自动控制原理试题库

1、5-2已知单位负反馈系统开环传递函数，试绘制系统开环对数频率特性曲线。解：750L( ) dB(1) G(s)=s(s+5)(s+15)40-20dB/dec10201501G(s)=1 s+1)( 1s+1)5s(-20-40dB/dec515 =15( )-60dB/dec20lgK=20dB 0o1=52 =0 ( )=-90-180-90 =o ( )=-270-270(3) G(s)=10L( ) dB(2s+1)(8s+1)20-20dB/dec解： 20lgK=20 dB20lgK00.1250.5-201=0.125 2=0.5-40dB/dec( ) =0 ( )=0o0o-

2、90 ( )=-180=-180解： (5) G(s)=10L() dB1 =1s(s-1)40-20dB/dec20lgK=200dB20-40dB/dec(o-201=0)=-270 ( ) = ()=- -180o0-90-180-27010(s+0.2)1.33(5s+1)(7) G(s)= s2 (s+0.1)(s+15) s 2(10s+1)(0.67s+1)dB=L( ) dB40-40dB/dec20-60dB/dec15解： 20lgK=2.501 =0.12 =0.23 =150.1 0.21-20o-40dB/dec=0) ( )=-180(-60dB/dec = 0o(

3、)=-270-90-180-2705-4 已知系统的开环幅频率特性曲线，写出传递函数并画出对数相频特性曲线。L()dB(a)20lgK=20K=102020lgK-20dB/dec(b) 20lgK=-20K=0.1L( ) dB0120101000.1s -20 20dB/decG(s)=G(s)=(0.1s+1)10c(0.05s+1)(d) 20lgK=48K=251L( ) dB(c)K=100L() dB48-20dB/dec-20dB/dec20lgK-40dB/dec1000110 50 10000.01-40dB/dec-60dB/dec-60dB/dec251G(s)=100

4、G(s)=s(100s+1) (0.01s+1)(s+1)(0.1s+1)(0.01s+1)(c) K=100L( ) dB-20dB/dec10000.01100G(s)= s-40dB/dec(100s+1)(0.01s+1)-60dB/dec(e) 由图可得：12得：L( ) dB20lgM r =4.58 dBM r =1.7 =-20dB/dec4.58dB= 0.942 1-0100=±±0.32=0.3r=45.3122-60dB/dec n n =50K=1001-20r =10021G(s)=222T =( n ) =0.022T =0.01s(0.02s

5、) +0.01s+1)5-7 已知奈氏曲线， p 为不稳定极点个数， 为积分环节个数，试判别系统稳定性。p=0ImIm=0(b)=2(a)p=0 =0+ -1 =0=00Re系统不稳定-10 Re系统稳定(c) p=0Im =0+ Im=2p=0(d)+ =0=3=00Re-1=0-10Re系统不稳定(e)p=0系统稳定Im =1 (f)Imp=1-1 =0=0=00Re0-1Re系统稳定 =0 +系统稳定(g)p=0Im(h)p=1Im=0=1=0-1=0Re-10Re0系统稳定=0 +系统不稳定5-17 已知系统开环幅频率特性曲线 (1) 写出传递函数。 (2)利用相位裕量判断稳定性 (3

6、)将对数幅频特性向右平移十倍频程，讨论对系统性能的影响。L( ) dB解： K=10G(s)= 10 s(10s+1)(0.05s+1)1021c =110 c=180 o +( c )oo-1-1=180-90 -tg10-tg0.05=90 o -84.3 o -2.9 o = 2.8 o-20dB/dec-40dB/dec2000. 1c10 -60dB/dec ( )0-90-1805-18 已知系统结构，试绘制系统的开环对数幅频特性曲线，并计算相角稳定裕量。R(s)51C(s)-20.5s +1s(0.02s +1)解： G(s)=10L( ) dBs(0.5s+1)(0.02s+1)40-20dB/dec10200.5 c 2 1c=4.51 2 c50=180 o o-1 ×0-90 -tg0.5 4.5-20-40dB/dec-tg -10.02 ×4.5=90 o -66o-2.6 o= 21.4 o-60dB/dec已知单位负反馈系统开环传递函数，采G0(s)=500K o6-1用超前校正满足K =100, 45o。s(s+5)解： K =K v=100100G 0(s)=L( ) dBs(0.2s+1)20lgK=40dB100210.2 cc=22.4°=12.6取=5.6 om = -=45o12