Lead/Lag Compensator Frequency Domain Properties and Design Methods


 Mervyn Blankenship
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1 Lectures 6 and 7 Lead/Lag Cmpensatr Frequency Dmain Prperties and Design Methds Definitin Cnsider the cmpensatr (ie cntrller Fr, it is called a lag cmpensatr s K Fr s, it is called a lead cmpensatr Ntatin When using such cmpensatrs in carrying ut pleplacement rt lcusbased design, we ften used the frm s K In the same way, mst tables f Laplace transfrm pairs express a first rder system as G s We s a chse the ntatin in Definitin because the units f thse parameters are, indeed, radians/sec Since in engineering the symbl is cmmnly used t represent frequency in radians/secnd, we have simply chsen t highlight this fact, when addressing these cmpensatrs in the frequency dmain Example Cnsider s K Because, this is a lead cmpensatr In relatin t the rt s lcus, this cmpensatr adds angle (r phase t the rt lcus angle criterin Furthermre, the zer lead s (r ccurs befre the ple n the negative real axis (a Fr K, this is a unity static gain cmpensatr Give its Bde plt Then verlay a straightline apprximatin The StraightLine Apprximatin Bde Plt The symmetry in bth plts is n accident The magnitude plt straightline apprximatin has turnn frequencies At the zer break frequency the slpe changes frm zer t /decade At the ple break frequency the  /decade slpe assciated with this ple turns n, and cancels that f the zer The straightline phase apprximatin is trickier By itself, the phase f the numeratr f (s ges frm t 9 as : Hence, it has nt nly a turnn frequency, but als a turnff frequency The frmer is at, and the latter is at Because this turnn/turnff interval spans tw decades, the slpe is 45 / decade ver the interval [, ]The denminatr term will clearly have a negative slpe ver the interval [, ] Because these intervals verlap, the slpes will cancel each ther in that regin
2 (b At what frequency will the phase cmpensatr cntribute the mst phase t the pen lp system At 3rad / s In fact, frm the symmetry f the magnitude plt, we can speculate that the frequency at which the phase is imum is the gemetric mean f [which is equivalent t lg( [lg( lg( ]/ ] (, : (c Estimate the amunt f amplificatin the cmpensatr gives at Then infer a general expressin fr it At 3rad / s the cmpensatr magnitude is abve the static gain f It is als belw the cmpensatr high frequency gain Hence we can infer that at the cmpensatr gain is higher that then static gain, where lg( / / (d Summarize the gain and phase relatins fr the lead cmpensatr lead is imum s K at the frequency where its phase s SUMMARY: The frequency f imum phase lead is, and the amplificatin re: static gain is where lg( / / (e Infer a similar summary fr the lag cmpensatr SUMMARY: The frequency f imum phase lag (ie negative phase is, and the attenuatin re: static gain is / where lg( / (ie this will be a negative value Example Suppse that we have experimentally arrived at the FRF at the right fr a given plant that is t be cntrlled (afr unity feedback prprtinal cntrl with K, estimate the CL GM and PM gc r / s PM 5 and pc r / s PM is undefined (b By incrprating the unity static gain cmpensatr in Example, determine hw will the OL Bde plt be affected at 3rad / s Fr s, the cmpensatr Bde plt gives M ( 3 and ( 3 55 Hence, at this frequency, the OL s magnitude will becme  + = , and the OL phase will becme = 5 (c Determine hw much we wuld need t gain up the cmpensatr in (b s that the pen lp gain crssver frequency is at 3 r/s [Nte: this wuld result in a ~75 CL PM] Answer: We wuld need K / K 3 6
3 3 (darrive at a transfer functin mdel fr the plant, using the plant Bde plt At lw frequencies, the magnitude slpe is /dec This can be mdeled by a /s term This lpe breaks t a 4/dec value at the frequency 5 r/s Hence, we will mdel the plant as G p s ( s 5 T find the value fr, we will use the lw frequency gain 46/ 46 M p ( M p ( (5 5 Hence, ur plant mdel is G p s ( s 5 (efr the cmpensatr achieved by (bc, and the plant mdel in (d, use the cmmand margin t verify the design The plt at right verifies the design (fplt the CL step respnse, and cmment n it The plt is at lwer right The CL has unity static gain, achieves steady state in ~ secnd, and has n versht Remark It shuld be nted that we did nt use the mdel in (d t design the cntrl system We used the experimental FRF fr the plant We nly used the mdel as a way f analytically verifying the design, since we d nt have a real plant t implement it n
4 4 Derivatin f Phase Cmpensatr Design Equatins fr Changing Clsed Lp Phase Margin Here we will derive the mathematical frmulas needed t design a unity static gain phaselead cmpensatr: s with s ( First, we write the cmpensatr FRF: i ( i i Hence: ( M G ( i and ( tan ( / tan ( / c (afind the frequency at which ( is imum Slutin: OK Here s the first f the tw tricks that I use t slve this prblem: Recall (ha! ha! the trig identity: tan( tan( tan( tan( tan( T use this, let tan ( / and tan ( / We then have tan[ ( ] ( / ( /, and hence: ( / ( / ( tan where d( T find the frequency ( achieves its imum value, we need t slve Nw, n the surface it d may seem like this is a real pain, due t the arctan functins Hwever recall that the arctan is a mntnically increasing functin n the interval ( /, / T see why this matters, recall the chain rule fr differentiatin (the secnd trick: g(x dg[ f ( ] d dg df df d If is strictly mntnic, that means that it derivative is strictly greater than (r less than zer Hence, if in this case df we set the abve equatin t zer, then it must be that Applying this secnd trick, we get d d( d d d ( ( ( ( Fr the rightmst quantity t equal zer, its numeratr must equal zer; that is: ( ( ( ( ( ( ( The slutin t this is = the frequency where ( achieves its imum (b Find the expressin fr the cntrller magnitude at Slutin: This is relatively easy:
5 5 M ( ( ( Hence: M ( Nw, the static gain f the cntrller is D ( and its high frequency gain is D ( If we cnvert these gains t decibels, then we have D ( lg, D( and Magnitude ( Bde Diagram D ( i lg In wrds, the cntrller achieves half f its ttal gain at T visualize this, let The Bde plt fr this cntrller is and Phase (deg Frequency (rad/sec shwn at right We see that the phase is imum at is, and the assciated gain is half way frm t (cderive the expressin fr ( Slutin: T this end, define / [NOTE: On p349 the term α used by the authrs is /! Hence, my α is the inverse f theirs] We then have ( tan ( tan (/ Applying the abve trig identity gives tan( Nw cnsider the fllwing right triangle at right We see that 4 ( Hence sin(, r sin( sin( We can nw summarize the design prcedure fr using a lead cntrller t increase the pen lp phase by a specified amunt : Having specified sin(, cmpute Since /, by knwing α we knw the rati f the sin( cntrller break frequencies Next, decide at what frequency this phase gain shuld ccur This decisin will ften be made in cnjunctin with ther clsed lp specificatins (eg errr cnstants, etc, and s n ne prcedure can sin( be described here Since and, we can slve these fr the unknwns & : sin( Frm the first, we have / Substituting this int the secnd gives: sin( Having this, we can nw slve fr / sin(
6 6 Remark The abve develpment is nt given in the bk (it is left as a hmewrk prblem Furthermre, expressins (639 and (64 n p365 invlve the quantities While I d briefly use the variable α, it is nt the same as the authrs On p366 the authrs use the quantities z & p While ne might presume these are the cmpensatr zer and ple, respectively, they are nt As evidenced by (64 they are the negatives f the zer and ple Finally, I chse t use cmpensatr break frequencies design These break frequencies are directly read frm an pen lp Bde plt T d & & fr the simple reasn that we are engaging in frequency dmainbased cntrller Implementatin f the Design Equatins There are tw standard methds f implementatin Methd : When the PM is specified but n value fr is specified In this case, fr a specified PM ne wuld first identify the current OL gain crssver frequency, current PM The difference PMPM wuld be raise the OL gain by / And s, except fr the fact that if we center the cmpensatr in rder t read the, it will will n lnger be the gain crssver frequency If we have the freedm t use a nnunity static gain cmpensatr, then we can simply make its static gain equal t / Then stature as the gain crssver frequency If we have n such freedm (and ften we d nt since by adjusting the cmpensatr static gain, we wuld alter steady state errr, then we have t resrt t the fllwing apprach: will retain its Identify the frequency at which the OL magnitude is currently /, and hpe that the OL phase has nt changed much in relatin t what it is at Typically, it will be less than what it was at Fr this reasn, we wuld add a cushin t PM PM at the utset The steps fr this methd are summarized belw Cmpute the current PM, and then cmpute the needed PM PM cushin A reasnable first chice fr the cushin is 5 sin( Cmpute an initial value fr sin( 3 Lcate the frequency, O L, where the current pen lp has G ( i / [This is because, if yu were t nw recenter the cmpensatr at, this wuld becme ( new ( 4 Cmpute the pen lp phase ( ld L O L ( O ] 5 Cmpute 8 ( new O L ( If this value is clse t the required PM, missin accmplished If this value is deemed t be unacceptably lw, then return t and add a larger cushin [If this value is deemed t be unacceptably large (which rarely happens!, then return t and reduce the cushin ] 6 Cmpute /, fllwed by (new gc
7 7 Example 3 [See bk Example 65 n p 367] Cnsider a unity feedback system with pen lp transfer functin G / s( s with FRF at right Suppse that we desire PM 45 withut changing the OL lw frequency gain Using margin(g we get PM 8 We will chse (ie a cushin Then a=(+sind(3/(sind(3 = 3546, r a=*lg(a =5 3 Fr GO L ( i / 5 5 Using the data cursr, we find 4r / s ( ld 4 Using the data cursr, we then find O L ( 67, hence ( new O ( L ( new 5 8 O L ( PM Missin accmplished (in a single try! 6 / 33r / s, fllwed by 758r / s 5 Hence, s s The clsed lp step s s 758 respnses fr the cntrllers G C and s 33 G C 35 are shwn at right Clearly, the s 758 inclusin f the PM specificatin resulted in a huge imprvement Befre leaving this example, in rder t better tie this design methd t the rt lcus ple placement methd, we ffer the rt lcus crrespnding t the pen lp system s 33 G K 35 s 758 s( s [Nte: K crrespnds t ur final OLTF] The rt lcus plt at the right reveals that fr K (required t satisfy the steady state errr specificatin, the cmplex clsed lp ples are imally damped [As an aside: Whereas the data cursr states that thse ples crrespnd t % versht, we see frm the abve step respnse, that the actual versht is ~% This increase is due t the clsed lp zer that the phase cmpensatr intrduced] QUESTION: In view f the abve, why didn t we simply design the cntrller via the rt lcus pleplacement methd? It seems like we are just learning a new methd that gives the same results we wuld have gtten using a methd we already knw? ANSWER: Recall that the frequency dmain methd never actually used direct knwledge f the plant transfer functin Even thugh we btained the plant FRF by assuming it was knwn, very ften a plant FRF is btained frm experimental
8 8 data Hence, the answer is: Because ften we d nt have a mathematical mdel fr the plant transfer functin, the frequency dmain methd can be mre apprpriate As reasnable as the abve answer is, there is yet anther additinal answer that is ften as, if nt mre imprtant T this end, suppse that there is a sec time delay between when the errr is cmputed and when it is received by the cntrller T incrprate this delay int the pen lp, st recall that fr e( t E( s, then e( t t e E( s And s nw the final pen lp transfer functin is: G( s s e s 758 s( s s The clsed lp unit step respnses withut and with this time delay included are shwn at right Clearly, the time delay had a destabilizing effect The term in the pen lp transfer functin means that it is n lnger a rati f plynmials, frm which ne simply btains ples and zers Hence, nw, the rt lcus methd des nt even apply i t Even s, this term is trivial t incrprate int the pen lp FRF This is because e has a magnitude f (and s des nthing t the pen lp magnitude, and cntributes angle t And s, at the new gain e st t ( crssver frequency 4r / s this time delay will reduce the pen lp phase by t ( (4(8 / 4 Hence, when the delay is included, the clsed lp PM 45 4 OUCH! Basically, we are back t where we were prir t incrprating the phase cmpensatr Yes, that IS true Hwever, had that cmpensatr NOT been incrprated, the clsed lp PW wuld be PM In layman s terms, we wuld have been SOL Methd : When the PM is specified and als a value fr is specified In sme ways this case is easier that Methd It typically requires the cmbined use f a lead and a lag cmpensatr The lead cmpensatr is use t adjust the phase at Then the lag cmpensatr is used as a high frequency attenuatr t frce the OL gain t be at cmpensatr needs t be used as a high frequency amplifier] [Nte: It can als happen that, as ppsed t a lag cmpensatr, a secnd lead Example 4 Cnsider again the plant in Example 5: / s( s The Bde plt is given again here fr cnvenience G p
9 9 Suppse that we require that the clsed lp have PM 6 at an pen lp gain crssver frequency 58 r gc / will first design a unity static gain lead cmpensatr with and r / s The abve design equatins give: & 8, and s G lead (8/ 5( s 5 /( s 8 The resulting OL Bde plt is shwn belw 5 c s We We see that the phase is what we want at r / s, but the OL gain is abut  S we will use a secnd lead cmpensatr as a high frequency amplifier G lead c ( / ( s /( s
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