2Objectives for desired response Improving transient responsePercent overshoot, damping ratio,settling time,peak timeImproving steady-state errorSteady state error
3Gain adjustmentHigher gain, smaller steady stead error, larger percent overshootReducing gain, smaller percent overshoot, higher steady state error
4Compensator Allows us to meet transient and steady state error. Composed of poles and zeros.Increased an order of the system.The system can be approx. to 2nd order using some techniques.
5Improving transient response Point A and B have the same damping ratio.Starting from point A, cannot reach a faster response at point B by adjusting K.We have pole at A on the root locus, but we want response like at B.Compensator is preferred.
6Compensator configulations CascadeCompensatorFeedbackCompensatorThe added compensator can change a pattern of root locus
7compensator Method of implementing compensator: 1. Proportional control systems: feed the error forward to the plant.2. Integral control systems: feed the integral of the error to the plant.3. Derivative control systems: feed the derivative of the error to the plant.Es gibt eine Reihe von verschiedenen Kompensationsanlagen, die eingesetzt werden können, um zu beheben bestimmte System-Metriken, die außerhalb eines richtigen Betriebsbereich sind werden. Am häufigsten sind die Phaseneigenschaften in der Notwendigkeit der Kompensation, insbesondere wenn der Amplitudengang ist konstant bleiben.
8Types of compensator Active compensator Passive compensator PI, PD, PID use of active components, i.e., OP-AMPRequire power sourcess error converge to zeroExpensivePassive compensatorLag, Lead use of passive components, i.e., R L CNo need of power sourcess error nearly reaches zeroLess expensive
9Improving steady-state error (PI) Placing a pole at the origin to increase system order; decreasing ss error as a result!!(a) Original system without compensation(b) Add a pole at the origin but angular contribution at point A is no longer 180
10Improving Steady-State Error (PI) Also add a zero close to thepole at the origin. As angularcontribution of thecompensator zero and polecancels out, point A is stillon the root locus, and thesystem type has beenincreased.
11Improving Steady-State Error (PI):Example Choose zero at -1Damping ratio = in both uncompensated and PI cases
14Improving Steady-State Error (PI) As shown in the figure, the step response of the PI compensated system approaches unity in the steady-state, while the uncompensated system response approaches 1−0.108 = The simulation shows that it takes 18 seconds for the compensated system to reach and stay within ±2% of the final value of unity, while the uncompensated system takes about 6 seconds to settle to within ±2% of its final value of This is because there is no pole-zero cancelation and the pole not canceled is very close to the origin.
16Finding an intersection between damping ratio line and root locus Damping ratio line has an equation:where a = real part, b = imaginary part of the intersection point,Summation of angle from open-loop poles and zeros to the point is 180 degrees
18Magnitude of open loop system is 1 Use the formula to get the real and imaginary part of the intersection point and getMagnitude of open loop system is 1No open loop zero
19Draw root locus with compensator (system order is up by 1--from 3rd to 4th) Needs complex poles corresponding to damping ratio of (K=158.2)From K, find the 3rd and 4th poles (at and )Pole at can do phase cacellation with zero at -1 (3th order approx.)Compensated system and uncompensated system have similar transient response (closed loop poles and K are aprrox. The same)
20PI ControllerA compensator with a pole at the origin and a zero close tothe pole is called an ideal integral compensator, or PIcontroller
21Lag CompensatorIdeal integral compensation: pole is in the origin, requires active network (costly).Real (passive) integral compensation: pole is close to origin (not in the origin), cheaper.(a) Type 1 uncompensated system(b) Type 1 compensated systemType is not increased.What about steady-state error
23Example With damping ratio of 0.174, add lag Compensator to improve steady-state errorby a factor of 10
24Step I: find an intersection of root locus and damping ratio line ( j3.926 with K=164.56)Step II: find Kp = lim G(s) as s0 (Kp=8.228)Step III: steady-state error = 1/(1+Kp)= 0.108Step IV: want to decrease error down to[Kp = (1 – )/ = ]Step V: require a ratio of compensator zero to poleas /8.228 =Step VI: choose a pole at 0.01, the correspondingZero will be at *0.01 = 0.111
253rd order approx. for lag compensator (= uncompensated system) makingSame transient response but 10 timesImprovement in ss response!!!
27If we choose a compensator pole at 0.001 (10 times closer to the origin), we’ll get a compensator zeroat (Kp=91.593)New compensator:4th pole is at -0.01(compared to )producing a longertransient response.
28SS response improvement conclusions Can be done either by PI controller (pole at origin) or lag compensator (pole closed to origin).Improving ss error without affecting the transient response.Next step is to improve the transient response itself.
29Improving Transient Response Objective is toDecrease settling timeGet a response with a desired %OS (damping ratio)Techniques can be used:PD controller (ideal derivative compensation)Lead compensator
30PD controller: Improving transient response System above controlled by a pure gain (P controller) in the forward path has its root locus going through point A for some value of gain K.• Our goal is to speed up the response at A to that at B, while keeping the percent overshoot unchanged.• The above root locus with a P controller cannot go through point B (sum of angles from the open-loop finite poles and zeros to point B is not an odd multiple of 180◦). A solution is to add a (nonzero) zero to the forward path (e.g., PD controller).
31PD controller: Improving transient response • Transfer function of the PD controller Gc(s) = K2 s + K1 = K2(s+K1/K2) = K(s+zc) introduces a zero at −zc Into the forward path.• Effect of the added zero: The added zero will contribute to make the sum of angles from the open-loop finite poles and zeros to the desired point (point B) be an odd multiple of 180◦.Note: an added zero has the effect of pushing the root locus to the left while an added pole has the effect of pushing it to the right.• The new root locus can meet the specific transient response (with shorter settling time) by going through point B for some value of gain K.
32Ideal Derivative Compensator So called PD controllerCompensator adds a zero to the system at –Zc to keep a damping ratio constant with a faster response
33(a) Uncompensated system, (b) compensator zero at -2 (d) compensator zero at -3, (d) compensator zero at -4Indicate peak timeIndicate settling time
34Settling time & peak time: (b)<(c)<(d)<(a) %OS: (b)=(c)=(d)=(a)ss error: compensated systems has lower value than uncompensated one cause improvement in transient response always yields an improvement in ss error
36Exampledesign a PD controller to yield 16% overshoot with a threefold reduction in settling time
37Step I: calculate a corresponding damping ration (16% overshoot = 0 Step I: calculate a corresponding damping ration (16% overshoot = damping ratio)Step II: search along the damping ratio line for an odd multiple of 180 (at ±j2.064) and corresponding K (43.35)Step III: find the 3rd pole (at -7.59) which is far away from the dominant poles 2nd order approx. works!!!
38More details in step II and III Characteristic equation:
39Step IV: evaluate a desired settling time: Step V: get corresponding real and imagine number of the dominant poles( and )
41Step VI: summation of angles at the desired pole location, -275 Step VI: summation of angles at the desired pole location, , is not an odd multiple of 180 (not on the root locus) need to add a zero to make the sum of 180.Step VII: the angular contribution for the point to be on root locus is =95.6 put a zero to create the desired angle
42Compensator: (s+3.006)Might not have a pole-zero cancellation forcompensated system
45Lead Compensation Zeta2-zeta1=angular contribution • A PD controller can be approximated with a lead compensator, which is implemented with a passive network.• If the lead compensator pole is farther from the imaginary axis than the compensator zero, the angular contribution of the compensator is still positive and thus approximates an equivalent single zero.• The advantages of a passive lead compensator over an active PD controller are that (1) no additional power supplies are required and (2) noise due to differentiation is reduced.
46Lead Compensation Zeta2-zeta1=angular contribution • The concept behind lead compensation: the difference between 180◦ and the sum of the angles from the uncompensated system’s poles and zeros to the design point (desired pole location) must be the angular contribution required of the compensator.That is,q2 −q1 −q3 −q4 +q5 = (2k+1)180◦where q2 −q1 = qc is the angular contribution of the lead compensator.
47• The angular contribution qc can be determined from the rays originating from the desired closed-loop pole and terminating at the compensator pole and zero. These rays can be rotated about the desired closed-loop pole and thus different pairs of compensator pole and zero can be used to meet the transient response requirement.• Different possible lead compensators: differences are in the values of the static error constants, the static gain, the difficulty in justifying a second-order approximation when the design is complete, and the ensuing transient response.• For design we arbitrarily select either a lead compensator pole and zero and find the angular contribution at the design point of this pole or zero with the uncompensated system’s open-loop poles and zeros. The difference between this angle and 180◦ is the required contribution of the remaining compensator pole or zero.
48Example Design three lead compensators for the system that has 30% OS and will reduce settling time downby a factor of 2.
49Step I: %OS = 30% equaivalent to damping ratio = 0.358, Ѳ= 69.02 Step II: Search along the line to find a point that gives 180 degree (-1.007±j2.627)Step III: Find a corresponding K ( )Step IV: calculate settling time of uncompensated systemStep V: twofold reduction in settling time (Ts=3.972/2 = 1.986), correspoding real and imaginary parts are:
50Step VI: let’s put a zero at -5 and find the net angle to the test point (-172.69) Step VII: need a pole at the location giving 7.31 degree to the test point.
52Note: check if the 2nd order approx Note: check if the 2nd order approx. is valid for justify our estimates of percent overshoot and settling timeSearch for 3rd and 4th closed-loop poles(-43.8, )-43.8 is more than 20 times the real part of the dominant poleis close to the zero at -5The approx. is then valid!!!