线性系统理论习题答案ans1to6

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(完整word版)《线性系统理论》试卷及答案

R C 2《线性系统理论》试卷及答案1、（20分）如图所示RLC 网络,若e(t ）为系统输入变量r （t)，电阻R 2两端的电压为输出量y(t),选定状态变量为 x 1（t)=v 1（t ），x 2(t ）=v 2(t)，x 3（t)=i （t)要求列写出系统的状态空间描述。

2、(15分)求出下面的输入输出描述的一个状态空间描述。

y （4)+4y （3）+3y （2）+7y （1）+3y=u (3）+ 2u （1）+ 3u3、（15分）计算下列线性系统的传递函数。

[]210X 13101X y -⎡⎤⎡⎤=+⎢⎥⎢⎥-⎣⎦⎣⎦=4、(10分）分析下列系统的能控性.0111X X u a b •⎡⎤⎡⎤=+⎢⎥⎢⎥-⎣⎦⎣⎦5、（10分）分析下列系统的能观性。

[]1110a X X y Xb •⎡⎤==-⎢⎥⎣⎦6、（15分)判断下列系统的原点平衡状态x e 是否大范围渐近稳定。

12221123x x x x x x==--7、（15分）已知系统的状态方程为221012000401X X u •--⎡⎤⎡⎤⎢⎥⎢⎥=-+⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦试确定一个状态反馈阵K,使闭环极点配置为λ1*=—2、λ2*=-3、λ3*=—4.答案：1、（20分）如图所示RLC 网络，若e （t ）为系统输入变量r （t ）,电阻R 2两端的电压为输出量y （t ），选定状态变量为 x 1(t)=v 1（t)，x 2(t ）=v 2（t ）,x 3（t)=i （t ）要求列写出系统的状态空间描述。

2、（15分）求出下面的输入输出描述的一个状态空间描述。

列出向量表示形式解出解出解出r x x x L R x x x rx LR x x x xx x C R x x x C xC x r x R x L L LL⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+--=-=+=+==++1321113211311132122222112211333113000xy x xLy （4）+4y (3）+3y （2)+7y (1）+3y=u （3）+ 2u （1)+ 3u[]得出了状态空间表达式列出向量表示形式，就求导，有选取状态变量令有令⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=++=⎪⎪⎩⎪⎪⎨⎧+----=========⎩⎨⎧++==++++++++=++++++===43211025233375y ~y ~x y ~x y ~...y ~x y ~x y ~3y ~2y ~y ~3y ~7y ~3y ~4y ~u 3734p 1y ~3734p 32p y d/dtp 4214321(4)43(2)22(1)1(3)4(1)21(1)(3)(1)(2)(3)(4)2342343x x x x x x x y u x x x x x x x x y u p p p u p p p p(完整word 版)《线性系统理论》试卷及答案3、(15分）计算下列线性系统的传递函数.[]Xy u X 10103112X =⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡--=[][][]计算得出传递函数⎥⎦⎤⎢⎣⎡⎥⎦⎤⎢⎣⎡-------=-=⎥⎦⎤⎢⎣⎡-------=⎥⎦⎤⎢⎣⎡--=--==⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡--==⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡--=----1021131)3)(2(110)()(21131)3)(2(13112)()()(1010311210103112X 1111s s s s B A Is C s G s s s s s s A Is BA Is C s G CB A Xy u X(完整word 版)《线性系统理论》试卷及答案4、（10分）分析下列系统的能控性。

线性系统分析_习题答案

线性系统分析_(吴大正_第四版)习题答案(总184页)-CAL-FENGHAI.-(YICAI)-Company One1-CAL-本页仅作为文档封面，使用请直接删除专业课习题解析课程西安电子科技大学844信号与系统2专业课习题解析课程第1讲第一章信号与系统（一）3专业课习题解析课程第2讲第一章信号与系统（二）451-1画出下列各信号的波形【式中)()(t t t r ε=】为斜升函数。

（2）∞<<-∞=-t et f t,)( （3）)()sin()(t t t f επ=（4）)(sin )(t t f ε= （5）)(sin )(t r t f = （7）)(2)(k t f kε= （10）)(])1(1[)(k k f kε-+=解：各信号波形为（2）∞<<-∞=-t et f t,)(（3）)()sin()(t t t f επ=6（4）)(sin )(t t f ε=（5）)(sin )(t r t f =7（7）)(2)(k t f k ε=（10）)(])1(1[)(k k f k ε-+=81-2 画出下列各信号的波形[式中)()(t t t r ε=为斜升函数]。

（1）)2()1(3)1(2)(-+--+=t t t t f εεε （2）)2()1(2)()(-+--=t r t r t r t f （5）)2()2()(t t r t f -=ε （8）)]5()([)(--=k k k k f εε （11）)]7()()[6sin()(--=k k k k f εεπ （12）)]()3([2)(k k k f k---=εε解：各信号波形为（1）)2()1(3)1(2)(-+--+=t t t t f εεε9（2）)2()1(2)()(-+--=t r t r t r t f（5）)2()2()(t t r t f -=ε10（8）)]5()([)(--=k k k k f εε（11）)]7()()[6sin()(--=k k k k f εεπ11（12）)]()3([2)(k k k f k ---=εε1-3 写出图1-3所示各波形的表达式。

山东大学2011级线性系统理论试题答案

x1 y 0 0 1 x2 x 3
(s 1)(s 2)(s 3) 0 1 1, 2 2, 3 3
k k k s4 1 2 3 k1 1.5, k2 2, k3 0.5 ( s 1)( s 2)( s 3) s 1 s 2 s 3
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故 {
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x1 y 1 1 1 x2 x 3
①
由 1
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② 由 2
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线性系统理论第一章(习题)

若 li 是 A 的特征值，试证 [1 li li 2 li n -1 ]T 是属于 li 的特征向量。 1—2 若 li 是 A 的一个特征值，试证 f (li ) 是矩阵函数 f (A) 的一个特征值。 1—3 试求下列矩阵的特征多项式和最小多项式
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y =
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29
u(t) 1 2 t 1 2 (b) t
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F(n, m ) 为何？
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n -1
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1 ù ú s+3ú 5s + 1 úú s + 2 úû
的实现，并画出其模拟图。 1—25 设{ A , B , C , D }和{ A , B , C , D }是两个线性时不变系统，其维数不一定相同。证明当且仅当

第一篇线性系统理论习题答案

⎡ s +1 ⎢s2 + s +1 ⎢ −1 = [1 0 1]⎢ 2 ⎢s + s +1 ⎢ 0 ⎢ ⎣
9－7 设有三维状态方程
⎡0 ⎤ ⎢1 ⎥ ⎢ ⎥ ⎢ ⎣1 ⎥ ⎦
1 s + s +1 s 2 s + s +1
2
0
⎤ 0 ⎥ ⎥ ⎡0 ⎤ s 2 + 2 s 1⎥ = 3 0 ⎥ ⎢ ⎢ s −1 ⎥ ⎥ 1 ⎥ ⎢ ⎣1⎥ ⎦ s − 1⎥ ⎦
⎡ R M ⎤ ⎡ R −1 ∵⎢ ⎥×⎢ ⎣0 T ⎦ ⎣ 0
− R −1 MT −1 ⎤ ⎡ R −1 ⎥=⎢ T− ⎦ ⎣ 0
⎡R M ⎤ ∴⎢ ⎥ ⎣0 T ⎦
9－10 解
−1
⎡ R −1 =⎢ ⎣ 0
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−1
对可控标准形 A 和 b ，计算 ( sI − A) b
+
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& 2 = x1 + y = x1 − C 2 x
写成矩阵形式为
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图 9－1 RLC 网络
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x1 ， x 2 有下列关系存在 x1 = x1 + x 2 x 2 = − x1 − 2 x 2
试求系统在 x 坐标中的状态方程。解 ①
&1 = x & = x2 x &2 = & & = −2 x1 − 3 x 2 + u x x

《线性系统理论》作业参考答案

x 11 e t x 21 , 21 0 , x
x11 ( t 0 ) 1 x 21 ( t 0 ) 0
，
x 12 e t x 22 , 22 0 , x
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解得
x12 e t e t 0 x11 1 ， x 21 0 x 21 1 1 (t ) x 0 e
( sI A )
1
s ( s 1) 0 2 det( sI A ) s ( s 1) 0 adj ( sI A ) 1
s 1 ( s 1) 0
2
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2
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。
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A 1 I 0 , ( A 1 I ) 2 0 , ( A 1 I ) 3 0 , ( A 1 I ) 4 0
At
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e 1 1 0
t
t t 1 e te t e 1 。 t e
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A det C B IA det D 0 BD A I D C
c
k 0
k
A
k
设 x 是属于 i 的一个非零特征向量，故
A x i x
.
2 2 因此 A x A Ax A i x i Ax i i x i x .

线性系统分析_(吴大正_第四版)习题答案

专业课习题解析课程西安电子科技大学844信号与系统可编辑word,供参考版！专业课习题解析课程第1讲第一章信号与系统（一）可编辑word,供参考版！专业课习题解析课程第2讲第一章信号与系统（二）可编辑word,供参考版！可编辑word,供参考版！1-1画出下列各信号的波形【式中)()(t t t r ε=】为斜升函数。

（2）∞<<-∞=-t et f t,)( （3）)()sin()(t t t f επ=（4）)(sin )(t t f ε= （5）)(sin )(t r t f = （7）)(2)(k t f kε= （10）)(])1(1[)(k k f kε-+=解：各信号波形为（2）∞<<-∞=-t et f t,)(（3）)()sin()(t t t f επ=（4）)fε=t)(sin(t（5）)tf=r(t)(sin可编辑word,供参考版！（7）)t(kf kε=)(2（10）)f kεk-=(k+(])1()1[可编辑word,供参考版！可编辑word,供参考版！1-2 画出下列各信号的波形[式中)()(t t t r ε=为斜升函数]。

（1）)2()1(3)1(2)(-+--+=t t t t f εεε （2）)2()1(2)()(-+--=t r t r t r t f （5）)2()2()(t t r t f -=ε （8）)]5()([)(--=k k k k f εε （11）)]7()()[6sin()(--=k k k k f εεπ（12）)]()3([2)(k k k f k---=εε 解：各信号波形为（1）)2()1(3)1(2)(-+--+=t t t t f εεε（2）)2()1(2)()(-+--=t rt rt rtf（5）)2()2()(ttrtf-=ε可编辑word,供参考版！可编辑word,供参考版！（8）)]5()([)(--=k k k k f εε（11）)]7()()[6sin()(--=k k k k f εεπ（12）)]()3([2)(kkkf k---=εε可编辑word,供参考版！1-3 写出图1-3所示各波形的表达式。

线性系统分析_(吴大正_第四版)习题答案

专业课习题解析课程/西安电子科技大学844信号与系统？专业课习题解析课程第1讲:第一章信号与系统（一）专业课习题解析课程{第2讲第一章信号与系统（二）1-1画出下列各信号的波形【式中)()(t t t r ε=】为斜升函数。

（2）∞<<-∞=-t e t f t,)( （3）)()sin()(t t t f επ=、（4）)(sin )(t t f ε= （5）)(sin )(t r t f =（7）)(2)(k t f kε= （10）)(])1(1[)(k k f kε-+=解：各信号波形为（2）∞<<-∞=-t et f t,)(（3）)()sin()(t t t f επ=（4）)fε=t)(sin(t（5）)tf=r(t)(sin}（7）)t(kf kε=)(2（10）)f kεk-=(k+(])1(1[)1-2 画出下列各信号的波形[式中)()(t t t r ε=为斜升函数]。

（1）)2()1(3)1(2)(-+--+=t t t t f εεε （2）)2()1(2)()(-+--=t r t r t r t f （5）)2()2()(t t r t f -=ε （8）)]5()([)(--=k k k k f εε （11）)]7()()[6sin()(--=k k k k f εεπ（12）)]()3([2)(k k k f k---=εε ；解：各信号波形为（1）)2()1(3)1(2)(-+--+=t t t t f εεε（2）)2()1(2)()(-+--=t rt rt rtf（5）)2()2()(ttrtf-=ε（8）)]5()([)(--=k k k k f εε（11）)]7()()[6sin()(--=k k k k f εεπ《（12）)]()3([2)(kkkf k---=εε1-3 写出图1-3所示各波形的表达式。

1-4 写出图1-4所示各序列的闭合形式表达式。

线性系统课后答案第4章

PROBLEMS OF CHAPTER 44.1 An oscillation can be generated by 一个振荡器可由下式描述：X X ⎥⎦⎤⎢⎣⎡-=0110 试证其解为：Show that its solution is )0(cos sin sin cos )(X t t t t t X ⎥⎦⎤⎢⎣⎡-=Proof: )0()0()(0110X eX e t X t At ⎥⎦⎤⎢⎣⎡-== ,the eigenvalues of A are j,-j;Let λββλ10)(+=h .If te h λλ=)(,then on the spectrum of A,thentj t e j j h t j t e j j h jt jt sin cos )(sin cos )(1010-==-=-+==+=-ββββ thenttsin cos 10==ββso ⎥⎦⎤⎢⎣⎡-=⎥⎦⎤⎢⎣⎡-+⎥⎦⎤⎢⎣⎡=+=t t t t t t A I A h cos sin sin cos 0110sin 1001cos )(10ββ )0(cos sin sin cos )0()0()(0110X t t t t X eX e t X t At ⎥⎦⎤⎢⎣⎡-===⎥⎦⎤⎢⎣⎡- 4.2 Use two different methods to find the unit-step response of 用两种方法求下面系统的单位阶跃响应：U X X ⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡--=112210[]X Y 32=Answer: assuming the initial state is zero state.method1:we use (3.20) to compute⎥⎦⎤⎢⎣⎡-+++=⎥⎦⎤⎢⎣⎡+-=---s s s s s s A sI 212221221)(211then t At e t t tt tt A sI L e ---⎥⎦⎤⎢⎣⎡--+=-=sin cos sin 2sin sin cos ))((11 and )22(5)22(5)()()(221++=++=-=-s s s s s s s BU A sI C s Y then t e t y tsin 5)(-= for t>=0method2:ttt tt At tt A tt A te t e t e t e t e C Be e CA B d e C d Bu e C t y --------=⎥⎦⎤⎢⎣⎡---+⎥⎦⎤⎢⎣⎡--=-===⎰⎰sin 51sin 3cos 1sin 2cos 015.01)()()(010)(0)(τττττfor t>=04.3 Discretize the state equation in Problem 4.2 for T=1 and T=π.离散化习题4.3中的状态方程，T 分别取1和π Answer:][][][][][]1[0k DU k CX k Y k BU d e k X e k X TA AT +=⎪⎭⎫ ⎝⎛+=+⎰ααFor T=1,use matlab: [ab,bd]=c2d(a,b,1) ab =0.5083 0.3096 -0.6191 -0.1108 bd =1.0471 -0.1821[]][32][][1821.00471.1][1108.06191.03096.05083.0]1[k X k Y k U k X k X =⎥⎦⎤⎢⎣⎡-+⎥⎦⎤⎢⎣⎡--=+for T=π,use matlab:[ab,bd]=c2d(a,b,3.1415926) ab =-0.0432 0.0000 -0.0000 -0.0432 bd =1.5648 -1.0432[]][32][][0432.15648.1][0432.0000432.0]1[k X k Y k U k X k X =⎥⎦⎤⎢⎣⎡-+⎥⎦⎤⎢⎣⎡--=+ 4.4 Find the companion-form and modal-form equivalent equations of 求系统的等价友形和模式规范形。

中国科学技术大学自动化专业《线性系统理论和设计》习题1-6章习题答案

1.7 证明：())()det(det )det(det )(det )det()det()(1111λλλλλλλA B A I T A I T T A I T AT T I B I AT T B B A ∆=-=⋅-⋅-=-=-=∆⇒=----相似，与设= 又因为特征值为特征方程()0λ∆=的根，故特征值也相同。

1.11 解：可以参照课本P18的例题1.12（1）,3,2,1)3)(2)(1()(,300020104132111===⇒---=∆⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=λλλλλλA A ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡==Λ∴⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⇒⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⇒=--3211000105411050140010)(1113211Q A Q Q q q q q A I ，，由λ（4）,2,1,1)2)(1)(1()(4344111432124==-==⇒-+-=∆⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--λλλλλλλA ，1241243111111()0,111122,()012,12,4822 2.P I A q q q u I A q q u λλλλλξλλη⎡⎤⎡⎤⎢⎥⎢⎥-⎢⎥⎢⎥==--=⇒==⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦==-=⎡⎤⎢⎥⎢⎥==<=⎢⎥⎢⎥⎣⎦===对于，，由对于的特征值，其代数重数由计算其对应的特征向量计算出一个特征向量，即几何重数个数小于代数重数，即标准型中存在一个对应的约当块，约当块的阶数即的指数可以利用[]4443434123414418 1.682,()001110111121,,44114412121181211212q I A q q q c q q Q q q q q Q A Q λλ-=-=⇒⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥-⎢⎥⎢⎥⎢⎥=⋅-=∴==⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦⎡⎤⎢⎥-⎢⎥∴Λ==⎢⎥⎢⎥⎣⎦的式计算的广义特征向量由取1.12 证明：12n 222112n n 1n-1n-112n 21n 121n 1221n n 1n-3n-3221n 21n-22n-2n-2221n n 1111(1110()()0()()(0()()λλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλλ-⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦⎡⎤--⎢⎥--⎢⎥--⎢⎥==⎢⎥--⎢⎥⎢⎥--⎣⎦后一行减去前一行的倍）n-221n n 123n 2131n 1n-2n-2n-223n j i 1i j n)()111()()()()()λλλλλλλλλλλλλλλλλλ≤<≤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎡⎤⎢⎥⎢⎥=---=⎢⎥⎢⎥⎣⎦=-∏同理2.6 解：(d) 令24231211y x y x yx y x ====，，，，则状态空间方程为： u m m k m k m k mk ⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎣⎡+⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡--=0010020100000200112211x xx y ⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=010*******y y (e) 令yx y x ==21，，则状态空间方程为： u e e t t ⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡--=-10102x x[]x y 01= 2.7 解：(c)非线性方程： ⎩⎨⎧==21221u-x xx x[]x y 01= (d) 设⎪⎩⎪⎨⎧+=⇒=+⋅++-=⇒=+⋅+ux sx x u)(x s u x x sx x s )x (u 333221122121112，则状态空间方程可为：u ⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡-=310312x x[]x y 01= 另法：先求出传递函数2323s G(s)s s +=+-，按2.6（b ）方法求解。

线性系统理论习题答案ans1to6

-1 l 0 M
0 -1 l M
an -1 an - 2 -1 l l 0 0 M M
= l2 0
L L L O L 0 -1 l M
an - 2 an - 3 an - 4 = l + a1ln -1 + L + an
n
l -1 0 L 0 l -1 L 0 0 =l 0 l L 0 0 + (-1) n +1 an (-1) n -1 M M M M O M l + a1 an -1 an - 2 an - 3 L l + a1 L 0 0 L 0 + l (-1) n an -1 (-1) n - 2 + an = L L O M L l + a1
而Q AB = BA \ e
A+ B
\ F (t , t0 ) = e B (t - t 0 )
= e A × eB = eB × e A
e - At e( A + B )(t - t 0 )e At 0 = e - At e A( t - t 0 ) e B ( t - t 0 ) e At 0 = e - At e At e - At 0 e B (t -t 0 )e At 0 = e - At 0 e B ( t - t 0 ) e At 0 = e - At 0 e At 0 e B (t - t 0 ) = e B ( t - t 0 ) = F (t , t0 )
1-6 证明：由 A Î R p ´ q , B Î R q ´ p 得令C = ê
éA ëIq
Ipù éB ,D = ê ú 0û ëI p
Iq ù é I p + AB 0 ù é BA + I q ，则 CD = ê , DC = ê ú ú - Aû Iq û ë B ë 0

(word完整版)线性系统理论考试试卷A答案

《线性系统理论基础》期末考试试卷（A ）北京工业大学电控学院日期：2011年6月17日姓名：___________ 学号：___________ 得分: ___________一、（20分)按如下要求建立系统的状态空间模型（1）已知系统的微分方程描述为 326y y y y u +-+=,写出其状态空间模型.（2）已知系统由图1所示的基本模块构成，写出其状态空间模型。

图 1解:(1）0121,2, 3.6a a a b ==-==010000101236[100]x x u y x⎡⎤⎡⎤⎢⎥⎢⎥=+⎢⎥⎢⎥⎢⎥⎢⎥--⎣⎦⎣⎦= (2）111123113222312223123331231232222234443102202114430[110]x x u u u u x x x x ux x u u x x x x x u y x x x x uy x x x x x xx x u y x=+==-=-+⎧⎧⎪⎪=-+=+⇒=--+=+⎨⎨⎪⎪=-+=+=+-⎩⎩-⎡⎤⎡⎤⎢⎥⎢⎥⇒=--+⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦=二、（20分）已知系统的状态空间模型为010341[20]x x u y x⎡⎤⎡⎤=+⎢⎥⎢⎥--⎣⎦⎣⎦= （1）画出系统的结构框图.（2）将该系统化为约当(或对角线)标准型.（3)计算标准型系统在t u e -=激励下的零初态响应和输出响应。

解：（1)因为 122121342x x x x x u y x=⎧⎪=--+⎨⎪=⎩所以系统的结构框图如下：2x(2）2121(4)343(1)(3)3413I A λλλλλλλλλλλ--==++=++=+++=-=-所以：交换矩阵:1319P --⎡⎤=⎢⎥⎣⎦ 131931*********P -⎡⎤--⎢⎥⎡⎤=-=⎢⎥⎢⎥--⎣⎦⎢⎥⎢⎥⎣⎦ 11003P AP --⎡⎤=⎢⎥-⎣⎦11216P B -⎡⎤-⎢⎥=⎢⎥⎢⎥⎢⎥⎣⎦[]31CP =--所以约旦标准型为: 11021036[31]x x u y x⎡⎤-⎢⎥-⎡⎤=+⎢⎥⎢⎥-⎣⎦⎢⎥⎢⎥⎣⎦=-- （3）0330233121610()030()100210061002110(1)026121()12t tt t tt t t t t t tx t e e d e e e e d t e e e te e e ee d τττττττττ⎡⎤⎢⎥⎢⎥⎣⎦--------------⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦⎡⎤⎡⎤-⎢⎥⎡⎤⎢⎥=⎢⎥⎢⎥⎢⎥-⎢⎥⎣⎦⎢⎥⎣⎦⎢⎥⎣⎦⎡⎤-⎡⎤⎢⎥⎡⎤⎢⎥=⎢⎥⎢⎥⎢⎥--⎢⎥⎣⎦⎣⎦⎢⎥⎣⎦⎡⎤-⎢⎥=⎢⎥⎢-⎢⎣⎦⎰⎰⎰⎥⎥ 331()()212t t t y t te e e ---=--三、(20分）给定系统状态空间模型如下[]210102000031102x x u y x-⎡⎤⎡⎤⎢⎥⎢⎥=-+⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦= （1）判断该系统的能控性，写出判断过程。

线性系统课后答案第3章

线性系统课后答案第3章3.1consider Fig3.1 ,what is the representation of the vector with respect to thebasis ? What is the representation of with respect ro ?图3.1中,向量关于的表⽰是什么? 关于的表⽰有是什么?If we draw from two lines in parallel with and , they tutersect at and as shown, thus the representation of with respect to the basis is , this can be verifiedfromTo find the representation of with respect to ,we draw from two lines inparallel with and , they intersect at -2 and , thus the representation of withrespect to , is , this can be verified from3.2 what are the 1-morm ,2-norm , and infinite-norm of the vectors , 问向量的1-范数,2-范数是什么?3.3 find two orthonormal vectors that span the same space as the two vectors in problem 3题3.2中的两个向量张成同⼀空间的两个标准正交向量.Schmidt orthonormalization praedure ,The two orthomormal vectors areIn fact , the vectors and are orthogonal because so we can only normalize thetwo vectors3.4 comsider an matrix A with , if all colums of A are orthonormal , then ,what can you say abort ? ⼀个阶矩阵A(), 如果A的所有列都是标准正交的,则问是怎么样?Let , if all colums of A are orthomormal , that isthenif A is a symmetric square matrix , that is to say , n=m for everythen3.5 find the ranks and mullities of the following matrices 求下列矩阵的秩和化零度3.6 Find bases of the range spaces of the matrices in problem 3.5 求题3.5中矩阵值域空间的基the last two columns of are linearly independent , so the set can be used as abasis of the range space of , all columns of are linearly independent ,so the setcan be used as a basis of the range spare oflet ,where denotes of are linearly independent , the third colums can beexpressed as , so the set can be used as a basis of the range space of3.7 consider the linear algebraic equation it has three equations and two unknowns does a solution exist in the equation ? is the solution unique ? doesa solution exist if ?线性代数⽅程有三个⽅程两个未知数, 问⽅程是否有解?若有解是否唯⼀?若⽅程是否有解?Let clearly and are linearly independent , so rank(A)=2 , is the sum of and,that is rank([A ])=2, rank(A)= rank([A ]) so a solution exists in A=Nullity(A)=2- rank(A)=0The solution is unique if , then rank([A ])=3 rank(A), that is to say theredoesn’t exist a solution in A=3.8 find the general solution of how many parameters do you have ?求⽅程的同解,同解中⽤了⼏个参数?Let we can readily obtain rank(A)= rank([A ])=3 so this lies in the rangespace of A and is a solution Nullity(A)=4-3=1 that means the dimension ofthe null space of A is 1 , the number of parameters in the general solution willbe 1 , A basis of the null space of A is thus the general solution of A= can be expressed an for any realis the only parameter3.9 find the solution in example 3.3 that has the smallest Euclidean norm 求例3中具有最⼩欧⽒范数的解,the general solution in example 3.3 is for any real andthe Euclidean norm of ishas the smallest Euclidean norm ,3.10 find the solution in problem 3.8 that has the smallest Euclidean norm 求题3.8中欧⽒范数最⼩的解,for any realthe Euclidean norm of is has the smallest Euclidean norm3.11 consider the equationwhere A is an matrix and is an column vector ,under what conditions on A andexist to meet the equation for any ?令A是的矩阵, 是的列向量,问在A和满⾜什么条件时,存在,对所有的,它们都满⾜⽅程write the equation in this formwhere is an matrix and is an column vector , from the equation we can see , exist to meet t equation for any ,if and only if under this condition , there will exist to meet the equation f 3.12 given what are the representations of A with respect to the basis and the basis , respectively? 给定请问A 关于和基的表⽰分别是什么?we have thus the representation of A with respect to the basis isthus the representation of A with respect to the basis is3.13 find Jordan-form representations of the following matrices写出下列矩阵的jordan 型表⽰:the characteristic polynomial of is thus the eigenvelues of are 1 ,2 , 3 , they are all distinct Jordan-form representation of will be diagonal .the eigenvectors associated with ,respectively can be any nonzero solution ofthus the jordan-form representation of with respect to isthe characteristic polynomial of is has eigenvalues the eigenvectorsassociated with are , respectively the we havethe characteristic polynomial of is theus the eigenvalues of are 1 ,1 and 2 , the eigenvalue 1 has multiplicity 2 , and nullity the has two tinearly independent eigenvectors associated with 1 , thus we havethe characteristic polynomial of is clearly has lnly one distinct eigenvalue 0with multiplicity 3 , Nullity(-0I)=3-2=1 , thus has only one independenteigenvector associated with 0 , we can compute the generalized , eigenvectorsof from equations below , then the representation of with respect to the basisis3.14 consider the companion-form matrix show that its characterisicpolynomial is given byshow also that if is an eigenvalue of A or a solution of then is an eigenvectorof A associated with证明友矩阵A的特征多项式并且,如果是的⼀个特征值, 的⼀个解, 那么向量是A 关于的⼀个特征向量proof:if is an eigenvalue of A , that is to say , then we havethat is to say is an eigenvetor oa A associated with3.15 show that the vandermonde determinant equals , thus we concludethat the matrix is nonsingular or equivalently , the eigenvectors are linearly independent if all eigenvalues are distinct ,证明vandermonde ⾏列式为, 因此如果所有的特征值都互不相同则该矩阵⾮奇异, 或者等价地说, 所有特征向量线性⽆关,proof: let a, b, c and d be the eigenvalues of a matrix A , and they are distinct Assuming the matrix is singular , that is abcd=0 , let a=0 , then we have and from vandermonde determinantso we can see ,that is to say are not distinetthis implies the assumption is not true , that is , the matrix is nonsingular let be the eigenvectors of A ,3.16 show that the companion-form matrix in problem 3.14 is nonsingular if and only if , under this assumption , show that its inverse equals 证明题3.14 中的友矩阵⾮奇异当且仅当, 且矩阵的逆为proof: as we know , the chacteristic polynomial is so let , we haveA is nonsingular if and only if3.17 consider with and T>0 show that is an generalized eigenvector of grade 3 and the three columns of constitute a chain of generalized eigenvectors of length 3 , venty矩阵A 中,T>0 , 证明是3 级⼴义特征向量, 并且矩阵Q 的3 列组成长度是3的⼴义特征向量链,验证Proof : clearly A has only one distinct eigenvalue with multiplicity 3 these two equation imply that is a generalized eigenvctorof grade 3 ,andthat is the three columns of Q consititute a chain of generalized eigenvectors of length 33.18 Find the characteristic polynomials and the minimal polynomials of the following matrices求下列矩阵的特征多项式和最⼩多项式,3.19 show that if is an eigenvalue of A with eigenvector then is an eigenvalue of with the same eigenvector证明如果是A 的关于的特征向量,那么是的特征值, 是关于的特征向量, proof let A be an matrix , use theorem 3.5 for any function we can definewhich equals on the spectrum of Aif is an eigenvalue of A , then we have andwhich implies that is an eigenvalue of with the same eigenvector3.20 show that an matrix has the property for if and only if A has eigenvalues 0 with multiplicity n and index m of less , such a matrix is called a nilpotent matrix证明的矩阵在当且仅当A的n 重0特征值指数不⼤于m ,这样的矩阵被称为归零矩阵,proof : if A has eigenvalues 0 with multiplicity n and index M or less then the Jordan-form representation of A is wherefrom the nilpotent property , we have so if andthenIf then where ,So we havewhich implies that A has only one distinct eigenvalue o with multiplicity n and index m or less ,3.21 given , find 求A的函数,the characteristic polynomial of A islet on the spectrum of A , we havethe we havethe computeto compute :3.22 use two different methods to compute for A1 and A4 in problem 3.13⽤两种⽅法计算题3.13 中A1和A4的函数method 1 : the Jordan-form representation of A1 with respect to the basis is method 2: the characteristic polynomial of is let on the spectrum of , we have the characteristic polynomial of is , let on the spectrum of ,we havethus3.23 Show that functions of the same matrix ; that is consequently we have 证明同⼀矩阵的函数具有可交换性，即因此有成⽴proof: let ( n is the order of A)let then we have3.24 let , find a B such that show that of for some I ,then B does not existlet , find a B such that Is it true that ,for any nonsingular c ,there exists a matrix B such that ?令证明若,则不存在B 使若 ,是否对任意⾮奇异C都存在B使, ?Let sowhere , if for some i , does not existfor we have , where then does mot exist , so B does mot exist , we can conclude that , it is mot true that , for any nonsingular C THERE EXISTS aB such that3.25 let and let m(s) be the monic greatest common divisor of all entries of Adj(Si-A) , Verify for the matrix in problem 3.13 that the minimal polynomial of A equals令, , 并且令m(s)是Adj(Si-A)的所有元素的第⼀最⼤公因⼦,利⽤题3.13 中验证A的最⼩多项式为verification : we can easily obtain that3.26 Define where are constant matrices theis definition is valid because the degree in s of the adjoint of (sI-A) is at most n-1 , verifywhere tr stands for the trase of a matrix and is defined as the sum of all its diagonal entries this process of computing is called the leverrier algorithm定义其中是A 的特征多项式是常数矩阵,这样定义是有效的, 因为SI-A 的伴随矩阵中S的阶次不超过N-1 验证其中矩阵的迹tr 定义为其对⾓元素之和, 这种计算和的程式被称为leverrier 算法.verification:’where3.27 use problem 3.26 to prove the cayley-hamilton theorem利⽤题3.26证明cayley-hamilton 定理proof:multiplying ith equation by yields ( )then we can see that is3.28 use problem 3.26 to show利⽤题3.26 证明上式,Proof: another : let3.29 let all eigenvalues of A be distinct and let be a right eigenvector of A associated with that is define and define,where is the ith row of P , show that is a left eigenvector of A associated with , that is 如果A的所有特征值互不相同, 是关于的⼀个右特征向量,即,定义并且其中是P 的第I ⾏, 证明是A的关于的⼀个左特征向量,即Proof: all eigenvalues of A are distinct , and is a right eigenvector of A associated with , and so we know thatThat isso , that is , is a left eigenvector of A associated with3.30 show that if all eigenvalues of A are distinct , then can be expressed as where and are right and left eigenvectors of A associated with 证明若A的所有特征值互不相同,则可以表⽰为其中和是A的关于的右特征值和左特征值,Proof: if all eigenvalues of A are distinct , let be a right eigenvector of A associated with , then is nonsingular , and where is aleft eigenvector of A associated with ,That is3.31 find the M to meet the lyapunov equation in (3.59) with what are the eigenvalues of the lyapunov equation ? is the lyapunov equation singular ? is the solution unique ?已知A,B,C,求M 使之满⾜(3.59) 的lyapunov ⽅程的特征值, 该⽅程是否奇异>解是否唯⼀?The eigenvalues of the Lyapunov equation areThe lyapunov equation is nonsingular M satisfying the equation3.32 repeat problem 3.31 for with two different C ,⽤本题给出的A, B, C, 重复题3.31 的问题,the eigenvalues of the lyapunov equation are the lyapunov equation is singular because it has zero eigenvalue if C lies in the range space of the lyapunov equation , then solution exist and are not unique,3.33 check to see if the following matrices are positive definite orsenidefinite 确定下列矩阵是否正定或者正半定,1. is not positive definite , nor is pesitive semidefinite2. it has a negative eigenvalue , so the second matrix is not positive definite , neither is positive demidefinte ,,3 the third matrix ‘s prindipal minors `that is all the principal minors of the thire matrix are zero or positive , so the matrix is positive semidefinite ,3.34 compute the singular values of the following matrices 计算下列矩阵的奇值,the eigenvalues of are 6 and 1 , thus the singular values of are and 1 ,the eigenvalues of are , thus the singular values of are3.35 if A is symmetric , what is the ralatimship between its eigenvalues and singular values ? 对称矩阵A 的特征值与奇异值之间有什么关系?If A is symmetric , then LET be an eigenvector of A associated with eigenvaue that is , thus we haveWhich implies is the eigenvalue of , (n is the order of A )So the singular values of A are where is the eigenvalue of A3.36 show证明上式成⽴let A is and B isuse (3.64) we can readily obtain3.37 show (3.65) 证明(3.65)proof: letthen we havebecausewe have det(NP)=det(QP)And3.38 Consider , where A is and has rank m is a solution ? if not , under what condition will it be a solution? Is a solution ?阶矩阵A 秩为m , 是不是⽅程的解? 如果不是, 那么在什么条件下,他才会成为该⽅程的解? 是不是⽅程的解?A is and has rank m so we know that , and is a square matrix of rankA=m , rank( ,So if ,then rank()If m=n , and rankA=m , so A is nonsingular , then we haverank()=rank(A)=m , and A=A that is is a solution ,RankA=MRank()=m is monsingular and exists , so we have , that is , is a solution of ,。

线性系统课后题答案

第一章数学基础1、加法不变性：R(S)中存在零元0，使得对()()S R s f ∈∀，都有()()s f s f =+0成立。

乘法不变性：R(S)中存在单位元1，使得对()()S R s f ∈∀，都有()()()s f s f s f =⋅=⋅11成立。

2、反证法证明：(1)加法不变性的唯一性假设在域F 中，存在0和0’，0≠0’，..t s αααα=+=+'0,0,对F ∈∀α成立。

以α+0=α为例，取α=0’，则0’+0=0’ 因为0’为零元，所以0’+0=0 所以0’=0，与假设矛盾。

（2）乘法不变性的唯一性假设在域F 中，存在1和1’，'11≠，..t s αααααα=⋅=⋅=⋅=⋅'1'1,11,对F ∈∀α成立。

以ααα=⋅=⋅11为例，取'1=α，则有'1'111'1=⋅=⋅ '1为单位元1'111'1=⋅=⋅∴'11=∴ 与假设矛盾3、试用反例证明你对下列问题的回答域交换环环 []R s 是是是 n n R *是是元素[]R s ∈的对角矩阵是是是 []p R s 是是是[]n np R s *是是其中：()p R s 是元素为常态的实有理分式（当s →∞，()R s 有界）；()n n p R s ⨯是元素属于()p R s 的n n ⨯矩阵证明：⑴[]R s 不是域。

如 ()1f +=s s ，显然()[]s R s f ∉-1。

（2）n nR* 不是交换环。

如⎥⎦⎤⎢⎣⎡=1010α，⎥⎦⎤⎢⎣⎡=0101β，显然22⨯∈R βα、。

但是βααβ≠。

（3）不是域。

如⎥⎦⎤⎢⎣⎡+=0001s α，1-α不存在。

（4）()p R s 不是域。

如∈+=1s 1α()p R s ，1-α=s+1.∞→∞→-1α时，s , 所以1-α∉()p R s 。

线性系统理论多年考题和答案

线性系统理论多年考题和答案2019级综合大题⎡400⎤⎡1⎤⎥x +⎢1⎥u x =⎢0-21⎢⎥⎢⎥⎢⎢⎣00-1⎥⎦⎣0⎥⎦y =[112]x1 能否通过状态反馈设计将系统特征值配置到平面任意位置？2 控规范分解求上述方程的不可简约形式？3 求方程的传递函数；4 验证系统是否渐近稳定、BIBO 稳定、李氏稳定；（各种稳定之间的关系和判定方法！）5 可能通过状态反馈将不可简约方程特征值配置到-2，-3？若能，确定K ，若不能，请说明理由；6 能否为系统不可简约方程设计全阶状态观测器，使其特征值为-4，-5； 7画出不可简约方程带有状态观测器的状态反馈系统结构图。

参考解答： 1.判断能控性：能控矩阵M =⎡⎣B可控，不能任意配置极点。

2按可控规范型分解AB⎡1416⎤⎢1-24⎥, rank (M ) =2. 系统不完全A 2B ⎤=⎦⎢⎥⎢⎣000⎥⎦⎡1⎢3140⎡⎤⎢1⎢⎥-1取M 的前两列，并加1与其线性无关列构成P =1-20，求得P =⎢⎢⎥⎢6⎢⎥⎢⎣001⎦⎢0⎢⎣2⎤⎡08⎢3⎥⎡1⎤⎢⎥1⎢⎥-1-1进行变换=PAP ⎢12-⎥, =PB =0, =cP =[222]⎢⎥⎢6⎥⎢⎢⎥⎣0⎥⎦001⎢⎥⎢⎥⎣⎦2⎤0⎥3⎥1-0⎥⎥6⎥01⎥⎥⎦⎧⎡08⎤⎡1⎤⎪x =⎢⎥x +⎢0⎥u12所以系统不可简约实现为⎨⎣⎦⎣⎦⎪y =[22]x ⎩3.G (s ) =c (sI -A ) -1B =4.2(s -1)(s +1) 2(s -1)=(s -4)(s +2)(s +1) (s -4)(s +2)det(sI -A ) =(s -4)(s +2)(s +1) ，系统有一极点4，位于复平面的右部，故不是渐近稳定。

G (s ) =c (sI -A ) -1B =2(s -1)，极点为4，-2，存在位于右半平面的极点，故系统不(s -4)(s +2)是BIBO 稳定。

系统发散，不是李氏稳定。

南航江驹线性系统理论习题

1-11 若系统的系统矩阵 S ( s ) 为
2s 1 s ( s 1) 2s 1 s ( s 1) 2
0 0 0 1 0 s 2 ( s 1) s( s 2) s S (s) 0 0 ( s 2) 1 0 0 1 0
1 0 x 0 0
0 1 1 0
0 0 5 0
0 0 0 0 x u, 0 1 6 1
y 1 0 0 1 x
将系统进行标准结构分解。 2-6 判断下列系统的输出可控性，输出函数可控性和输入函数可观测性
0 1 0 （1） A 0 0 1 , 3 2 1 0 1 0 （2） A 1 2 0 , 0 0 3 0 1 0 （3） A 1 2 0 , 0 0 3 0 1 0 （4） A 1 2 0 , 0 0 3
det e At ei t
i 1
n
1-14 若图 1-11 两个反馈链接的子系统，其传递函数阵分别为
1 s 1 G1 ( s ) 0
1 s 2 , s 1 s 2
1 s 3 G2 ( s ) 1 s 1
2-13 给定单变量线性定常系统
y 0 1 x
Ax bu , y cx x
已知 ( A, b) 为可控，问是否存在 C 使得 ( A, C ) 总是可观测。请加以论证，并举例说明之。 2-14 已知系统的传递阵为
( s 3) 1 s 1 ( s 2)( s 1) （1） G ( s ) ( s 2) 1 s 1 ( s 3)( s 1)

【理论】北航线性系统理论完整版答案

【关键字】理论1-1 证明：由矩阵可知A的特征多项式为若是A的特征值，则所以是属于的特征向量。

1-7 解：由于，可知当时，，所以系统不具有因果性。

又由于，所以系统是时不变的。

1-8 解：容易验证该系统满足齐次性与可加性，所以此系统是线性的。

由于而，故，所以系统是时变的。

又因为而，故，所以系统具有因果性。

1-11 解：由题设可知，随变化的图如下所示。

随变化的图如下所示。

从上述两图及所描述的系统，分析如下：当，且即时，有；当时，；当时，有；当时，有；当时，有；综上所示，该松弛系统在上述输入而激励的输出为：1-15 解：由上述齐次方程，可得两线性无关的解向量为：，所以即其基本矩阵为；状态转移矩阵为：1-17 证明：由题设我们可知故，得证。

1-19 证明：由题设可知：由上式可推出又由及习题1-17的结论可推出由以上两个结论，我们可得到所以得证。

即得证。

1-20 解：设其等价变换为，则可知: 由于P 是非奇异矩阵，所以。

1-24 解：易知，其中为严格真有理函数矩阵，进行下列计算：，则所以因此，可得一个实现如下：其模拟图如下所示。

1-25 证明：由题设知同理可知若要使得两系统零状态等价，则要满足，即满足，得证。

2-2 解： a,由题设可知：[]315 1 7- 1 1 1-7- 1 1 1- 1 0 1 1- 10 0 1 B A AB B 2=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=rank rank ，所以系统可控； 30 2 2 8- 14- 8-1- 3- 2-4 4 2 1 2 1 1- 10 2=⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡rank CA CA C rank ，所以系统可观。

b,[]x c c c y u x x 0 1 1 0 0 1 1 0 0 0 1 0 0 1 1 321=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡+⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=•由题设可知：[]30 1 0 1 1 0 1 0 1 1 01 A B 2=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡==rank B rank rankB ，所以系统可控；（1）若0321===c c c ，则系统不可观；（2）若321c c c ，，中至少有一个不等于零，则3 2 CA CA C 321132113212≠⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡++=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡c c c c c c c c c c c rank rank ，所以系统不可观；总之，该系统不可观。