Feedback control systems
This presentation is the property of its rightful owner.
Sponsored Links
1 / 38

Feedback Control Systems PowerPoint PPT Presentation

  • Uploaded on
  • Presentation posted in: General

Feedback Control Systems. Dr. Basil Hamed Electrical & Computer Engineering Islamic University of Gaza. Root Locus. PROBLEM DEFINITION.

Download Presentation

Feedback Control Systems

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

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.   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= 0.707. 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).

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= 0.707 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

Block Diagram

Part 1)

Part 2)

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.

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 120752. Again we use the equation (-real(pole))/ωn 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

Block Diagram

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).

a) Root locus of unity feedback system

b) The output step response of the system

c) From MATLAB results it can be seen gain K for damping ratio 

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

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

d) Obtained OS% and Tp for gain K=41.8962 calculated in part (c)



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

e) The resulting output step response of the elevator system


Problem 3


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

Block Diagram

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= 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 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).

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

K1= 1.56, K2 =0.7475

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

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

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

Using the final value theorem the steady-state error is


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

The piloted output step response (switch closed) for K1=2.0602 and K2=1.5665 is


  • Login