自控第三章习题

3-1设温度计需要在一分钟内指示出响应值的98%，并且假设温度计为一阶系统，求时间常数T。如果将温度计放在澡盆内，澡盆的温度以10oC/min的速度线性变化，求温度计的误差。第三章习题课(3-1)解:

c(t)=c(∞)98%t=4T=1min

r(t)=10t

e(t)=r(t)-c(t)

c(t)=10(t-T+e)-t/T=10(T-e)-t/Tess=lim

t→∞

e(t)=10T=2.5T=0.253-2电路如图，设系统初始状态为领．第三章习题课(3-2)解:求系统的单位阶跃响应,及uc(t1)=8

时的t1值．R1Cs+1R1/R0G(s)=

R0=20kΩ

R1=200kΩ

C=2.5μF

uc(t)=K(1–etT-)KTs

+

1=T=R1C=0.5

K=R1/R0=10

=10(1–e-2t)8=10(1–e-2t)0.8=1–e-2te-2t=0.2

t=0.8

-∞++R1R0Curucg(t)=e-t/TTKt1=0.8

=4

(2)求系统的单位脉冲响应，单位斜坡响应,及单位抛物响应在t1时刻的值．第三章习题课(3-2)解:uc(t)=K(t-T+Te-t/T)=4

R(s)=1s2R(s)=1R(s)=1s3T2=K(ss+1/T+Ts2-1s3-T2)=1.2

1s3KTs

+

1Uc(s)=-0.5t+0.25-0.25e-2t)12t2uc(t)=10(3-3已知单位负反馈系统的开环传递函数，求系统的单位阶跃响应。4s(s+5)G(s)=第三章习题课(3-3)解:=s2+5s+4

C(s)R(s)4s(s+1)(s+4)C(s)=4R(s)=s11/3s+41s+=4/3s+1-13c(t)=1+-4t-t43-ee3-4已知单位负反馈系统的开环传递函数，求系统的上升时间tr、峰值时间tp、超调量σ%和调整时间ts。1s(s+1)G(s)=第三章习题课(3-4)解:=s2+s+1

C(s)R(s)12

=1ωn2ωnζ=1ζ=0.5=1ωn=0.866dω

=ωn2ζ1-

=60o-1ζ=tg

β21-

ζtr=dωπβ-

=3.14-3.14/30.866=2.42tp=dωπ3.140.866

==3.63σ%=100%e-ζζπ1-2=16%-1.8ets=ζ3ωn=6ts=ζ4ωn=83-6已知系统的单位阶跃响应:c(t)=1+0.2e-60t-10t-1.2e(1)求系统的闭环传递函数。(2)求系统的阻尼比和无阻尼振荡频率。第三章习题课(3-6)解:0.2s+601s+C(s)=1.2s+10-s(s+60)(s+10)=600=s2+70s+600

C(s)R(s)600R(s)=s12

=600ωn2ωnζ=70ζ=1.43=24.5ωn3-7设二阶系统的单位阶跃响应曲线如图，系统的为单位反馈，求系统的传递函数。tc(t)010.11.3解:第三章习题课(3-7)tp==0.121-

ζπωn=0.3e-ζζπ1-2eζζπ1-2=3.3ωn2ζ1-

3.140.1==31.421-

ζπ/ζ=ln3.3=1.1921-

ζπ)2/ζ(=1.429.862ζ=1.42-1.422ζζ=0.35=33.4ωnωnωnζs(s+2)G(s)=21115.6s(s+22.7)=3-8已知单位负反馈系统的开环传递函数，求系统K、T值以满足动态指标：σ%≤30%，ts≤0.3(5%)。Ks(Ts+1)G(s)=第三章习题课(3-8)解:=Ts2+s+K

C(s)R(s)Ks2+=T1s+TKTKT12ωnζ=ts=ζ3ωn≤0.3ωnζ≥10T≤0.05≤0.3e-ζζπ1-2ζ≥0.35≥28.6ωn2

=ωnTK=817.96K≥40.93-11已知闭环系统的特征方程式，试用劳斯判据判断系统的稳定性。第三章习题课(3-11)(1)s3+20s2+9s+100=0解：劳斯表如下：s1

s0

s3

s2

19201004

100系统稳定。(3)s4+8s3+18s2+16s+5=01185s4

s3

816劳斯表如下：s2

165s1

216

16

s0

5系统稳定。3-12已知单位负反馈系统的开环传递函数，试确定系统稳定时K值范围。第三章习题课(3-12)G(s)=s(s+1)(0.5s2+s+1)K(0.5s+1)解:0.5s4+1.5s3+2s2+s++0.5Ks+K=00.52Ks4

s3

1.51+0.5Ks2

b31

b31=

1.5*2-0,5(1+0.5K)

1.5

3-0,5-0.25K>0

0.25K<2.5

=1.67-0.167Ks1

K

b41

b41=

(1.67-0.167K)(1+0.5K)-1.5K1.67-0.167K3-13已知系统结构如图，试确定系统稳定时τ值范围。第三章习题课(3-13)-１s1+τs-10s(s+1)R(s)C(s)解:G(s)=s2+s+1010(1+sτ)s1=s(s2+s+1010(s+1)sτ)Φ(s)=s3+s2+1010(s+1)s2+10s+10τ110s3

s2

(1+10

τ)10s1

b31

s0

10

b31=

10(1+10

τ)-101+10

τ>0τ>03-14已知系统结构如图，试确定系统稳定时τ值范围。第三章习题课(3-14)R(s)-τs+1s10s(s+1)C(s)解:G(s)=s2(s+1)10(τs+1)Φ(s)=s3+s2+10s+10τ10(τs+1)s3

s2

110

110τs1

b31

s0

10

b31=

10

τ-101

>0τ>1r(t)=I(t)+2t+t23-16已知单位反馈系统的开环传递函数，试求Kp、Kv和Ka

．并求稳态误差ess．第三章习题课(3-16)解：(0.1s+1)(0.2s+)(1)G(s)=20Kp=20

υ=0ess1=R01+Ks2R(s)=1s2+s32+=121Kυ=0ess2=∞Ka=0ess3=∞

ess=∞υ=1Kυ=10Ka=0ess3=∞

ess=∞s(s+2)(s+10)(2)G(s)=200s(0.5s+1)(0.1s+1)=10Kp=∞

ess1=0ess2=K2=210s2(s2+4s+10)(3)G(s)=10(2s+1)υ=2Kp=∞

ess1=0s2(0.1s2+0.4s+1)=(2s+1)Kυ=∞

ess2=0Ka=1

ess3=2

ess=23-17已知系统结构如图。第三章习题课(3-17)解:-R(s)-K1τs１s2C(s)单位阶跃输入:G(s)=s2+K1sτK1Φ(s)=s2+K1s+K1τK1确定K1和τ值

。

σ%=20%ts

=1.8(5%)2ωnζ=K1τ2

=K1ωn=0.2e-ζζπ1-2ts=ζ3ωn=1.8ζ=0.45ωn31.8*0.45==3.72

ωnK1==13.7τ=0.24(2)求系统的稳态误差：r(t)=I(t),t,12t2

解:G(s)=s2+K1sτK1=s(s+1)τK11τ1υ=1Kp=∞

ess1=0R(s)=1sR(s)=s21Kυ=Kess2=τ=0.24R(s)=s31Ka=0ess3=∞第三章习题课(3-17)3-18已知系统结构如图。为使ζ=0.7时单位斜坡输入的稳态误差ess=0.25第三章习题课(3-18)解:确定K和τ值

。-τs-Ks(s+2)R(s)C(s)G(s)=s2+2s+KsτK=s(s+1)τ2+K1τ2+KKΦ(s)=s2+(2+K)s+KτK2ωnζ=2+Kτ2

=Kωn

=2*0.7K

ess=

2+K

τK

=0.25=

0.25K-2τK

K=31.6τ=0.186+R(s)G(s)F(s)C(s)+-D1(s)D2(s)E(s)3-19系统结构如图。第三章习题课(3-19)解:

r(t)=d1(t)=d2(t)=I(t)求r(t)作下的稳态误差．essr=lims·1+G(s)F(s)s→0s1=

1+G(0)F(0)1(2)求d1(t)和d2(t)同时作用下的稳态误差．essd=lims-F(s)1+G(s)F(s)s→0-11+G(s)F(s)+s1[]Ed(s)=-G2(s