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爱校园() 课后答案网() 淘答案()旨在为广大学生朋友的自主学习提供一个分享和交流的平台。Khdaw团队一直秉承用心为大家服务的宗旨,以关注学生的学习生活为出发点,最全最多的课后习题参考答案,尽在课后答案网()!大学答案 - 中学答案 - 考研答案 - 考试答案课后答案网,用心为你服务!电位器放大器电动机减速器阀门水箱浮子 杠杆_电位器放大器电动机绞盘位置大门_11-5 解:系统的输出量:电炉炉温 给定输入量:加热器电压 被控对象:电炉仓库大门自动控制开(闭)的职能方框图门实际开(闭)门 的位置工作原理:系统的被控对象为大门。被控量为大门的实际位置。输入量为希望的大门位置。当合上开门开关时,桥式电位器测量电路产生偏差电压,经放大器放大后,驱动电动机带动绞盘转动, 使大门向上提起。同时,与大门连在一起的电位器电刷上移,直到桥式电位器达到平衡,电动机停转,开 门开关自动断开。反之,当合上关门开关时,电动机带动绞盘反转,使大门关闭。受控量:门的位置测量比较元件:电位计1-4 解:受控对象:门。 执行元件:电动机,绞盘。 放大元件:放大器。水位自动控制系统的职能方框图h chr出水 电动机通过减速器使阀门的开度减小(或增大),以使水箱水位达到希望值 hr 。当 hc = hr 时,电位器电刷位于中点位置,电动机不工作。一但 hc hr 时,浮子位置相应升高(或降低),通过杠杆作用使电位器电刷从中点位置下移(或上移),从而给电动机提供一定的工作电压,驱动电压 ur 相对应,此时电位器电刷位于中点位置)。rc(与电位器设定工作原理:系统的被控对象为水箱。被控量为水箱的实际水位 h 。给定值为希望水位 h测量元件:浮子,杠杆。放大元件:放大器。执行元件:通过电机控制进水阀门开度,控制进水流量。比较计算元件:电位器。c被控量:水箱的实际水位 h受控对象:水箱液面。1-1(略)1-2(略)1-3 解:习题第一章自动控制原理(非自动化类)习题答案电位器电压 放 大功 率 放 大电机 加热器电炉热 电偶K1K21s 2 + s1TsK3K21Ts1s2 + sK1K3-21 3Ts3 + (T + 1)s2 + s + K K,C (s) / R(s) =K1K3X5(s)X4(s)X3(s)X2(s)R(s)C(s)_N1(s)+X1(s)N2(s)将方块图连接起来,得出系统的动态结构图:X5(s)-X4(s)C(s)X5(s)X4(s)X3(s)N2(s)X5(s)C(s)-X2(s)X1(s)X3(s)X2(s)X1(s)+R(s) 3 5绘制上式各子方程的方块图如下图所示:N1(s)K X (s) = s2C (s) + sC (s) X 5 (s) = X 4 (s) K2 N2 (s)TsX 4 (s) = X 3 (s) X 2 (s) = K1 X1 (s) X 3 (s) = X 2 (s) X 5 (s)2-1 解:对微分方程做拉氏变换: X1 (s) = R(s) C (s) + N1 (s)习题第二章炉温给定 炉温放大元件:电压放大器,功率放大器,减速器比较元件:电位计 测量元件:热电偶 职能方框图:1s + 1Ks1Ts + 1sTTs+1s1s + 11Ts + 1K-31 31 42 32 4(b)R(s)1 + G G G G + G G G G(a)=R(s)ms2 + fs + KG1 + G2C (s) =1C(s)2-3 解:(过程略)0N (s) =C (s)(s + 1)(Ts + 1)1 +Ts2 + (T + 1)s + (K + 1)kR(s)C (s)= (s + 1)(Ts + 1)(s + 1)(Ts + 1) =K + s+K sX4(s)X3(s)X1(s)R(s) C(s)X5(s)X2(s)N(s)将方块图连接得出系统的动态结构图:C(s)X4(s)X4(s)X3(s)X5(s)N(s)N(s)X5(s)C(s)-X3(s)X1(s)X2(s)R(s)X1(s)R(s)X2(s) X 5 (s) = (Ts + 1) N (s)绘制上式各子方程的方块如下图:C (s) = X (s) N (s)4(Ts + 1) X 4 (s) = X 3 (s) + X 5 (s) X 2 (s) = sR(s)(s + 1) X 3 (s) = X1 (s) + X 2 (s)2-2 解:对微分方程做拉氏变换 X1 (s) = KR(s) C (s)1 3Ts3 + (T + 1)s2 + s + K K2C (s) / N (s) = K2 K3TsC (s) / N1 (s) = C (s) / R(s) ,三个回路均接触,可得 = 1 La = 1 + G1G2 + 2G14 La = L1 + L2 + L3 = G1G2 G1 G1a =13(b)(1)系统的反馈回路有三个,所以有R1 + G1G2G5 + G2G3G4 G4G2G5G1G2G3 + 1C =三个回路两两接触,可得 = 1 La = 1 + G1G2G5 + G2G3G4 G4G2G5(2)有两条前向通道,且与两条回路均有接触,所以P1 = G1G2G3 , 1 = 1P2 = 1, 2 = 1(3)闭环传递函数 C/R 为 La = L1 + L2 + L3 = G1G2G5 G2G3G4 + G4G2G5a =132-5 解:(a)(1)系统的反馈回路有三个,所以有K1K2nG (s) =Kn s1 2 3Ts2 + s + K K K(2)要消除干扰对系统的影响C (s) / N (s) = K n K3 s K1K2 K3Gn = 0Ts + 1s11 2 3K231 +sTs2 + s + K K KKKnn 1C (s) / N (s) = (K G KK3K2 ) Ts + 1 = K n K3 s K1K2 K3Gn求 C/N,令 R=0,向后移动单位反馈的比较点1 2 31 + G(s)Ts2 + s + K K K=C (s) / R(s) =G(s)K1K2 K3s(Ts + 1)2-4 解 :(1)求 C/R,令 N=0G(s) = K1K2 K3R(s)1 + G1G2 + G2G3 + G3G4 + G1G2G3G4(e)G1G2G3G4C (s) =R(s)1 G2G3R(s)1 + G1 + G2G1(d)(c)C(s) = G1 G2C(s) = G2 + G1G25n1 2= 0.1t p =13-2 解:系统为欠阻尼二阶系统(书上改为“单位负反馈”,“已知系统开环传递函数”) % = e / 1 100% = 1.3 1 100%2H1 + 10K= 10HK= 0.9H 1 + 10K0= 10 K= 1010K0要使过渡时间减小到原来的 0.1 倍,要保证总的放大系数不变,则:(原放大系数为 10,时间常数为 0.2)1 + 10KHHs + 10.2R(s)0 1 + G(s)K=1 + 10K HG(s) (s) = C (s) = K10K0采用 K0 , K H 负反馈方法的闭环传递函数为0.2s + 1)3-1 解:(原书改为 G(s) =10习题第三章N3 (s)N3 (s)N2 (s)N2 (s)1 + G1G2G3 + G2= 1=E(s) = C (s)(1 + G2 )G3E (s) = C (s)N1 (s)N1 (s)1 + G1G2G3 + G2R(s)1 + G1G2G3 + G2E(s) = C (s) = G2G3 G1G2G3E(s) = 1 + G2 G2G3N3 (s)1 + G1G2G3 + G2N2 (s)1 + G1G2G3 + G2=C (s)C (s) = 1 (1 + G1G2G3 + G2 ) = 1(1 + G2 )G3N1 (s)R(s)1 + G1G2G3 + G2= C (s) / R(s)C (s)C (s) = G1G2G3 + G2G32-6 解:用梅逊公式求,有两个回路,且接触,可得 = 1 La = 1 + G1G2G3 + G2 ,可得1 + G1G2 + 2G11 + G1G2 + 2G1RG1G2 + G2C = G1G2 + G1 + G2 G1 =(2)有四条前向通道,且与三条回路均有接触,所以P1 = G1G2 , 1 = 1P2 = G1 , 2 = 1P3 = G2 , 3 = 1P4 = G1 , 4 = 1(3)闭环传递函数 C/R 为6nc. = 0.1, = 1s1 时,ns= 3.5st =3.52 % = e / 1 100% = 72.8%nb. = 0.1, = 10s1 时,ns= 7st =3.52 % = e / 1 100% = 72.8%na. = 0.1, = 5s1 时,n2n = 10解得:n = 14.14, = 0.354, %=30%, t p = 0.238结论,K 增大,超调增加,峰值时间减小。3-4 解:(1) 2 = 200s2 + 10sG(s) =200(2) K = 20s1 时:n2n = 10解得:n = 10, = 0.5, % = 16.3%, t p = 0.363 2 = 100s2 + 10sG(s) =1003-3 解:(1) K = 10s1 时:s(s + 24.1)s(0.041s + 1)=G(s) =47.11136所以,开环传递函数为:n = 33.71 = 0.358解得:7系统不稳定。(b)用古尔维茨判据5210203104.73.25532s4 s3 s2 s1s0系统稳定。(2)(a)用劳思判据= 8000D3 =001001009202010= 80D1 = 20, D2 =1009201系统稳定。(b)用古尔维茨判据910001204100s3 s2 s1s0则 % 减小, ts 减小3-5 解:(1)(a)用劳思判据(3) 讨论系统参数: 不变, % 不变; 不变,n 增加,则 ts 减小;n 不变, 增加,ns= 1.4st =3.52 % = e / 1 100% = 16.3%n(2) = 0.5, = 5s1 时,ns= 35st =3.52 % = e / 1 100% = 72.8%8劳斯表:s3 + 21s2 + 10s + 10(a) 系统传递函数:10(s + 1)3-7 解:3解得 K 44若系统稳定,则:K 1 0, K 03K40.20.83 K 1K 1Ks3s2s1s0劳思表0.2S 3 + 0.8S 2 + (K 1)s + K = 04(2)系统闭环特征方程为若系统稳定,则: K 1 0, K 0 。无解4K0.210.8K K 1s3s2s1s0系统不稳定。3-6 解:(1)系统闭环特征方程为0.2S 3 + 0.8S 2 s + K = 0劳思表2= 3060D4 =0300002151510310(其实 D4 不必计算,因为 D3 0, 0 K 0, K 0 时系统稳定当 2 0,2 0.01Ks02K1K0.0122 0.01Ks3s2s1劳思表:0.01s3 + 2 s2 + s + K = 0系统稳定。3-8 解:系统闭环特征方程为:100110110s2s1s0劳思表:s2 + 101s + 10(b) 系统传递函数:10系统稳定。101000121200 / 2110s3 s2 s1s0101 +s(s + 4)(s2 + 2s + 2)s 0s 0s2s2 7(s + 1)sss= 8 / 711输入 r (t ) = t 1(t) 时, R(s) = 1 , e = lim sE = lim ss(s + 4)(s2 + 2s + 2)s 0s 0ss1 + 7(s + 1)sss1 = 01当输入 r (t) = 1(t ) 时, R(s) = 1 , e = lim sE = lim s1 + G(s)sE i RR(s)(s)R(s) =E = 1系统稳定。1071507167.59.47s4 s3 s2 s1s0劳思表:s4 + 6s3 + 10s2 + 15s + 7 = 0解法二、系统的闭环特征方程为:K7ssss= 。1当 r (t ) = t 1(t) 时, e = 8 = 1.14 ;当 r (t ) = t 2 1(t ) 时, e8ss(2)解法一、因为 = 1 ,属于型无差系统,开环增益 K = 7 ,故当 r (t ) = 1(t ) 时, e = 0 ;s(0.1s + 1)(0.5s + 1)1 +s0s 0s3s3 10sss= 11输入 r (t ) = t 2 1(t ) 时, R(s) = 2 , e = lim sE = lim ss(0.1s + 1)(0.5s + 1)1 +s 0s 0s2 10s2sss= 0.111输入 r (t ) = t 1(t) 时, R(s) = 1 , e = lim sE = lim ss(0.1s + 1)(0.5s + 1)s 0sss 0 101 +sss1 = 01当输入 r (t) = 1(t ) 时, R(s) = 1 , e = lim sE = lim s1 + G(s)sE i RR(s)(s)R(s) =E = 1系统稳定。11s2输入 r (t ) = 10t, R(s) =10调节时间 ts = 4T = 1min, T = 0.25 minTs + 1R(s)为一阶惯性环节3-10 解:系统传递函数为= G(s) =1C (s)1 +s2 (0.1s + 1)s 0s0s3s3 8(0.5s + 1)sss= 0.2521输入 r (t ) = t 2 1(t ) 时, R(s) = 2 , e = lim sE = lim s1 +s2 (0.1s + 1)s 0s 0s2 8(0.5s + 1)s2sss= 011输入 r (t ) = t 1(t) 时, R(s) = 1 , e = lim sE = lim ss2 (0.1s + 1)s 0s0ss1 + 8(0.5s + 1)sss1 = 01当输入 r (t) = 1(t ) 时, R(s) = 1 , e = lim sE = lim s1 + G(s)sE i RR(s)(s)R(s) =E = 1系统稳定。0.14183.28s3 s2 s1s0劳思表:0.1s3 + s2 + 4s + 8 = 0解法二、系统的闭环特征方程为:K当 r (t ) = t 1(t) 时, ess = 0 ;当 r (t ) = t 1(t) 时, ess = 0.25 。22(3)解法一、因为 = 2 ,属于型无差系统,开环增益 K = 8 ,故当 r (t) = 1(t ) 时, ess = 0 ;1 +s(s + 4)(s2 + 2s + 2)s0s 0s3s3 7(s + 1)sss= 11输入 r (t ) = t 2 1(t ) 时, R(s) = 2 , e = lim sE = lim s12在扰动点之后引入积分环节 1/s,s 0所以对输入响应的误差, ess = lim sE(s) = 0 。s(0.05s + 1)(s + 5) + 2.5Kss(0.05s + 1)(s + 5) + 2.5KE (s) = s(0.05s + 1)(s + 5) 2.5s(0.05s + 1) 1 = (0.05s + 1)(s + 5) 2.5(0.05s + 1)s(0.05s + 1)(s + 5)s(0.05s + 1)(s + 5) + 2.5KN (s)1 + 2.5KE i N 2.5(0.05s + 1)s= E (s) = s + 5 =2.5s(0.05s + 1)(s + 5)s(0.05s + 1)(s + 5) + 2.5KR(s)1 + 2.5KE i R =s(0.05s + 1)(s + 5)1= E(s) =(3)在扰动点前的前向通道中引入积分环节 1/s,s 05 + 2.5Kss= 0.0455 。比较说明,K 越大,稳态误差越小。e = lim sE(s) =2.5(2)当 K=20 时s 0s0(0.05s + 1)(s + 5) + 2.5Ks5 + 2.5Kss= 0.0238e2.5= lim sE(s) = lim s (0.05s + 1)(s + 5) 2.5(0.05s + 1) 1 =ss(1)当 K=40 时输入 R(s) =, N (s) =11(0.05s + 1)(s + 5) + 2.5KsE (s) = (0.05s + 1)(s + 5) 2.5(0.05s + 1) 1(0.05s + 1)(s + 5)1 +N (s)2.5KE i N = E (s) = s + 5 2.5(0.05s + 1)(s + 5)1 +R(s)2.5KE i R 1= E(s) =s 03-11 解:用梅森公式:ess = lim sE (s) = 2.5(C )D稳态误差:ss (0.25s + 1)2210E (s) = R(s) C (s) = 10 T1s + 21s(T2 + K )s + 5 + k C(s) s +113E i N s3ssnE i R s3ssre= = lim ss 0= , e= lim ss 011s3s3令 R(s) =, N (s) =11E i N s2ssnE i R s2ssr= = lim ss 0e= 2(K + 5) ,= lim ss0e11s2s2令 R(s) =, N (s) =11E i N sssnE i R sssre= lim ss 0= lim ss01 = 21 = 0 ,ses令 R(s) =, N (s) =1121 s2 + Ks + 5s)Ts + 2(T s2 + K211 +N (s)s(T s + 2)(T s + K s + K + 5) + ( s + 1)1E i N ( s + 1)= E(s) =( s + 1)(T2 s + 2)s(T2 s + 5) + Ks( s + 1)( s + 1)21Ts + 2(T s2 + K s2 + Ks + 5s)211 +s(T s + 2)(T s + K s + K + 5) + ( s + 1)R(s)1 ( s + 1)E i R =s(T1s + 2)(T2 s + K s + K + 5)1= E (s) =系统开环 = 1 ,故对 R 为型,干扰 N 作用点之前无积分环节,系统对 N 为 0 型解法二、用梅森公式R(s)N(s)3-12 解:解法一、原系统结构图变换为s 0所以对输入响应的误差, ess = lim sE(s) = 。K1s(0.05s + 1)(s + 5) + 2.5K sE (s) = R(s)E i R + N (s)E i N =(0.05s + 1)(s2 + 5s 2.5) 1(0.05s + 1)(s + 5)sss(0.05s + 1)(s + 5) + 2.5KN (s)1 + 2.5KE i N 2.5(0.05s + 1)= E(s) = s + 5 1 =2.5(0.05s + 1)(s + 5)ss(0.05s + 1)(s + 5) + 2.5KR(s)1 + 2.5KE i R =s(0.05s + 1)(s + 5)1= E (s) =14nn n 1 +s2 + 2 ss 0s 0s2 2ssse = lim sE R(s) = lim s11 = 2s2(2) 输入 r (t) = 1(t), R(s) =1n n 1 +s2 + 2 ss0s0s 2ssse = lim sE R(s) = lim s1 = 01s(1)输入 r (t) = 1(t), R(s) =1n1 + G(s)s2 + 2 ssE i RR(s)(s)R(s) =G(s) = n ,误差传递函数 E = 1 23-14 解:开环传递函数为s0根据定义 e = r c , ess = essr + essn = essn = lim sEn (s) = 0.1 。si0.5s2 + s + 200(b)系统开环 = 1 ,为型系统,故 essr = 0 ;又 En (s) = N (s)iC i N =2000.1信号 essn = 0 ,从而有 ess = essr + essn = 0 。r (t ) = t 1(t) 时,essr = 0 ,又在 n(t)作用点以前原系统串联了一个积分环节,故对阶跃干扰R(s)s2 + s + 1,因为分子分母后两项系数对应相等,故系统为无差,在解法二、=s 0ss + 1s2C (s)输入 R(s) =, N (s) =,所以 ess = lim sE(s) = 011E(s) = R(s) C(s) = R(s) (R(s)iC i R + N (s)iC i N )N (s)s(s + 1) + 1R(s)s(s + 1) + 1C i N C i R =,=(a) 解法一、解得,s(s + 1)C (s)s + 1C (s)系统对 r(t)为型,对 n(t)为 0 型。3-13:154-2 解:4-1 解:习题第四章16作图测得 = 0.5 的阻尼线与根轨迹交点 s1,2 = 0.33 j0.58 ,根据根之和法则,9所以,无超调时 K 的取值范围为 0 K = 0.1925 。392333d 1 + 1 ( 1) + 1 =1当 0K3 时系统稳定, K =31333 = 2, K = 3与虚轴交点:1 + GH = s(s + 1)(0.5s + 1) + K = 0.5s3 + 1.5s2 + s + K 0.5( j )3 + 1.5( j )2 + j + K = 02分离角为 3(k = 1) 331 ,(k = 1),分离点坐标 s = =3(2k + 1)3(k = 0)3 条渐近线与实轴夹角3a渐近线交点为 =(0 1 2) = 1160D4-3 解:根轨迹如图极点 P1 = 0, P2 = 1, P3 = 2 ,共有三条渐近线17由根轨迹可以看出适当增加零点可以改善系统稳定性,使本来不稳定的系统变得稳定。 % = 25%4-5 解:(题目改为单位负反馈)n0.01s2 + 0.08s + 1= 0.88s , n = 10 , = 0.4 , ts =系统可以 看作 (s) =3.510.590.67231与零点 z=构成偶极子,所以主导极点为 s , s ,即(2)由于极点为 s1 = 110.67s + 1s,t = 3T = 2s, % = 0即 (s) =10.6712,3= 4 j9.2 ,主导极点为 s ,系统看成一阶系统。1.5 , s4-4 解:(1) s1 = 1n = 0.5 ,n = 0.667 ,其阶跃响应下的性能指标为 % = 16.3% , ts = = 10.5s 。13.51(s s )(s s2)s2 + 0.667s + 0.445,从而得到=主导极点,系统近似为二阶,即 (s) =0.445s1s2 s1 + s2 + s3 = p1 + p2 + p3 ,求得 s3 = 2.34 。s3 对虚轴的距离是 s1,2 的 7 倍,故认为 s1,2 是18s(s + 8)(1) G(s) =1605-3 解:s2 + 44.37s + 986.96G(s) =986.96 R = 44959()10 106 R= 0.7081 2100 2 106= 1013(H ) L =11041 100 2 L 106 + 10 106 Rj,1LCs2 + RCs + 1G(10 j) =设 G(s) =15超过10D ,所以不满足要求。5-2 解: = 2 f = 2 5 = 10 , G(10 j) = 3.54 = 0.708, G(10 j) = 90D5-1 解: 0 = arctan T = arctan 2 f T = arctan 2 10 0.01 = 32.14 ,相位差D第五章习题答案19s(s + 0.5)(s2 + 3.2s + 64)(3) G(s) =64(s + 2)s(s + 1)(s + 20)(2) G(s) =100(s + 2)20L(k ) = 20 lg K 20 lg k = 0 , K = k = 100 ,由图可知r = 45.3 ,nn 21s(s2 + 2 s + 1)由一个放大环节、一个积分环节、一个振荡环节组成(c) G(s) =K80s( 1 s + 1)40 K = 40G(s) =80Tccc1= 0 ,1 = 1 = 80 T =,穿越频率 = 40 , L( ) = 20 lg K 20 lg s(Ts + 1)由一个放大环节、一个积分环节、一个惯性环节组成(b) G(s) =K0.1s + 1G(s) =10T20 lg K = 20, K = 10 ;1 = 10 T = 0.11Ts + 1由一个放大环节、一个惯性环节组成K5-4 解:(a) G(s) =s(s2 + s + 1)(s2 + 4s + 25)(4) G(s) =s(s + 0.1)21伯德图:50s (s + 1)s (s + 50)221, 20 lg K = 14=52505-5 解:(1) G(s) =s(0.25s2 + 0.2s + 1)G(s) =10(s + 1)L = 20 lg K = 20 , K = 102,在 1 = 1 处, = 8 = 0.21 = 1 = 1 , 2 = 2 n = 2 , 20 lg1nn 2s(s2 + 2s + 1)1(e) G(s) =K ( s + 1)101s2 (s + 1)2K = 0.1990 ,G(s) = 0.1990(10s + 1) )2c( 或 者采用 精 确表示 : L( ) = 20 lg K + 20 lg 102 + 1 20 lg12 20 lg(1 + 1) = 0 ,s2 (s + 1)2ccL( ) = 20 lg K + 20 lg10 20 lg( 2 + 1) = 0 , K = 0.2 ,G(s) = 0.2(10s + 1)1 = 0.1 得 = 10 ; 2 = 1 得T = 1组成s2 (Ts + 1)2由一个放大环节、一个微分环节、两个积分环节、两个惯性环节(d) G(s) =K ( s + 1)s(s2 + 30s + 2.5 103 )G(s) =2.5 103 1001 2 22 1 2nn20 lg,得到 50 , = 0.3 (0.954 舍去)。= 4.85 , =1r22P = 0, N = 0 无穿越,故 Z = P 2 N = 0 稳定155s( s + 1)(s + 1)11, 20 lg K = 10.46(2) G(s) =s(s + 5)(s + 15) 3 250有一次负穿越, P = 0 , Z = P 2N = 2 故不稳定1023s(0.5s + 1)(0.02s + 1)5-6 解: G(s) =,1 = 2,2 = 50, 20 lg K = 2010P = 0, N = 0 无穿越,故 Z = P 2 N = 0 稳定155s2 ( s + 1)(s + 1)s (s + 5)(s + 15)211, 20 lg K = 10.46=(3) G(s) = 3 250(s + 1)10 (s + 1)24 n Lh = 20 lg G( jg ) = 20 1 g= 180 ,得 = 102g求g , 90 arctanDDn2 gn = 10, = 0.05s(0.01s2 + 0.01s + 1)5-7 解: G(s) =K10(5 j + 1)(0.2 j + 1)gh 20 lg 5 = 13.98 dBL = 20 lg G( j ) = 20 lg10DDDDDDD = 180 arctg (0.5c ) arctg (0.02c ) 90 = 180 66 5 90 = 190.01= 10得 0.5g i0.02g = 1 g =1(90 + arctg (0.5g ) + arctg (0.02g ) = 180DD20 lg K 20 lg c 20 lg(0.5c ) = 0 ,得c 20 = 2 5 4.47 (精确解 4.2460);25(2)= 70.36D= 180D 90D + 78.69D 84.29D 14.04D= 180 90 + tg (5c1 ) tg (10c1 ) tg (0.25c1 )111DD1 = 180 + G( jc1 )Ds(10s + 1)(0.25s + 1)G(s) =2(5s + 1)110 1c1= 1 K = 2 = 1 K 5得 2 = 5, T1 = 10,T3 = 0.25T3T1 2(1)= 0.2,= 0.1,= 4111s(T1s + 1)(T3 s + 1)5-8 解: G(s) =K ( 2 s + 1) 10 1 = 0.1 902因为c = 0.1 , = 180 90 arctgDDD0.01 0.1s(0.01s2 + 0.01s + 1)G(s) =0.1ggg(0.01 )2 + (1 0.01 2 )2c= 0.1,得 K = 0.1 = 0.1K26= 1系统稳定性不变= 180 90 + tg (5c1 ) tg (10c1 ) tg (0.25c1 )111DD111DD= 180 90 + tg (0.5c 2 ) tg (c 2 ) tg (0.025c 2 ) 2 = 180 + G( jc 2 )Ds(s + 1)(0.025s + 1)G(s) =20(0.5s + 1) K = 2010 10= 1c 2 = 10 = 10c1 K 0.5 10s(s + 1)(0.025s + 1)G(s) =K (0.5s + 1)(3)右移 10 倍频程,则T1 = 1, 2 = 0.5, T3 = 0.025P = 0, N = 0 (无穿越) Z = P 2N = 0 (稳定)27少为 45D 13D = 32D ,引入超前网络后,新的截止频率会增大,从而存在相角裕度损失,因 s + 1c(2)为了将系统相角裕度提高到 45D ,使用超前校正网络 G (s) = s + 1 ,所需超前相角至不能满足要求。s(0.1s + 1)c(1) 原系统开环传递函数 G(s) =,计算得到 = 44s1 50s1 , = 13D 50s1 ,从而得到 =1 + 1m
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