LAG LEAD COMPENSATOR 7 6 Root Locus Design

LAG LEAD COMPENSATOR

7. 6 Root Locus Design Example. Consider rigid satellite control system A(s) + compensator K motor R(s) The RL is along the j axis, then the system will always oscillate regardless the value of K. To obtain an acceptable design the following system will be employed – compensator R(s) + K – R(s) j + KV The open loop-function is now: –j motor

7. 6 Root Locus Design And the RL is j –j The rate feedback changes the gain to KKv and add a zero at s = - 1/Kv. The system is seen to be stable for all K and Kv positive. Since there are 2 parameters K and Kv, the poles theoretically can be placed anywhere in the s plane. However it must be remembered that the physical system always has limitation. The rate feedback is actually a PD compensation with TF A problem with PD compensator is that its gain increases with frequency. If high frequency noise presents another compensator (phase lead) can be used to limit the high frequency gain.

PHASE LEAD DESIGN The TF of 1 st order phase lead compensator is given by The angle criteria is (all zero angles) – (all pole angles) = r consider the RL of the uncompensated is j Where |z 0| < |p 0|, and Kc, z 0, and p 0 are to be determined to satisfy the design criteria. First before we present the design procedure we consider the CE of compensated system The product of KKc can be considered as single gain parameter 2 poles –j Adding the lead compensation the root locus will shift the RL to the left. Suppose S 1 is on the RL then θz – θp- 2θ 1=-180 j θz θp p 0 z 0 θ 1 –j 2 poles

Analytical Phase Lead Design For this procedure it is convenient to express lead compensator as We can choose K=1 leaving 3 unknown. ao can be chosen arbitrarily, a 1 and a 2 is computed using (7. 53) from the reference book The object of the design is to choose a 0, a 1, and b 1 such that given s 1 That is we design the compensator places the pole at s 1. Here we have 4 unknowns with only 2 equations (magnitude and angle).

Controller Design Using Computer The following material can be downloaded from http: //www. engin. umich. edu/group/ctm/rlocus. html Using computer the design will be much easier. MATLAB®, give a convenient way to design. Consider an open loop system which has a transfer function of Say our design criteria are 5% overshoot and 1 second rise time. First we have to draw the root locus using MATLAB® program: num=[1 7]; den=conv([1 0], [1 5]), conv([1 15], [1 20])); rlocus(num, den) axis([-22 3 -15 15])

Controller Design Using Computer The program will produce the root locus chart as follows

Choosing a value of K from the root locus In our problem, we need an overshoot less than 5% (which means a damping ratio Zeta of greater than 0. 7) and a rise time of 1 second (which means a natural frequency Wn greater than 1. 8). Enter in the Matlab command window: zeta=0. 7; Wn=1. 8; sgrid(zeta, Wn) This command will give chart as follows

To meet the requirement the pole must be located in the shaded area. Note: the blue shaded area is added to the original chart for clarity