自控原理习题解答(第六章)

第六章习题解答现测得其频率响应为：当ω=1rad／s时，幅频M(1)=12√2，相频Φ(1)=-π/4，求此环节的放大系数K和时间常数T。6-1设某环节的传递函数为1Tsk)s(G241T212k14tgT4TtgTtg2121Tk1Tke1Tk)j(G16211222Tjtg221-解：答6-2已知一蒸汽过热器的传递函数为1)(45s1)(30s3.2)s(G2若输入信号为x(t)=0.8sin0.1t，求过热器输出y(t)的稳态响应。6.2201t.0056sin.047.7713.1431t.0sin25.211056.21.045tg1.0302tg1t.0sin11.0202511.09008.02.3jG1t.08sin.0jGtye1202519002.3jG26112245tg302tgj22211解：答•6-3设单位反馈控制系统的开环传递函数为G0(s)=5/(s+1)，当把下列输入信号作用在闭环系统上时，试求系统的稳态输出。(1)x(t)=sin(t+30°)；(2)x(t)=2cos(2t-45°)；(3)x(t)=sin(t+30°)-2cos(2t-45°)。54.20t822sin.0ty54.203046.9;46.961tgjG822.03753615jGMe3656j5jG;6s51s511s5sG1s5sG30tsintx]136[12121-116jtg201-；已知：）（答57.262t58sin.1ty57.264543.18;43.1862tgjG1.584053645jGMe3656j5jG;6s51s511s5sG452t2sin452t902sin452t902sintx,90sincos1s5sG,452t2costx]236[12121-226jtg201-根据已知：）（答57.262t58sin.154.20t822sin.0tytyty452t2cos30tsintx]336[21根据叠加原理：；已知：）（答。s2es1)s(H)s(G)3()1s(s12s.0)s(H)s(G)2(;1)1)(2s1)(0.2s1s.(0)1s(5)s(H)s(G)1(6-4闭环系统的开环传递函数如下：试分别画出它们的奈奎斯特图并判别各闭环系统的稳定性。有凹凸。因有零点，故奈氏曲线稳定终止于原点，切于终点：起点：型系统典型画法答,000NPZ180901390mn0,50,k012s12s.011s.01s5sHsG.1146jImRe123456ω0-118090909090,0M:0,5M:02tg2.0tg-1.0tgtgjHjG212.011.0115jHjG12s12s.011s.01s5sHsG.214611112222大致画法答den,numnyquist;13.262.004.0den;1500num13s.262s.004s.015ssHsG3.MATLAB14623画法答不稳定。内。象限至落在相角比较：与并终止于原点。，终点：切于）负实轴无穷远处起点：（型系统典型画法答,22-0NPZ2270180,tg2.0tg1s110.2s180901390mn18021ss10.2ssHsG.1246112闭环系统不稳定。大致画法答,2NPZ180,0M:180,M:0tg9022.0tg1104.0M1ss10.2ssHsG.2246112222den,numnyquist;0011den;10.200numss10.2ssHsG3.MATLAB24623画法答,0M32.147,1M190,M0j01jHjG180147.32--147.3257(rad).21221,111,Me1jHjG,es1sHsG34690js：：：）点，闭环系统稳定。，曲线未包围（故答-1RejIm•6-5试求图6-68所示电路的频率特性G(jω)=E0(jω)/Ei(jω)，并画出其对数幅频渐近线。eie0R1R1R2CCR2eie0R1C1C2R2eie0(a)(b)(c)22112121212112122121212112121221221112121122i0T1T11k,TTRRCRRTC,RT,RRRk1sT11sTk1sT1sTk1sRRCRR1CsRRRRRRCsRRRCsRRCsRCs1RRCs1RRRCs1RCs1RRsEsEa56答12dBL19045045-90-222112122121212122i0T1T1CRRTC,RT1sT1sT1CsRR1CsRCs1RRCs1RsEsEb56）（）（答R1R2Ceie0(b)12dBL19045045-90-2baba21ba2121221ba21bac21ba22b11ababa2baba212211222112211211222221111112222i0cTTTTTTTTTT2121sTTsTT1sT1sTsT1sT11sT1sTsGCRTT,CRT,CRT11sTTTTsTT1sT1sT1sCRCRCRsCRCR1sCR1sCRsCR1sCR1sCR1sCR1sCRRsC1RsC1sC1RsC1RsEsEsGc56，式和比较答络的超前部分网后部分，后面的分式为前面的分式为网络的滞则有：其中设答sT11TsT11sTsT1sT11T1sTsGTT,TT,1TT,1TTTT,TTTTTTTT,TTTTc56bbaababacb2a1a1b21aa1baba21ba211adBL19045045-90-b22ba1b2bb2a1aa1babbaa1,;,1T1;,1T1;TT,T1,T1设ab2•6-6已知一些元件的对数幅频特性曲线如图6-69所示，试写出它们的传递函数，并计算出各参数值。型。说明系统为与横轴平行对数幅频渐近线起始段则组成，且和惯性环节由比例答011s.0101TsksG,1k1Ts1k1.01011T,1010k2020lgk20lgMa6611s.01s1k1TsksG,1k,020lgk1.01011T,10b66则，由于由比例和微分环节组成答分环节。起始段向上斜说明有微则时，或当即时的幅值为当起始段的斜率为组成，和惯性环节、微分由比例答105s.00.1s1TskssG1,.0k,020lg10ks/10rad,1.0k,k-1,10lgk-20,20lgkL20lgk,120dB/dec,05.02011T1Ts1skc661-10.01s1100ss1001sT1sTsksG100,kk120dB/dec,0.0110011T100,0.0111T1sT11sT1s1kd6621v221121则轴的交点其渐近线的延长线与横型系统，为起始段的斜率为组成，和、两个惯性环节、积分由比例答2005.1991010k,4620lgk20dB20lgk,0dBL10,20dBL10110.0033s10.1s1s101sT1sT1sTksG,10k,20dB20lgk20lgk0,20lgk0dB/dec,0.003330011T0.11011T1,111T1sT11sT11sT1ke663.22046221pp322211321。设故仍时，而降低，变化到从从图中看到：则型系统，为高度为起始段的斜率为，组成，和、、三个惯性环节由比例答22.7558s.4s227.577.477s.448.02s77.410s2sksG4.773.5-148.00,159.559.1,559.559.5,1159.5,1-12182.1182.110-121,0725.02025.1-121lg25.12025.21-12120lg20lgM,-121M68610k2020lgks2skf6622222nn22nn2nr24242220725.0222r2r2nn22n，则，解得：由解得：等式两边平方得得：由，组成，由和二阶环节由比例答•6-7设有开环传递函数G(s)H(s)的奈氏曲线图如图6-70所示，图中P是G(s)H(s)分母中实部为正根的数目，试说明闭环系统是否稳定，为什么?闭环系统稳定；：闭环系统稳定；：闭环系统不稳定；：答,022NPZc,000NPZb,220NPZa76•6-8画出下列开环传递函数的对数幅频特性和对数相频特性，并判别闭环系统的稳定性。)254ss)(1ss(s0.1)8(s)s(H)s(G)3(;)16s)(1ss(s2)s(H)s(G)2(;)18s)(12s(2)s(H)s(G)1(2222闭环系统稳定。知