现代控制理论习题 附答案

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1、*%+&'")((!#,)1.qjf6vn−a1−a2−a31x˙(t)=100x(t)+0u(t),0100y(t)=[b1,b2,b3]x(t).Wqjh#s^-b$qjh#s^ns+a1a2a3f(s)=det[sI−A]=−1s00−1s=s3+as2+as+a.123qj-bn−1s+a1a2a31G(s)=c[sI−A]−1b=[b,b,b]−1s001230−1s0s2∗∗11=[b1,b2,b3]s∗

2、∗032s+a1s+a2s+a31∗∗0b1s2+b2s+b3=.s3+a1s2+a2s+a3[a<Æ*n{d2.Wr
<Æh#s^
−a1−a2···−an−1−an10···00Ac=01···00∈Rn×n,...............00···1000···0−an10···0−an−1Aco=01···0−an−2∈Rn×n,...............00···1−a112−a110···0−a201···0.....

3、.....n×nAo=.....∈R,−an−100···1−an00···0010···0001···0..........n×nAob=.....∈R.000···1−an−an−1−an−2···−a1$h#s^nf(s)=sn+asn−1+asn−2+···+a.12n3.qj-bn2s+3G(s)=,(s+1)(s+2)u"
,%v$2s+3G(s)=,s2+3s+2AV,%vn"#"#−a1−a21x˙(t)=x(t)+u(t)100"#"#−3−21=x(t)+u(t

4、),100y(t)=[2,3]x(t).Ay,%vn"#"#0−a21x˙(t)=x(t)+u(t)1−a10"#"#0−21=x(t)+u(t),1−30"#−T10y(t)=[b1,b2]x(t)a11"#−T10=[2,3]x(t)=[2,−3]x(t).31+ÆV,%vn"#"#−a11b1x˙(t)=x(t)+u(t)−a20b23"#"#−31b1=x(t)+u(t),−20b2y(t)=[1,0]x(t).+Æy,%vn"#"#−1"#0110b1x˙(t)=x(t)+u(t)−a2−a1a11b2"#"#−1"#0110b1=x(t)+u(t)

5、−2−331b2"#"#"#01102=x(t)+u(t)−2−3−313"#"#012=x(t)+u(t)−2−3−3y(t)=[1,0]x(t).4.qj-bn6s2+22s+18G(s)=,(s+1)(s+2)(s+3)u"
,%v$6s2+22s+18G(s)=,s3+6s2+11s+6AV,%vn−6−11−61x˙(t)=100x(t)+0u(t)0100y(t)=[6,22,18]x(t).Ay,%vn00−61x˙(t)=10−11x(t)+0u(t)01−60

6、−T100y(t)=[6,22,18]610x(t)11611−625=[6,22,18]01−6x(t)=[6,−14,36]x(t).0014+ÆV,%vn−6106x˙(t)=−1101x(t)+22u(t)−60018y(t)=[1,0]x(t).+Æy,%vn0106x˙(t)=001x(t)+−14u(t)−6−11−636y(t)=[1,0,0]x(t).5.qj-bn1G(s)=,(s+1)(s+2)uqj"7w$11G(s)=−,

7、s+1s+2T

8、"7"wn#"#−101x˙(t)=x(t)+u(t)0−21y(t)=[1,−1]x(t).6.qj-bn6s2+22s+18G(s)=,(s+1)(s+2)(s+3)uqj"7w$123G(s)=++,s+1s+2s+3T

9、"7wn−1001x˙(t)=0−20x(t)+1u(t)00−31y(t)=[1,2,3]x(t).7.qj-bn5s2+13s+9G(s)=,(s2+2s+1)(s+2)uqjw$123G(s)=++,(s+1)2s+1s+25T

10、wn