自动控制原理胡寿松第四版课后答案

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1、13解：系统的工作原理为：当流出增加时，液位降低，浮球降落，控制器通过移动气动阀门的 开度，流入量增加，液位开始上。当流入量和流出量相等时达到平衡。当流出量减小时，系 统的变化过程则相反。授课：XXX希望液位流出量授课：XXX高度液位高度控制器气动阀水箱流入量浮球图一14（1） 非线性系统（2） 非线性时变系统（3） 线性定常系统（4） 线性定常系统（5） 线性时变系统（6） 线性定常系统授课：XXX2-1解：显然，弹簧力为 kx(t ) ，根据牛顿第二运动定律有：授课：XXXF (t ) kx(t) = m移项整理，得机械系统的微分方程为：d 2 x(t )dt 2授课：XXX2m d x(

2、t ) + kx(t ) = F (t )dt 2对上述方程中各项求拉氏变换得：ms 2 X (s) + kX (s) = F (s)授课：XXX所以，机械系统的传递函数为：G(s) =X (s) =F (s)1ms 2 + k授课：XXX2-2解一：授课：XXX由图易得：i1 (t )R1 = u1 (t ) u2 (t ) uc (t ) + i1 (t )R2 = u2 (t ) duc (t ) 授课：XXXi1 (t ) = Cdt由上述方程组可得无源网络的运动方程为：授课：XXX授课：XXXC ( R + R ) du2 (t ) u (t ) = CRdu1 (t ) u (t

3、) 授课：XXX12dt+ 22+ 1dt授课：XXX对上述方程中各项求拉氏变换得：C (R1 + R2 )sU 2 (s) + U 2 (s) = CR2 sU1 (s) + U1 (s) 所以，无源网络的传递函数为：授课：XXXG(s) = U 2 (s) =U1 (s)1 + sCR21 + sC(R1 + R2 )授课：XXX解二（运算阻抗法或复阻抗法）：授课：XXXU (s) 1+ R21 + R Cs授课：XXX 2 = Cs = 2 授课：XXXU (s) R + 1 + R1 + ( R + R )Cs授课：XXX1121Cs22-5解：按照上述方程的顺序，从输出量开始绘制系统

4、的结构图，其绘制结果如下图所示：依次消掉上述方程中的中间变量 X 1 , X 2 , X 3 , 可得系统传递函数为：授课：XXXC(s) =R(s)G1 (s)G2 (s)G3 (s)G4 (s)1 + G2 (s)G3 (s)G6 (s) + G3 (s)G4 (s)G5 (s) + G1 (s)G2 (s)G3 (s)G4 (s)G7 (s) G8 (s)授课：XXX2-6解：授课：XXX 将 G1 (s) 与 G1 (s) 组成的并联环节和 G1 (s) 与 G1 (s) 组成的并联环节简化，它们的等效传递函数和简化结构图为：G12 (s) = G1 (s) + G2 (s)G34 (

5、s) = G3 (s) G4 (s) 将 G12 (s), G34 (s) 组成的反馈回路简化便求得系统的闭环传递函数为：授课：XXX2-7解：C(s) =R(s)G12 (s)1 + G12 (s)G34 (s)=G1 (s) + G2 (s)1 + G1 (s) + G2 (s)G3 (s) G4 (s)授课：XXX授课：XXX由上图可列方程组：E (s)G1 (s) C (s)H 2 (s)G2 (s) = C (s)授课：XXXR(s) H1(s) C (s)G2 (s)= E (s)授课：XXX联列上述两个方程，消掉 E (s) ，得传递函数为：授课：XXXC(s) =R(s)G1

6、(s)G2 (s)1 + H1 (s)G1 (s) + H 2 (s)G2 (s)授课：XXX联列上述两个方程，消掉 C (s) ，得传递函数为：授课：XXXE(s) =R(s)1 + H 2 (s)G2 (s)1 + H1 (s)G1 (s) + H 2 (s)G2 (s)授课：XXX2-8解：授课：XXX将反馈回路简化，其等效传递函数和简化图为：0.41G (s) = 2s + 1 =1 + 0.4 * 0.52s + 115s + 3授课：XXX将反馈回路简化，其等效传递函数和简化图为：1授课：XXX2 2G (s) = s + 0.3s + 1 =5s + 32授课：XXX231 +

7、0.45s + 4.5s+ 5.9s + 3.4授课：XXX(s + 0.3s + 1)(5s + 3)授课：XXX授课：XXX将反馈回路简化便求得系统的闭环传递函数为：0.7 * (5s + 3) o (s) = 5s 3 + 4.5s 2 + 5.9s + 3.4 =3.5s + 2.1授课：XXXi (s)1 + 0.7 * Ks(5s + 3)5s 3+ (4.5 + 3.5K )s 2+ (5.9 + 2.1K )s + 3.4授课：XXX5s 3-3解：该二阶系统的最大超调量：授课：XXX p = e /1 2*100%授课：XXX授课：XXX当 p= 5% 时，可解上述方程得：授

8、课：XXX = 0.69授课：XXX当 p= 5% 时，该二阶系统的过渡时间为：授课：XXX授课：XXXt s 3wn授课：XXX授课：XXX所以，该二阶系统的无阻尼自振角频率 wn3-4解： 3t s=30.69 * 2= 2.17授课：XXX由上图可得系统的传递函数：10 * (1 + Ks)授课：XXXC (s) =R(s)s(s + 2)1 + 10 * (1 + Ks)s(s + 2)=10 * (Ks + 1)2s + 2 * (1 + 5K )s + 10授课：XXX授课：XXX所以 wn =10 ，wn = 1 + 5K授课：XXX授课：XXX 若= 0.5 时， K 0.11

9、6授课：XXX授课：XXX所以 K 0.116 时，= 0.5授课：XXX 系统单位阶跃响应的超调量和过渡过程时间分别为：授课：XXX p = e /1 2*100% = e0.5*3.14 /10.52*100% 16.3%授课：XXXts =3wn=3授课：XXX0.5 * 1.910授课：XXX 加入 (1 + Ks ) 相当于加入了一个比例微分环节，将使系统的阻尼比增大，可以有效地减小原系统的阶跃响应的超调量；同时由于微分的作用，使系统阶跃响应的速度（即变授课：XXX化率）提高了，从而缩短了过渡时间：总之，加入 (1 + Ks ) 后，系统响应性能得到改善。3-5解：授课：XXX由上图

10、可得该控制系统的传递函数：C(s) =110K1授课：XXXR(s)二阶系统的标准形式为：C (s)R(s)s 2 + (10 + 1)s + 10Kw 2= n s 2 + 2w s + w2授课：XXXnn所以w2n = 10K12wn = 10 + 1由授课：XXXp= e /1 2*100%授课：XXXt p =wn1 2授课：XXX p = 9.5%t p = 0.5可得 = 0.6授课：XXX2wn = 10K1 = 0.6wn = 7.85授课：XXX由和2wn = 10 + 1wn = 7.85可得：授课：XXXK1 = 6.16 = 0.84授课：XXXt s 3wn授课：X

11、XX= 0.64授课：XXX3-6解： 列出劳斯表为：因为劳斯表首列系数符号变号 2 次，所以系统不稳定。 列出劳斯表为：因为劳斯表首列系数全大于零，所以系统稳定。 列出劳斯表为：因为劳斯表首列系数符号变号 2 次，所以系统不稳定。3-7解：系统的闭环系统传递函数：K (s +1)授课：XXXC (s) =R(s)=s(2s +1)(Ts +1)=1 + K (s +1)s(2s +1)(Ts +1)K (s +1)K (s +1)s(2s +1)(Ts +1) + K (s +1)授课：XXX2Ts3 + (T + 2)s 2 + (K +1)s + K列出劳斯表为：s32TK +1s2T

12、+ 2Ks1(K +1)(T + 2) 2KT T + 2s0K授课：XXX授课：XXXT 0 ，T + 2 0 ， (K + 1)(T + 2) 2KT T + 2 0 ， K 0授课：XXXT 0K 0 ， (K + 1)(T + 2) 2KT 0授课：XXX(K +1)(T + 2) 2KT = (T + 2) + KT + 2K 2KT= (T + 2) KT + 2K = (T + 2) K (T 2) 0K (T 2) (T + 2)3-9解：授课：XXX由上图可得闭环系统传递函数：C (s) =KK2 K3授课：XXX2 3 2 3 2 3R(s)(1 + KK K a)s2 K

13、K K bs KK K代入已知数据，得二阶系统特征方程：(1 + 0.1K )s2 0.1Ks K = 0授课：XXX列出劳斯表为：s21 + 0.1K Ks1 0.1Ks0 K授课：XXX可见，只要放大器10 K 0 ，系统就是稳定的。授课：XXX3-12解：系统的稳态误差为：授课：XXXess= lim e(t ) = lim sE (s) = limsR(s)授课：XXXt s0s 0 1 + G0 (s)授课：XXX授课：XXX G0 (s) =10s(0.1s + 1)(0.5s + 1)授课：XXX系统的静态位置误差系数：授课：XXXK= lim G(s) = lim10= 授课：

14、XXXps 00s 0 s(0.1s + 1)(0.5s + 1)授课：XXX系统的静态速度误差系数：授课：XXXK = lim sG(s) = lim10s= 10授课：XXXvs 00s 0 s(0.1s + 1)(0.5s + 1)授课：XXX系统的静态加速度误差系数：授课：XXXK = lim s 2 G(s) = lim10s 2= 0授课：XXXas00s0 s(0.1s + 1)(0.5s + 1)授课：XXX当 r (t ) = 1(t ) 时， R(s) = 1s授课：XXXess= lims* 1 = 0授课：XXX当 r (t ) = 4t 时， R(s) =s0 10s

15、1 +s(0.1s + 1)(0.5s + 1)4授课：XXXs 2e= lims* 4 = 0.4授课：XXXsss 0 s 2授课：XXX当 r (t ) = t 2 时， R(s) =1 +10s(0.1s + 1)(0.5s + 1)2s 3授课：XXXess= lims 01 +s* 2 = 10s 3s(0.1s + 1)(0.5s + 1)授课：XXX当 r(t) = 1(t) + 4t + t 2 时， R(s) = 1 + 4 + 2ss 2s 3授课：XXX3-14解：ess = 0 + 0.4 + = 授课：XXX由于单位斜坡输入下系统稳态误差为常值=2，所以系统为 I

16、型系统授课：XXX设开环传递函数 G(s) =Ks(s2 + as + b)K = 0.5 b授课：XXX授课：XXX闭环传递函数(s) =G(s)=K1 + G(s)s3 + as2 + bs + K授课：XXXQ s = 1 j 是系统闭环极点，因此授课：XXXs3 + as2 + bs + K = (s + c)(s2 + 2s + 2) = s3 + (2 + c)s2 + (2c + 2)s + 2c授课：XXXK = 0.5bK = 2cb = 2c + 2a = 2 + cK = 2a = 3b = 4c = 1授课：XXX授课：XXX所以 G(s) =2。s(s2 + 3s +

17、 4)授课：XXX授课：XXX4-1js js 授课：XXX授课：XXXk k = 0k k = 00 k = 0k k k = 00授课：XXX(a)(b)j s 授课：XXXjs 0 授课：XXX0(c)(d)4-2j s 授课：XXXp 3 = 10 p 1 = 0 p 2 = 0授课：XXX授课：XXXp1 = 0,p2 = 0,p3 = 1授课：XXX授课：XXX1. 实轴上的根轨迹(, 1)(0, 0)1授课：XXX2. n m = 33 条根轨迹趋向无穷远处的渐近线相角为授课：XXX= 180(2q + 1) = 60,180a 3(q = 0,1)授课：XXX渐近线与实轴的交点

18、为n m pi zi授课：XXXi =1j =10 0 1 1授课：XXX a =3. 系统的特征方程为n m= 33授课：XXX授课：XXX1+G(s) = 1 +K= 0s2 (s +1)授课：XXX即K = s2 (s +1) = s3 s2授课：XXXdK = 3s2 2s = 0dss(3s + 2) = 0授课：XXX根s1 = 0（舍去）s2 = 0.667授课：XXX授课：XXX4. 令 s = j代入特征方程1+G(s) = 1 +K= 0s2 (s +1)授课：XXXs2 (s +1) + K =0( j )2 ( j +1) + K =0 2 ( j +1) + K =0

19、K 2 j =0K 2 =0 = 0授课：XXX=0（舍去）授课：XXX与虚轴没有交点，即只有根轨迹上的起点，也即开环极点p1,2 = 0在虚轴上。授课：XXX授课：XXX2授课：XXX授课：XXX5-1G(s) =50.25s +1G( j ) =50.25 j +1授课：XXX授课：XXXA( ) =5 (0.25 )2 +1() = arctan(0.25)授课：XXX授课：XXX输入 r(t) = 5 cos(4t 30) = 5 sin(4t + 60)=4授课：XXX授课：XXXA(4) =5(0.25 * 4)2 +1= 2.5 2(4) = arctan(0.25 * 4) =

20、 45授课：XXX系统的稳态输出为c(t ) = A(4) * 5 cos4t 30 + (4)= 2.5 2 * 5 cos(4t 30 45)= 17.68 cos(4t 75) = 17.68 sin(4t +15)sin = cos(90 ) = cos( 90) = cos( + 270)授课：XXX5-3或者，c(t ) = A(4) * 5 sin4t + 60 + (4)= 2.5 2 * 5 sin(4t + 60 45)= 17.68 sin(4t +15)11授课：XXX（2）G(s) =(1 + s)(1 + 2s)G( j ) =(1 + j )(1 + j 2 )授

21、课：XXX授课：XXXA( ) =1(1 + 2 )(1 + 4 2 )() = arctan arctan 2授课：XXX() = arctan arctan 2 = 90arctan + arctan 2 = 90授课：XXX = 1/(2) 2 = 1/ 2A( ) =1=(1 +1 / 2)(1 + 4 *1/ 2)2 = 0.473授课：XXX与虚轴的交点为（0，-j0.47）授课：XXXjY()0 = -j0.47 = 01X ()授课：XXX1授课：XXX授课：XXX（3） G(s) =1s(1 + s)(1 + 2s)G( j ) =1j (1 + j )(1 + j2 )授课

22、：XXX授课：XXXA( ) =1(1 + 2 )(1 + 4 2 )() = 90 arctan arctan 2授课：XXX() = 90 arctan arctan 2 = 180arctan + arctan 2 = 90授课：XXX = 1/(2) 2 = 1/ 2A( ) =11/2 (1 +1/ 2)(1 + 4 *1/ 2)= 2 = 0.673授课：XXX与实轴的交点为（-0.67，-j0）授课：XXX-0.670 = 0.707 = 0jY () = X ()授课：XXX授课：XXX（4） G(s) =1s2 (1 + s)(1 + 2s)G( j ) =1( j )2 (

23、1 + j )(1 + j 2 )授课：XXX授课：XXXA( ) = 21(1 + 2 )(1 + 4 2 )( ) = 180 arctan arctan 2授课：XXX() = 180 arctan arctan 2 = 270arctan + arctan 2 = 90授课：XXX = 1/(2) 2 = 1/ 2A( ) =1= 2 (1/ 2) (1 +1/ 2)(1 + 4 *1/ 2)32 = 0.94授课：XXX与虚轴的交点为（0，j0.94）授课：XXX = 0.707 = 00.940jY() = X ()2授课：XXX5-4（2）1 = 0.5 ，2 = 1 ， k =

24、 1 ， = 0L ( ) ( d B )授课：XXX00.01-20dB0.10.5-20dB /dec1 10-40dB /dec授课：XXX-40dB（3）1 = 0.5 ，2 = 1 ， k = 1 ， = 1授课：XXXL ( ) ( d B )-20dB /dec授课：XXX20dB -40dB /dec0授课：XXX0.010.10.51 10授课：XXX授课：XXX-20dB-40dB-60dB /dec授课：XXX（4）1 = 0.5 ，2 = 1 ， k = 1 ， = 2L ( )(d B )授课：XXX60dB-40dB /dec授课：XXX40dB20dB -60dB

25、 /dec0授课：XXX0.010.10.51 10授课：XXX授课：XXX-20dB-40dB-80dB /dec授课：XXX授课：XXX5-6G(s) =1s 1是一个非最小相位系统授课：XXX3授课：XXX授课：XXXG( j ) =1=1(1 j ) =1e j ( 180o +arctg )授课：XXXj 11 + 21 + 2授课：XXXG(s) =1s +1是一个最小相位系统授课：XXXG( j ) =1=1(1 j ) =1e jarctg授课：XXXj +1 1 + 21 + 2授课：XXX5-8(a) = 0 授课：XXX = -10X ( )授课：XXX = 0 +系统开

26、环传递函数有一极点在 s 平面的原点处，因此乃氏回线中半径为无穷小量 的半圆弧 对应的映射曲线是一个半径为无穷大的圆弧： ：0 0+ ； ：90 0 90； () ：90 0 90N=P-Z， Z=P-N=0-(-2)=2闭环系统有 2 个极点在右半平面，所以闭环系统不稳定（b）jY ( ) 授课：XXX = 0 = 0+ = -1 0X ( ) 授课：XXX4授课：XXX系统开环传递函数有 2 个极点在 s 平面的原点处，因此乃氏回线中半径为无穷小量 的半圆弧对应的映射曲线是一个半径为无穷大的圆弧： ：0 0+ ； ：90 0 90； () ：180 0 180N=P-Z， Z=P-N=0-

27、0=0闭环系统有 0 个极点在右半平面，所以闭环系统稳定授课：XXX5-10KK2.28K授课：XXX（1）G(s)H (s) =Ts +1()()12.28s +1=s + 2.28授课：XXX1 = 2.280授课：XXX90 ( )授课：XXX授课：XXXG s H s = K1= K1=2.28K授课：XXX（2）( )( ) ( )()s Ts +1s12.28s +1s(s + 2.28)授课：XXX901 = 2.28授课：XXX授课：XXX180 ( )授课：XXX授课：XXXK s +11K 0.5s +14K (s + 0.5)授课：XXX（3）G(s)H (s) =s T

28、s +1 s1=s (s + 2)授课：XXX222s +12L ( )( d B )-40dB /dec-20dB /deca授课：XXXb 00.512-40dB /dec授课：XXX授课：XXX5授课：XXX授课：XXX20 lg 1= a 20 lg K + 20 lg 1= 40 lg 120 lg K = 20 lg 1授课：XXX0.520 lg(K )1 = 20 lg 20.50.5K = 1/ 2 = 0.50.5授课：XXXG(s)H (s) = 4K (s + 0.5) = 2(s + 0.5)授课：XXXs2 (s + 2)s2 (s + 2)授课：XXX授课：XXX

29、90()()1 = 0.52 = 2授课：XXX授课：XXX180 ( )授课：XXX授课：XXX5-11 = 0jY ()授课：XXX授课：XXX = +0 = (-1,j0)X ()授课：XXX = 0+授课：XXXG(s)H (s) =Ks(s +1)(3s +1) G( j )H ( j ) =Kj ( j +1)(3 j +1)授课：XXX( ) = 90 arctan arctan 3 = 180arctan + arctan 3 = 90授课：XXX = 1/(3) 2 = 1/ 3A( ) =K1 /3 (1 +1 / 3)(1 + 9 *1/ 3)= 3 K = 14授课：X

30、XXKc = 4/3 = 1.33授课：XXX6授课：XXX6-2 (1)6 2G(s) = n 授课：XXXnns(s2 + 4s + 6)s(s2 + 2 s + 2 )授课：XXX授课：XXX 2 = 6 =6 =2.45,2 =4 =4= 2= 0.816授课：XXXn n n2n 6授课：XXX授课：XXXK = 1所以，c = 120lgK = 0授课：XXX授课：XXX 2 / ( ) = 90 arctgcn 2 * 0.816 *1/ 2.45 = 90 arctg授课：XXXc 1 2 / 2 1 1/ 2.452授课：XXXcn = 90 arctg 2 * 0.816

31、*1 / 2.45 = 90 arctg 0.666 = 90 arctg 0.7995授课：XXX1 1 / 2.452 0.833 授课：XXX= 90 38.64 = 128.64 = 180 + (c ) = 180 128.64 = 51.36L( )(dB)授课：XXX50403020100-10-20-30-400.01-20dB /dec0.1n12.4510-60dB /dec授课：XXX授课：XXX(2)1 = 1,2 =1/0.2=5授课：XXX授课：XXX 2 / ( ) = 90 arctgcn + arctgc arctg c授课：XXXc 1 2 / 2 授课：X

32、XXcn 1 2 授课：XXX15 = 128.64 + arctg 1 arctg 1 = 128.64 + 45 11.31 = 94.95 = 180 + (c ) = 180 94.95 = 85.05授课：XXX1授课：XXX课后答案网L() (dB )授课：XXX50403020100-10-200.01-20dB /dec0.1n12.4520dB /decG c5 10授课：XXX-30-40-40dB /dec -60dB /dec-60dB /dec授课：XXX授课：XXX6-5 (1)G(s) =10s(0.5s +1)(0.1s +1)授课：XXX授课：XXX = 1,

33、 20 lg K =20lg10=20dB1 = 1/ 0.5 = 2,2 = 1 / 0.1 = 10授课：XXX授课：XXX1 = 2时， L(1 ) = 20 20(lg 2 lg1) = 20lg10 20 lg 2 = 20lg5 = 14dB授课：XXX授课：XXX2 = 10时， L(2 ) = 14 40(lg10 lg 2) = 13.96dB授课：XXX所以，1 c 2L(1 ) = 40(lg c lg 2) = 40(lg c / 2) = 14dBc = 4.48 (c ) = 90 arctg 0.5c arctg 0.1c = 90 arctg 2.24 arct

34、g 0.448= 90 65.94 24.13 = 180.07 = 180 + (c ) = 180 180.07 = 0.07L ( )(dB)授课：XXX50403020100-10-20-30-400.1-20dB /dec1 2-40dB /dec c 10-60dB /dec100授课：XXX授课：XXX2授课：XXX授课：XXX(2)G(s)Gc (s) =10(0.33s +1)s(0.5s +1)(0.1s +1)(0.033s +1)授课：XXX = 1, 20 lg K =20lg10=20dB授课：XXX1 = 1 / 0.5 = 2,2 = 1/ 0.33 = 3,3

35、 = 1 / 0.1 = 10,4 = 1/ 0.033 = 30授课：XXX授课：XXX2 = 3时， L(1 ) L(2 ) = 40(lg 2 lg 1 ) 14 L(2 ) = 40(lg 4.35 lg 2)授课：XXXL(2 ) = 7dBL(3 = 10) L(2 = 3) = 20(lg 3 lg 2 ) = 3.37dB所以2 c 2 3L(2 ) = 20(lg c 2 lg 2 ) = 20(lg c 2 / 3) = 7dBc 2 = 6.72 (c ) = 90 arctg 0.5c 2 arctg 0.1c 2 + arctg 0.33c 2 arctg 0.033

36、c 2= 90 arctg 3.36 arctg 0.672 + arctg 2.22 arctg 0.222= 90 73.43 33.90+ 65.7512.52 = 144.1 2 = 180 + (c 2 ) = 180 144.1 = 35.9L( )(dB)授课：XXX504030-20dB /dec20100-40dB /decc 220dB /decG c10授课：XXX-10-200.1123c130G cG100授课：XXX-30-40-20dB /dec-40dB /dec-60dB /dec授课：XXX-60dB /dec校正环节为相位超前校正，校正后系统的相角裕量增加，系统又不稳定变为稳定，且有一定的稳定裕度，降低系统响应的超调量；剪切频率增加，系统快速性提高；但是高频段增益提 高，系统抑制噪声能力下降。3 （注：可编辑下载，若有不当之处，请指正，谢谢!） 授课：XXX