计算机控制系统：综合大作业1

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1、计算机控制系统综合习题1已知D(£)解:(1)试用Z变换、一阶向后差分、向前差分、零极点匹配、Tustin变换和预修正的Tustin(设关键频率=4)变换等方法将D⑸离散化，采样周期分别取为0.1s和0.4s；①Z变换法:4/(,)=L-(—)=4^,D(z)=4zz-e^74z当T=0.1s时Q(z)=——z-0.674z当T=0.4s时,D(z—'7z-0.2019②一阶向后差分法:D(z)=D(s)4Tz1-z-1s=T(4T+l)z-l立Zz.2857Z当T=O.ls吋,D(z)=;')z-0.7143当T=0・4s时,D(z)=;'7z-0.3846③一阶向前差分

2、法:D(z)=D(s)z-l4Tz+(4Jl)04当T=0.1s时,D(z)=——:——;z-0.6当T=0.4s时,£)(z)=———;z+0.6④零极点匹配法：4因为D(y)二s+4所以z=evT,D(z)=k,(Z+J)(零点匹配到J)当T=0.1s时'£)(z)=0」65(z+l)z-0.67当Z4s时Q(z)」399(z+1)'7z-0.2019⑤Tustin变换法:D(z)=D(s)27z+2T2z-l(2T+l)z+(2T-l)当皿时,％)亠67(2+1)')z-0.667当TE时Q(z)=凹凹'7z-0.111⑥修正Tustin变换法,关键频率(丿二4rad/s4

3、tan(^)(z+l)D(z)=D(s)z-i2[4tan(^)+©]z+[4tan(%^)—®]当T=0.1s时,£>(z)=0.1686(z+l)z-0.6629当T=0.4s时,£)(z)=0.5073(z+l)z+0.0146(2)将D(z)的零极点标在Z平面图上®Z变换法i)T=0.1s-0.608-10500.51RealPart8o6o4o202••o-OAJPUaelu-4-Oii)T=0.4s202•・oo□050RealPart0.5②一阶冋后差分法:i)T=O.ls10.50RealPart-0.5-18642024••••••oooo-ogUEdXJeu_

4、6elu-68a6■•ii)T=0.4s0.50RealPart864202468••••••••oooo£999treaAJeuguJ-②一阶向前差分法:i)T=0・ls-1-0.50RealPart0.518642024•・••••oooo-O-O806-Oii)T=0.4s864202468••••••••oooo-o-o-o-oRealPart④零极点匹配法:i)T=0.1s-10RealPart0.51864202oo.ci0o.ii)T=0.4s匸OJdAJWU一6BE-.50.42o24aa00tredajbueeuj-0.510RealPart05■0.50Rea

5、lPart42024••••oo-O-OcedAJeu_6eE-⑤Tustin变换法:i)T=0.1sii)T=0.4s⑥修止Tustin变换法:i)T=O.lsii)T=0.4s（3）波特图①连续域10.80.60.4202468a0000redAJeu&euJ-4202•••oootred一-0.4•0.6■0.8-1-0.500.51RealPartBodeDiagram-520■30■(8P)Opau6e乏45■(6OP)Oss•901011010°101Frequency(rad/s)②z变换i)T=0.1BodeDiagram2520System:sysFrequen

6、cy(rad/s):4Magnitude(dB):18.730■(6ep)ose匸dSystem:sysFrequency(rad/s):4Phase(deg):-34.210°Frequency(rad/s)1015BodeDiagram(8P)(Dpacbel/M100-10-201③一阶向后差分法T=0.1(6(dp)Ossr1T—r—iii1—i—i*•iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiI*T-

7、1111

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9、iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiilir111111

10、iiiiSystem:sys~iiiiiiii■丄■■■i[l一」__iiiiiiiii丄Frequency(rad/s):4Phase(deg):-11.3丄L“J.」1iiiiiiiiiiiiiiiiiiiiiiiii1iiiiiIIIiiiiiiiii1iiiiiIIIIiIiiiill111111111111111111]-千』10°Frequency(rad/s)101BodeDiagra