課件第八章非線性系統(tǒng)2describing function ll

1、18.3 Describing function method(Harmonic Linearizing method) Describing function method is first proposed by P.J. Daniel in1940. Basic ideaFor the nonlinear systemTypical structure of the nonlinear systems NG(s)( )e t( )x t)(tc0)( trH(s)nonlinearlinear: ( )sinIfe tAt x(t), maybe it is not a sinusoid

2、al but a periodic function, can be expressed as a Fourier series.a sinusoidal input,2Assumption: The harmonic of x(t) could be neglected, then:11( )sin()x txtoutput frequency is equal toinput frequency approximately.Similar to the frequency analysis of linear system, we can perform frequency analysi

3、s for the nonlinear system based on the assumption.Describing function3 The describing function method is mainly used to analyze the stability and self-oscillation of the nonlinear system without external excitation. Advantage: It is not confined by the order of the system. Disadvantages: It is an a

4、pproximate analysis method. It can only be used to study the system frequency characteristics. 4 1. The linear part and the nonlinear part can be separated.NLG-Fig. 8-9 Typical structure of nonlinear systemry8.3.1 Concept of describing functionThe describing function method can be applied to nonline

5、ar systems with the following features:Shown in Fig. 8-9，NL is a nonlinear part，G is the transfer function of the linear part.xe52. The system has an odd-symmetric nonlinearity, and the input-output relationship of the nonlinear part is static (without energy storage elements). If so, sinusoidal inp

6、ut periodic output The output can be expanded into Fourier series with a zero D. C. component.f(x)=-f(-x)3. The linear part is a good low-pass filterWe can suppose the higher-oder harmonic is filtered out.There is only a fundamental component in the the output .6Definition of the describing function

7、 If all the conditions above are satisfied, we can describe the nonlinear components by the frequency response like as that we did in the linear systems. So we have: The describing function N(A) of the nonlinear element is: the complex ratio of the fundamental component of the output y(t) and the si

8、nusoidal input e(t), that is: 1111 ( )sin, ( )cossin sin()For e tAtx tAtBtxt11( )jx eN AA7Assume the input of nonlinear is sinusoidal Normally, the output is periodic, which can be expressed as a Fourier series: 10sincos2)(nnntnBtnAAtx The nonlinearity is odd-symmetric(奇對稱奇對稱). 20)(cos)(1tdtntxAn 20

9、)(sin)(1tdtntxBntAte sin)( A008For the fundamental component, we have2101( )cos()Ax tt dt2101( )sin()Bx tt dtThus, the fundamental component is 11111( )cossinsin()x tAtBtxtwhere22111xAB111AarctgB9The describing function is then given by11( )jxN AeAObviously, the describing function is a function of

10、the input amplitude A. So we can regard it as a variable gain amplifier. AAjABeABAANBAarctgj11212111)( Replacing the nonlinear part by N(A), we can extend the frequency response method of linear system to the nonlinear system so as to analyze the frequency characteristics of nonlinear system.10Norma

11、lly, the describing function N is a function of the amplitude and the frequency of input signal, it should be expressed as .Remarks：),( ANIn most of the nonlinear components, there are no energy storage elements. The frequencies of output and input are then independent. So the describing function N

12、of common nonlinear components is only a function of the amplitude of input, which can be expressed as N(A)。11If the nonlinearity is single-valued odd-symmetric01 A0, 011 BARemarks： (cont.)ABAN1)( The output x(t) is an odd function.The describing function is a real function of input amplitude A.If t

13、he nonlinearity is not single-valued odd-symmetricThe output x(t) is neither an odd nor even.The describing function is a complex function of input amplitude A.128.3.2 Describing function of typical nonlinear characteristic 1. SaturationtAte sin)( Assume the input isFig 8-10 Input and output wavefor

14、m of a saturation nonlinearity x(t)t ke(t)e(t)x(t)t a13when , the output x(t) is ,sin,sin)(tKAKatKAtx ttt0where，AaaA1sin,sin Because the output is odd， A10 (single-valued odd-symmetric)01111 BAtg aA x(t) ke(t)e(t)x(t)t a14 01)(sin)(2tdttxB 211sin2AaAaAaKA 022)(sin)(sin)(sin2ttdKAttdKattdKAThe descri

15、bing function of saturation nonlinearity is： 211sin2AaAaAaK ABAN1)( N(A) is a nonlinear real function of input amplitude A. It can be regarded as a variable gain amplifier. 152. Dead-zoneAssume the input is ，tAte sin)( x(t)x(t)e(t)e(t)k .Fig8-11 input and output waveforms of dead zone nonlinearityt

16、at a 16When ，the output of dead zone nonlinearity is , 0),sin(, 0)(atAKtx ttt0The output is oddA10， 01 01)(sin)(2ttdtxBwhere，aA AaaA1sin,sin )(sinsin2ttdatAK17The describing function of dead zone nonlinearity is 2111sin22)(AaAaAaKABAN 2111sin22AaAaAaKAB (2) N(A) decreases as becomes larger(1) When i

17、s very small, i.e., the non-sensible zone is small, N(A) is approximately equal to K；/a ANote:/1( )0a AN A(3)/a A18Assume the input is ，3. GaptAte sin)( Fig 8-12 input and output waveforms of gap nonlinearityx(t)x(t)e(t)e(t)kt 2at a 22 A19 tatAKtaAKtatAKtx),sin(2),(20),sin()(From the mathematical de

18、scription of gap nonlinearity, the x(t) is given bywhere，AaAaAA2sin,2)sin(1 20 2)(cos)(2ttdaAK )(cos)sin(2ttdatAK AaAaKA24 01)(cos)(2tdttxA 20)(cos)sin(2 ttdatAK 212122sin2AaAAaAAaAKA 20)(sin)sin(2 ttdatAK 01)(sin)(2tdttxB21So, we can obtain the describing function N(A) of gap nonlinearity as follow

19、s: 111212( )2224()sin12( )jBAN AjAAKAaAaAaKa aAjAAAAN A e222124()222( )sin12Ka aAKAaAaAaN AAAAA21121()4222sin12a aAAtgAaAaAaAAA224. RelayAssume the input is ，tAte sin)( Fig 8-13 input and output waveforms of relay nonlinearity with dead zone and hysteresis ringx(t)x(t)e(t)e(t)kt at 2 1 bmama a 2 1 2

20、2 1 23 , 0, 0)(btx ttt22110The output of the relay characteristic is :where，AmamaAAaaA122111sin,)sin(sin,sin 24Amabb )1(2)sin(sin212 2212112)cos(cos2AaAmabb 21)(cos21 ttdbA)(sin2211ttdbB 2511112121)()(BAjtgjeABAAeANAN 22422)1(1122)(AamAamAamAbAN 2221111)1( AaAamAamtg The describing function N(A) of

21、relay nonlinearity with dead zone and hysteresis ring is26When a0, we can obtain the describing function of a ideal relay nonlinearityAbAN 4)( x(t)e(t)-bbFig 8-14 Ideal relay nonlinearity Corollary：27When m1 and a0, we can obtain the describing function of a relay with dead zone214)( AaAbAN x(t)e(t)

22、b-ba-a Fig 8-15 A relay with dead zone28When m1, we can obtain the describing function of a relay with hysterics 2114)( AaAajtgeAbAN x(t)e(t)b-b-aaFig 8-16 A relay with hysteresis Summary: Nonlinear system analysis by the Describing Function Method:1. Draft of x-y, x-t, y-t;2. Decide the odd/even qu

23、ality of y(t);3. Decide the symmetry property of y(t);4. Calculate A1, B1 - by integration;5. Calculate N(A).29308.3.3 Describing function of multiple nonlinearitiesThe describing function of a cascaded nonlinear system is not equal to the product of the two describing functions of each nonlinear el

24、ements. NL1NL2xyzFig 8-17 cascaded nonlinear system1. Cascaded nonlinear system)()()(21ANANAN 31assume NL1 is a dead zone nonlinearity, NL2 is a saturation nonlinearity, the composite nonlinearity of the cascaded nonlinear system is shown in Fig.8-18.xyzxyk1a1yzk2b1 (a) cascaded nonlinear system Fig

25、 8-18 the cascaded nonlinear system and its composite nonlinearity xz111kba (b) composite nonlinearity zx1a21kk322. Parallel nonlinear system NL1NL2xy1y2yFig 8-19 Parallel nonlinear systemN(A) = N1(A) + N2(A) +According to the definition of describing function, the describing function N(A) of the ou

26、tput y and the input x is equal to the sum of two describing functions： Example 1 Obtain the describing function of the nonlinear system shown in the following figure.Solution： 31210120303103233xxxxxxxxxy yy=x23x1x2x0+h-hFig 8-20 multiple nonlinearities1sinassume xAt10A)()()()()(4321ANANANANAN thenN

27、L1 is an ideal relay nonlinearity when a=0,3334Obtain N1(A)： 03314)(sin2httdhBAhABAN 3114)( Obtain N2(A)： 02213)(sinsin32AhttdtAhB223)(hAN hAAN8)(3 Obtain N4(A)： 0331)(sinsin2ttdtAB343A 0221)(sinsin32ttdthAB 28hA Obtain N3(A)： tsuppose 033)(cossin2dA 022033cossin3cossin2dA35362443)(AAN 22343834)(AhA

28、hAhAN Then, the describing function of the multiple nonlinearity is37Fundamental component of the output x(t)Amplitude of the sinusoidal input e(t)N(A) = It can only reflect partial dynamic characteristics of the system.If a self-excited oscillation occurs in the system, any input x(t) can be regard

29、ed as a sinusoidal signal because the linear part is a low-pass filter. NLG-Typical structure of nonlinear systemr=0yxeThe describing function method is then applicable.8.3.4 Analyze nonlinear system with describing function method38Assume the input of a nonlinear system is zero,N(A) is the describi

30、ng function of a nonlinear element, analyze the condition for self-exited oscillation.22sinXAtAssume Fig. 8-21 nonlinear system1122()()()sin()XGjGjH jAt Thenwhere: )()()(21 jHjGjG G1N(A)G2HX1X1X2X2Yr=0-39 jeANAN)()( )sin()()()()()(2212 tAjHjGjGANtx1)()()()(21 jHjGjGAN )12( nAssume:tAx sin22 thenthe

31、condition of self-oscillation is22( )( )xtx tThe self-oscillation occurs.Note:Considering we have ,40Suppose the transfer function of linear part satisfies )()()()(21sHsGsGsG We can conclude that the condition of self-excited oscillation is )(1)(ANjG 1( ) ()0N A G jor1)()()()(21 jHjGjGAN )12( n41010

32、jjGjGsystemtheofquationteristic eThe charac)( )(1 : unstableRe1Im c 1/Kgstable Critical stability)( jG G(s)(tc)(trFor the linear system:),( )( :is system stable the of condition sufficient and necessary the function,transfer phase minimum a is )( If jpointircle the does not cjGsG142nonlinear system

33、N(A) G(s)( )e t( )x t)(tc)(tr NonlinearLinear G(s)(tc)(trLinear systemTransfer function of the system:)(1)()()()( jGjGjRjCj Characteristic equation:1)(0)(1 jGjGplane )( the In jGA point()( ) ()()()1( ) ()C jN A G jjR jN A G j 1( )()01 () ( )N A G jG jN A A curve43 Because the describing function N(A

34、) actually is a linearized frequency response, we can expand the Nyquist criterion to the nonliear system:1 () dont circle the ( )curve, the nonlinear system is stable;1 () circle the curve, ( )the nonlinear system is unstable;1 () intersect with the ( )curve, there wiG jN AG jN AG jN All be a self-

35、oscillation in the nonlinear system .(1)(3)(2) stability. critical the in is system the , ), 1(point the withintersect )( jjG stable; is system the , ), 1(point the circlet don )( jjG (1)unstable; is system the , ),( point the circle )( jjG1(2)(3)temlinear systhcompare wi(For example the minimum pha

36、se system)44ImRe)( jG)(1AN (a)ImRe)(1AN )( jG(b) do not circle 1/N(A) (Stable)( jG circle 1/N(A) (Unstable)( jG45)( jG(c)ImRe)( jG)(1AN UnstableImRe(d)aaabbb)( jG)(1AN intersect with 1/N(A) (Self-oscillation)ImReaaabbb)( jG)(1AN Analysis：(1) Self-oscillation occurs at point a and point b. (2) The se

37、lf-oscillation at point a is unstable while the self- oscillation at point b is stable. (3) There is only one stable self-oscillation in an actual physical system.1)()( ANjG The amplitude and frequency of the self-oscillation can be obtained by solving4647Example 2 A relay control system structure i

38、s shown in Fig. 8-23. Suppose a1, b3. (1) Is there a self-oscillation in the system? If there is, obtain the amplitude and frequency of the oscillation. (2) How to adjust the parameters if you want to eliminate the self-oscillation？Solution：the describing function of a relay nonlinearity with dead z

39、one is214)( AaAbAN )1)(15 . 0(2 sssbar=0Fig.8-23 relay control sytem-48There is an extreme value of function on the real axis.214)(1 AabAAN 1( )1( )whenAaN AwhenAN A ，1( )N A10( )ddAN AaA2 Substitute a1, b3 into above equations, we have2 A21 2()0aA4952. 06)(12 AAN)125. 125. 0()5 . 01(2)125. 125. 0(3

40、)(24224 jjGSubstituting into the real part, we have2 66. 05 . 11)(Re2 jG2 Set the imaginary part to zero, we haveImRe)( jG)(1AN )1)(15 . 0(2)( ssssGWe can obtain two amplitude: A11.11 ，A22.3 There is a self-oscillation in the system with the amplitude 2.3 and the frequency .25 . 111112)(12 AAAN Let（

41、not exist in the reality）5051(2) To eliminate the self-oscillation, let 5 . 1114)(122 aAAabAAN We can get the ratio of relay parameters 36. 225 . 1 abAdjust the ratio of a and b to b2a, we can eliminate the self-oscillation. 52Example 3 A multi-loop control system is shown in Fig. 8-25. If G1(s)=1, the natural oscillation frequency and damping ratio of system is and when working on the linear area of the saturation. If we suppose , try to find the minimal ratio T1/T2 when the system is stable.2 n 1 ssG811)(1 G1(s) 2010 Ts s0 02 T145 r = 0iEoEFig.