Design a Lead controller using Bode techniques for the plant,

Question

such that the following performance specifications are met:



System is stable.



What will be an ideal response?

Answer 1

The plant spec Bode plots are shown below.



Again, a lead compensator is used to add phase at specific frequencies and gain compensation to move the LM plot vertically without affecting the phase. In this problem we need to add more phase than a single first order lead can supply. Two first order leads could be used but the frequency range necessary to obtain that much phase is excessively wide compared to the range of a second order lead. The transfer function for the second order lead is:



The following table shows the relationship between the lead ratio, and the phase produced by a second order lead compensator for two values of damping, and 0.8.



For this problem we need to add 140° of phase. Using the second order lead with this corresponds to a lead ratio of and because the lead is to be applied at 10 rad/sec,



and



The second order lead compensator becomes,



The Bode plot of the lead compensator plus plant is shown below.



At this point we will apply the corrective gain necessary to obtain a 10 rad/sec crossover frequency. From the LM plot this gain is 42 dB or 125.



The final control system is shown in the following figure.





The unit step response is presented below.





Notice that the response “trails” off towards zero as time increases. This is true because we did not address the steady state error as part of the specs, our only focus was on the transient response of the system. An additional PI or Lag compensator could be used to satisfy this additional spec had it existed originally.