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。希 望 液 位流出量高度 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22-1 解:显然,弹簧力为 kx(t ) ,根据牛顿第二运动定律有:F (t ) kx(t) = m移项整理,得机械系统的微分方程为:d 2 x(t )dt 2m d x(t ) +
2、 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)所以,机械系统的传递函数为:G(s) = X (s) =F (s) 1ms 2 + k2-2 解一:由图易得:i1 (t )R1 = u1 (t ) u2 (t ) uc (t ) + i1 (t )R2 = u2 (t ) duc (t ) i1 (t ) = C dt由上述方程组可得无源网络的运动方程为:C ( R + R ) du2 (t ) u (t ) = CR du1 (t ) u (t ) 1 2 dt + 2 2 + 1dt对上述方程中各项求拉氏变换得:C
3、 (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:G(s) = U 2 (s) =U1 (s)1 + sCR21 + sC(R1 + R2 )解二(运算阻抗法或复阻抗法) :U (s) 1 + R2 1 + R Cs 2 = Cs = 2 U (s) R + 1 + R 1 + ( R + R )Cs1 1 21 Cs 22-5 解: 按照上述方程的顺序, 从输出量开始绘制系统的结构图, 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:C(s) =R(
4、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)2-6 解: 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 (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:2-7 解:
5、C(s) =R(s)G12 (s)1 + G12 (s)G34 (s) =G1 (s) + G2 (s)1 + G1 (s) + G2 (s)G3 (s) G4 (s)由上图可列方程组:E (s)G1 (s) C (s)H 2 (s)G2 (s) = C (s)R(s) H1(s) C (s)G2 (s)= E (s)联列上述两个方程,消掉 E (s) ,得传递函数为:C(s) =R(s)G1 (s)G2 (s)1 + H1 (s)G1 (s) + H 2 (s)G2 (s)联列上述两个方程,消掉 C (s) ,得传递函数为:E(s) =R(s)1 + H 2 (s)G2 (s)1 + H1
6、(s)G1 (s) + H 2 (s)G2 (s)12 22 32-8 解:将反馈回路简化,其等效传递函数和简化图为:0.4G (s) = 2s + 1 =1 + 0.4 * 0.52s + 115s + 3将反馈回路简化,其等效传递函数和简化图为:1G (s) = s +0.3s + 1 = 5s + 321 + 0.4 5s + 4.5s + 5.9s + 3.4(s + 0.3s + 1)(5s + 3)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 + 2.1i (s) 1 + 0.7
7、 * Ks(5s + 3) 5s 3 + (4.5 + 3.5K )s 2 + (5.9 + 2.1K )s + 3.425s 3-3 解:该二阶系统的最大超调量: p = e / 1 2 *100%当 p = 5% 时,可解上述方程得: = 0.69当 p = 5% 时,该二阶系统的过渡时间为:t s 3wn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由上图 可得系统的传递函数:10 * (1 + Ks)C (s) =R(s)s(s + 2)1 + 10 * (1 + Ks)s(s + 2)= 10 * (Ks + 1)s + 2 *
8、(1 + 5K )s + 10所以 wn = 10 , wn = 1 + 5K 若 = 0.5 时, K 0.116所 以 K 0.116 时, = 0.5 系 统单位阶跃响应的超调量和过渡过程时间分别为: p = e / 1 2 *100% = e0.5*3.14 / 10.52 *100% 16.3%ts = 3wn =30.5 * 1.910 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w212p化率)提高了,从而缩短了过渡时间:总之,加入 (1 + Ks
9、) 后,系统响应性能得到改善。3-5 解:由上图可得该控制系统的传递函数:C(s) = 10K1R(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 + w2n n所以n = 10K12wn = 10 + 1由 = e /1 2 *100%t p =wn 1 2 p = 9.5%t p = 0.5可得 = 0.6wn = 10K1 = 0.6wn = 7.85由 和2wn = 10 + 1 wn = 7.85可得:K1 = 6.16 = 0.84t s 3wn= 0.643-6 解: 列出 劳斯表为:因 为劳斯表首列系数符
10、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)C (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)2Ts3 + (T + 2)s 2 + (K +1)s + K列出劳斯表为:s3 2T K +1s2 T + 2 Ks1 (K +1)(T + 2) 2KT T + 2s0 K2 3
11、 2 3 2 3T 0 , T + 2 0 , (K + 1)(T + 2) 2KT T + 2 0 , K 0T 0 K 0 , (K + 1)(T + 2) 2KT 0(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:由上图可得闭环系统传递函数:C (s) = KK2 K3R(s) (1 + KK K a)s2 KK 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列出劳斯表为:s2 1 +
12、 0.1K Ks1 0.1Ks0 K可见,只要放大器 10 K 0 ,系统就是稳定的。3-12 解:系统的稳态误差为:ess = lim e(t ) = lim sE (s) = lim s R(s)t s 0 s 0 1 + G0 (s) G0 (s) = 10s(0.1s + 1)(0.5s + 1)系统的静态 位置误差系数:K = lim G (s) = lim 10 = ps 0 0 s 0 s(0.1s + 1)(0.5s + 1)系统的静态速度误差系数:K = lim sG (s) = lim 10s = 10vs 0 0 s 0 s(0.1s + 1)(0.5s + 1)系统的静
13、态加速度误差系数:K = lim s 2 G (s) = lim 10s 2= 0as 0 0 s 0 s(0.1s + 1)(0.5s + 1)当 r (t ) = 1(t ) 时, R(s) = 1sess = lim s * 1 = 0当 r (t ) = 4t 时, R(s) =s 0 10 s1 +s(0.1s + 1)(0.5s + 1)4s 2e = lim s * 4 = 0.4sss 0 s 2当 r (t ) = t 2 时, R(s) =1 + 10s(0.1s + 1)(0.5s + 1)2s 3ess = lims 0 1 +s * 2 = 10 s 3s(0.1s
14、+ 1)(0.5s + 1)当 r(t) = 1(t) + 4t + t 2 时, R(s) = 1 + 4 + 2s s 2 s 33-14 解:ess = 0 + 0.4 + = 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 型系统设开环传递函数 G(s) = Ks(s2 + as + b) K = 0.5 b闭环传递函数 (s) = G(s) = K1 + G(s) s3 + as2 + bs + KQ s = 1 j 是系统闭环极点,因此s3 + as2 + bs + K = (s + c)(s2 + 2s + 2) = s3 + (2 + c)s2 + (2c + 2)
15、s + 2cK = 0.5bK = 2cb = 2c + 2 a = 2 + cK = 2a = 3b = 4c = 1所以 G(s) = 2 。s(s2 + 3s + 4)4-1j s j s k k = 0k k = 00 k = 0 k k k = 0 0(a) (b)j s j s 0 0(c) (d)4-2j s p 3 = 10 p 1 = 0 p 2 = 0p1 = 0, p2 = 0, p3 = 11. 实轴上的 根轨迹 ( , 1) (0, 0)1= 2. n m = 33 条根轨 迹趋向无穷远处的渐近线相角为 180(2q + 1) = 60,180a 3 (q = 0,1
16、)渐近线与实轴的交点为n m pi zii =1 j =1 0 0 1 1 a =3. 系统的特 征方程为n m = = 3 31+G(s) = 1 + K = 0s2 (s +1)即 K = s2 (s +1) = s3 s2dK = 3s2 2s = 0dss(3s + 2) = 0根 s1 = 0 (舍去 ) s2 = 0.6674. 令 s = j 代入特征方程 1+G(s) = 1 + K = 0s2 (s +1)s2 (s +1) + K =0( j )2 ( j +1) + K =0 2 ( j +1) + K =0K 2 j =0K 2 =0 = 0=0 (舍去)与虚轴没有交点
17、,即只有根轨迹上的起点,也即开环极点 p1,2 = 0 在虚轴上。25-1 G(s) = 50.25s +1 G( j ) = 50.25 j +1A( ) = 5 (0.25 )2 +1() = arctan(0.25)输入 r(t) = 5 cos(4t 30) = 5 sin(4t + 60) =4A(4) = 5(0.25 * 4)2 +1= 2.5 2 (4) = arctan(0.25 * 4) = 45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
18、 sin(4t +15)sin = cos(90 ) = cos( 90) = cos( + 270)5-3或者,c(t ) = A(4) * 5 sin4t + 60 + (4)= 2.5 2 * 5 sin(4t + 60 45)= 17.68 sin(4t +15)1 1(2 ) G(s) =(1 + s)(1 + 2s) G( j ) =(1 + j )(1 + j2 )A( ) = 1(1 + 2 )(1 + 4 2 )() = arctan arctan 2() = arctan arctan 2 = 90 arctan + arctan 2 = 90 = 1/(2) 2 = 1/
19、 2 A( ) = 1 =(1 +1 / 2)(1 + 4 *1/ 2)2 = 0.473与虚轴的交点为(0 ,-j0.47 )jY()0 = -j0.47 = 01X ()1(3 ) G(s) = 1s(1 + s)(1 + 2s)G( j ) = 1j (1 + j )(1 + j2 )A( ) =1(1 + 2 )(1 + 4 2 )() = 90 arctan arctan 2() = 90 arctan arctan 2 = 180 arctan + arctan 2 = 90 = 1/(2) 2 = 1/ 2 A( ) = 11/2 (1 +1/ 2)(1 + 4 *1/ 2) =
20、 2 = 0.673与实轴的交点为(-0.67 ,-j0 )-0.670 = 0.707 = 0jY () = X ()(4 ) G(s) = 1s2 (1 + s)(1 + 2s)G( j ) = 1( j )2 (1 + j )(1 + j2 )A( ) = 21(1 + 2 )(1 + 4 2 )( ) = 180 arctan arctan 2() = 180 arctan arctan 2 = 270 arctan + arctan 2 = 90 = 1/(2) 2 = 1/ 2 A( ) = 1 = 2 (1/ 2) (1 +1/ 2)(1 + 4 *1/ 2) 3 2 = 0.9
21、4与虚轴的交点为(0 ,j0 .94) = 0.707 = 0 0.940jY() = X ()25-4(2 ) 1 = 0.5 , 2 = 1 , k = 1 , = 0L ( ) ( d B )0 0.01-20dB0.1 0.5-20dB /dec1 10-40dB /dec-40dB(3 ) 1 = 0.5 , 2 = 1 , k = 1 , = 1L ( ) ( d B ) -20dB /dec20dB -40dB /dec0 0.01 0.1 0.5 1 10-20dB-40dB-60dB /dec(4 ) 1 = 0.5 , 2 = 1 , k = 1 , = 2L ( )(d
22、B )60dB-40dB /dec40dB20dB -60dB /dec0 0.01 0.1 0.5 1 10-20dB-40dB-80dB /dec5-6G(s) = 1s 1是一个非最小相位系统3G( j ) = 1 = 1 (1 j ) = 1 e j ( 180o +arctg )j 11 + 2 1 + 2G(s) = 1s +1是一个最小相位系统G( j ) = 1 = 1 (1 j ) = 1 e jarctgj +1 1 + 2 1 + 25-8(a) = 0 = -1 0 X ( ) = 0 +系统开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
23、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+ ; :9 0 0 90; () :9 0 0 9 0N=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 ( ) = 0 = 0+ = -1 0 X ( ) 4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+ ; :9 0 0 90; () :1 80 0 180N=P-Z, Z=P-N=0-0=0闭环系统有 0 个极点在右 半平面,所以闭环系统稳定5-10K K 2.28K
24、(1 ) G(s)H (s) = =Ts +1()()12.28s +1=s + 2.281 = 2.280 90 ( )G s H s = K 1 = K 1 = 2.28K(2 ) ( ) ( ) ( )()s Ts +1 s 12.28s +1 s(s + 2.28)901 = 2.28180 ( )K s +11K 0.5 s +1 4K (s + 0.5)(3 ) G(s)H (s) = =s Ts +1 s 1=s (s + 2)2 2 2s +12L ( )( d B )-40dB /dec-20dB /decab 0 0.5 1 2 -40dB /dec520 lg 1 = a
25、 20 lg K + 20 lg 1 = 40 lg 1 20 lg K = 20 lg 10.520 lg(K )1 = 20 lg 20.5 0.5K = 1/ 2 = 0.50.5G(s)H (s) = 4K (s + 0.5) = 2(s + 0.5)s2 (s + 2) s2 (s + 2)90()() 1 = 0.5 2 = 2180 ( )5-11 = 0jY () = +0 = (-1,j0)X () = 0+G(s)H (s) = Ks(s +1)(3s +1) G( j )H ( j ) = Kj ( j +1)(3 j +1)( ) = 90 arctan arctan
26、3 = 180 arctan + arctan 3 = 90 = 1/(3) 2 = 1/ 3 A( ) = K1 /3 (1 +1 / 3)(1 + 9 *1/ 3) = 3 K = 14Kc = 4/3 = 1.336n n1 5 6-2 (1)6 2G(s) = = n s(s2 + 4s + 6) s(s2 + 2 s + 2 ) 2 = 6 = 6 =2.45, 2 =4 = 4 = 2 = 0.816n n n 2n 6K = 1 所以 , c = 1 20lgK = 0 2 / ( ) = 90 arctg c n 2 * 0.816 *1/ 2.45 = 90 arctgc 1
27、 2 / 2 1 1/ 2.452 c n = 90 arctg 2 * 0.816 *1 / 2.45 = 90 arctg 0.666 = 90 arctg 0.7995 1 1 / 2.452 0.833 = 90 38.64 = 128.64 = 180 + (c ) = 180 128.64 = 51.36L( )(dB)50403020100-10-20-30-400.01-20dB /dec0.1n1 2.4510-60dB /dec(2) 1 = 1, 2 =1/0.2=5 2 / ( ) = 90 arctg c n + arctg c arctg cc 1 2 / 2 c
28、n 1 2 = 128.64 + arctg 1 arctg 1 = 128.64 + 45 11.31 = 94.95 = 180 + (c ) = 180 94.95 = 85.051课后答案网L() (dB )50403020100-10-200.01-20dB /dec0.1n1 2.4520dB /decG c5 10-30-40-40dB /dec -60dB /dec-60dB /dec6-5 (1)G(s) = 10s(0.5s +1)(0.1s +1) = 1, 20 lg K =20lg10=20dB 1 = 1/ 0.5 = 2, 2 = 1 / 0.1 = 101 =
29、2 时, L(1 ) = 20 20(lg 2 lg1) = 20lg10 20 lg 2 = 20lg5 = 14dB2 = 10 时, L(2 ) = 14 40(lg10 lg 2) = 13.96dB所以 , 1 c 2L(1 ) = 40(lgc lg 2) = 40(lgc / 2) = 14dBc = 4.48 (c ) = 90 arctg 0.5c arctg 0.1c = 90 arctg 2.24 arctg 0.448= 90 65.94 24.13 = 180.07 = 180 + (c ) = 180 180.07 = 0.07L ( )(dB)5040302010
30、0-10-20-30-400.1-20dB /dec1 2-40dB /dec c 10-60dB /dec1002(2)G(s)Gc (s) = 10(0.33s +1)s(0.5s +1)(0.1s +1)(0.033s +1) = 1, 20 lg K =20lg10=20dB1 = 1 / 0.5 = 2, 2 = 1/ 0.33 = 3, 3 = 1 / 0.1 = 10, 4 = 1/ 0.033 = 302 = 3 时, L(1 ) L(2 ) = 40(lg2 lg 1 ) 14 L(2 ) = 40(lg 4.35 lg 2)L(2 ) = 7dBL(3 = 10) L(2
31、= 3) = 20(lg 3 lg 2 ) = 3.37dB所 以 2 c 2 3L(2 ) = 20(lgc 2 lg 2 ) = 20(lgc 2 / 3) = 7dBc 2 = 6.72 (c ) = 90 arctg 0.5c 2 arctg 0.1c 2 + arctg 0.33c 2 arctg 0.033c 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)504030 -20dB /dec20100-40dB /decc 220dB /dec G c10 -10-200.1 1 2 3 c1 30G cG100-30-40-20dB /dec -40dB /dec -60dB /dec-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