Stability analysis of feedback systems

Chapter 1 Stability analysis of feedback systems 1.1 Introduction A control system must be asymptotically stable. A method for determination of the stability property of a control system will be described in the following. The method is based on the frequency response ofthelooptransferfunction 1,L(jω),anditisdenotedNyquist sstability criterion. This is a graphical analysis method. There are also algebraic analysis methods, as the Routh s stability criterion which is based on the coefficients of the characteristic polynomial of the control system. Such stability analysis methods are not described in this book since they have limited practical importance(and the mathematical operations become quite complicated except for the simplest models). 1.2 Stability analysis of dynamic systems in general This section defines the different stability properties that a dynamic systemcanhaveintermsofimpulseresponseofthesystem. Then,the corresponding transfer function pole locations in the complex plane are derived. In Section 1.3, and subsequent sections, these results are applied loop. 1 The loop transfer function is the product of all the transfer functions in the closed 1

2 to feedback(control) systems, which are a special case of dynamic systems. 1.2.1 Stability properties and impulse response The different stability properties can be defined in several ways. I have chosentousetheimpulseresponse ofthesystemasthebasisfordefinition of the stability properties. The impulse response is the time-response in outputvariableofthesystemduetoanimpulseontheinput. Usingthe impulse response makes it relatively simple to relate stability properties to thepoles ofthesystem(thisisbecausetheimpulseresponseandthepoles are closely connected), as you will see soon. Some words about the impulse signal: It is a time-signal which in principle has infinitely short duration and infinite amplitude, but so that the integral ofthesignal theintegralisequaltotheareaunderthetime-functionof thesignal isfinite. Thisareaisalsocalledthestrength oftheimpulse. Animpulseofstrengthoneiscalledaunitimpulse,δ(t). Thesquarepulse infigure1.1approachesanimpulsefunctionas goestozero. A t Figure1.1: Thesquarepulseapproachesanimpulsefunctionas goestozero. Herearethestabilitydefinitions: Adynamicsystemhasoneofthe following stability properties: Asymptotically stable system: The stationary impulse response,

3 h(t), is zero: lim h(t)=0 (1.1) t Marginally stable system: The stationary impulse response is different from zero, but limited: 0< lim t h(t)< (1.2) Unstable system: The stationary impulse response is unlimited: lim h(t)= (1.3) t A system is stable if it is either asymptotically stable or marginally stable. Figure 1.2 depicts the different stability properties. Asymtotically stable system h(t) 0 t Impulse, System Impulse response y(t)=h(t) Marginally stable system 0 t Unstable system Figure 1.2: Different stability properties Inthedefinitionaboveitisassumedthattheimpulseintheinputustarts fromzeroandthattheimpulseresponseintheoutputyhaszeroinitial value. Inpracticetheseinitialvaluesmaybedifferentfromzeroifthe system initially is some other operating point than the zero operating point.

4 Oneproblemwiththeidealimpulsefunctionisitcannotbegenerated fullyinpractice,butinpracticethereishardlyeveraneedtoperform impulse response experiments to determine stability properties. It is more useful as a conceptual definition of stability, cf. the next section. 1.2.2 Stability properties and poles Inmanycasesitwouldbequiteimpracticaliftheonlywaytodetermine thestabilitypropertyofasystemwastodoexperimentsortorun simulations, to obtain the impulse response(or the step response). Fortunately, we can conclude about the stationary part of the impulse response, and hence conclude about the stability property, by just analyzing the mathematical model of the system. This is because the stationary impulse response is a function of the poles of the system. The connection between the impulse response and the poles is derived in the following. Letusassumethatthemathematicalmodelofthesystemisatransfer function, H(s), from input signal u to output signal y: y(s) = H(s)u(s) (1.4) Theinputuwillbeaunitimpulse,thatis,u(t)=δ(t). Itcanbeshown thatthelaplacetransformofδ(t)is1. Letusdenotetheimpulseresponse h(t). The Laplace transform of h(t) is h(s)=h(s)l{δ(t)}=h(s) 1=H(s) (1.5) Thus, the Laplace transform of the impulse response equals the transfer function of the system. We need the impulse response time-function h(t) since it is the basis for the definitions of the different stability properties. With results from Laplace transform theory, h(t) can be calculated using the following formula: h(t) = 1 d m 1 lim s p i (m 1)! i ds m 1 (s p i ) m H(s) e st (1.6) }{{} h(s) = lim s p i i (s p i)h(s) e st ifm=1 (1.7) }{{} {p i }isthesetofpolesinh(s),andhencetherootsofthedenominatorof H(s). m is the multiplicity of the poles(so-called simple poles have h(s)

5 m = 1). The denominator polynomial of H(s) is called the characteristic polynomial,, a(s). The poles are the roots of a(s). Consequently, the poles are the solutions of the characteristic equation An example: Transfer function H(s)= a(s) = 0 (1.8) 1 (s+2)(s+3) hasthepolesp 1 = 2andp 2 = 3,andthecharacteristicpolynomialis (1.9) a(s)=(s+2)(s+3)=s 2 +5s+6 (1.10) We will now use(1.6) and(1.7) to connect the different stability properties tothepoleplacementinthecomplexplane. Letusfirstassumethatthe polesofh(s)aresimple. Thenm=1,andh(t)isgivenby(1.7). Apoleis generally a complex number: p i =a i +jb i (1.11) wherea i istherealpartandb i istheimaginarypartofthepole. (1.7) impliesthath(t)isequaltothesumoftheipartialresponsesofthetotal impulse response: h(t)= h i (t) (1.12) i where h i (t)=k i e p it =k i e (a i+jb i )t =k i e a it e jb it (1.13) Herek i issomeconstant. Theterme jb it isacomplexnumberontheunity circleandthereforeithasabsolutevalueequalto1. 2 Thus,itistheterm e a it whichdeterminesthesteady-state(t )absolutevalueofthe partialresponseh i (t)accordingtothefollowinganalysis: Supposethattherealpart,a i,ofthepoleisstrictlynegative,thatis, a i <0,whichmeansthatthepoleliesinthelefthalfplane. This impliese a it 0,andthereforeh i (t) 0ast. Supposethattherealpart,a i,ofthepoleiszero,thatisa i =0, whichmeansthatthepoleliesontheimaginaryaxis. Thisimplies e a it =1,andthereforeh i (t)goestowardsaconstantvaluedifferent fromzeroast. 2 Ifapolehasaimaginarypartbdifferenfromzero,theremustbeacomplexconjugate pole with imaginary part b. However, this fact does not influate the conclusions of the analysis we are performing.

6 Supposethattherealpart,a i,ofthepoleisstrictlypositive,thatis, a i >0,whichmeansthatthepoleliesintheleftrightplane. This impliese a it,andthereforeh i (t) ast. From the above analysis we can conclude as follows for transfer functions havingpolemultiplicityone: (1)Ifeachofthepolesliesinthelefthalf plane, the system is asymptotically stable, because then each of the partial impulseresponseterms,h i (t),goestowards0ast. (2)Ifonepolelies ontheimaginaryaxiswhiletherestofthepolesliesonthelefthalfplane, thesystemismarginallystable,becausethenoneoftheh i (t)-termsgoes towardsaconstantvaluedifferentfrom0ast. (3)Ifatleastoneof thepolesliesintherighthalfplane,thesystemisunstable,becausethen atleastonetermh i (t)goesto ast. Multiple poles: It would have been nice to conclude about stability and poleplacementnow,butwehavetolookcloseratthecaseofmultiple polesofh(s). Theimpulseresponseh(t)isgivenby(1.6). Supposethat thatthemultiplicityofthepolep i ism=2. Thecorrespondingpartial impulse response becomes { d [ h i (t)= lim (s p i ) 2 H(s)e st]} (1.14) s pi ds Here,theterm d ds (est )isequaltote st,whichmeansthath i (t)willcontain termsaste p it. Byperformingthesameanalysisasforsimplepoles,wewill findthefollowing: (1)h i (t) 0forapolewithnegativerealpart(since te p it goestowardszerobecausee p it decreasesfasterthantincreases). (2) h i (t) forapoleontheimaginaryaxis(te p it equalst). (3)h i (t) forapolehavingpositiverealpart. Wewillgetthesameresultsifthe multiplicity m is greater than two. Now we can conclude by stating the following correspondence between stability and pole placement: Asymptotically stable system: Each of the poles of the transfer function lies strictly in the left half plane(has strictly negative real part). Marginallystable system: Oneormorepolesliesonthe imaginaryaxis(haverealpartequaltozero),andallthesepolesare distinct. Besides, no poles lie in the right half plane. Unstable system: Atleastonepoleliesintherighthalfplane(has realpartgreaterthanzero). Or: Therearemultiplepolesonthe imaginary axis.

7 Figure 1.3 gives a illustration of the relation between stability property and pole placement. Left half plane Im Right half plane Asymptotically stable pole area Unstable pole area Re Figure 1.3: The relation between stability property and pole placement Example 1.1 Stability property of some simple dynamic systems The first order transfer function H 1 (s)= 1 s+1 haspolep= 1whichliesinthelefthalfplane. Thus,thesystemis asymptotically stable. (1.15) The transfer function H 2 (s)= 1 s (1.16) (whichisthetransferfunctionofanintegrator)haspolep=0,whichlies ontheimaginaryaxisandhasmultiplicityone. So,thesystemis marginally stable. The transfer function H 3 (s)= 1 s 2 (1.17) havepolesp=0,whichisontheimaginaryaxiswithmultiplicitytwo. The system is therefore unstable. The transfer function H 4 (s)= 1 s 1 (1.18)

8 haspolep=+1,whichliesintherighthalfplane. Thesystemistherefore unstable. [End of Example 1.1] Example 1.2 Stability property of a mass-spring-damper Figure 1.4 shows a mass-spring-damper-system.y is position. F is applied F [N] m K [N/m] D [N/(m/s)] 0 y [m] Figure 1.4: Mass-spring-damper force. D is damping constant. K is spring constant. Newton s 2. Law gives the following mathematical model: mÿ(t)=f(t) Dẏ(t) Ky(t) (1.19) ThetransferfunctionfromtheforceF topositionyis H(s)= y(s) F(s) = 1 ms 2 +Ds+K (1.20) Assumethatm=20kg,D=4N/(m/s),andK=2N/m. Whatisthe stability property of the system? The characteristic polynomial becomes which has roots a(s)=ms 2 +Ds+K=20s 2 +4s+2 (1.21) p 1, p 1 = 4± 4 2 4 20 2 2 20 = 0.1 ± j0.3 (1.22) which are the poles of H(s). Both these poles have strictly negative real parts( 0.1). The system is therefore asymptotically stable. Figure 1.5 shows the poles(marked as crosses) in the complex plane. Figure 1.6 shows the impulse response. AssumenowthatthedamperisremovedsothatD=0. Thenthe characteristic polynomial is a(s)=ms 2 +K (1.23)

9 Figure 1.5: The poles of a mass-spring-damper plotted in the complex plane. Thepolesarep 1,2 = 0,1±j0,3. Figure 1.6: The impulse response for a mass-spring-damper with m = 20 kg, D=4N/(m/s)andK f =2N/m. Thesystemetisasymptoticallystable. andthepolesare K p 1, p 1 =±j =±j0.32 (1.24) m which lies on the imaginary axis, and they have multiplicity one. The system is then marginally stable. Figure 1.7 shows the impulse response. [End of Example 1.2]

10 Figure 1.7: The impulse response of the mass-spring-damper with m = 20 kg, D=0N/(m/s)ogK f =2N/m. Thesystemismarginallystabile. 1.3 Transfer functions used for stability analysis of feedback systems Which is the transfer function to be used to determine the stability analysis of feedback systems, e.g. control systems? Figure 1.8 shows a block diagram(transfer function based block diagram) of a control system. Wemustselectatransferfunctionfromoneoftheinputsignalstothe closedlooptooneoftheoutputsignalsfromtheloop. Letusselectthe transferfunctionfromthesetpointy msp totheprocessmeasurementy m. This transfer function is y m (s) y msp (s) = H c(s)h u (s)h m (s) 1+H c (s)h u (s)h m (s) = L(s) =T(s) (1.25) 1+L(s) which is the tracking transfer function of the control system. (If we had selected some other transfer function, for example the transfer function from the disturbance to the process output variable, the result of the analysis would have been the same.) L(s) is the loop transfer function of the control system: L(s)=H c (s)h u (s)h m (s) (1.26) Figure1.9showsacompactblockdiagramofacontrolsystem. The transferfunctionfromy msp toy m isthetrackingfunction: s T(s)= L(s) 1+L(s) = n L (s) d L (s) 1+ n L(s) d L (s) = n L (s) d L (s)+n L (s) (1.27) wheren L (s)andd L (s)arethenumeratoranddenominatorpolynomialsof L(s), respectively. The characteristic polynomial of the tracking function is c(s)=d L (s)+n L (s) (1.28)

11 Process disturbance Process d(s) Reference or setpoint y r (s) H sm (s) Reference in measurement unit y mr (s) Model of sensor with scaling Manipulating Control error variable e m (s) Hc (s) u(s) Controller Measurement y m (s) H u (s) H d (s) Actuator transfer function H s (s) Sensor (measurement) with scaling Disturbance transfer function y(s) Process output variable Figure 1.8: Block diagram(transfer function based) of a control system The stability of the control system is determined by the placement of the roots of(1.28) in the complex plane. Example 1.3 Stability of a feedback control system See Figure 1.8 which shows a block diagram(transfer function based) of a control system. We assume that the individual transfer functions with parameter values are as follows: H sm (s)=k sm =1 (1.29) y msp L(s) y m Figure1.9: Compactblockdiagramofacontrolsystemwithsetpointy msp as inputvariableandprocessmeasurementy m asoutputvariable

K p =1(asympt. stable) K p =2(marg. stable) K p =4(unstable) p 1 = 1.75 p 1 = 2 p 1 = 2.31 p 2 = 0.12+j0.74 p 2 =j p 2 =0.16+j1.31 p 3 = 0.12 j0.74 p 3 = j p 3 =0.16 j1.31 Table1.1: PolesofthetrackingtransferfunctionforvariousK p -values 12 H s (s)=k s =1 (1.30) H c (s)=k p (proportionalcontroller) (1.31) 1 H u (s)= (s+1) 2 (1.32) s 1 H d (s)= (s+1) 2 (1.33) s The stability property of the control system can be determined from the placementofthepolesofthetrackingtransferfunction,t(s),whichisthe transferfunctionfromthereferencey SP totheprocessoutputvariabley. The tracking transfer function is Inserting(1.29) (1.33) gives T(s)= y(s) y SP (s) = H sm(s)h c (s)h u (s) 1+H s (s)h c (s)h u (s) H yr,y(s)= The characteristic polynomial of T(s) is (1.34) K p s 3 +2s 2 +s+k p (1.35) a(s)=s 3 +2s 2 +s+k p (1.36) Table1.1showsthepolesforthreedifferentK p -values. 3 Figure1.10showsthestepresponseinyforthethreeK p -values(itisa unitstepinr). [End of Example 1.3] 1.4 Nyquist s stability criterion The Nyquist s stability criterion will now be derived. We start with a little rewriting: Therootsof(1.28)arethesameastherootsof d L (s)+n L (s) =1+ n L(s) =1+L(s)=0 (1.37) d L (s) d L (s) 3 The poles can be calculated using the MATLAB-function roots or pzmap or pole or using the LabVIEW-function Complex Polynomial Roots.

13 Figure 1.10: Example 1.3: Step response in the process output variable y for threedifferentk p -values which, therefore, too can be denoted the characteristic equation of the control system. (1.37) is the equation from which the Nyquist s stability criterion will be derived. In the derivation we will use the Argument variation principle: Argument variationprinciple: Givenafunctionf(s)wheresisa complexnumber. Thenf(s)isacomplexnumber,too. Aswithall complexnumbers,f(s)hasanangleorargument. Ifsfollowsa closed contour Γ(gamma) in the complex s-plane which encircles a numberofpolesandanumberofzerosoff(s),seefigure1.11,then the following applies: arg Γ f(s)=360 (numberofzerosminusnumberofpolesoff(s)insideγ) (1.38) wherearg Γ f(s)meansthechangeoftheangleoff(s)whenshas followed Γ once in positive direction of circulation(i.e. clockwise).

14 Γ contour Im(s) Right half plane Re(s) P CL poles of closed loop system in right half plane P OL poles of open loop system in right half plane Positive direction of circulation Figure 1.11: s shall follow the Γ contour once in positive direction (counter clockwise). Forourpurpose,weletthefunctionf(s)intheArgumentvariation principle be f(s)=1+l(s) (1.39) TheΓcontourmustencircletheentirerighthalfs-plane,sothatweare certainthatallpolesandzerosof1+l(s)areencircled. Fromthe Argument Variation Principle we have: arg Γ d L (s)+n L (s) [1+L(s)] = arg Γ d L (s) = 360 (numberofrootsof(d L +n L )inrhp (1.40) minusnumberrootsofd L inrhp) (1.41) = 360 (numberpolesofclosedloopsysteminrhp minusnumberpolesofopensysteminrhp) = 360 (P CL P OL ) (1.42) whererhpmeansrighthalfplane. By opensystem wemeanthe (imaginary)systemhavingtransferfunctionl(s)=n L (s)/d L (s),i.e.,the

15 original feedback system with the feedback broken. The poles of the open systemaretherootsofd L (s)=0. Finally, we can formulate the Nyquist s stability criterion. But before we dothat,weshouldremindourselveswhatweareafter,namelytobeable todeterminethenumberpolesp CL oftheclosedloopsysteminrhp.it those poles which determines whether the closed loop system(the control system)isasymptoticallystableornot. If P CL =0theclosedloopsystem is asymptotically stable. Nyquist sstabilitycriterion: LetP OL bethenumberofpolesofthe opensystemintherighthalfplane,andletarg Γ L(s)betheangular changeofthevectorl(s)asshavefollowedtheγcontouroncein positivedirectionofcirculation. Then,thenumberpolesP CL ofthe closedloopsystemintherighthalfplane,is P CL = arg ΓL(s) 360 +P OL (1.43) IfP CL =0,theclosedloopsystemisasymptoticallystable. Letustakeacloserlookatthetermsontherightsideof(1.43): P OL are therootsofd L (s),andthereshouldnotbeanyproblemcalculatingthese roots. Todeterminetheangularchangeofthevector1+L(s). Figure1.12 showshowthevector(orcomplexnumber)1+l(s)appearsinanyquist diagram foratypicalplotofl(s). ANyquistdiagramissimplya CartesiandiagramofthecomplexplaneinwhichLisplotted. 1+L(s)is thevectorfromthepoint( 1, 0j),whichisdenotedthecriticalpoint,to the Nyquist curve of L(s). More about the Nyquist curve of L(jω) LetustakeamoredetailedlookattheNyquistcurveofLassfollowsthe Γ contour in the s-plane, see Figure 1.11. In practice, the denominator polynomial of L(s) has higher order than the numerator polynomial. This impliesthatl(s)ismappedtotheoriginofthenyquistdiagramwhen s =. Thus,thewholesemicircularpartoftheΓcontourismappedto the origin. TheimaginaryaxisconstitutestherestoftheΓcontour. Howisthe mappingofl(s)assrunsalongtheimaginaryaxis? Ontheimaginary axiss=jω,whichimpliesthatl(s)=l(jω),whichisthefrequency responseofl(s). Aconsequenceofthisisthatwecaninprinciple

16 The critical point 1 Im L(s) Negative ω Infinite ω 0 Re L(s) 1 + L(s) Nyquist curve of L(s) Positive ω Decreasing ω Figure1.12: TypicalNyquistcurveof L(s). Thevector1+L(s)isdrawn. determine the stability property of a feedback system by just looking at the frequency response of the open system, L(jω). ωhasnegativevalueswhens=jωisonthenegativeimaginaryaxis. For ω < 0 the frequency response has a mathematical meaning. From general properties of complex functions, L( jω) = L(jω) (1.44) and L( jω) = L(jω) (1.45) ThereforetheNyquistcurveofL(s)forω<0willbeidenticaltothe Nyquistcurveofω>0,butmirroredabouttherealaxis. Thus,weonly needtoknowhowl(jω)ismappedforω 0. TherestoftheNyquist curvethencomesbyitself! ActuallyweneednotdrawmoreoftheNyquist curve(forω>0)thanwhatissufficientfordeterminingifthecritical point is encircled or not. WemustdosomeextraconsiderationsifsomeofthepolesinL(s),which arethepolesoftheopenloopsystem,lieintheorigin. Thiscorrespondsto pure integrators in control loop, which is a common situation in feedback control systems because the controller usually has integral action, as in a PI or PID controller. If L(s) contains integrators, the Γ contour must go outside theorigo. Buttotheleftortotheright? Wechoosetotheright, seefigure1.13. (Wehavetherebydecidedthattheoriginbelongstothe

17 Im(s) Infinetely small radius Im L(s) Infinitely large radius from origin 1 0 Re(s) 0 Re L(s) 1 + L(s) Figure 1.13: Left diagram: If L(s) has a pole in origin, the Γ contour must pass the origin along an arbitrarily small semicircle to the right. Right diagram: AtypicalNyquistcurveof L. lefthalfplane. ThisimpliesthatP OL doesnotcountthesepoles.) The radius of the semicircle around origin is arbitrarily small. The Nyquist curvethenbecomesasshowninthediagramtotherightinthesame figure. The arbitrarily small semicircle in the s-plane is mapped to an infinitelylargesemicircleinthel-plane. Theisbecauseass 0,theloop transfer function is approximately L(s) K s (ifweassumeonepoleintheorigin). Onthesmallsemicircle, s=re jθ (1.46) which gives L(s) K r e jθ (1.47) Whenr 0andwhensimultaneouslyθgoesfrom+90 via0 to 90, the Nyquist curve becomes an infinitely large semicircle, as shown. The Nyquist s stability criterion for non-rational transfer functions The Nyquist s stability criterion gives information about the poles of feedback systems. So far it has been assumed that the loop transfer

18 function L(s) is a rational transfer function. What if L(s) is irrational? Here is one example: L(s)= 1 s e τs (1.48) wheree τs representstimedelay. InsuchcasesthetrackingratioT(s)will alsobeirrational,andthedefinitionofpolesdoesnotapplytosuch irrational transfer functions. Actually, the Nyquist s stability criterion can be used as a graphical method for determining the stability property on basis of the frequency response L(jω). Nyquist s special stability criterion Inmostcasestheopensystemisstable,thatis,P OL =0. (1.43)then becomes P CL = arg Γ[L(s)] 360 (1.49) This implies that the feedback system is asymptotically stable if the Nyquist curve does not encircle the critical point. This is the Nyquist s special stability criterion or the Nyquist s stability criterion for open stable systems. The Nyquist s special stability criterion can also be formulated as follows: The feedback system is asymptotically stable if the Nyquist curve of L has thecriticalpointonitsleftsideforincreasing ω. Another way to formulate Nyquist s special stability criterion involves the amplitudecrossoverfrequency ω c andthephasecrossoverfrequency ω 180. ω c isthefrequencyatwhichthel(jω)curvecrossestheunitcircle,while ω 180 isthefrequencyatwhichthel(jω)curvecrossesthenegativereal axis. In other words: L(jω c ) =1 (1.50) and argl(jω 180 )= 180 (1.51) See Figure 1.14. Note: The Nyquist diagram contains no explicit frequency axis. We can now determine the stability properties from the relation between these two crossover frequencies: Asymptoticallystableclosedloopsystem:ω c <ω 180 Marginallystableclosedloopsystem:ω c =ω 180 Unstableclosedloopsystem:ω c >ω 180

19 Im L(s) L(jω 180 ) j Unit circle 1 0 Re L(s) L(jω c ) Decreasing ω Positive ω Figure1.14: Definitionofamplitudecrossoverfrequencyω c andphasecrossover frequencyω 180 The frequency of the sustained oscillations There are sustained oscillations in a marginally stable system. The frequencyoftheseoscillationsis ω c =ω 180.Thiscanbeexplainedas follows: Inamarginallystablesystem,L(±jω 180 )=L(±jω c )= 1. Therefore,d L (±jω 180 )+n L (±jω 180 )=0,whichisthecharacteristic equationoftheclosedloopsystemwith±jω 180 insertedfors. Therefore, thesystemhas±jω 180 amongitspoles. Thesystemusuallyhave additionalpoles,buttheylieinthelefthalfplane. Thepoles±jω 180 leads tosustainedsinusoidaloscillations. Thus,ω 180 (orω c )isthefrequencyof the sustained oscillations in a marginally stable system. 1.5 Stability margins 1.5.1 StabilitymarginsintermsofgainmarginGM and phasemarginpm An asymptotically stable feedback system may become marginally stable if the loop transfer function changes. The gain margin GM and the phase margin PM [radiansordegrees]arestabilitymargins whichintheirown

20 ways expresses how large parameter changes can be tolerated before an asymptotically stable system becomes marginally stable. Figure 1.15 shows the stability margins defined in the Nyquist diagram. GM is the Im L(s) L(jω 180 ) 1/GM j Unity circle 1 PM 0 Re L(s) L(jω c ) Figure1.15: GainmarginGM andphasemarginpm definedinthenyquist diagram (multiplicative, not additive) increase of the gain that L can tolerate at ω 180 beforethelcurve(inthenyquistdiagram)passesthroughthe critical point. Thus, L(jω 180 ) GM =1 (1.52) which gives GM= 1 L(jω 180 ) = 1 ReL(jω 180 ) (Thelatterexpressionin(1.53)isbecauseatω 180,ImL=0sothatthe amplitudeisequaltotheabsolutevalueoftherealpart.) (1.53) Ifweusedecibelastheunit(likeintheBodediagramwhichwewillsoon encounter), then GM [db]= L(jω 180 ) [db] (1.54) ThephasemarginPM isthephasereductionthatthelcurvecantolerate atω c beforethelcurvepassesthroughthecriticalpoint. Thus, argl(jω c ) PM= 180 (1.55) which gives PM =180 +argl(jω c ) (1.56)

21 Wecannowstateasfollows: Thefeedback(closed)systemis asymptotically stable if GM>0dB =1andPM >0 (1.57) This criterion is often denoted the Bode-Nyquist stability criterion. Reasonable ranges of the stability margins are and 2 6dB GM 4 12dB (1.58) 30 PM 60 (1.59) Thelargervalues,thebetterstability,butatthesametimethesystem becomes more sluggish, dynamically. If you are to use the stability margins as design criterias, you can use the following values(unless you have reasons for specifying other values): GM 2.5 8dBandPM 45 (1.60) Forexample,thecontrollergain,K p,canbeadjusteduntiloneofthe inequalitiesbecomesanequality. 4 Itcanbeshown 5 thatforpm 70,thedampingofthefeedbacksystem approximately corresponds to that of a second order system with relative damping factor ζ PM 100 (1.61) Forexample,PM =50 ζ=0.5. 1.5.2 Stability margins in terms of maximum sensitivity amplitude, S(jω) max An alternative quantity of a stability margin, is the minimum distance fromthel(jω)curvetothecriticalpoint. Thisdistanceis 1+L(jω),see Figure1.16. So,wecanusetheminimalvalueof 1+L(jω) asastability margin. However,itismorecommontotaketheinverseofthedistance: Thus,astabilitymarginisthemaximum valueof1/ 1+L(jω). And 4 Butyoushoulddefinitelycheckthebehaviourofthecontrolsystembysimulation,if possible. 5 The result is based on the assumption that the loop transfer function is L(s) = ω 2 0/[(s+2ζω 0)s] which gives tracking transfer function T(s) = L(s)/[1 + L(s)] = ω 2 0/ [ s 2 +2ζω 0 s+ω0] 2. ThephasemarginPM canbecalculatedfroml(s).

22 Im L(s) 1 0 Re L(s) 1+L(jω c ) min = S(jω c ) max L(jω) Figure 1.16: The minimum distance between the L(jω) curve and the critical point can be interpreted as a stability margin. This distance is 1+L min = S max. since1/[1+l(s)]isthesensitivityfunctions(s),then S(jω) max represents a stability margin. Reasonable values are in the range 1.5 3.5dB S(jω) max 3.0 9.5dB (1.62) Ifyouuse S(jω) max asacriterionforadjustingcontrollerparameters,you can use the following criterion(unless you have reasons for some other specification): S(jω) max =2.0 6dB (1.63) 1.6 Stability analysis in a Bode diagram ItismostcommontouseaBodediagramforfrequencyresponsebased stability analysis of closed loop systems. The Nyquist s Stability Criterion says: The closed loop system is marginally stable if the Nyquist curve(of L)goesthroughthecriticalpoint,whichisthepoint( 1,0). Butwhereis the critical point in the Bode diagram? The critical point has phase (angle) 180 andamplitude1=0db.thecriticalpointtherefore constitutestwolinesinabodediagram: The0dBlineintheamplitude diagramandthe 180 lineinthephasediagram. Figure1.17shows typical L curves for an asymptotically stable closed loop system. In the figure,gm,pm,ω c andω 180 areindicated. Example 1.4 Stability analysis of a feedback control system

23 [db] L(jω) 0 db ω c GM ω (logarithmic) [degrees] arg L(jω) -180 PM ω 180 Figure 1.17: Typical L curves of an asymptotically stable closed loop system withgm,pm,ω c andω 180 indicated Given a feedback control system with structure as shown in Figure 1.18. The loop transfer function is 1 L(s)=H c (s)h p (s)= K p }{{}(s+1) 2 = s }{{} H c(s) H p(s) K p (s+1) 2 s =n L(s) d L (s) (1.64) We will determine the stability property of the control system for different valuesofthecontrollergaink p inthreeways: Poleplacement,Nyquist s Stability Criterion, and simulation. The tracking transfer function is T(s)= y m(s) y msp (s) = L(s) 1+L(s) = n L (s) d L (s)+n L (s) = The characteristic polynomial is K p s 3 +2s 2 +s+k p (1.65) c(s)=s 3 +2s 2 +s+k p (1.66) Figures1.19 1.21showthestepresponseafterastepinthesetpoint,the poles,thebodediagramandnyquistdiagramforthreek p valueswhich result in different stability properties. The detailed results are shown below.

24 y msp Controller H c (s) Process with measurement and scaling H p (s) y m Figure 1.18: Example 1.4: Block diagram of feedback control system K p =1: Asymptoticallystablesystem,seeFigure1.19. Fromthe BodediagramwereadoffstabilitymarginsGM=6.0dB=2.0and PM =21. weseealsothat S(jω) max =11dB=3.5(alargevalue, butitcorrespondswiththesmallthephasemarginofpm =20 ). K p =2: Marginallystablesystem,seeFigure1.20. FromtheBode diagram,ω c =ω 180. TheLcurvegoesthroughthecriticalpointin thenyquistdiagram. S max hasinfinitelylargevalue(sincethe minimumdistance,1/ S max,between L andthecriticalpointis zero). LetuscalculatetheperiodT p oftheundampedoscillations: Since ω 180 =1.0rad/s,theperiodisT p =2π/ω 180 =6.28s,whichfitswell with the simulation shown in Figure 1.20. K p =4: Unstablesystem,seeFigure1.21. FromtheBodediagram, ω c >ω 180. FromtheNyquistdiagramweseethattheLcurvepasses outside the critical point. (The frequency response curves of M and N havenophysicalmeaninginthisthecase.) [End of Example 1.4] 1.7 Robustness in term of stability margins Per definition the stability margins expresses the robustness of the feedback control system against certain parameter changes in the loop transfer function: Thegainmargin GM ishowmuchtheloopgain,k,canincrease beforethesystembecomesunstable. Forexample,isGM =2when

25 K=1.5,thecontrolsystembecomesunstableforK largerthan 1.5 2=3.0. Thephasemargin PM ishowmuchthephaselagfunctionofthe loop can be reduced before the loop becomes unstable. One reason of reducedphaseisthatthetimedelayincontrolloopisincreased. A changeofthetimedelayby τ introducesthefactore τs inl(s) andcontributestoarglwith τ ω[rad]or τ ω 180 π [deg]. L ishowevernotinfluencedbecausetheamplitudefunctionofe τs is1, independentofthevalueofτ. Thesystembecomesunstableifthe timedelayhaveincreasedby τ max suchthat 6 PM= τ max ω c 180 π [deg] (1.67) which gives the following maximum change of the time delay: τ max = PM ω c π 180 (1.68) IfyouwanttocalculatehowmuchthephasemarginPM isreduced ifthetimedelayisincreasedby τ,youcanusethefollowing formula which stems from(1.67): PM = τ ω c 180 π [deg] (1.69) For example, assume that a given control system has ω c =0.2rad/minandPM=50. Ifthetimedelayincreasesby1min, thephasemarginisreducedby PM=1 0.2 180 π =11.4,i.e. from 50 to38.6. 6 RememberthatPM isfoundatω c.

Figure 1.19: Example 1.4: Step response(step in setpoint), poles, Bode diagramandnyquistdiagramwithk p =1. Thecontrolsystemisasymptotically stable. 26

Figure 1.20: Example 1.4: Step response (step in setpoint), poles, Bode diagram and Nyquist diagram with K p = 2. The control system is marginally stable. 27

Figure 1.21: Example 1.4: Step response(step in setpoint), poles, Bode diagramandnyquistdiagramwithk p =4. Thecontrolsystemisunstable. 28