Active control of suspension

Active control of suspension

One of the most potentially beneficial applications of an automotive control system is the active control of the suspension system. One feedback control system uses a shock absorber consisting of a cylinder filled with a compressible fluid that provides both spring and damping forces. The cylinder has a plunger activated by a gear motor, a displacement-measuring sensor, and a piston. Spring force is generated by piston displacement, which compresses the fluid. During piston displacement, the pressure imbalance across the piston is used to control damping.

The plunger varies the internal volume of the cylinder. This feedback system is shown in figure below.

  1. a) Assume the physical system (piston and plunge) is modeled as Gp=1/(s+1), the transfer function of gear motor is found from Table 2.5 in the textbook as Gg= 2/[s(s+20)(s+10)], and the controller transfer function is Gc=Ks, K is a control parameter. Sensor feedback H=1. Find the transfer function for this device with K in it. Use a single block diagram to describe the device.
  2. b) Determine control parameter K value range to yield a stable system.
  3. c) Plot the actual piston travel in the system with labels and title when K=20, if system input of desired piston travel is a unit step function and all initial conditions are zero.

Solution

  1. a) Diagram block

The open loop is:

If H = 1, the close loop is calculated by:

b)applying the Routh’s Stability Criterion we have:

Then, the conditions that all roots have negative real parts (stable) are:

So, the values range to yield a stable system is

  1. c) Using the equation (1) in Matlab the Code with K=20, we have

SimulationCode.m

clc

clearall

closeall

% Oppen loop

K=20;

Num = [2*K];

Den = [1 31 230 200];

Gol = tf(Num,Den);

% Close Loop

Gcl = feedback(Gol,1);

%Step

step(Gcl)

title(‘Step response with K = 20’)