INC341 Design Using Graphical Tool (continue) Lecture 10

Improving Both Steady-State Error and Transient Response PI, Lag improve steady-state error PD, Lead improve transient response PID, Lead-lag improve both (PID = Proportional plus Intergal plus Derivative controller)

PID Controller

PID controller design Evaluate the performance of the uncompensated system Design PD controller to meet transient response specifications Simulate and Test, redesign if necessary Design PI controller to get required steady-state error Find K constant of PID

Example Design PID controller so that the system can operate with a peak time that is 2/3 of uncompensated system, at 20% OS, and steady-state error of 0 for a step input

Step 1 %OS = 20%  damping ratio = 0.456  Ѳ = 62.87 Search along ther line to find a point of 180 degree (-5.415±j10.57) Find a correspoding K=121.51 Then find the peak time

Step 2 Decrease peak time by a factor of 2/3  get imaginary point of a compensator pole: To keep a damping ratio constant, real part of the pole will be at The compensator poles will be at -8.13±j15.867

Sum of the angles from uncompensated poles and zeros to the test point (-8.13±j15.867) is -198.37 The contribution angle for the compensator zero is then 180-198.371 = 18.37 ดPD controller is (s+55.92)

Step 3 Simulate the PD compensated system to see if it reduces peak time and improves ss error

Step 4 design PI compensator (one pole at origin and a zero near origin; at -0.5 in this example) Find a new point along the 0.456 damping ratio line (-7.516±j14.67), with an associate gain of 4.6

Step 5 Evaluate K1, K2, K3 of PID controller Compare to

Step 6

Lead-Lag Compensator Design Same procedures as in designing PID: Begin with designing lead compensator to get the desired transient response design lag compensator to improve steady-state error

Example Design lead-lag compensator so that the system can operate with 20% OS, twofold reduction in settling time, and tenfold improvement in steady-state error for a ramp input

Step 1 %OS = 20%  damping ratio = 0.456  Ѳ = 62.87 Search along ther line to find a point of 180 degree (-1.794±j3.501) Find a correspoding K=192.1 Then find the settling time

Step 2 Decrease settling time by a factor of 2  get a real part of a compensator pole: To keep a damping ratio constant, imaginary part of the pole will be at The compensator poles will be at -3.588±j7.003

Select the compensator zero at -6 to coincide with the open-loop pole Sum of the angles from uncompensated poles and zeros to the test point (-3.588±j7.003) is -164.65 The contribution angle for the compensator zero is then 180-164.65 = 15.35 Lead compensator is

Then find a new K at the design point (K=1977)

Step 3 Simulate the lead compensated system

Step 4 Originally the uncompensated system has the transfer function:

After adding the lead compensator, the system has changed to Static error constant, Kv, is then 6.794 (lead compensator has improved ss error by a factor of 6.794/3.201=2.122) So the lag compensator must be designed to improve ss error by a factor of 10/2.122=4.713

Step 5 Pick a pole at 0.01, then the associated zero will be at 0.04713 Lag-lead compensator Lag-lead compensatated open loop system

Step 6

Conclusions

Feedback Compensation Put a compensator in the feedback path

Tachometer Popular feedback compensator, rate sensor Tachometer generates a voltage output proportional to input rational speed

rate feedback

Example Design a feedback compensator to decrease settling time by a factor of 4 and keep a constant %OS of 20

Step 1 %OS = 20%  damping ratio = 0.456  Ѳ = 62.87 Search along the daping ratio line to get a summation of angle of 180 degrees at -1.809±j3.531 Find the corresponding K from the magnitude rule settling time

Step 2 Reduce the Settling time by a factor of 4 A new location of poles is at -7.236±j14.123

At dominant pole -7.236±j14.123, KG(s)H(s) has a net angle of = -277.33  needs an additional angle from zero of 277.33-180 = 97.33 Find the corresponding K to the pole at -7.236+j14.123 using the magnitude rule: K = 256.819

Feedback block is 0.185(s+5.42)

Physical System Realization PI Compensator

Lag Compensator

PD Compensator

Lead Compensator

PID Compensator