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Time-Domain Solution of LTI State Equations. 2 State-Variable Response of Linear Systems
- Abigayle Tate
- 1 years ago
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1 2.14 Analysis and Design of Feedback Control Systems Time-Domain Solution of LTI State Equations Derek Rowell October 22 1 Introduction This note examines the response of linear, time-invariant models expressed in the standard state -equation form: ẋ Ax + Bu (1) y Cx + Du. (2) that is, as a set of coupled, first-order differential equations. The solution proceeds in two steps; first the state-variable response x(t) is found by solving the set of first-order state equations, Eq. (1), and then the state response is substituted into the algebraic output equations, Eq. (2) in order to compute y(t). As in the classical solution method for ordinary differential equations with constant coefficients, the total system state response x(t) is considered in two parts: a homogeneous solution x h (t) that describes the response to an arbitrary set of initial conditions x(), and a particular solution x p (t) that satisfies the state equations for the given input u(t). The two components are then combined to form the total response. The solution methods used in this note rely heavily on matrix algebra. In order to keep the treatment simple we attempt wherever possible to introduce concepts using a first-order system, in which the A, B, C, andd matrices reduce to scalar values, and then to generalize results by replacing the scalars with the appropriate matrices. 2 State-Variable Response of Linear Systems 2.1 The Homogeneous State Response The state-variable response of a system described by Eq. (1) with zero input, u(t), and an arbitrary set of initial conditions x() is the solution of the set of n homogeneous first-order differential equations: ẋ Ax. (3) To derive the homogeneous response x h (t), we begin by considering the response of a firstorder (scalar) system withstate equation ẋ ax + bu (4) 1
2 withinitial condition x(). In this case the homogeneous response x h (t) has an exponential form defined by the system time constant τ 1/a, or: x h (t) e at x(). (5) The exponential term e at in Eq. (5) may be expanded as a power series, to give: ( x h (t) 1+at + a2 t 2 + a3 t ak t k ) +... x(), (6) 2! 3! k! where the series converges for all finite t>. Let us now assume that the homogeneous response x h (t) ofthestatevectorofahigher order linear time-invariant system, described by Eq. (3), can also be expressed as an infinite power series, similar in form to Eq. (6), but in terms of the square matrix A, thatiswe assume: ( x h (t) I + At + A2 t 2 + A3 t Ak t k ) +... x() (7) 2! 3! k! where x() is the initial state. Each term in this series is a matrix of size n n, andthe summation of all terms yields another matrix of size n n. To verify that the homogeneous state equation ẋ Ax is satisfied by Eq. (7), the series may be differentiated term by term. Matrix differentiation is defined on an element by element basis, and because eachsystem matrix A k contains only constant elements: ( ẋ h (t) + A + A 2 t + A3 t Ak t k 1 ) 2! (k 1)! +... x() ( A I + At + A2 t 2 + A3 t Ak 1 t k 1 ) +... x() 2! 3! (k 1)! Ax h (t). (8) Equation (8) shows that the assumed series form of the solution satisfies the homogeneous state equations, demonstrating that Eq. (7) is in fact a solution of Eq. (3). The homogeneous response to an arbitrary set of initial conditions x() can therefore be expressed as an infinite sum of time dependent matrix functions, involving only the system matrix A. Because of the similarity of this series to the power series defining the scalar exponential, it is convenient to define the matrix exponential of a square matrix A as e At I + At + A2 t 2 + A3 t Ak t k +... (9) 2! 3! k! which is itself a square matrix the same size as its defining matrix A. The matrix form of the exponential is recognized by the presence of a matrix quantity in the exponent. The system homogeneous response x h (t) may therefore be written in terms of the matrix exponential x h (t) e At x() (1) 2
3 which is similar in form to Eq. (5). The solution is often written as x h (t) Φ(t)x() (11) where Φ(t) e At is defined to be the state transition matrix 1 5. Equation (11) gives the response at any time t to an arbitrary set of initial conditions, thus computation of e At at any t yields the values of all the state variables x(t) directly. Example 1 Determine the matrix exponential, and hence the state transition matrix, and the homogeneous response to the initial conditions x 1 () 2, x 2 () 3 of the system withstate equations: ẋ 1 2x 1 + u ẋ 2 x 1 x 2. Solution: The system matrix is A From Eq. (9) the matrix exponential (and the state transition matrix) is Φ(t) e At ( I + At + A2 t 2 + A3 t Ak t k ) ! 3! k! t 2 + t ! 8 t ! t + 4t2 8t ! 3! +t 3t2 + 7t t + t2 2! 3! 2! t3 3! (i) The elements φ 11 and φ 22 are simply the series representation for e 2t and e t respectively. The series for φ 21 is not so easily recognized but is in fact the first four terms of the the expansion of e t e 2t. The state transition matrix is therefore e Φ(t) 2t e t e 2t e t (ii) 3
4 and the homogeneous response to initial conditions x 1 () and x 2 () is x h (t) Φ(t)x() (iii) or x 1 (t) x 1 ()e 2t (iv) x 2 (t) x 1 () ( e t e 2t) + x 2 ()e t. (v) With the given initial conditions the response is x 1 (t) 2e 2t (vi) x 2 (t) 2 ( e t e 2t) +3e t 5e t 2e 2t. (vii) In general the recognition of the exponential components from the series for each element is difficult and is not normally used for finding a closed form for the state transition matrix. Although the sum expressed in Eq. (9) converges for all A, in many cases the series converges slowly, and is rarely used for the direct computation of Φ(t). There are many methods for computing the elements of Φ(t), including one presented in Section 4.3, that are much more convenient than the direct series definition. 1,5,6 2.2 The Forced State Response of Linear Systems We now consider the complete response of a linear system to an input u(t). Consider first a first-order system witha state equation ẋ ax + bu written in the form ẋ(t) ax(t) bu(t). (12) If bothsides are multiplied by an integrating factor e at, the left-hand side becomes a perfect differential e at ẋ e at ax d ( e at x(t) ) e at bu (13) dt which may be integrated directly to give t d ( e aτ x (τ) ) t dτ e at x (t) x () e aτ bu (τ) dτ (14) dτ and rearranged to give the state variable response explicitly: x (t) e at x () + t 4 e a(t τ) bu (τ) dτ. (15)
5 The development of the expression for the response of higher order systems may be performed in a similar manner using the matrix exponential e At as an integrating factor. Matrix differentiation and integration are defined to be element by element operations, therefore if the state equations ẋ Ax + Bu are rearranged, and all terms pre-multiplied by the square matrix e At : e At ẋ (t) e At Ax (t) d ( e At x (t) ) e At Bu(t). (16) dt Integration of Eq. (16) gives t d dτ ( e Aτ x (τ) ) dτ e At x(t) e A x() t e Aτ Bu(τ)dτ (17) and because e A I and e At 1 e At the complete state vector response may be written in two similar forms t x(t) e At x() + e At e Aτ Bu(τ)dτ (18) x(t) e At x() + t e A(t τ) Bu(τ)dτ. (19) The full state response, described by Eq. (18) or Eq. (19) consists of two components: the first is a term similar to the system homogeneous response x h (t) e At x() that is dependent only on the system initial conditions x(). The second term is in the form of a convolution integral, and is the particular solution for the input u(t), withzero initial conditions. Evaluation of the integral in Eq. (19) involves matrix integration. For a system of order n and with r inputs, the matrix e At is n n, B is n r and u(t) isanr 1 column vector. The product e A(t τ) Bu(τ) is therefore an n 1 column vector, and solution for eachof the state equations involves a single scalar integration. Example 2 Findtheresponseofthetwostatevariablesofthesystem ẋ 1 2x 1 + u ẋ 2 x 1 x 2. to a constant input u(t) 5 for t>, if x 1 (), and x 2. Solution: This is the same system described in Example 1. The state transition matrix was shown to be e Φ(t) 2t e t e 2t e t 5
6 Withzero initial conditions, the forced response is (Eq. (18)): t x(t) e At e Aτ Bu(τ)dτ. Matrix integration is defined on an element by element basis, so that x1 (t) e 2t t e 2τ 5 x 2 (t) e t e 2t e t e τ e 2τ e τ dτ e 2t t e t e 2t e t 5e2τ dτ t 5eτ 5e 2τ dτ e 2t 5 2 5e t e 2t (i) (ii) (iii) (iv) 3 The System Output Response For either the homogeneous or forced responses, the system output response y(t) may be found by substitution of the state variable response into the algebraic system output equations y Cx + Du. (2) In the case of the homogeneous response, where u(t), Eq. (2) becomes y h (t) Ce At x(), (21) while for the forced response substitution of Eq. (19) into the output equations gives t y(t) Ce At x() + C e A(t τ) Bu(τ)dτ + Du(t). (22) Example 3 Find the response of the output variable y 2x 1 + x 2 in the system described by state equations ẋ 1 2x 1 + u ẋ 2 x 1 x 2. 6
7 Scalar exponential: e at 1+at + a2 t 2 e a 1 e at 1 e at 2! + a3 t 3 3! Matrix exponential: +... e At I + At + A2 t 2 e A I e At e At 1 e a(t 1+t 2 ) e at 1 e at 2 e A(t 1+t 2 ) e At 1 e At 2 2! + A3 t 3 3! +... e (a 1+a 2 )t e a1t e a 2t e (A 1+A 2 )t e A1t e A2t only if A 1 A 2 A 2 A 1 d dt eat ae at e at d a dt eat Ae At e At A t e at dt 1 e at 1 t e At dt A 1 e At I e At I A 1 a if A 1 exists. Otherwise defined by the series. Table 1: Comparison of properties of the scalar and matrix exponentials. to a constant input u(t) 5 for t>, if x 1 (), and x 2. Solution: This is the same system described in Example 1 with the same input and initial conditions as used in Example 2. The state-variable response is (Example 2): 5 x1 (t) e 2t 5 x 2 (t) 2 5e t + 5 (i) 2 e 2t The output response is y(t) 2x 1 (t)+x 2 (t) e 2t 5e t. (ii) 4 The State Transition Matrix 4.1 Properties of the State Transition Matrix Table 1 shows some of the properties that can be derived directly from the series definition of the matrix exponential e At. For comparison the similar properties of the scalar exponential e at are also given. Although the sum expressed in Eq. (9) converges for all A, in many cases the series converges slowly, and is rarely used for the direct computation of Φ(t). There are many methods for computing the elements of Φ(t), including one presented in the next 7
8 section, that are much more convenient than the series definition. The matrix exponential representation of the state transition matrix allows some of its properties to be simply stated: (1) Φ() I, which simply states that the state response at time t is identical to the initial conditions. (2) Φ( t) Φ 1 (t). The response of an unforced system before time t may be calculated from the initial conditions x(), x( t) Φ( t)x() Φ 1 (t)x() (23) and the inverse always exists. (3) Φ(t 1 )Φ(t 2 )Φ(t 1 + t 2 ). With this property the state response at time t may be defined from the system state specified at some time other than t, for example at t t. Using Property (2), the response at time t is x() Φ( t )x(t ) (24) and using the properties in Table 1, x(t) Φ(t)x() Φ(t)Φ( t )x(t ) (25) or x h (t) Φ(t t )x(t ). (26) (4) If A is a diagonal matrix then e At is also diagonal, and eachelement on the diagonal is an exponential in the corresponding diagonal element of the A matrix, that is e aiit. This property is easily shown by considering the terms A n in the series definition and noting that any diagonal matrix raised to an integer power is itself diagonal. 4.2 System Eigenvalues and Eigenvectors In Example 1 eachelement of Φ(t) e At was shown to be a sum of scalar exponential terms, and the resulting homogeneous response of each of the two state variables is therefore a sum of exponential terms. In general, the homogeneous response of linear systems is exponential in form, containing components of the form e λt,whereλ is a root of the characteristic equation. For example, the first-order homogeneous output response is of the form y(t) Ce λt where C is determined from the initial condition C y(), and the second-order response consists of two exponential components y(t) C 1 e λ 1t + C 2 e λ 2t where the constants C 1 and C 2 are determined by a pair of initial conditions, usually the output and its derivative. It is therefore reasonable to conjecture that for an nth order system the homogeneous response of eachof the n state variables x i (t) isaweightedsumofn exponential components: n x i (t) m ij e λ jt j1 (27) 8
9 where the m ij are constant coefficients that are dependent on the system structure and the initial conditions x(). The proposed solution for the complete state vector may be written in a matrix form or x 1 (t) x 2 (t). x n (t) m 11 m m 1n m 21 m m 2n m n1 m n2... m nn e λ 1t e λ 2t x h (t) M. e λnt where M is an n n matrix of the coefficients m ij. To determine the conditions under which Eq. (29) is a solution of the homogeneous state equations, the suggested response is differentiated and substituted into the state equation. From Eq. (27) the derivative of each conjectured state response is e λ 1t e λ 2t. e λnt (28) (29) or in the matrix form x 1 x 2. x n dx n i dt λ j m ij e λ jt j1 λ 1 m 11 λ 2 m λ n m 1n λ 1 m 21 λ 2 m λ 2 m 2n λ 1 m n1 λ 2 m n2... λ n m nn e λ 1t e λ 2t. e λnt (3). (31) Equations (28) and (31) may be substituted into the homogeneous state equation, Eq. (3), λ 1 m 11 λ 2 m λ n m 1n e λ 1t m 11 m m 1n e λ 1t λ 1 m 21 λ 2 m λ 2 m 2n e λ 2t A m 21 m m 2n e λ 2t (32) λ 1 m n1 λ 2 m n2... λ n m nn e λnt m n1 m n2... m nn e λnt and if a set of m ij and λ i can be found that satisfy Eq. (32) the conjectured exponential form is a solution to the homogeneous state equations. It is convenient to write the n n matrix M in terms of its columns, that is to define a set of n column vectors m j for j 1, 2,...,n from the matrix M m j 9 m 1j m 2j. m nj
10 so that the matrix M may be written in a partitioned form Equation (32) may then be written λ1 m 1 λ 2 m 2... λ n m n M m 1 m 2... m n. (33) e λ 1t e λ 2t. e λnt A m 1 m 2... m n and for Eq.(34) to hold the required condition is λ1 m 1 λ 2 m 2... λ n m n A m 1 m 2... m n e λ 1t e λ 2t. e λnt (34) Am 1 Am 2... Am n. (35) The two matrices in Eq. (35) are square n n. If they are equal then the corresponding columns in eachmust be equal, that is λ i m i Am i i 1, 2,...,n. (36) Equation (36) states that the assumed exponential form for the solution satisfies the homogeneous state equation provided a set of n scalar quantities λ i,andasetofn column vectors m i can be found that each satisfy Eq. (36). Equation (36) is a statement of the classical eigenvalue/eigenvector problem of linear algebra. Given a square matrix A, the values of λ satisfying Eq. (36) are known as the eigenvalues, or characteristic values, of A. The corresponding column vectors m are defined to be the eigenvectors, or characteristic vectors, of A. The homogeneous response of a linear system is therefore determined by the eigenvalues and the eigenvectors of its system matrix A. Equation (36) may be written as a set of homogeneous algebraic equations λ i I A m i (37) where I is the n n identity matrix. The condition for a non-trivial solution of such a set of linear equations is that (λ i )detλ i I A. (38) which is defined to be the characteristic equation of the n n matrix A. Expansion of the determinant generates a polynomial of degree n in λ, and so Eq. (38) may be written λ n + a n 1 λ n 1 + a n 2 λ n a 1 λ + a (39) or in factored form in terms of its n roots λ 1,...,λ n (λ λ 1 )(λ λ 2 )...(λ λ n ). (4) For a physical system the n roots are either real or occur in complex conjugate pairs. The eigenvalues of the matrix A are the roots of its characteristic equation, and these are commonly known as the system eigenvalues. 1
11 Example 4 Determine the eigenvalues of a linear system with state equations: ẋ 1 1 x 1 ẋ 2 1 x 2 + u(t). ẋ x 3 1 Solution: The characteristic equation is det λi A or det λ 1 λ λ +4 (i) λ 3 +4λ 2 +9λ + 1 (ii) (λ +2)(λ +(1+j2)) (λ +(1 j2)) (iii) The three eigenvalues are therefore λ 1 2, λ 2 1+j2, and λ 2 1 j2. For each eigenvalue of a system there is an eigenvector, defined from Eq. (37). If the eigenvalues are distinct, the eigenvectors are linearly independent, and the matrix M is non-singular. In the development that follows it is assumed that M has an inverse, and therefore only applies to systems without repeated eigenvalues. An eigenvector m i associated witha given eigenvalue λ i is found by substituting into the equation λ i I A m i. (41) No unique solution exists, however, because the definition of the eigenvalue problem, Eq. (36), shows that if m is an eigenvector then so is αm for any non-zero scalar value α. The matrix M, which is simply a collection of eigenvectors, defined in Eq. (33) therefore is not unique and some other information must be used to fully specify the homogeneous system response. Example 5 Determine the eigenvalues and corresponding eigenvectors associated with a system having an A matrix: 2 1 A
12 Solution: The characteristic equation is det λi A or λ +2 1 det 2 λ +3 λ 2 +5λ +4 (λ +4)(λ +1). (i) The two eigenvalues are therefore λ 1 1, and λ 2 4. To find the eigenvectors these values are substituted into the equation λ i I A m i. For the case λ 1 1 thisgives m11 m 21. (ii) Both of these equations give the same result; m 11 m 21. The eigenvector cannot be further defined, and although one particular solution is 1 m 1 1 the general solution must be defined in terms of an unknown scalar multiplier α 1 m 1 α1 α 1 (iii) provided α 1. Similarly, for the second eigenvalue, λ 2 4, the equations are 2 1 m m 22 whichbothstate that 2m 12 m 22. The general solution is α m 2 2 2α 2 (iv) (v) for α 2. The following are all eigenvectors corresponding to λ 2 4: Assume that the system matrix A has no repeated eigenvalues, and that the n distinct eigenvalues are λ 1,λ 2,...,λ n.definethemodal matrix M by an arbitrary set of corresponding eigenvectors m i : M m 1 m 2... m n. (42) 12
13 From Eq. (29) the homogeneous system response may be written e λ 1t e λ 2t x h (t) M. e λnt α 1 m 1 α 2 m 2... α n m n for any non-zero values of the constants α i. The rules of matrix manipulation allow this expression to be rewritten: α 1 e λ 1t x h (t) α 2 e λ 2t m 1 m 2... m n. α n e λnt e λ 1t... α 1 e λ 2t... α 2 m 1 m 2... m n e λnt α n e λ 1t e λ 2t. e λnt (43) Me Λt α (44) where α is a column vector of length n containing the unknown constants α i,ande Λt is an n n diagonal matrix containing the modal responses e λit on the leading diagonal e λ 1t... e Λt e λ 2t (45)... e λnt At time t all of the diagonal elements in e Λ are unity, so that e Λ I is the identity matrix and Eq. (44) becomes x() MIα. (46) For the case of distinct eigenvalues the modal matrix M is non-singular, and the vector α may be found by pre-multiplying eachside of the equation by M 1 : α M 1 x() (47) specifying the values of the unknown α i in terms of the system initial conditions x(). The complete homogeneous response of the state vector is x h (t) Me Λt M 1 x(). (48) 13
14 Comparison with Eq. (11) shows that the state transition matrix may be written as leading to the following important result: Φ(t) Me Λt M 1 (49) Given a linear system of order n, described by the homogeneous equation ẋ Ax, and where the matrix A has n distinct eigenvalues, the homogeneous response of any state variable in the system from an arbitrary set of initial conditions x() is a linear combination of n modal components e λ it,wheretheλ i are the eigenvalues of the matrix A. 4.3 A Method For Determining the State Transition Matrix Equation (49) provides the basis for the determination of the state transition matrix for systems withdistinct eigenvalues: (1) Substitute numerical values into the system A matrix, and compute the system eigenvalues, a modal matrix M, anditsinversem 1. A computer based linear algebra package provides a convenient method for doing this. (2) Form the diagonal matrix e Λt by placing the modal components e λ it on the leading diagonal. (3) The state transition matrix Φ(t) isφ(t) Me Λt M 1. Example 6 Determine the state transition matrix for the system discussed in Example 2, and find its homogeneous response to the initial conditions x 1 () 1, and x 2 () 2. Solution: The system is described by the matrix 2 1 A 2 3 and in Example 2 the eigenvalues are shown to be λ 1 1, and λ 2 4, and a pair of corresponding eigenvectors are 1 1 m 1, m A modal matrix M is therefore M
15 and its inverse M 1 is M The matrix e Λt is found by placing the modal responses e t and e 4t on the diagonal: e e Λt t e 4t. (i) The state transition matrix Φ(t) is Φ(t) Me Λt M e t e 4t e t + e 4t e t e 4t 3 2e t 2e 4t e t +2e 4t. (ii) The homogeneous response for the given pair of initial conditions is x h (t) Φ(t)x() or or x1 (t) x 2 (t) 1 3 2e t + e 4t e t e 4t 2e t 2e 4t e t +2e 4t 1 2 (iii) x 1 (t) 4 3 e t 1 3 e 4t (iv) x 2 (t) 4 3 e t e 4t. (v) 4.4 Systems with Complex Eigenvalues The modal components of the response are defined by the eigenvalues of the matrix A, as found from the roots of the characteristic equation det λi A which is a polynomial of degree n in λ, withconstant and real coefficients. Any polynomial equation with real coefficients must have roots that are real, or which appear in complex conjugate pairs. The eigenvalues of A are therefore either real, or occur as pairs of the form λ i,i+1 σ ± jω, wherej 1. In suchcases there are modal response components in 15
16 the matrix e Λt of the form e (σ±jω)t e σt e ±jωt. However, when the state transition matrix is computed the conjugate components combine, and the Euler relationships sin ωt 1 ( e jωt e jωt) 2j (5) cos ωt 1 ( e jωt + e jωt) 2 (51) may be used to express the elements of the state transition matrix in a purely real form e σt sin (ωt) ore σt cos (ωt). Example 7 Determine the state transition matrix for the undamped mechanical oscillator, witha mass m 1 Kg suspended on a spring withstiffness K 1 N/m, as shown withits linear graphin Fig. 1. Figure 1: Simple spring-mass mechanical oscillator. Solution: The state equations for this system are F K v m K 1/m FK v m + 1/m F s (t). (i) With the values given the state equations become F K v m 1 1 FK v m + 1 F s (t). (ii) The characteristic equation det λi A is λ (iii) 16
17 so that the eigenvalues are purely imaginary: λ 1 j1 and λ 2 j1. A pair of corresponding eigenvectors is found by substituting these values back into the equation λ i I A m i and choosing an arbitrary scaling constant: m 1 1.1j, m 2 1.1j The modal matrix M and its inverse are 1 1 M M 1 1.1j 1.1j.1j.2j.1j 1 and the state transition matrix is e j1t.1j 1 Φ(t).2j.1j.1j e j1t.1j 1 e j1t + e j1t 1 ej1t e j1t 2 2j.1 ej1t e j1t e j1t + e j1t 2j 2 cos(1t) 1sin(1t). (iv).1sin(1t) cos(1t) The homogeneous response to arbitrary initial values of F K and v m is therefore FK (t) cos(1t) 1sin(1t) FK () (v) v m (t).1sin(1t) cos(1t) v m () and the individual state responses are F K (t) F K () cos(1t)+1v m () sin(1t) (vi) v m (t).1f K () sin(1t)+v m () cos(1t) (vii) which are purely real responses, despite the imaginary system eigenvalues Systems with Repeated Eigenvalues The method for deriving the state transition matrix presented in the previous section is dependent on the existence of the inverse of the modal matrix M, thatism must be nonsingular. In general, if there are two or more eigenvalues with the same value, the eigenvectors are not linearly independent, and M 1 does not exist. It is possible to extend the method to handle repeated eigenvalues, as described in references on linear system theory. 17
18 When there is a pair of repeated eigenvalues, instead of a linear combination of simple exponential terms, the state response may have a pair of components e λit and te λit corresponding to the two identical eigenvalues. This does not imply that the state transition matrix Φ(t) does not exist for systems withrepeated eigenvalues; it may be computed by other methods. It may however lose the simple exponential form assumed throughout this chapter. We do not discuss this case further in this introductory note, and readers are referred to more advanced texts Stability of Linear Systems In state-space terminology, a system is said to be asymptotically stable if the homogeneous response of the state vector x(t) returns to the origin of the state-space x from any arbitrary set of initial conditions x() as time t. This definition of stability is equivalent to stating that the homogeneous response of all state variables must decay to zero in the absence of any input to the system, or lim x i(t) (52) t for all i 1,...,n. This condition may be rewritten in terms of the state transition matrix lim Φ(t)x() (53) t for any x(). All of the elements of the state transition matrix are linear combinations of the modal components e λ it, therefore the stability of a system depends on all such components decaying to zero with time. For real eigenvalues this requires that λ i <, since any positive eigenvalue generates a modal response that increases exponentially with time. If eigenvalues appear in complex conjugate pairs λ i,i+1 σ ± jω the state homogeneous response contains components of the form e σt sin(ωt) ore σt cos(ωt). If σ>thesecomponentsgrowexponentially with time and the system is by definition unstable. The requirements for system stability may be therefore summarized: A linear system, described by state equations ẋ AX + Bu, is asymptotically stable if and only if all eigenvalues of the matrix A have negative real parts. Three other separate conditions should be considered: 1. If one or more eigenvalues, or pair of conjugate eigenvalues, has a positive real part there is at least one corresponding modal component that increases exponentially without bound from any finite initial condition, violating the definition of stability. 2. Any pair of conjugate eigenvalues that are purely imaginary, λ i,i+1 ±jω witha real part σ, generate an undamped oscillatory component in the state response. The magnitude of the homogeneous system response neither decays or grows, but continues to oscillate for all time at a frequency ω. Sucha system is defined to be marginally stable. 18
19 3. An eigenvalue witha value λ generates a modal component e t that is a constant. The system response neither decays nor grows, and again the system is defined to be marginally stable. Example 8 Discuss the stability of an inverted pendulum consisting of a mass m on the end of a long light rod of length l which is mounted in a rotational bearing that exhibits a viscous rotational drag, as shown in Fig. 2. Figure 2: An inverted pendulum. Solution: The system may be treated as a rotational system. The moment of inertia J of the mass is J ml 2, and when the rod is displaced from the vertical, gravity exerts a torque mgl sin θ about the bearing as shown. This system is inherently non-linear because of the angular dependence of the torque, but it may be linearized by using the small angle approximation sin θ θ for small θ. Then the restoring torque after a displacement θ is mglθ. This is the constitutive relationship of an ideal T-type element, and allows the gravitational effect to be modeled as a torsional spring witha negative spring constant K mgl. Let the frictional drag about the bearing be represented by the damping coefficient B. The linear graph for the system is shown in Fig. 2. Define the states to be the torque in the spring T K and the angular velocity of the shaft Ω. The linearized state equations for this homogeneous system are: T K Ω J K 1/J B/J TK Ω J. (i) The characteristic equation is det λi A det λ K 1/Jλ + B/J (ii) 19
20 or λ 2 + B J λ + K J λ 2 + B J λ g R. (iii) The quadratic equation may be used to determine the two eigenvalues λ 1,2 B 2J ± 1 2 The following may be noted directly from Eq. (v). (B ) 2 + 4g R. (iv) 1. The quantity under the radical is positive, therefore both eigenvalues are real. 2. (B ) g 2 J R > B (v) 2J so that there is always a positive eigenvalue. We conclude, therefore, that the inverted pendulum is an unstable system and in its linearized form will exhibit an exponential growth in angular displacement from the vertical for any finite initial offset or angular velocity. J 4.7 Transformation of State Variables The choice of a set of state variables used to represent a system is not unique. In Chapter operational methods to transfer between representations are discussed. It is possible to define an infinite number of different representations by transforming the state vector by linear operations. If a system is described by a state vector x, a new set of state variables x may be generated from any linearly independent combination of the original state variables x i (t), for example: x i(t) p i1 x 1 (t)+p i2 x 2 (t)+ + p in x n (t) (54) where the p ij are constants. This linear transformation may be written in matrix form x Px (55) where P is a non-singular n n square matrix. (With this definition the elements p ij in Eq. (52) above are elements of P 1.) The state equations in the new state variables become ẋ Pẋ APx + Bu (56) 2
21 and pre-multiplying eachside by the inverse of P, the new set of state equations gives The output equation must be similarly transformed ẋ ( P 1 AP ) x + ( P 1 B ) u. (57) y (CP) x + Du. (58) The system is now represented by modified A, B, andc matrices. The state variable representation is an internal system representation that is made observable to the system environment through the output equations. The input-output system dynamic behavior should be independent of the internal system representation. For the transformed state equations the characteristic equation is defined in terms of the transformed A matrix If the substitution I P 1 IP is made: det λi ( P 1 AP ). (59) det λp 1 P P 1 AP (6) det P 1 λi A P. (61) Since the determinant of a product is the product of the determinants, the characteristic equation is det P 1 det λi AdetP (62) and because P is not singular, det P anddetp 1, the transformed characteristic equation is det λi A (63) which is the same as that of the original state equations, leading to the following important conclusion: The characteristic equation, and hence the eigenvalues λ i and the modal response components e λ it, are invariant under a linear transformation of the system state variables. These quantities are properties of the system itself and are not affected by the choice of state variables used to describe the system dynamics. If a transformation of the state variable representation is made, it is necessary to similarly transform any initial conditions to the new representation using x () P 1 x(). (64) 21
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Friday, 19 March 2010
Cabbage Rolls
I know that Cabbage Rolls are not traditional English food. I think they are probably something that originated in Eastern Europe, but they are something that we have enjoyed for many years in my family.
When I was a child, it was always a real treat when my mom had Cabbage Rolls on the menu. She made hers quite differently than most. They were cooked in a huge Wearever Aluminum pot on top of the stove all day, and she always added potatoes, carrots, and wedges of cabbage to the pot along with the cabbage rolls.
My dad never liked the cabbage. I never liked the meat . . . but we made a deal with each other every time my mom made them. I gave him my meat, and he gave me his cabbage. We were both happy.
Oh how delicious those vegetables were on my plate . . . the potatoes lightly mashed with my fork, and then all of them spread with cold butter and sprinkled with salt and pepper. The butter melting and mingling with the vegetable juices and tomatoes . . .
Oh my . . . just the thought of them makes my mouth water.
I make mine quite differently than my mother did. She never put rice in hers, but I like rice in mine. I also add minced streaky bacon to the meat mixture, which gives it an added layer of flavour.
She only ever used tinned tomatoes to cook hers in. I have developed a delicious and slightly sweet and sour tomatoey sauce to cook mine in.
Her's were cooked on top of the stove. Mine are baked in the oven, so that those tasty little rolls aborb all the delicious flavours from the sauce and become meltingly tender.
Both quite different from the other . . . but equally as delicious. I do hope you will give these a try before the weather becomes too warm for oven meals.
That beautiful Spring Cabbage that is in the shops right now, (the pointed sweetheart variety) is just perfect for these!!
*Cabbage Rolls*
Makes 12
Printable Recipe
This is our favourite cabbage roll recipe. Delicious, moist and meaty with a fabulous sweet and sour sauce. These always go down a real treat. I often double and triple the recipe when I take them to pot luck suppers. I always bring home an empty dish.
1 pound extra lean minced beef
1/4 cup raw long grain rice
1 large egg, beaten
1 medium onion, peeled and grated
1 carrot, peeled and grated
1 fat clove of garlic, peeled and crushed
1 tsp salt
1/2 tsp freshly ground black pepper
1/2 pound smoked streaky bacon, minced
12 wilted cabbage leaves, thick veins trimmed to thin
For the sauce:
80g soft light brown sugar, (1/2 cup packed)
2 ounces fresh lemon juice
8 ounces passata
Place the meat into a bowl and mix together with the rice, egg, onion, carrot, garlic, salt, pepper and bacon. Mix well. Shape into 12 equal sized ovals. Place each oval at the wide end of a wilted cabbage leafe. Roll up, tucking in the sides to completely encase the meat. Place, folded side down, into a greased baking dish. Mix together the brown sugar, lemon juice and tomato sauce. Pour this sauce over the rolls. Cover tightly with a lid. Preheat the oven to 190*C/375*F. Place the casserole into the heated oven and bake for 1 hour. Uncover at the end of that time and bake for 20 minutes longer. Serve hot.
21 comments:
Ok, you're gonna think I'm stupid but I once tried making these and stuffed cabbage with already cooked meat. No wonder why it was so hard to shape the cabbage into a perfect roll! :) Thanks for the pictures. I'll have to try it again!ReplyDelete
I LOVE cabbage rolls, Marie! So glad you shared this recipe tonight!ReplyDelete
How's the moving preparations going? I hope you're not working yourselves too hard! Are there people that are helping you? I sure hope so!
Keeping you in my prayers, as always...much love sent your way this lovely night...see you tomorrow!
- Anonymous19 March 2010 at 06:23
Marie, this food is turkish.. however, here in Greece is very popular and we use oil, lemon and eggs as a sauce..ReplyDelete
have a nice day..
ps: happy i can finally post a comment..:)
I think I must be about the only person in the world never to have had a cabbage roll. They look pretty impressive. Are they a main meal in their own right or do they get served along side something else?ReplyDelete
Chele, I like to serve them with some steamed rice. The sauce tastes so good with it. I also serve some vegetables on the side. The other day we had roasted carrots and parsnips and they were very good together.ReplyDelete
Interesting to read they're Turkish. it makes sense though as that was a great influence in Italy and my grandmothers both made them. Yours look incredible as alwaysReplyDelete
xoxo Pattie
They look just right..Everything the way I like:)ReplyDelete
Great pics Marie~
Haven't cabbage rolls in a long time! You have the talent of making them look good and sound good. Now I'm hungry for cabbage rolls.ReplyDelete
Oh, I love cabbage rolls and need to make them again after seeing this! They look delicious!ReplyDelete
Never heard of these Marie, they look very tasty.ReplyDelete
Wishing you well, sending you my love and best and hope you are doing ok. Lucie xx
This is perfect. I bought mince and cabbage today in my supermarket shop, so this may be appearing on my menu this weekend!ReplyDelete
My mom makes them this way too. maybe it is a german or russian thing.ReplyDelete
Fabulous recipe, Marie. My impression, too, is that these are distinctly Eastern European. Romanians, for example, love this stuff! My family, quite Romanian on my mom's side, cooked up a huge pan of these for every big gathering while I was growing up. Your recipe looks so similar to what my mom would have made, I am making it a point to save it. I can almost smell them! (Another beauty of a recipe from Marie!)ReplyDelete
I used to have something similar to this when I was growing up. I always knew them as "gołabki" (from the Polish side of my family).ReplyDelete
Ive always wanted to try making a veggie version of cabbage rolls. The sauce theyre in is a great idea also.ReplyDelete
- Teresa21 March 2010 at 12:56
Marie, I made these cabbage rolls last night for dinner and yes! they went down a real treat! I didn't have the bacon, but that it is the only deviation that I made to your recipe. And.....I have leftovers to warm up for today....YAY!ReplyDelete
Well, to me they're Polish!! They are called 'glembokie' - pronounced something like 'gerwompki' in English and I learnt them at my Polish mother's knee, as she did at hers - but I am sure the recipe has spread across from Russia and moving west and south, as they serve them in Moscow too!!! I don't suppose we'll ever know where they originated.....ReplyDelete
- Anonymous15 September 2010 at 23:50
They sure are Polish right along with pierogis and Keilbasa and saurkraut! I grew up on this kind of food! Yummy!ReplyDelete
- Anonymous3 March 2011 at 10:37
i think they are one of those turkish foods that made it all across eastern europe! they are very popular here in canada too due to our large eastern european population especially in the prairies - and thank goodness for that!ReplyDelete
marie - here is a trick to really make the cabbage rolls sweet and sour - my mother always layered home-made sauerkraut over the cabbage rolls. to this day everyone always wonders why my cabbage rolls taste so much better than everyone else's!!!
Oh yes, I have seen that done anonymous and I love it! You are making me hungry for some cabbage rolls again!ReplyDelete
Gosh they look so good. I could ravage a cabbage right now. ツReplyDelete
Thanks for stopping by. I love to hear from you so do not be shy! Please don't attempt to leave spam. It will be deleted immediately.


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Which Joe gave his name to 'sloppy joes'? We look at five interesting sandwiches and their lexical origins.
Definition of advice note in English:
advice note
(Chiefly in commercial contexts) a note giving formal notification of something; especially a note recording the contents of a consignment of goods.
Mid 19th century; earliest use found in The Liverpool Mercury.
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Pet Sitter in Mount Dora
My name is Jennifer Lindsey! I am a per owner presently a blue Merle Pomeranian named Weechee 1 years old and a Pompoo named Bear 3 months old I have had ownership of cats domestic Siamese and two wild I have had goats horses pigs and cows to care for since childhood I love caring for animals training them. Lavishing them with affection ( a necessity for every one) I can provide phone references and a resume from previous positions held
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Most listings are in the $10's per square foot per year. The Springfield submarket is seeing a an increase in rates. In the last 3 months, the median rate increased $2 per sqft per year. That's a 9% increase in rent.
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Even though the median size is 2,740 sqft, there are still 0 new listings below 2,000 sqft. Most properties in this region are close to 2 floors. Most buildings in this area were built in 1988 and are Class A.
Trends and status of the commercial real estate market in Springfield.
Market Report
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OCR B - halogenoalkanes
- Created by: Former Member
- Created on: 09-03-13 13:53
First 249 words of the document:
It is an alkane with at least one halogen atom in
place of a hydrogen atom.
Naming halogenoalkanes- the prefixes are listed in alphabetical order.
Bonds are PERMANENT-PERMANENT DIPOLE because carbon
and halogen together is a polar molecule. Their boiling points depends
on the size and number of halogen atom present; the bigger the halogen atom and
the more halogen atoms there are higher the boiling point. For e.g. halogenoalkane
with a Cl or Br atom has higher boiling point than F.
larger Halogenoalkanes can cause GREAT ENVIRONMENTAL DAMAGE.
For e.g. chlorofluorocarbons (CFC') are Halogenoalkanes where all the hydrogen
atoms are replaced by chlorine and fluorine atom. They are used for lots of things
because they are very Unreactive. They are used for dry cleaning and degreasing
but this causes damage to the ozone layer.
Key terms
Nucleophile (Nu-) molecule or negatively charged ion with a lone pair of e-s that
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Substitution when one group is replaced by another
Mechanism Shows the stages of a reaction
Curly arrows show movement of e-s to make/break bonds
single headed - 1 e-
double headed pair of e-s
Reactions of Halogenoalkanes involve breaking C-H bonds. The bond can
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Oak Lodge Medical Centre parking
Driveway on Milford Gardens, HA8
Parking Space on Playfield Road, HA8
Parking Space on Littlefield Road, HA8
Driveway on Garratt Road, HA8
Driveway on Deans Lane, HA8
Parking Space on Fairfield Crescent, HA8
Parking Space on Clarinet Court, , HA8
Driveway on Hale Drive, NW7
Driveway on Manor Park Gardens, HA8
Information about parking near Oak Lodge Medical Centre
Looking for parking near Oak Lodge Medical Centre? You've come to the right place! Parking's what we do best - and we have a handy selection of 250 spaces conveniently located just a stone's throw from the hospital entrance.
Hospital parking can be a nightmare. Not only are spaces in limited supply, but it's often incredibly overpriced too . Plus you always run the risk of getting a ticket for overstaying, and no one wants that - especially after visiting hospital.
We think hospital trips are stressful enough without the added pain of finding a parking space that isn't a rip-off. That's why we offer a cheaper, hassle-free alternative to hospital car parks- in fact, you can reserve your very own guaranteed parking space near Oak Lodge Medical Centre for just £4.00. Not bad, eh?
With JustPark, you'll never be late to an appointment at because of parking problems ever again. And you'll certainly never be punished with a parking fine for being late back to your car. If you overstay with us, you just pay a bit more at the same rate to cover the extra time.
We're taking the pain out of parking for 1 million drivers across the UK. But don't just take our word for it - check out our latest customer reviews for Oak Lodge Medical Centre parking below:
"Great thanks! Easy to find from M1 J4, nice and clean, easy access, lots of room (parked a Ford Transit Custom no problem). Very close to Edgware station. " — Driveway on Manor Park Gardens, HA8
"Perfect little spot for get to Edward station.. and a superb price. Will use again." — Driveway on Manor Park Gardens, HA8
None of our delicious JustPark spaces tickle your fancy? Don't worry - we've got your back. There are plenty of other parking options available near Oak Lodge Medical Centre - here's the best of the rest:
Off-Street Parking
Here's a selection of nearby off-street parking options:
|Watling Avenue
|London Borough of Barnet
|£1.00 / 2 hours
|9am - 5pm,
Mon - Fri
|The Broadwalk Shopping Centre
|The Broadwalk Centre
|£1.00 / 3 hours
|12am - 11pm,
Mon - Sat
|Queensbury Station
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|12am - 11pm,
Mon - Fri
|Kingsbury Road
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|£1.50 / hour
|8am - 6pm,
Mon - Sat
|Canons Park Station
|£5.20 / day
|12am - 11pm,
Mon - Fri
On-Street Parking
Here's a selection of nearby on-street parking options:
|Max Stay
|High Street, Edgware, On-Street
|£0.50 / 20 minutes
|2 hours
|Mollison Way, Edgware, On-Street
|£0.10 / 20 minutes
|2 hours
|Mead Road, Edgware, On-Street
|£0.50 / 20 minutes
|2 hours
|Bacon Lane, On-Street
|£0.10 / 20 minutes
|2 hours
|High Street, Edgware, On-Street
|£0.50 / 20 minutes
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Nearby Transport
- Burnt Oak Station 0.29 miles away
- Edgware Station 0.72 miles away
- Queensbury Station 0.91 miles away
- Mill Hill Broadway Station 1.17 miles away
- Hendon Station 2.14 miles away
- Wealdstone Station 2.31 miles away
Nearby Hospitals
- Oak Lodge Medical Centre 0 miles away
- The Bacon Lane Surgery 0.24 miles away
- Watling Clinic 0.61 miles away
- Grahame Park Health Centre 1.15 miles away
- Honeypot Lane Centre 1.29 miles away


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DescriptionCreated as a luxurious private residence, villa Jiva Puri embodies the most magical elements of Bali, wrapping you up in a sublime, never-forgotten experience. It is a divine five bedroom retreat family or friends, and the perfect party venue for up to 40 guests should you wish to entertain.
As you enter the cloistered realm of Jiva Puri, the outside world simply drops away. An apparently endless array of employees attends you - dedicated entirely to your party's comfort, entertainment and relaxation. The talented kitchen staff, led by the former Executive Chef of the Grand Hyatt, can whip up anything from Balinese specialities to home favourites.
Searching out the perfect spot on the unspoilt southwest coast of Bali to build this two acre estate was the beginning of a dream come true for the owner. Jiva Puri is an architect-designed collection of buildings inspired by Balinese culture and featuring every modern comfort. The serenity is enhanced by myriad pools and frangipani trees, but if you need to break the spell by watching TV in bed, you'll need to ask for one.
Guests sleep in individual pavilions, gathering in the beautiful communal lounge building for chill-out time. Delicious chef-cooked meals are served up at the enormous banqueting table which is surrounded by vistas of the tropical gardens and the tranquil sea. Relaxation levels are raised even higher by the round-the-clock security staff protecting guests within this luxury villa's walled grounds.
Once you've become accustomed to this pampered lifestyle, you may want to explore your new kingdom. Why not venture in to the gym for a workout, or down to the private beach? The famous Nirwana golf course is 15 minutes away and there are tennis courts nearby. Clinging to the rocks beyond Jiva Puri's black sand beach is a striking Balinese temple which can be visited by special arrangement.
But if that all sounds too energetic, there are plenty of ways to relax in the gentle Indonesian heat. Just lie back, sipping a cocktail on the sun deck beside the pool, or have a massage and beauty treatment in your own private spa. Heavenly.
What's IncludedPlease call us on +44 (0)203 432 7784 for rates and the latest availability.
Rates include:
Firefly pre-arrival concierge service
Airport transfers to and from Jiva Puri
Daily breakfast (choice of Asian, American or continental)
Chef service as required
Daily housekeeping
Villa concierge service
All food and beverage will be charged for separately.
RoomsVilla Jiva Puri has 4 individual sleeping pavilions plus one master pavilion, providing accommodation for a total of 10 people.
The 4 guest bedrooms each have a double bed and a private bathroom with bath and shower, stocked with designer toiletries.
The master bedroom is more spacious than the others, and has an en-suite bathroom with separate bath and shower. It also has an additional study area with desk.
All the sleeping pavilions have a safe in the room, wireless internet, air conditioning and a Bose sound system, and TVs can also be installed on request. They each have a private outdoor area.
An additional room can also be made available for an accompanying domestic helper.
EatingIf you want to feast on genuine Balinese cuisine, then Jiva Puri is the place. Having written the definitive book about Balinese Cooking, the Executive Chef is an expert traditional ingredients, flavours and techniques. As well as dining on mouth-watering dishes featuring ingredients such as peanut, lemongrass and coconut, you can learn Chef's secrets in a Balinese cookery class.
Food at Jiva Puri is not limited to local specialities however. The chefs are also adept at classic Western dishes and are used to specific dietary requirements and special requests. All food except for breakfast is charged at an additional cost, so you just need to let the chef know whether you'd like lunches, dinners or just the occasional meal. A range of menus can be suggested, and catering for events is no problem.
Soft drinks are included, but all wines and spirits are charged as extras. The service at this luxury villa is very much a la carte and bespoke to your own needs. So whether you have complicated requests, want to eat at unusual times or throw a party, we can make it happen.
ReviewsThere are currently no reviews for Jiva Puri.
KidsThe unique beauty of this part of Bali is bound to make an everlasting impression on children. From the fascinating temples to the tropical wildlife, wild forests and inviting beaches, there's plenty to explore.
Older children might like trips out to the markets of Ubud and Seminyak, or perhaps a bike ride through the countryside. A visit to an elephant safari park or a monkey sanctuary is easily arranged, and the Jiva Puri staff can arrange traditional dance and music performances at the villa.
The 30m swimming pool is ideal for kids to use up some energy before flopping in front of a DVD in the lounge pavilion. However, there are water features and ponds all around the villa grounds, so very young children will need to be supervised for their own safety.
If you need any equipment such as cots or highchairs, just let us know in advance and they can be provided.


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The goal is to develop job and social skills through on-site internship experiences that lead to competitive employment in the community.
Building Independence: Project SEARCH Overview
Meet the Class of 2016 Video
Project SEARCH benefits young adults through:
We're excited to announce Milwaukee's newest Project SEARCH location—the Milwaukee County Zoo. This unique and diverse transition experience will be available to young people motivated to work in the community.
Internship rotations at the Milwaukee County Zoo include:
|Project SEARCH Application
|Project SEARCH Eligibility Criteria
|Project SEARCH Brochure
Mission Support Center
Tel: (414) 847-4200
James O. Wright Center forWork & Training
Tel: (414) 353-6400
Metropolitan Chicago -
Tel: (312) 994-1440