1、Measurement and control technology for Mechanical Engineering (I)1.Block diagram 2.The transfer function of linear systems 3. Feedback Control SystemCharacteristicsstabilityContent4. Frequency Domain The frequency response Frequency response plotsStability in the Frequency Domain Compensation and co

2、mpensator校正器校正器1.Block diagram 1.block diagram reduction1.a thermotank automatic control system is shown followingTry to sketch the block diagram.P1.1 many luxury(奢侈,華貴) automobiles have thermostatically controlled air-conditioning system（恒溫空調系統） for the comfort of the passengers. Sketch(繪制) a block

3、 diagram of an air-condition temperature on a dashboard panel（儀表盤）. Identify the function of each element of the thermostatically controlled cooling system P1.1keyThermostatically controlled air-conditioning systemAutomobile cabinTemperature sensor+-Desired temperature set by driverAutomobile cabin

4、temperaturemeasurementcontrollerprocessMeasured temperatureP1.8 P1.8 The student-teacher learning process is P1.8 The student-teacher learning process is inherentlyinherently(天性地,固有地) a feedback process intended to a feedback process intended to reduce the system error to a minimum .With the aid red

5、uce the system error to a minimum .With the aid of Figure 1.3,construct a feedback model of the of Figure 1.3,construct a feedback model of the learning process and identifylearning process and identify(識別,確定) each block of each block of the system.the system.P1.8keyteacherstudentexam+-Desired knowl

6、edgeknowledge outputmeasurementcontrollerprocessMeasured knowledgelectureError3.a system is shown following, Try to find the transfer function by block diagram reduction.2.The transfer function of linear systems Differential Equations of physical systems The transfer function Definition :The transfe

7、r function of a linear system is defined as the ratio of the Laplace transform of the output variable to the Laplace transform of the input variable, with all initial conditions assumed to be zero.The Laplace transform ;atMetf)(0)()()(dtetftfLsFstIf :The Laplace inverse transform; 0,)(21)()(1tdsesFj

8、sFLtfjjstTo find response)(1)()(1)(sFsdttfLsFsdttfLnn)()()()()()(222sFsdttfdLsFsdttfdLssFdttdfLnnnqProperties of Laplace transformLaf1(t)+bf2(t)=aLf1(t)+bLf2(t)Laf(t)=aLf(t)1.A spring-mass-damper system is shown following Write the differential equation and transfer function Xo(s)/Xi(s) describing t

9、he system. the differential equation：0)(0)()(2 1ooioxxbxKmxxxbxxKmK2 xmx”0()b xxThe transfer function Xo(s)/Xi(s) describing the system is0)()()()(0)()()()(221sXsXbssXKsXmssXsXbssXsXKooio)()(22sXKbsmsbssXo)()()()(22122221sXsbKbsKbsmsKbsmsKsXio22122221)()()()(sbKbsKbsmsKbsmsKsXsXio2.as shown in Fig .

10、Determine the transfer function X1(s)/F(s). mv 1xmv)(21xxb)(211xxkF22xk mt 2xmt)(21xxb)(211xxk22xb111211221212212122( )()( )()()0vtm xb xxk xxF tm xb xxb xk xxk x2111112212212111()( )()( )( )()( )()( )0vtm sbsk X sbsk XsF sm sbsb skk Xsbsk X s112121212( )( )()tb skXsXsm sbb skk21212122212121111()( )

11、( )( )()()()ttvm sbb skkX sG sF sm sbb skkm sbskbsk3. Feedback Control System Steady-state error)(lim)(lim0ssEteestssCalculating steady-state error)(lim)(teeetssCharacteristicsE(s)=1-T(s)R(s)()(1)()(SHsGsGsTE(s) = Yr(s) Y(s）； E(s) = R(S) Y(s) when h(s)=1Disturbance:Ed(s)=0-Td(s)D(s)Error: Error cons

12、tant for the standard test inputsPosition error constant (Unit step input)位置誤差系數)0()0()()(lim0HGsHsGKspVelocity error constant (Unit ramp input)速度誤差系數pssssKsHsGsRsHsGse11)()(11lim)()()(11lim00vssssKsHssGssRsHsGse1)()(1lim)()()(11lim00)()(lim0sHssGKsv閉環？開環？開環！Acceleration error constant (Unit parabol

13、ic input)加速度誤差系數)()(lim20sHsGsKsaassssKsHsGsssRsHsGse1)()(1lim)()()(11lim2200432( )24(1)( )623Y ssR sssss 432()2 4 (1 )()1()6232 4 (1 )TssGsTssssss02 42 4l i m()32 42 1psKGs12 113s speK4324320062324(1) 121lim (1( )( )lim6233sssssssssesT sR sssssss E6.11 A system with a transfer function Y(s)/R(s) is

14、Determine the steady-state error to a unit step input. Is the system stable?1) method 1:method 2:A necessary and sufficient condition for a feedback system to be asymptotically stable is that ALL the poles of the system transfer function have negative real parts (i.e. lie in the left-hand s-plane).S

15、tabilityThe Routh-Hurwitz stability criterion kraitirin (標準標準) The number of roots with positive real parts of the charac-teristic equation is equal to the number of changes in sign of the first column of the Routh array. i.e. the necessary and sufficient condition for a feedback system to be stable

16、 is that there is no changes in sign in the first column of the Routh array of the system characteristic equation.NYQUIST STABILITY CRITERIONA feedback system is stable if and only if the contour L in the L(s)-plane does not encircle the (-1, j0) point when the number of poles of L(s) in the right-h

17、and s-plane is zero (P=0)A feedback system is stable if and only if, for the contour L, the number of counterclockwise encirclements of the (-1, j0) point is equal to the number of poles of L(s) with positive real parts.To compare: Routh array (陣列) 0)(0111asasasasDnnnnsnan an-2 an-4 an-6 sn-1an-1 an

18、-3 an-5 an-7 Characteristic equation:The first two lines of the Routh array(陣列) are scheduled as follows.snan an-2 an-4 an-6 sn-1an-1 an-3 an-5 an-7 sn-2bn-1bn-3bn-5bn-7 sn-3cn-1cn-3cn-5cn-7 s2en-1en-3s1fn-1s0gnnnnnnnnnnnaaaaaaaaaab5141154131nnnnnnnnnnnaaaaaaaaaab7161176151nnnnnnnnnnnaa

19、aaaaaaaab1.A cassette tape storage device has been designed for mass-storage. It is necessary to control accurately the velocity of the tape. The transform function of the speed control of the tape drive is given by T(s) as following.Determine the limiting gain for a stable system.Determine a suitab

20、le gain so that the overshoot to a step command is approximately 5%.2321010( )(100)(40400) 10140440040000 10KKT ssssKsssK3s2s1s1(140440040000 10)140K0s4 0 0 05 7 6 0 0KAnswer:a) the Routh array: 1 4400 140 40000+10K 40000+10Kthe system is stable if . .5%0.69PO223222()(1.38)(1.38)(1.38)nnnnnnsb sssb

21、sb sb32140440040000 100sssK104.225.93003nbKb) so the desired characteristic polynomial isthe actual characteristic polynomial isso the suitable gain is k=3003 dominate poles.The relative stability of feedback control systems aSSa : the desired minimum distance of the poles to the j-axisExample 10: f

22、ind an appropriate gain K so that the real parts of the poles are less than 1. 018189)(23KssssDShift the j-axis left to 1, let the new complex variable1 ss010183618) 1(18) 1(9) 1()(2323KsssKssssDRouth array Find KHow to find stability by Nyquist criterion?G(s)H(s)G(jw)H(jw)A(w),(w)Nyquist plotZ=N+P4

23、. Frequency Domain The frequency responseThe frequency response of a system is defined as the steady-state response of the system to a sinusoidal (adj正弦) input signal. The sinusoid(正弦曲線) is a unique(唯一的) input signal, and the resulting output signal for a linear system, as well as signals throughout

24、 the system, is sinusoidal in the steady state; it differs from the i n p u t wav e fo r m o n l y i n a m p l i t u d e( 幅 值 ) a n d phase angle(相位角). Definition:Magnitude response:)()()()()()(jRjYjRjYjTAPhase angle response:)()()()(jRjYjT Frequency response of a system)(Im)(Re)()(jTjjTsTjTjs傳遞函數與頻

25、率響應的關系輸入、輸出、系統關系Chapter 8: Frequency Response Methods Department of Mechanical Engineering, Southeast University Frequency Response Plots of primary factors v Proportional factorKsG)(Transfer functionFrequency response0)(jKeKjGReal partKR)(Imaginary part0)(XLogarithmic gainKLlog20)(Phase angle0)(Cha

26、pter 8: Frequency Response Methods Department of Mechanical Engineering, Southeast Universityv Inertial(慣性) factor11)(TssGTransfer functionFrequency responsejarctgTeTTjjG221111)(magnitude2211)(TAPhase anglearctgT)(如果符號為負，幅值？相位？Chapter 8: Frequency Response Methods Department of Mechanical Engineerin

27、g, Southeast Universityv First-order differential factor1)( ssGTransfer functionFrequency responsejarctgejjG2211)(Logarithmic gain221log20)(LPhase anglearctg)(Chapter 8: Frequency Response Methods Department of Mechanical Engineering, Southeast Universityv Integral factor0ReIm =0 =Polar DiagramssG1)

28、(211)(jejjGlog20)(L90)(0)(P1)(A1)(QChapter 8: Frequency Response Methods Department of Mechanical Engineering, Southeast Universityv Ideal differential factorssG)(2)(jejjG0)(P)(Qlog20)(L)(A() = 900ReIm =0 =Polar diagramChapter 8: Frequency Response Methods Department of Mechanical Engineering, South

29、east Universityv Oscillation factorTransfer function：10,2121)(22222nnnssTssTsGFrequency characteristic：nnnnnjjjG2112)(2222Chapter 8: Frequency Response Methods Department of Mechanical Engineering, Southeast University222211)(nnA212)(nnarctg2222211)(nnnP222212)(nnnQ Polar plot (Nyquist Plot) Frequen

30、cy response plots)()()()()()()(Im)(Re)(jjGjeAejGjQPjGjjGjGWhere )()()()()()(22PQarctgQPAv Magnitude plot x-coordinate a logarithmic coordinate for in terms of the logarithm to the base 10units: rad/sec or Hz Bode plot (Logarithmic (對數) Plot)y-coordinate the linear coordinate for the logarithm of the

31、 magnitude represented in terms of the logarithm to the base 10units: decibels(dB) 分貝L()=20logA()v Phase plot x-coordinate same as magnitude ploty-coordinate the linear coordinate for ()units: degreeq Basic steps for drawing the polar diagram Express the transfer function as the multiplication of pr

32、imary factors)()()()(21sGsGsGsGn Obtain the frequency characteristic)()(2)(1)()()()()()(21njnjjjeAeAeAeAjG)()()(2121)()()(njneAAAdrawing the frequency response diagram Obtain A(0)、(0)；A()、(), add some characteristic point, (such as the intersection with the coordinate axes) q Basic steps for drawing

33、 the Bode diagram Express the transfer function as the multiplication of primary factors Obtain the corner frequency of each primary factor) 12)(1() 1() 12)(1() 1()()(112211112211sTsTsTsTsssssKsHsGqqqqvpppp,2121TTand mark them in logarithmic frequency axis. Calculate 20lgK，draw a line with a slope o

34、f -20v dB/dec at the point (1, 20lgK), and extends the line left to the left of all the corner frequencies to obtain the asymptote in the lowest frequency region. Extend the asymptote in the lowest frequency region right, and change the slope at every corner frequency of the corresponding primary fa

35、ctors. Correcting the asymptote to get the exact magnitude curve Obtain the phase characteristic by adding the phase of each primary factor.Chapter 8: Frequency Response Methods Department of Mechanical Engineering, Southeast University Example 1: Find the transfer function of a minimum phase system

36、-200-20-40200.1120 (rad/s)L()pHow to find a transform function from a Bode diagram1201lg1 . 0lglg2020KKOr: when w=0.1, 20lgA(w)=20lg A(0.1)20lg(k/w)=20 =K=1ttr25)(2.the minimum phase open-loop asymptotic logarithmic magnitude plot of a unity feedback control system is shown following1) Find the open

37、-loop transfer function of the system. 2) Determine the steady-state error of the feedback control system for an input 111111 . 0)(4321sssssKsG10040lg20KK解：1）開環積分環節數v = 0。 = 0.1處，斜率變化量為20dB/dec，為一階微分環節；=1、2、3 、4 處，斜率變化量均為-20dB/dec，為慣性環節。系統傳遞函數具有下述形式確定轉折頻率1、2、3 、4 60lg100lg5040lglg30520lglg5030201 .

38、0lglg405043423154.8257.19957. 1316. 04321154.82157.191957. 11316. 011 . 0100)(ssssssG 2) steady-state error is infinite.q Relationship between the Nyquist and Bode diagram.Bode diagram.)()()(,)()()(222111sDsNKsHsDsNKsG)()()()()()()()(212121sDsKNsDsDsNsNKKsHsGKK)()()()()()()(1)(sDsDsDsDsKNsHsGsFKBKKK

39、)()()()()()()()()(1)()(211211sDsDsNKsKNsDsDsNKsHsGsGsTBKKObviously the zeros of F(s) is the poles of T(s). Stability in the Frequency Domain1.the bode diagram of a minimum phase open-loop transfer function is shown following1) try to draw the polar diagram corresponding to this bode diagram. 2) if t

40、he open-loop gain K is 50, Determine the parameter K region so that the closed-loop is stable. 2）K5 or 50/3K100/3 or 62.5k100Cauchys theorem Principle of the argument（復數的幅角）If a contour s in the s-plane encircles Z zeros and P poles of F(s) and does not pass through any poles or zeros of F(s) and th

41、e traversal is the clockwise direction along the contour, the corresponding contour F in the F(s)-plane encircles the origin of the F(s)-plane N=Z-P times in the clockwise direction.Note: if N0, i.e. N+ - N-, is equal to P/2, the half of the number of the unstable poles of the loop transfer function

42、 L(s).1.a open-loop transfer function of a feedback system is shown following210( )610sGH sssdetermine whether the feedback system is stable utilizing the Nyquist criterion.22222 2210 10(10)6 (10)60 ()(10)6(10)36jjGH jj2(10)6000;70210(1070)6 70( 70)0.17(1070)36 70A 解：解： find intersection with the re

43、al axis: the polar plot passes through the 1 point and P=0, the system is marginally stable1036)010(06)010(10)0(2ARelative Stability and the Nyquist CriterionGain Margin G.M.| )()(|1)(1.gggjHjGAMGThe phase crossing frequency g at which the phase angle reaches -180.)()(log20)(1log20.gggLAAMGmethode 2

44、:Phase Margin P.M.pm 180 (c)c gain crossing frequency A(c) = 1， L(c) = 0dB( )(1)(2)KG ss ss()901802gggarctgarctg 9 02gga rctga rctg2gdBjGKggggg06326log20416log20| )(|log20221.a open-loop transfer function of a unity feedback system is shown followingIf K=6, determine the gain margin. find the gain m

45、argin:解：find the phase crossover frequency )4)(2() 1()(sssKsGtttr3cos55sin3)(2.a open-loop transfer function of a negative unity feedback system is shown following1) select a gain K, so that the phase margin is 45 . 2) if K=8 and the input signal is , try to write the steady-state response of the fe

46、edback control system. 13542360cccarctgarctgarctg22542cccarctgarctgarctg22254cccarctgarctgarctg2121141452cccc2)211 (45cc0192cc53. 847or. 0cc解：解：1）6 . 959. 842. 976. 81164222cccK1164110222ccc45.802c97. 8cThe gain crossover frequency is 8.53rad/sselect K=10When K=10, the gain crossover frequency is :4

47、2360)(ccccarctgarctgarctg133664 .776 .83360P.M=47162) 1(6)(2ssssT22224)16(16)(A2162)(arctgarctg27. 2)5(A3 .53910)5(arctgarctg2) The closed-loop transfer function is 06. 2)3(A31763)3(arctgarctgtttr3cos55sin3)()313cos(3 .10)3 .535sin(81. 6)3(3cos()3(5)5(5sin()5(3)(tttAtAtyif the input signal is , the

48、steady-state response of the feedback control system isq Steps for design phase-lead compensator adjust the open-loop gain to guarantee保正 the steady-state performance Evaluate the uncompensated system performance when the steady-state error are satisfied Find the m, , c (m)155(.)()(.MPMPMPMPccm0log2