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1、5.1Is the network shown in Fing5.2 BIBO stable ? If not , find a bounded input that will excite an unbound output图5.2 的网络BIBO 是否稳定?如果不是.请举出 一般激励无界输出的有界输入.From Fig5.2 .we can obtain thatAssuming zero initial and applying Laplace transform then we havebecause=1+=1+=1+=1+2=Which is not bounded .thus th
2、e network shown in Fin5.2 is not BIBO stable ,If u(t)=sint ,then we havey(t)=we can see u(t)is bounded ,and the output excited by this input is not bounded5.2consider a system with an irrational function .show that a necessary condition for the system to be BIBO stable is that |g(s)| is finite for a
3、ll Res 0一个系统的传递函数是S的非有理式。证明该系统BIBO 稳定的一个必要条件是对所有Res0. |g(s)|有界。Proof let s=thenIf the system is BIBO stable ,we have . for some constant Mwhich implies shat is convergent .For all Res=We have|g(t)|g(t0so and are also convergent , that is .where N and L are finite constant. Then| is finite ,for all R
4、es .Another Proof, If the system is BIBO stable ,we have for some constant M .And.For all Res . We obtain |.That is . | is finite for all Res5.3Is a system with impulse g(t)=BIBO stable ? how about g(t)=t for t脉冲响应为个g(t)= 的系统是否BIBO 稳定? 的系统又如何?Because the system with impulse response g(t)= is not BIB
5、O stable ,while the system with impulse response g(t)= for is BIBO stable.54 Is a system with transfer function BIBO stable一个系统的传递函数为 ，问该系统是否BIBO 稳定。Laplace transform has the property .If then We know . Thus So the impulse response of the system is . for the system is BIBO stable5.5 Show that negati
6、ve-feedback shown in Fig . 2.5(b) is BIBO stable if and only if and the gain a has a magnitude less than 1.For a=1 ,find a bounded input that will excite an unbounded output.证明图2.5(b) 中的负反馈系统BIBO 稳定当且增到a 的模小于1。对于a=1情况，找出一有界输入r(t), 她所产生的输出是无界的。Poof,If r(t)= .then the output is the impulse response of
7、 the system and equal=the impulse is definded as the limit of the pulse in Fig.3.2 and can be considered to be positive .thus we haveand which implies that the negative-feedback system shown in Fig.2.5(b) is BIBO stable if and only if the gain a has magitude less than 1.For .if we choose clearly it
8、is bounded the output excited by isAnd y(t) is not bounded5.6 consider a system with transfer function 。 what are the steady-state responses byu ( t )=3.for .and by u ( t )=sin2t. for 一个系统的传递函数是，分别由 和所产生的稳态响应是什么？The impulse response of the system is 。And we haveso the system is BIBO stableusing Theo
9、rem 5.2 ,we can readily obtain the steady-state responsesif u ( t )=3for .,then as approaches if u ( t )=sin2t for then ,as approaches5.7 considery= -2 3 is it BIBO stable ?问上述状态方程描述的系统是否BIBO稳定？The transfer function of the system is=The pole of is 1.which lies inside the left-half s-plane, so the sy
10、stem is BIBO stable5.8 consider a discrete-time system with impulse response for is the system BIBO stable ?一个离散时间系统的脉冲响应序列是，。问该系统是否BIBO 稳定？Because =Gkis absolutely sumable in . The discrete-time system is BIBO stable .5.9 Is the state equation in problem 5.7 marginally stable? Asymptotically stable
11、 ? 题5.7中的状态方程描述的系统是否稳定？是否渐进稳定？The characteristic polynomialthe eigenralue Ihas positive real part ,thus the equation is not marginally stable neither is Asymptotically stable.5.10 Is the homogeneous stable equation .marginally stable ? Asymptotically stable?齐次状态方程，是否限界稳定？The characteristic polynomia
12、l is And the minimal polynomial is .The matrix has eigenvalues 0 , 0 . and 1 . the eigenvalue 0 is a simple root of the minimal .thus the equation is marginally stableThe system is not asymptotically stable because the matrix has zero eigenvalues5 。11 Is the homogeneous stable equation marginally st
13、able ? asymptotically stable ?齐次状态方程 是否限界稳定？是否渐进稳定？The characteristic polynomial .And the minimal polynomial is .the matrix has eigenvalues 0. 0. and 1 . the eigenvalue 0 is a repeated root of the minimal polymial so the equation is not marginally stable ,neither is asymptotically stable .5.12 Is th
14、e discrete-time homogeneous state equation marginally stable ?Asymptotically stable ?离散时间齐次状态方程 是否限界稳定？是否渐进稳定？The characteristic of the system matrix is .and the minimal polynomial is the equation is not asymptotically stable because the matrix has eigenvalues 1. whose magnitudes equal to 1 . the eq
15、uation is marginally stable because the matrix has all its eigenvalues with magnitudes less than or to 1 and the eigenvalue 1 is simple root of the minimal polynomial . . 5.13 Is the discrete-time homogeneous state equation marginally stable ? asymptotically stable ? 离散时间齐次状态方程 是否稳定？是否渐进稳定？Its chara
16、cteristic polynomial is and its minimal polynomial is the matrix has eigenvalues 1 .1 .and 0.9 . the eigenvalue 1 with magnitude equal to 1 is a repeated root of the minimal polynimial . .thus the equation is not marginally stable and is not asymptotically stable . 5.14 Use theorem 5.5 to show that
17、all eigenvalues of have negative real parts .用定理 5.5 证明A 的特征值都具有负实部. Poof :For any given definite symmetric matrix N where with a0 .and The lyapunov equation can be written as that we have thus is the unique solution of the lyapunor equation and M is symmetric and we see is positive definite because
18、 for any given positive definite symmetric matrix N, The lyapunov equation has a unique symmetric solution M and M is positive definite , all eigenvalues of A have negative real parts ,5.15 Use theorem 5.D5show that all eigencalues of the A in Problem 5.14 magnitudes less than 1.Poof : For any given
19、 positive definite symmetric matrix N , we try to find the solution of discrete Lyapunov equation M Letwhere a0 and and M is assumed as ,then this is the unique symmetric solution of the discrete Lyapunov equation And we have we see M is positive definite .use theorem 5.D5,we can conclude that all e
20、igenvalues of A have magnitudes less than 1.5.16 For any distinct negative real and any nonzero real ,show that the matrix is positive definite . 正明上定义的矩阵M是正宗的 （ 为相异的负实数， 为非零实数）。Poof : are distinct negatire real numbers and are nonzero real numbers , so we have (2) . = ) 0(3) From (1), (2) and (3) ,
21、 We can conclude that M is positive definite 5.17 A real matrix M (not necessarily symmetric ) is defined to be positive definite if for any non zero .Is it true that the matrix M is positive definite if all eigenvalues of Mare real and positive or you check its positive definiteness ?一个实矩阵M被定义为正定的，
22、如果对任意非零都有。如果M的所有特征值都是正实数，是不是就有M正定/ 或者如果M的所有主子式都是正，M是不是正定呢？如果不是，要如何确定它的正定性呢？If all eigenvalues of M are real and positive or if all its leading pricipal minors are positive ,the matrix M may not be positive definite .the exampal following can show it .Let , Its eigenvalues 1,1 are real and positive a
23、nd all its leading pricipal minors are positive , but for non zero we can see . So M is not positive definite .Because is a real number . we have , and This 0, that is , M is positive and the matrix is symmetric , so we can use theorem 3.7 to cheek its positive definiteness .5.18 show that all eigen
24、values of A have real parts less than if and only if ,for any given positive definite symmetric matrix N , the equation has a unique symmetric solution M and M is positive definite .证明A的所有特征值的实部小于 当且对任意给定的正定对称阵N。方程有唯一解M，且M 是正定的。Proof : the equation can be written as Let , then the equation becomes .
25、 So all eigenvalues of B have megative parts if and only if ,for any given positive definite symmetric matrix N, the equation has a unique symmetric solution m and is positive definitive .And we know , so , that is , all eigenvalues of A are the eigenvalues of B subtracted . So all eigenvalues of A
26、are real parts less than .if and only all eigenvalues of B have negative real parts , eguivalontly if and only if , for any given positive symmetric matrix N the equation has a unique symmetric solution M and M is positive definite .5.19 show that all eigenvalues of A have magnitudes less than if an
27、d only if , for any given positive definite symmetric matrix N ,the equation . Has a unique symmetric solution M and M is positive definite .证明A的所有特征值的模都小于 当且仅当对任意给定的对称正定N ， 方程有唯一解M ，且M 是对称正定阵。Proof : the equation can be written as let , then that is ,all eigenvalues of A are the eigenvalues B multi
28、plied by ,so all eigenvalue of B have msgritudes less than 1 . equivalently if and only if , for any given positive definite symmetric matrix N the equation has a unique symmetric solution M and M is positive definite .520 Is a system with impulse response , for, BIBO stable ? how about 脉冲响应为， 的系统是否
29、BIBO 稳定？ 的系统是否BIBO 稳定？ 、 Both systems are BIBO stable .5.21 Consider the time-varying equation is the equation stable ? marginally stable ? asymptotically stable ?时变方程 是否BIBO 稳定 限界稳定？ 渐进稳定？is a fundamental matrix its stable transition matrix is Its impulse response is so the equation is BIBO stable
30、. is not bounded ,that is . there does not exist a finite constant M such that so the equation is not marginally stable and is not asymptotically stable .5.22 show that the equation in problem 5.21 can be transformed by using with , into is the equation BIBO stable ?marginally stable ? asymptoticall
31、y ? is the transformation a lyapunov transformation ?证明题5。21方程是可通过with 变换成 该方程是否BIBO 稳定？界限稳定？渐进稳定？该变换是不是lyapunov 变换 ?proof : we have found a fundamental matrix of the equation in problem 5.21 .is so from theorem 4.3 we know , let and , then we have and the equation can be transformed into the impuls
32、e response is invariant under equivalence transformation , so is the BIBO stability . that is marginally stable . is a fundamental matrix the state transition matrix is so the equation is marginally stable . (不趋于零)so the equation is not asymptotically stable . from theorem 5.7 we know the transforma
33、tion is mot a lyapunov transformation 5.23 is the homogeneous equation marginally stable ? asymptotically stable? 齐次方程 是否限界稳定？渐进稳定？is a fundamental matrix of the equation . is the state transition of the equation. for all and . the equation is marginally stable .if is fixed to some constant , then (不趋于零)。 so the equation is not asymptotically stable . for all t and with . we have . every entry of is bounded , so the equation is marginally stable .because the (2,2)th entry of does not approach zero as the equation is not asymptotically .