pid controller designing for active suspension system of bike

1. Control Systems of Active
Suspension of Bike
Term Project of Control System
NS AbdullahBin Masood(2009-204)
PC Mubashar Sharif (2009-462)
08/01/2014

2. Abstract
The shock absorberof a bike isa verysignificant element when we are considering the driver comfort.
Active suspension system is a type of shock absorber in which there is an actuator which gives a
controlled force to our system in order to absorb the shock and to settle the system in a desired/pre-
defined settling time.
In our projectwe have mathematicallymodeledthe active suspension system of bike taking in account
the jerksand bumpsthata bike passesoverwhichare basicallythe “step Input” to the system and then
a controls system has been designed in order to absorb that shock in a particular time span (settling
time) and have proposed a PD Compensator to serve the purpose.
We have designedthe control systemforshockabsorberof frontwheel only,assumingthatsame will be
the system for rear wheel as well.
Whenwe simulatedoursystem(uncompensated),there wasnostead-stateerror,i.e. zero steady-state
error,in the systemresponse.Therefore,ouronlyjobwastocontrol the settling-time of system. So, we
have designedthe PD(proportional-differential)Compensatoronlyinsteadof PID(proportional-integral-
differential) Compensator which is used when you have to minimize the steady-state error along with
the response time.
Actuator:
The actuator consideredishydraulicactuator.The PDcompensatordesignedwill control the force input
from actuator and apply it to the system in order to settle the system response in respective settling
time. Transfer function of hydraulic actuator is:
𝑇 =
60
𝑠 + 60
We are assumingthe shockto be a steep/slope orabump,the inputwill be stepinput. Say, the bump is
of 12 centimeters.

3. Design Conditions:
Settling time of uncompensated system = 8 seconds
Desired settling time = 2 seconds
% O.S. = 5 %
→ Damping ratio = ᶓ = 0.69

4. MATLAB Code of UncompensatedSystem
k1=3500;
c1=350;
k2=4000;
m1=100;
m2=20;
c2=70;
s = tf('s');
W1 = ((m1+m2)*s^2+c2*s+k2)/((m1*s^2+c1*s+k1)*(m2*s^2+(c1+c2)*s+(k1+k2))-
(c1*s+k1)*(c1*s+k1));
W2 = (-m1*c2*s^3-m1*k2*s^2)/((m1*s^2+c1*s+k1)*(m2*s^2+(c1+c2)*s+(k1+k2))-
(c1*s+k1)*(c1*s+k1));
W3=((-m1*c2*s^3-m1*k2*s^2)/((m1+m2)*s^2+c2*s+k2));
W4= 0.12 * W3;
W5= 60/(s+60);
W6=W4+W5;
total= W6* W1;
T1=-0.42 *(s+61.83) *(s+55.31) *(s+0.2897) *(s-0.2883);
T2=(s+60)*(s^2 + 1.026*s + 18.43)*(s^2 + 23.47*s + 379.7);
T=T1/T2;

5. Time Response ofUNCOMPNESATEDSystem:
Root Locus of UNCOMPNESATEDSystem:

6. Root Locus of UNCOMPNESATEDSystemwith damping lines and gain point:

7. MATLAB Code of Compensated System
k1=3500;
c1=350;
k2=4000;
m1=100;
m2=20;
c2=70;
s = tf('s');
W1 = ((m1+m2)*s^2+c2*s+k2)/((m1*s^2+c1*s+k1)*(m2*s^2+(c1+c2)*s+(k1+k2))-
(c1*s+k1)*(c1*s+k1));
W2 = (-m1*c2*s^3-m1*k2*s^2)/((m1*s^2+c1*s+k1)*(m2*s^2+(c1+c2)*s+(k1+k2))-
(c1*s+k1)*(c1*s+k1));
W3=((-m1*c2*s^3-m1*k2*s^2)/((m1+m2)*s^2+c2*s+k2));
W4= 0.12 * W3;
W5= 60/(s+60);
W6=W4+W5;
total= W6* W1;
T1=-0.42*15.2*(s+61.83) *(s+55.31) *(s+0.2897) *(s-0.2883)*(s+2.44);
T2=(s+60)*(s^2 + 1.026*s + 18.43)*(s^2 + 23.47*s + 379.7);
T=T1/T2;

8. Time Response ofCOMPNESATED System:
Root Locus of COMPNESATED System:

9. Time Response ofCOMPNESATED System:
Conclusion:
Ourexpectedsettlingtime was2secondsfrom8 seconds.Whereas,due tosome errorthe settlingtime
has notreducedto exact2 secondsbutto approx.5 seconds.