13控制原理课件wa第五章频域分析

1、1By Hui WFTP:9/1682Outline of Chapter 5IntroductionBode Plots (Logarithmic plots)Direct Polar PlotsNyquists Stability CriterionPhase Margin and Gain Margin and Their Relation to Stability 3Frequency Response2. Bode plots (Logarithmic plots ) The use of semilog paper eliminates the need to

2、take logarithms of very many numbers and also expands the low-frequency rang, which is of primary importance. The advantages of logarithmic plot are following1)The mathematical operations of multiplication and division are transformed to addition and subtraction;2)The work of obtaining the transfer

3、function(fall into several categories) is largely graphical instead of analytical.Preliminary design-using the straight-line asymptotic approximations to obtain approximate performance characteristicsSpecified design-correcting the straight-line curves for greater accuracy.Logarithmic plotspolar plo

4、tsenough data4Basic definition of logarithmic terms:Logarithm: The logarithm of a complex number is itself a complex number. The abbreviation “log” is used to indicate the logarithm to the base 10)(434. 0)(loglog)(log)(log)()(jjGejGejGjjDecibel (dB): the unit of the logarithm of the magnitudeIt is o

5、mitted in rest of this bookFrequency Response2. Bode plots (Logarithmic plots )When logarithms of transfer functions are used, the input and output variables are not necessarily in same units.5Log magnitude. The logarithm of the magnitude of a transfer function G(j) (幅频特性) expressed in decibels isdB

6、jG)(log20The quantity is called the log magnitude, abbreviate Lm. Thus dBjGjLmG)(log20)( Since the transfer function is a function of frequency, the Lm is also a function of frequency.Frequency Response2. Bode plots (Logarithmic plots )Octave (倍频倍频). An octave is a frequency band from f1 to f2, wher

7、e f2/f1=2. For example: the frequency band from 1 to 2 Hz is 1 octave in width, and the frequency band from 17.4 to 34.8 Hz is also 1 octave in width.6Decade(十倍频十倍频). There is an increase of 1 decade from f1 to f2 when f2/f1=10. The frequency band from 1 to 10 Hz or from 2.5 to 25 Hz is 1 decade in

8、width. 1 2 8 10 20 80 100 decades decades decades Frequency Response2. Bode plots (Logarithmic plots )The dB values of some common numbers are given in P.248 Table 8.1.Note that the meaning of “decibel”. 7Frequency Response 2. Bode plots General frequency-transfer-function relationshipsThe angle equ

9、ation:2221)(121)1 ()()1 ()1 ()(jjTjjmTjrTjKjGnnam 180/0The frequency transfer function:)(/1 ()/2(1)1 ()()1)(1 ()(2211jjTjjTjTjKjGnnamrm(8.11)The log magnitude:2221)(121)1 ()()1 ()1 ()(jjLmTjLmjmLmTjrLmTjLmLmKjLmGnnam(8.12)11tanT221/1/2tannn90m8Frequency Response2. Bode plots: Drawing the Bode plots)

10、(/1 ()/2(1)1 ()()1)(1 ()(2211jjTjjTjTjKjGnnamrmThe generalized form of the transfer functionThe basic types of factors:pnnrmmjjTjjK221211 These curves for each factor can be added together graphically to get the curves for the complete transfer function, especially when asymptotic straight lines are

11、 used.122111211jjTjjKnnm9Constants:dBKLmKmmlog20The angle is zero as long as Km is positive. Angle0 Frequency Response2. Bode plots: Drawing the Bode plots: ConstantsLm0.11 The log magnitude diagram is a horizontal straight line.The constant raises or lowers the Lm curve of the complete transfer fun

12、ction by a fixed amount.20logKm10j Factors: there are 2 forms: (j) dBjjLmlog20log2011 the log magnitude curve is a straight line with a negative slope of 6dB/octave or 20dB/decade.The angle is constant and equal to -90.dBjjLmlog20log20 the log magnitude curve is a straight line with a positive slope

13、 of 6dB/octave or 20dB/decade.The angle is constant and equal to +90.Lm10.120-20AngleFrequency Response2. Bode plots: Drawing the Bode plots: j Factors(j)-1-90(j)-1j90j111+jT Factors: there are 2 forms (1+jT) , too.dBTTjTjLm22111log201log201For very small values of , that is, T1dBTTjTjLmlog20log2011

14、1For values of 1/T this function is a straight line with 20dB/decade slopeLm(dB)201/T-2010/T The corner frequency cf (转折频率)： the frequency at which the asymptotes to the log magnitude curve intersect. The corner frequency for the function (1+jT)-r= (1+j(/cf)-r : cf=1/TFrequency Response2. Bode plots

15、: Drawing the Bode plots: 1+jT Factors13The phase angle: at zero frequency , the angle is 0; at the corner frequency = cf , the angle is -45; at the infinite frequency , the angle is -90.Lm(dB)201/T-20(1+jT)-110/TAngle90-90-4545(1+jT)-11/TFrequency Response2. Bode plots: Drawing the Bode plots: 1+jT

16、 Factors14Angle90-90-4545(1+jT)-11/T The corner frequency is the same, and the angle varies from 0 to 90 as the frequency increases from zeros to infinity.Lm and angle curves for the function (1+j T) are symmetrical about the abscissa to the curves for (1+jT)-1.dBTTjLm221log201Similarly, the factor

17、1+jT:Lm1/T20-20(1+jT)-110/T(1+jT)(1+jT)Question:error?Frequency Response2. Bode plots: Drawing the Bode plots: 1+jT Factors15/cfLm120-20(1+jT)-1(1+jT)10The exact curve and the asymptotes line At the corner frequency: 3dBOne octave above and below the corner frequency : 1dBTwo octaves from the corner

18、 frequency: 0.26dBThe error between the exact curve and the asymptotes line are as following01.02.03.04.0/cf11/21/41/102104Corner frequency cf Frequency Response2. Bode plots: Drawing the Bode plots: error16ssG1 . 011)(1ssG1 . 01)(2Frequency Response2. Bode plots: Drawing the Bode plots: Examples 5-

19、1 17Quadratic Factors: 1 122121jjnnFor 1 the quadratic can be factored into two first-order factors with real zeros that can be plotted in the manner shown previously. Thus211221111121TjTjjjnn211221111121TjLmTjLmjjLmnn211221111121TjAngleTjAnglejjAnglennLm1/T1-201/T2(1+jT2)-1(1+jT1)-1(1+jT1)-1 (1+jT2

20、)-1Angle-90-1801/T11/T2Frequency Response2. Bode plots: Drawing the Bode plots: Quadratic Factors18122121jjnn For very small values of the low-frequency asymptote is represented by Lm=0dB.The high-frequency asymptotes has a slope of 40dB/decade. The asymptotes cross at the corner frequency cf = n2/1

21、222212221log20121nnnnjjLm221122/1/2tan121nnnnjjAngleFor 1 the quadratic contains conjugate factors. The Lm and the angle curves can be plotted by:nnnnjjLmlog40log2012122122 For very high values of frequency, the Lm is approximatelyLm-40cf = n10nFrequency Response2. Bode plots: Drawing the Bode plots

22、: Quadratic FactorsQuadratic Factors: 1 19The phase-angle curve : At zero frequency, the angle is 0 At the corner frequency, the angle is 90 At infinite frequency, the angle is -180122121jjnnAngle-180-90nFrequency Response2. Bode plots: Drawing the Bode plots: Quadratic FactorsQuadratic Factors: 1 2

23、022121jjnnFor the quadratic factor appears in the numerator22121jjnnLm and angle curves for (1+j2/n+(j/n)2) are symmetrical about the abscissa to the curves for (1+j2/n+(j/n)2)-1.Angle-180-90n90180122121jjnnLmn-4010n122121jjnn40Frequency Response2. Bode plots: Drawing the Bode plots: Quadratic Facto

24、rs22121jjnn22121jjnn21From the Equation2/1222212221log20121nnnnjjLm A resonant condition exists in the vicinity of = n, where the peak value of the Lm is greater than 0 dB. 122121jjnn222222411)(nnjG041)(222222nnddjGdd221nm2102121)(mmjGMQuadratic Factors: 1 Frequency Response2. Bode plots: Drawing th

25、e Bode plots: Quadratic Factors22122121jjnn2221121nmmM The Lm1+j2/n+(j/n)2-1 with 0.707 has a peak value. The magnitude of this peak value and the frequency at which it occurs are following:M()1.0MmmThe peak value Mm depends only on the damping ratio .The curve of M vs. has a peak value greater than

26、 unity only for 0.707. The frequency at which the peak value occurs depend on both the damping ratio and the undamped frequency n.Quadratic Factors: 1 Frequency Response2. Bode plots: Drawing the Bode plots: Quadratic Factors23A family of curves for several values of 1 is plottedLog magnitude diagra

27、m122121jjnnSee P255Frequency Response2. Bode plots: Drawing the Bode plots: Quadratic Factors24Phase diagram122121jjnnSee P255Frequency Response2. Bode plots: Drawing the Bode plots: Quadratic Factors25214 . 011)(sssG224 . 01)(sssG122121jjnnFrequency Response2. Bode plots: Drawing the Bode plots: Ex

28、amples 5-2 =0.2, n=1=0.2, n=126 The Lm curves for poles and zeros lying in the right-half (RH) s plane are the same as those for poles and zeros located in the Left-half (LH) s plane. However, the angle curve are different. For example: factor (1-jT) The angle varies from 0 to -90 as varies from zer

29、os to infinityLm1/T10/TAngle-90-45(1-jT)1/T20Frequency Response 2. Bode plotsDrawing the Bode plots (Nonminimum-phase system)(1jT)9045(1+jT)27Example 5-3 ssG1 . 01)(1ssG1 . 01)(2Frequency Response 2. Bode plotsDrawing the Bode plots (Nonminimum-phase system)28 If is negative, the quadratic factor 1-

30、j2/n+(j/n)2-1 contains RH s plane poles and zeros. Its angle varies from -360 at =0 to 180 at =Lm40Anglen-180-90122121jjnn-360-270122121jjnnFrequency Response 2. Bode plotsDrawing the Bode plots (Nonminimum-phase system)n-4010n122121jjnn29Frequency Response 2. Bode plotsDrawing the Bode plots (Nonmi

31、nimum-phase system)11)(22sssG= -0.5 Example 5-4 11)(21sssG=0.5, n=130The block diagram of a feedback control system: See P256R(j) +)125. 0(05. 01)21 ()5 . 01 (42jjjjjC(j) -C(j) The open-loop transfer function:)125. 0(05. 01)21 ()5 . 01 (4)(2jjjjjjGBasic factors:1211)125. 0(05. 01 )21 ()()5 . 01 (4jj

32、jjjFrequency Response2. Bode plots: Drawing the Bode plots: Examples 5-5 311)(j1)21 (j)5 . 01 (j82 . 0)125. 0(05. 01 12njj)125. 0(05. 01)21 ()5 . 01 ( 42jjjjjFrequency Response2. Bode plots: Drawing the Bode plots: Examples 5-5 FactorCorner frequency cfLog magnitudeAngle charecteristic4None Constant

33、 magnitude of 12 dBConstant 0NoneConstant slope of 20 dB/decade(0dB at =1)Constant -90 1=0.50 slope below corner frequency; -20 dB/decade slope above corner frequencyVaries from 0 to -90 2=2.00 slope below corner frequency; 20 dB/decade slope above corner frequencyVaries from 0 to 90 3=8.00 slope be

34、low corner frequency; 40 dB/decade slope above corner frequencyVaries from 0 to -1803202040-20-400.1110Lm0.52.0841)(j1)21 (j)5 . 01 (j12)125. 0(05. 01 jj-20dB/dec-40dB/dec-20dB/dec-60dB/dec)125. 0(05. 01)21 ()5 . 01 ( 42jjjjjFrequency Response2. Bode plots: Drawing the Bode plots: Examples 5-5 33020

35、40-20-400.1110Lm0.52.081)(j412)5 . 01 (j1)21 (j12)125. 0(05. 01 jjFrequency Response2. Bode plots: Drawing the Bode plots: Examples 5-5 -20dB/dec-40dB/dec-20dB/dec-60dB/dec)125. 0(05. 01)21 ()5 . 01 ( 42jjjjj34090-90-1800.1110Angle 0.52.0841)(j1)21 (j)5 . 01 (j12)125. 0(05. 01 jj-270)125. 0(05. 01)2

36、1 ()5 . 01 ( 42jjjjjFrequency Response2. Bode plots: Drawing the Bode plots: Examples 5-5 35)125. 0(05. 01)21 ()5 . 01 ( 42jjjjjPlotted by MatlabFrequency Response2. Bode plots: Drawing the Bode plots: Examples 5-5 36Determination of the phase-angle curve of G(j) can be simplified by using the follo

37、wing procedure (See P259):1) For the (j)-m term, draw a line at the angle of (-m)90.2)For each (1+jT)1 term, locate the angles at the corner frequency, an octave above and an octave below the corner frequency, and a decade above and below the corner frequency. Then draw a curve through these points

38、for each (1+jT)1 term.3)For each 1+j2/n+(j/n)2 term: (a) locate the 90 point at the corner frequency =n (b) for the respective , obtain a few points to draw the phase plot for each term. The angle at =0.707n may be sufficient and can be evaluated easily as tan-12.828Frequency Response2. Bode plots:

39、Drawing the Bode plots: Review374) Once the phase plot of each term of G(j) has been drawn, the composite phase plot of G(j) is determined by adding the individual phase curves. (a) Use the line representing the angle equal to 90)()(lim0mjGAs the base line, where m is the system type. Add or subtrac

40、t the angles of each factor from this reference line. (b) At a particular frequency on the graph, measure the angle for each single-order and quadratic factor. Add and/or subtract them from the base line until all terms have been accounted for. The number of frequency points used is determined by th

41、e desired accuracy of the phase plots. (c) At = the phase is 90 times the difference of the orders of the numerator and denominator -(n-w)90.Frequency Response2. Bode plots: Drawing the Bode plots: Review38The relationship between the steady-state error of a closed-loop system and the system type an

42、d the gain:The Lm curveThe system type and gain001KRess11KRess22KRessFrequency Response 2. Bode plotsSystem type and gain as related to log magnitude curvesThe Type of system(v)The steady-state error coefficientStep InputR0u-1(t)Ramp InputR1tu-1(t)Parabolic Input1/2R2t2u-1(t)K0K1K2Position errorVelo

43、city errorParabolic error0K0 00R0/(1+K)K1 00R1/KK2 00R2/K00039Type 0 system:A type 0 system has a transfer function of the formaTjKjG1)(0For a Type 0 system the characteristic are as follows1) The corner frequency 1=1/Ta2) At low frequencies, 1Frequency Response 2. Bode plotsSystem type and gain as

44、related to log magnitude curvesK 01 ?402 . 015 . 01)(02jTjKjGa02. 65 . 0log20log200K2 . 0121)(01jTjKjGa02. 62log20log200KFrequency Response 2. Bode plotsSystem type and gain as related to log magnitude curvesssG2 . 015 . 0)(2Example 5-6:ssG2 . 012)(141Type 1 system:A type 1 system has a transfer fun

45、ction of the formaTjjKjG1)(1At low frequency, 1/Ta, LmG(j)Lm(K1/j)=LmK1-Lm(j), which has a slope of 20dB/decade. At =K1, Lm(K1/j)=0. The low-frequency portion of the curve of slope 20dB/decade crosses the 0-dB axis at a value of x=K1 At =1, Lm(j)=0 Lm(K1/j)=1=20logK1For Ta K1Lm1-20 dB/dec-20logK1-40

46、dB/decx=K1=10111KTaFrequency Response 2. Bode plotsSystem type and gain as related to log magnitude curves42 The low-frequency portion of the curve of slope 20dB/decade may be extended until it does cross the 0-dB axis. The value of the frequency at which the extension crosses the 0-dB axis is x=K1.

47、 The plot Lm(K1/j) crosses the 0-dB value at x=K1. For the corner frequency 1=1/Ta 1 this value is a point on the extension of the initial slope.Lm1-20 dB/dec20logK1-40dB/decx=K1=10Lm(K1/j)111KTaThe frequency x is smaller or larger than unity according as K1 is smaller or larger than unityFrequency

48、Response 2. Bode plotsSystem type and gain as related to log magnitude curves43For a Type 1 system the characteristic are as follows1) The slope at low frequencies is 20 dB/decade.2) The intercept of the low-frequency slope of 20 dB/decade (or its extension) with the 0-dB axis occurs at the frequenc

49、y x, where x=K1.3) The value of the low-frequency slope of 20 dB/decade (or its extension) at the frequency =1 is equal to 20logK1.4) The gain K1 is the steady-state ramp error coefficient.Conclusion:Frequency Response 2. Bode plotsSystem type and gain as related to log magnitude curves44Example 5-7

50、:2 . 0121)(12jjTjjKjGa2 . 015 . 01)(11jjTjjKjGa02. 65 . 0log20log201K02. 62log20log201KFrequency Response 2. Bode plotsSystem type and gain as related to log magnitude curves45Type 2 systemA type 2 system has a transfer function of the form aTjjKjG1)(22At low frequency, 1, the quantity 20logK2 is po

51、sitive, and if K21, the quantity 20logK2 is negative.2KLm1-40 dB/dec20logK2-60dB/decy=10211KTa22KyLmK2/(j)2Lm1-40 dB/dec20logK2-60dB/decy=10LmK2/(j)2211KTa22KyFrequency Response 2. Bode plotsSystem type and gain as related to log magnitude curves47For a Type 2 system the characteristic are as follow

52、s1) The slope at low frequencies is 40 dB/decade.2) The intercept of the low-frequency slope of 40 dB/decade (or its extension) with the 0-dB axis occurs at the frequency y, where y2=K2.3) The value of the low-frequency slope of 40 dB/decade (or its extension) at the frequency =1 is equal to 20logK2

53、.4) The gain K2 is the steady-state parabolic error coefficient.Conclusion:Frequency Response 2. Bode plotsSystem type and gain as related to log magnitude curves48 2 . 015 . 01)(2221jjTjjKjGaExample 5-8: 2 . 0121)(2222jjTjjKjGa02. 65 . 0log20log202K02. 62log20log202KFrequency Response 2. Bode plots

54、System type and gain as related to log magnitude curves49The magnitude and angle of the ratio of the output to the input can be obtained experimentally for a steady-state sinusoidal input signal at a number of frequencies.These data are used to obtain the exact Lm and angle diagramThe asymptotes are

55、 drawn on the exact Lm curve.using the fact that their slopes must be multiples of 20dB/decadeThe system type and the approximate time constant are determinedFrequency Response 2. Bode plotsExperimental determination of transfer function50Care: Whether any zeros of the transfer function are in the R

56、H s planeMinimumphase system : a system that has no open-loop zeros in the RH s plane. All factors have the form 1+Ts and/or 1+As+Bs2.Nonminimumphase system : a system that has open-loop zeros in the RH s plane. All factors have the form 1-Ts and/or 1-As+Bs2. The angular variation for poles or zeros

57、 in the RH s plane is different from those in the LH s plane. Therefore, care must be exercised in interpreting the angle plot to determined whether any factors of the transfer function lie in the RH s plane. Many practical systems are in the minimum-phase categoryFrequency Response 2. Bode plotsExp

58、erimental determination of transfer function51The Lm curves for minimum-phase system is as following, determine the transfer function.At low-frequencies the slope of asymptotes is zero.log20)(mLa0m=0.1, The variation of slope is 20, the factor 1+jT, 1/T=0.1, T=10= 1, The variation of slope is -20, t

59、he factor (1+jT1)-1= 2, The variation of slope is -20, the factor (1+jT2)-1= 3, The variation of slope is -20, the factor (1+jT3)-1=100, The variation of slope is -20, the factor (1+jT4)-1Lm40203050200-20-40-600.11002134Frequency Response 2. Bode plotsExperimental determination of transfer function5

60、2The form of the transfer function:1111101)(4321sTsTsTsTsKsG30log20K62.31KThe equation of straight linebabmamkLLlglg)()(If a= 1 and b=0.1 then k=20, Lm( 1)=40 and Lm(0.1)=30 316. 0101 . 0203040116. 3111TIf a= 4 and b=100 then k=-60, Lm( 4)=5 and Lm(100)=0 54.821010060054012. 0144TLm40203050200-20-40