3已知模拟调节器的传递函数为()s.s.sD08501701+=试写出相应数字控制器的位置型PID算法和增量型PID控制算式，设采样周期T=0.2s。解：因为())11(17011208501701sTsTKs.s.s.sDdip++=+=+=）（所以07.102===dipTTK，，。故位置型PID控制器∑∑∑===+=⎥⎦⎤⎢⎣⎡+=⎥⎦⎤⎢⎣⎡−−++=kikiDkiIPiekeiekeTkekeTieTTkeKku000)(7.10.40)(2)(7.10.20)(2)1()()()()(故增量型PID控制器[][][])1(2)(35.4)1()(7.10.40)1()(2)1()2()1(2)()()1()()1()()1()(−−+−≈+−−+−=−+−−++−−+−=Δ+−=kekekukekekekukekekeKkeKkekeKkukukukuDIP7设有限拍系统如图，试设计分别在单位阶跃输入及单位速度输入作用下，不同采样周期的有限拍无波纹的D(z)，并计算输出响应y(k)、控制信号u(k)和误差e(k)，画出它们对时间变化的波形已知：（1）采样周期分别为①T=10s②T=1s③T=0.1s（2）对象模型为G(s)=)1(5+ss（3）0H(s)=seTs−−1解：（1）当采样周期T=10s时广义对象的脉冲传递函数为：HG(z)=￡[)1(51+•−−ssseTs]=(1—Z1−)￡[)1(52+ss]=5()11−−z￡[sss11112−++]=5()11−−z[112111111)1(*−−−−−−−−+−zzezzTT]T=10时000045.010==−−eeTT=1时368.01==−−eeTT=0.1时905.01.0==−−eeT①当T=10时，0000045.0≈=−Te因此：]1)1(*10)[1(5)(112111−−−−−−+−−=zzzzzzHG=1111111211)09.01(551)11(51555−−−−−−−−−−−=−−=−−zzzzzzzzz②当T=1时，368.0=−Te因此：]11368.011)1()[1(5)(112111−−−−−−−−+−−=zzzzzzHG)368.01()1()1)(368.01()1()368.01()1(51211121111−−−−−−−−−−−−−−+−−=zzzzzzzz)368.01)(1()718.01(84.1)368.01)(1()718.01(368.0511111111−−−−−−−−−−+=−−+=zzzzzzzz有限拍无波纹由上可知：有一个单位圆内零点（z=—0.718）))(718.01()(111−−−++=bzazzzGc))(1()1()(121−−+−=zfzzGe利用Gc(z)=1-Ge(z)可得：321718.0)718.0()(−−−+++=bzzbaazzGc321)12()2(1)(−−−+−−−+=fzzfzfzGe解得：⎪⎩⎪⎨⎧−=−=+−=fbfbafa718.012718.02⎪⎩⎪⎨⎧==−=592.0407.1826.0fab则)592.01)(1()368.01)(587.01(764.0)()()()(1111−−−−+−−−=⋅=zzzzzGezHGzGczD21111111)1()718.01(84.1)587.01)(718.01(407.1)()()()()()()()()(−−−−−−−−⋅⋅+−+=⋅=⋅⋅=⋅=zTzzzzzzzRzHGzGczRzGezDzEzDzU.....2.02.004.076.04321++++=−−−−zzzz......43407.1)587.01)(718.01(407.1)1()()()(432111211+++=−+⋅−=⋅=−−−−−−−−zzzzzzzTzzGczRzY8.某工业加热炉通过飞升曲线实验法测的参数为：K=1，τ=30s,T1=320s，即可用带纯滞后的一阶惯性环节模型来描述。若采用零阶保持器，取采样周期T=6s，试用大林算法设计工业炉温度控制系统的数字控制器D（z），一直Tτ=120s。解：由题：用带纯滞后的一阶惯性环节模型来描述：已知：τ=30sT=6sN=τ/T=5K=1.2Tτ=120s05.01206==τTT9512.005.0=−e019.032061==TT98.0019.0=−e)05.095.01(024.0)98.01(05.0])95.01(95.01)[98.01(2.1)98.01)(9.01()(611611−−−−−−−−−=−−−−−−=zzzzzzszD)05.095.01()98.01(08.2611−−−−−−=zzz9.已知被控对象传递函数为)1)(12()(++=−ssesGs，采样周期T=1，试用大林算法设计D（z），判断是否会出现振铃现象？如何消除？解：由题：τ=+1sT1=2sT2=1sT=1sK=1sn=τ/T=116.0)37.021.1(1)2(21111211=−−=−−+=−−eeς092.0)607.0735.0(22.0)2(211211)12/1(2=−−=−−+=−−+−eeeς])1(1)[092.016.0()368.01)(61.01)(1(])1(1)[62.016.0()1)(1)(1()(211111112111111211−−−−−−−−−−−−−−−−−−−−−−−−=−−−−−−−=zezezzzezeezzezeezDTTTTTTττττττ0553.1575.012)(/112/1/12/1//>−−=++−=++−=−−−−−−τττTTTTTTETeeeeeeeCCAR存在振铃现象]1)1(1)[575.01()368.01)(61.01)(1(25.6])1(1)[575.01(16.0)368.01)(61.01)(1()(11/1111/12/11/1111/1−−−−−−−−−−−−−−−−+−++−−−=−−−+−−−=zzezzzezezezzzezDTTTTTτττττ有两个在左...
