 results from the condition of light descent

Results shown here indicate that a gain of K2 = 120752 for the aircraft in the medium-cruise condition, will yield a damping ratio of approximately 0.707.

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 120752. Again we use the equation (-real(pole))/Ï‰n to calculate the damping ratio.

Controller Transfer function:120752 s^2 + 483008 s + 483008------------------------------------------------s^2 + 110 s + 1000

Actuator Transfer function: 10------------s + 10

Aircraft Dynamics Transfer function: 0.02 s + 0.2--------------------------s^2 + 1.5 s + 6.25

Closed-loop system Transfer function: 2.415e004 s^3 + 3.381e005 s^2 + 1.063e006 s + 966016-------------------------------------------------------------------------------------s^5 + 121.5 s^4 + 2.644e004 s^3 + 3.52e005 s^2 + 1.091e006 s + 1.029e006

poles = 1.0e+002 *

-0.537 + 1.483i ,-0.537 - 1.4833i, -0.10, -0.0198 + 0.0048i, -0.0198 -0.0048i

required zeta =0.7902

Problem 2

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

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).

is equal to K=41.8962. This gain is calculated by MATLAB code for interactively selected point where the root locus crosses

OS%=0.0152=1.52%

Tp=7.7927sec

Comparing the calculated and simulated (see step output characteristic for K=41.8962) values for OS% and Tp we can notice very small difference

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

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 toÂ Î¶= 0.707 .

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 suitableÂ value based on the root locus.

e) PlotÂ closed loop system step responses when the pilot control loop is open and when the pilot loop is added (switch closed).

from the MATLAB results we can see that the values of steady-state errors are:ess1=Â 3.8552for interactively selected K2=1.5665ess1=Â 5.9386for interactively selected K2=0.7475

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)

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,Â K2=0.7475