Chapter Six Transient And Steady State Responses

270TRANSIENT AND STEADY STATE RESPONSES Im{s}λd¡ Re{s}0λdFigure 6.6: Complex conjugate dominant system polesThis analysis can be also justified by using the closed-loop system transferfunction. Consider, for example, a system described by its transfer function as B E § " ª] E § « ! E 3 E « « " ³² Since the poles at –60 and –70 are far to the left, their contribution to the systemresponse is negligible (they decay very quickly to zero as !µ ¶ · and µº¹ · ). Thetransfer function can be formally simplified as follows » E §Ã E ½ 3 " 3 " 3 « " ¼ « « ² « ¾À¿« Á ¾w¿ ¶ ·¹ ·« « Á(6.21)§7 IÄ " Example 6.2: In this example we use MATLAB to compare the stepresponses of the original and reduced-order systems whose transfer functions "Æ Ä Æ and Åare given in Figureare given in (6.21). The results obtained for Å6.7. It can be seen from this figure that step responses for the original andreduced-order (approximate) systems almost overlap.

271TRANSIENT AND STEADY STATE RESPONSES0.250.2(1):y(t), (2):yr(t)(1)0.15(2)0.10.05000.511.522.5time t [sec]33.544.55Figure 6.7: System step responses for the original(1) and reduced-order approximate (2) systemsThe corresponding responses are obtained by the following sequence ofMATLAB functionsz -1;p [-3 —10 —60 —70];k 12600;[num,den] zp2tf(z,p,k);t 0:0.05:5;[y,x] step(num,den,t);zr -1;pr [-3 —10];kr 3;[numr,denr] zp2tf(zr,pr,kr);[yr,xr] step(numr,denr,t);plot(t,y,t,yr,’- —’);xlabel(’time t [sec]’);ylabel(’(1):y(t), (2):yr(t)’);grid;text(0.71,0.16,’(1)’);

272TRANSIENT AND STEADY STATE RESPONSESÇtext(0.41,0.13,’(2)’);Similarly one can neglect the complex conjugate non-dominant poles, as isdemonstrated in the next example.Example 6.3: Consider the following transfer function containing two pairsof complex conjugate polesÈÊÉ"Ë ÌÀÍÉ"Ë ÐÌÎ ÏÎÉEË Ð³Ñ ÒÔÓmÌ É Ë Ð³Ñ ÐÕÓ(Ì É Ë Ð ÑÒ ÓyÑ Ì É Ë Ð³ÑÖÐÕÓyÑ ÌÏÏÏÏand the corresponding approximate reduced-order transfer function obtained byÈ»ÉEË Ì ÍÉ"Ë! cÐÍÉ Ë ÐÎÎË ÐË ÐÎ ÏÎÎÉ"Ë ÐÌ ÉEË ÐÎ ÏÌÉEË ÐÌÎÎË ÐÎ ÏÌÎ Ï Ï Ø Ù Û Ú ÛÎ Ï ÏÌÉEË Ð Û Ð Ñ ÜÞÝ Û Û ÙÐÑÏÉ"Ë cÐÎÎÌË ÐÎÌÍ}Èàß ÉEË ÌThe step responses of the original and approximate reduced-order systems arepresented in Figure 6.8.0.120.1(1):y(t), (2):yr(t)0.08(2)(1)0.060.040.02000.511.522.5time t [sec]33.544.55Figure 6.8: System step responses for the original (1) andapproximate (2) systems with complex conjugate poles

273TRANSIENT AND STEADY STATE RESPONSESIt can be seen from this figure that a very good approximation for the stepresponse is obtained by using the approximate reduced-order model.áHowever, the above technique is rather superficial. In addition, for multiinput multi-output systems this procedure becomes computationally cumbersome.In that case we need a more systematic method. In the control literature one isable to find several techniques used for the system order reduction. One of them,the method of singular perturbations (Kokotović and Khalil, 1986; Kokotovićet al., 1986), is presented below. The method systematically generalizes thepreviously explained idea of dominant poles.The eigenvalues of certain systems (having large and small time constants,or slow and fast system modes) are clustered in two or several groups (seeFigure 6.9). According to the theory of singular perturbations, if it is possibleto find an isolated group of poles (eigenvalues) closest to the imaginary axis,then the system response will be predominantly determined by that group ofeigenvalues.Im{s}0Re{s}dominantslow modesfast modesFigure 6.9: System eigenvalues clustered in two disjoint groupsThe state space form of such systems is given byâ3äLã åä ã æ çÖèLâLé åå ê éPëðèòñé æåê Pé ì çå ä åâ ä åí³ñäLæ çCíæ ä æâLî ååê îæ(çºï(6.22)

274TRANSIENT AND STEADY STATE RESPONSESwhere ó is a small positive parameter. It indicates that the time derivatives forstate variables ô õ are large, so that variables ô õ change quickly, in contrast tovariables ô ö , which are slow. If the state variables ô õ are asymptotically stable,then they decay very quickly, so that after the fast dynamics disappear ô ø õ ù ú3û ,we get an approximation for the fast subsystem as ü õ ú'ù üPý ôLö þ ÿ ÿô õ þ ÿ ÿ(6.23)From this equation we are able to find ô õ þ ÿ ÿ (assuming that the matrix ü in nonsingular, which is the standard assumption in the theory of singularperturbations; Kokotović et al., 1986) asô õ þ ÿ ÿ ù üö õ Lû üPý ô ö þ ÿ ÿ(6.24)Substituting this approximation in (6.22), we get an approximate reduced-orderslow subsystem as ôô ø ö þ ÿ ÿ ù ü üù ü ù ôþ ÿ ÿ ù ö àüÖõ üöö þ ÿ ÿö Cõ ü ö þ ÿ ÿ üPýöü ý 'öù ö õIüÖõjüö ùõ ü(6.25)õFrom the theory of singular perturbations it is known that ôLö û is close toô ö þ ÿ ÿ û for every ! , and þ ÿ ÿ û is a good approximation for û for " ö # ! , where " ö indicates the fact that this approximation becomesvalid shortly after the fast transient disappears (Kokotović et al., 1986).Example 6.4: Consider a mathematical model of a singularly perturbed fluidcatalytic cracker considered in Arkun and Ramakrishnan (1983). The problemmatrices are given byüù %% (' ) *,'-'%%ú6T* ú6'' 9 *,'-'& :9 . * . )16* 1@2 */.-07' ) *50-0ú1324*51úúúú:9-. *5)úú) 0 *,'ú(' ) *5932(' ú323* 11 *5132 .A BBB ' 1 */8 2 B23'-*;2 1-. 1 *?' ' C(' ú-16* 0-0ú

275TRANSIENT AND STEADY STATE RESPONSESD7EGFIH3J-JLK?J M N O K P6J :N M J-K Q6J NSR O K PN7J M K5P O K5O-PTTWXF H T T T T JT J T T TYUP-Q6K?JTVUThe eigenvalues of this system areZ\[ ] F N:M6K a-R b N:c3K;cLa b Ndcfe\K O M b NSa MLK a P bgN7J M Q K5T-a hwhich indicates that the system has two slow (–2.85 and –7.78) and three fastmodes. The small parameter i represents the separation of system eigenvaluesinto two disjoint groups. It can be roughly estimated as i j c4Kkc a-l c e\K O-M j T K,J(the ratio of the smallest and largest eigenvalues in the given slow and fast] D Wsubsets). We use MATLAB to partition matrices b bas followseps 0.1;A1 A(1:2,1:2);A2 A(1:2,3:5);A3 A(3:5,1:2)*eps;A4 A(3:5,3:5)*eps;B1 B(1:2,1:2);B2 B(3:5,1:2)*eps;C1 C(1:2,1:2);C2 C(1:2,3:5);The slow subsystem matrices, obtained from (6.25), are given by] m FIHnNoepK Tqe4R M J M K/e e c e b D m IF HnJ P K a O-M6JT K,J RLe3a N:a K M T-O RrUR K5T-O-M TW m F H T K5T J J P T Kkc Tqe4P bIt m F H T6K P-Q OTJ UTvZ[] mF The eigenvalues of the slow subsystem matrix areN7J M K PO6K O-PsUTuT UNSO6K P MLe3R bnN:a K5P-Mqe4O h.This reflects the impact of the fast modes on the slow modes so that the originalslow eigenvalues located at –2.85 and –7.78 are now changed to –3.6245 and–8.6243. In Figure 6.10 the outputs of the original (solid lines) and reduced(dashed lines) systems are presented in the time interval specified by MATLABas t 0:0.025:5. It can be seen that the output responses of these systems arewremarkably close to each other.

276TRANSIENT AND STEADY STATE RESPONSES1.41.2y(t), yr(t)10.80.60.40.2000.511.522.5time t [sec]33.544.55Figure 6.10: Outputs of the original fifth-order system and reducedsecond-order system obtained by the method of singular perturbationsModels of many real physical linear control systems that have the singularlyperturbed structure, displaying slow and fast state variables, can be found inGajić and Shen (1993).A MATLAB laboratory experiment involving system order reduction andcomparison of corresponding system trajectories and outputs of a real physicalcontrol system by using the method of singular perturbations is formulated inSection 6.7.6.4 Steady State ErrorsThe response of an asymptotically stable linear system is in the long run determined by its steady state component. During the initial time interval the transientresponse decays to zero, according to the asymptotic stability requirement (6.2),so that in the remaining part of the time interval the system response is represented by its steady state component only. Control engineers are interested inhaving steady state responses as close as possible to the desired ones so that we

277TRANSIENT AND STEADY STATE RESPONSESdefine the so-called steady state errors, which represent the differences at steadystate of the actual and desired system responses (outputs).Before we proceed to steady state error analysis, we introduce a simplifiedversion of the basic linear control system problem defined in Section 1.1.Simplified Basic Linear Control ProblemAs defined in Section 1.1 the basic linear control problem is still very difficultto solve. A simplified version of this problem can be formulated as follows.Apply to the system input a time function equal to the desired system output.This time function is known as the system’s reference input and is denoted byx y z { . Note that x y z}{Y y z}{ . Compare the actual and desired outputs by feedingback the actual output variable. The difference y z { x y z { qy z}{ represents theerror signal. Use the error signal together with simple controllers (if necessary)to drive the system under consideration such that Ly z { is reduced as much aspossible, at least at steady state. If a simple controller is used in the feedbackloop (Figure 6.11) the error signal has to be slightly redefined, see formula (6.26).In the following we use this simplified basic linear control problem in orderto identify the structure of controllers (feedback elements) that for certain typesof reference inputs (desired outputs) produce zero steady state errors.Consider the simplest feedback configuration of a single-input single-outputsystem given in Figure 6.11. E(s)U(s) R(s) - Controller - Plant G(s)Y(s)H(s)Feedback ElementFigure 6.11: Feedback system and steady state errorsLet the input signal d q q represent the Laplace transform of the desiredoutput (in this feedback configuration the desired output signal is used as an

278TRANSIENT AND STEADY STATE RESPONSESinput signal); then for L : , we see that in Figure 6.11 the quantity q represents the difference between the desired output q Y d q and the actualoutput q . In order to be able to reduce this error as much as possible, we allowdynamic elements in the feedback loop. Thus, q as a function of has to bechosen such that for the given type of reference input, the error, now defined by q 7 n L ¡ L q (6.26)is eliminated or reduced to its minimal value at steady state.From the block diagram given in Figure 6.11 we have q q u q q q so that the expression for the error is given by q q § q q (6.27)The steady state error component can be obtained by using the final value theoremof the Laplace transform as q ª ª «? d ? ² ³ ª«? d µv¶ L ·7 ª«? ¹ : µ»º § L n ¼ n L 6½(6.28)This expression will be used in order to determine the nature of the feedbackelement L such that the steady state error is reduced to zero for differenttypes of desired outputs. We will particularly consider step, ramp, and parabolicfunctions as desired system outputs.Before we proceed to the actual steady state error analysis, we introduce oneadditional definition.Definition 6.1 The type of feedback control system is determined by thenumber of poles of the open-loop feedback system transfer function located atthe origin, i.e. it is equal to ¾ , where ¾ is obtained fromÀLÁ và à Ã} o§ ÀÂÄ q q gŠƧ ¿ Ç n Á n§ n §ÈÇpÉ pà à Ã} o§ÈÇ Ê3Ë Å (6.29)Now we consider the steady state errors for different desired outputs, namelyunit step, unit ramp, and unit parabolic outputs.

TRANSIENT AND STEADY STATE RESPONSES279Unit Step Function as Desired OutputAssuming that our goal is that the system output follows as close as possiblethe unit step function, i.e. ÌdÍ Î Ï Ð Ñ7ÍnÎLÏ ÐÓÒ ÔLÎ , we get from (6.28)ÎÒ ÐÒÒÕ Ö Ö ÐØÖ ;Ù Û:Ú ÜÞÝÐÒÆß à ÍnÎLÏ á Í ÎqÏ Î â ÒÆßãÖ ,Ù Û:Ú Üuä à Í ÎLÏná Í ÎqÏnå Ò ß æ çwhereæ ç(6.30)is known as the position constant and from (6.30) is given byæ ç(Ð Ö ,Ù Û:Ú Ü ä à Í ÎqÏ á Í ÎqÏ å(6.31)It can be seen from (6.30) that the steady state error for the unit step referenceis reduced to zero for æ ç ÐXè . Examining closely (6.31), taking into account(6.29), we see that this condition is satisfied for é ê Ò .Thus, we can conclude that the feedback type system of order at least oneallows the system output at steady state to track the unit step function perfectly.Unit Ramp Function as Desired OutputIn this case the steady state error is obtained asÕ Ö Ö Ð Ö ?Ù Û:Ú Ü ä Î ë Í ÎqÏ åìÐ Ö ?Ù¹Û:Ú Ü Ý Ò ß à Í Î ÎLÏ á Í ÎLÏ Î Ò îí â Ð Ö ?Ù¹Û:Ú Üuä Î à Ò Í ÎqÏ á ÍnÎLÏ}å Ð æ Ò ï(6.32)whereæ ïdÐ Ö ?Ù¹Û:Ú Ü ä Î à Í ÎqÏ á ÍnÎLÏ}å(6.33)is known as the velocity constant. It can be easily concluded from (6.29) and(6.33) that æ ïSÐØè , i.e. Õ Ö Ö Ð ð for é êãñ . Thus, systems having two and morepure integrators ( ÒLÔLÎ terms) in the feedback loop will be able to perfectly trackthe unit ramp function as a desired system output.Unit Parabolic Function as Desired OutputFor a unit parabolic function we have Ñ7Í ÎqÏ Ðòñ@Ô ÎÂó so that from (6.28)ñÕ Ö Ö Ð Ö ,Ù Û:Ú Ü Ý Ò ß à Í Î ÎqÏ á Í ÎqÏ Î ñ â Ð Ö ,Ù Û:Ú Ü ä Î íÂà ñ Í ÎqÏ á ÍnÎLÏ}å Ð æ ôóæ ô , is defined byõæõôdÐ Ö ?Ù Û:Ú Ü ö Î í à Í ÎqÏ á ÍnÎLÏø (6.34)where the so-called acceleration constant,(6.35)

280TRANSIENT AND STEADY STATE RESPONSESFrom (6.29) and (6.35), we conclude that ù ú7û ü for ý þ ÿ , i.e. the feedbackloop must have three pure integrators in order to reduce the corresponding steadystate error to zero.Example 6.5: The steady state errors for a system that has the open-looptransfer function as areù dû üù#"dûùõú:û û ! û û oû ü % '&)( *# ,( &)(.- / 02143. Since the open-loop transfer function of this system has one integrator the outputof the closed-loop system can perfectly track only the unit step.5Example 6.6: Consider the second-order system whose open-loop transferfunction is given by û 67 8ÿThe position constant for this system issteady state error is û 9 ' :ù ûù û87 ûso that the corresponding; The unit step response of this system is presented in Figure 6.12, from which it; can be clearly seen that the steady state output is equal to; hence the steady@? ; A ; B state error is equal toû.5Note that the transient analysis and the study of steady state errors can beperformed for discrete-time linear systems in exactly the same way as was usedfor continuous-time systems. The steady state errors for discrete-time systems

281TRANSIENT AND STEADY STATE RESPONSESare obtained by using the final value theorem of thethe same procedure as in Section 6.4.C-transform and 22.5time t [sec]33.544.55Figure 6.12: System step response for Example 6.66.5 Response of High-Order Systems by MATLABFor high-order systems, analytical expressions for system step responses are quitecomplex. However, we are still able to determine approximately the responseparameters in many cases. In this section, we plot the unit step response of ahigh-order control system by using MATLAB and determine approximately fromthe graph obtained some of transient response parameters and the correspondingsteady state error.Consider the mathematical model of a synchronous machine connected toan infinite bus. The matrix D of this seventh-order system is given in Problem3.28. The remaining matrices are chosen asEGFIH JPQFQHSRRRJRJRJRJJR!MUTKLJNM2OVWF7JFor a system represented in the state space form, the step response is obtainedby using the MATLAB function [y,x] step(A,B,C,D,1,t), where 1

282TRANSIENT AND STEADY STATE RESPONSESindicates that the step signal is applied to the first system input and t representstime. The step response of this system is given in Figures 6.13 and 6.14.1.41.21y(t)0.80.60.40.2000.511.522.5time t [sec]33.544.55Figure 6.13: Step response of a synchronous machine for X8Y[Z \L] Figure 6.13 is obtained for the initial time interval of a bdc e;f)g h . It shows the actualresponse shape, but it is hard to draw conclusions about the transient responseparameters from this figure. However, if we plot the system step response fortime interval aibQcjekf l e h , then a response shape very similar to that in Figure6.4 is obtained. It is pretty straightforward to read from Figure 6.14 that thepeak time is a monIgqp , the overshot is approximately equal to e;rBl , the rise timeis a stnQuvp , and the settling time is roughly equal to w9uvp . By using MATLAB,it is obtained that xLy y{z w}rjeLu}u. so that the steady state error is 9y'ytz e;rAeLuLu} .This can be obtained either by finding x; 'a for some a long enough or by usingthe final value theorem of the Laplace transform as x y'y8 2 y . 2 y ! 9 { y' w e (6.36)where is the system closed-loop transfer function, which can be obtainedby using MATLAB as [num,den] ss2tf(A,B,C,D,1). Then, for this

283TRANSIENT AND STEADY STATE RESPONSESparticular example of order seven, we have yss num(1,8)/den(1,8). Notethat num(1,8) 5048.8 and den(1,8) 4937.2.1.41.21y(t)0.80.60.40.2005101520time t [sec]25303540Figure 6.14: Step response of a synchronous machine for 8 [ } U . 6.6 Control System Performance SpecificationsControl systems should satisfy certain specifications such that systems underconsideration have the desired behavior for both transient and steady state responses. If the desired specifications are not met, controllers should be designedand placed either in the forward path or in the feedback loop such that the desired specifications are obtained. The desired specifications include the requiredvalues (or upper and/or lower limits) of already defined quantities such as phaseand gain margins, settling time, rise time, peak time, maximum percent overshoot, and steady state errors. Additional specifications can be defined in thefrequency domain like control system frequency bandwidth, resonant frequency,and resonance peak, which will be presented in Chapter 9.Of course, it is impossible to meet all the specifications mentioned above.Sometimes some requirements are contradictory and sometimes some of themare not affordable. Thus, control engineers have to compromise while trying to

284TRANSIENT AND STEADY STATE RESPONSESsatisfy all of imposed control system requirements. Fortunately, we are able toidentify the most important ones. First of all, systems must be stable; hence themain goal of controller design is to stabilize the system under consideration, inother words, the system phase and gain stability margins should be handled withincreased care. Secondly, systems shoul

steady state response, that is (6.1) The transient response is present in the short period of time immediately after the system is turned on. If the system is asymptotically stable, the transient response disappears, which theoretically can be recorded as "! (6.2) However, if the system is unstable, the transient response will increase veryFile Size: 306KBPage Count: 29Explore furtherSteady State vs. Transient State in System Design and .resources.pcb.cadence.comTransient and Steady State Response in a Control System .www.electrical4u.comConcept of Transient State and Steady State - Electrical .electricalbaba.comChapter 5 Transient Analysis - CAUcau.ac.krTransient Response First and Second Order System .electricalacademia.comRecommended to you b