Linear Time-Invariant Dynamical Systems

Transcription

1 Linear Time-Invariant Dynamical Systems CEE 629. System Identification Duke University, Fall Linearity and Time Invariance A system G that maps an input u(t) to an output y(t) is a linear system if and only if (α 1 y 1 (t) + α 2 y 2 (t)) = G[α 1 u 1 (t) + α 2 u 2 (t)] (1) where y 1 = G[u 1 ], y 2 = G[u 2 ] and α 1 and α 2 are scalar constants. If (1) holds only for α 1 + α 2 = 1 the system is called affine. The function G(u) = Au + b is affine, but not linear. A system G that maps an input u(t) to an output y(t) is a time-invariant system if and only if y(t t o ) = G[u(t t o )]. (2) Systems described by ẋ(t) = Ax(t) + Bu(t), x() = x o (3) y(t) = Cx(t) + Du(t) (4) are linear and time-invariant. variable description dimension x state vector n by 1 u input vector r by 1 y output vector m by 1 A dynamics matrix n by n B input matrix n by r C output matrix m by n D feedthrough matrix m by r

2 2 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 2 Example: a spring-mass-damper oscillator An externally-forced spring-mass-damper oscillator is described by m r(t) + cṙ(t) + kr(t) = f(t), r() = d o, ṙ() = v o. (5) Assuming a solution of the form r(t) = re λt, (mλ 2 + cλ + k) r =. This equation is valid for r =, the trivial solution, and λ = c 2m ± c 2 4m k m Note that λ can be complex, λ = σ ± iω. Now, defining the natural frequency ω 2 n = k/m, and the damping ratio ζ = c/(2 mk), we find c/(2m) = ζω n, so, and if ζ < 1, the root may be written λ = ζω n ± ζ 2 ω 2 n ω 2 n = ζω n ± ω n ζ2 1 (6) λ = ζω n ± iω n 1 ζ2. (i = 1) Note that if ζ > then σ < and the system is stable. The second-order ordinary differential equation (5) may be written as two first-order ordinary differential equations, by defining a state vector of the position and velocity, x = [r ṙ] T. d ṙ dt r = 1 k/m c/m ṙ r + 1/m f(t), r() ṙ() = d o v o (7) In terms of a desired response from this system, we may be interested in the force on the foundation, f F, and the acceleration of the mass, both of which can be computed directly through a linear combination of the states and the input. f F r = k c k/m c/m ṙ r + 1/m f(t) (8)

3 Linear Time Invariant Systems 3 A single degree of freedom oscillator and all other linear dynamical systems may be described in a general sense using state variable descriptions, ẋ(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t). x() = x o 3 Free State Response The free state response x(t) of ẋ = Ax to an initial state x o is where e At is called the matrix exponential. In Matlab, x(:,p)=expm(a*t(p))*xo; x(t) = e At x o (9) The j-th column of the matrix of free state responses X(t) = e At I n is the set of responses of all states x i, i = 1,..., n from an initial state x oj = 1 and x ok = for all k j. The i-th row of the matrix of free state responses X(t) = e At I n is the set of responses of the i-th state, from each initial state x oj = 1 and x ok = for all k j, and j = 1,..., n.. 4 Free Output Response The free output response y(t) of ẋ = Ax to an initial state x o is In Matlab, y(:,p)=c*expm(a*t(p))*xo; y(t) = Ce At x o, (1) The j-th column of the matrix of free output responses Y (t) = Ce At I n is the set of responses of all outputs from an initial condition x oj = 1 and x ok = for all k j. The i-th row of the matrix of free output responses Y (t) = Ce At I n is the set of responses of the i-th output, from each initial condition x oj = 1 and x ok = for all k j, and j = 1,..., n..

4 4 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 5 Forced Response The forced response of a linear time invariant system is y(t) = Ce At x o + C t ea(t τ) Bu(τ) dτ + Du(t). (11) An efficient method for computing y(t) for a arbitrary inputs u(t) is provided in the last sections of this document. 6 Unit Impulse Response Function The impulse response is a special case of the free and forced response, in which x o = and each u j (t) = δ(t), where δ(t) is the symmetric unit Dirac delta function and is applied equally and independently to all r inputs of the system. Each Dirac delta function is zero for t < ɛ and t > ɛ and has the following properties: ɛ δ(t) dt = 1 ɛ ɛ ɛ ɛ δ(t) dt = 1 2 ɛ δ(t) dt = 1 2 δ() = δ(t τ) f(τ) dτ = f(t) ɛ δ(t τ) f(τ) dτ = 1 2 f(t) ɛ δ(t τ) f(τ) dτ = 1 2 f(t) From the arbitrary forced response of an LTI system, equation (11), the unit impulse response matrix of any LTI system initially at rest is, H(t) = C t ɛ ea(t τ) B δ(τ) dτ From the commutative property of convolution and the properties of the Dirac delta function, above, the unit impulse response for all t > ɛ is H(t) = Ce At B. (12)

5 Linear Time Invariant Systems 5 The responses H(t) of the system are called the unit impulse response function The i, j element of H(t) is the response of output i due to a unit impulse at input j. Note that the impulse response is a special case of the free response. In other words, if there is a vector v such that x o = Bv, the free response and the impulse response are equivalent. In other words, the input matrix B forms a basis for the initial condition that produces the same free response as the unit impulse response. Note that there can be initial conditions x o which do not equal Bv because B does not necessarily span R n. The set of free responses can therefore be much richer than the set of impulse responses. In Matlab, H(p,:,:)=C*expm(A*t(p))*B; 7 The Matrix Exponential The matrix exponential is defined for A R n n as e A = A k /k! = I + A + AA/2 + AAA/6 + AAAA/24 + (13) k= Properties of the matrix exponential: If A = A T then e A >. If A = A T then [e A ][e A ] T = I e At e Bt = e (A+B)t [e A ] T = e AT [e A ] 1 = e A e T 1AT = T 1 e A T for any square invertible matrix T R n n. e T 1 AT = I + T 1 AT T 1 AT T 1 AT T 1 AT T 1 AT T 1 AT + e T 1 AT = T 1 T + T 1 AT T 1 AAT T 1 AAAT + = T 1 [I + A + AA/2 + AAA/6 + ]T = T 1 e A T. (14)

6 6 CEE 629. System Identification Duke University Fall 217 H.P. Gavin det(e At ) = e trace(at) e At A 1 = A 1 e At rank(e At ) = n for any A R n n, regardless of the rank of A. d dt eat = Ae At e At = I + At AAt AAAt AAAAt4 + d dt eat = A + AA AAAt AAAAt3 + = A[I + At AAt AAAt AAAAt4 + ] d dt eat = Ae At (15) A t e Aτ dτ = e At I A t A eaτ dτ = t A t t d dτ eaτ dτ = e Aτ t = e At e A = e At I eaτ dτ = e At I (16) e Aτ dτ = e At I (17)

7 Linear Time Invariant Systems 7 t e A(t τ) dτ = A 1 (e At I) t t ea(t τ) dτ = eat e Aτ dτ = e At t e Aτ dτ = e At ( A 1 ) ( e At I ) = e At A 1 e At + e At A 1 = A 1 Ae At A 1 e At + A 1 Ae At A 1 = A 1 e AAA 1t e At + A 1 e AAA 1 t = A 1 e At e At + A 1 e At = A 1 + A 1 e At = A 1 ( e At I ) (18) = t eaτ dτ (19) The solution to ẋ(t) = Ax(t) + Bu(t), x() = x o, for t is x(t) = e At x o + t ea(t τ) Bu(τ) dτ. (2) As a special case of the above, if the input to the system is vδ(t) where v is a m by 1 vector and δ(t) is the Dirac delta function, and if x o =, then the output y(t) is called the unit impulse response function, h(t). h(t) = Ce At Bv. (21) With definitions of the natural frequency, ω n = k/m, the damping ratio, ζ = c/(2 mk), and the damped natural frequency, ω d = ω n ζ2 1, let A = 1 k/m c/m = 1 2ζω n ω 2 n. (22) For this dynamics matrix, the matrix exponential depends on the damping ratio, ζ, as follows:

8 8 CEE 629. System Identification Duke University Fall 217 H.P. Gavin damping damping ratio e At undamped ζ = e At = [ cos ω n t 1 ω n sin ω n t ω n sin ω n t cos ω n t ] under-damped < ζ < 1 e At = e ζωnt cos ω dt + ζ sin ω 1 1 ζ 2 dt ω d sin ω d t ω d sin ω 1 ζ 2 dt cos ω d t ζ sin ω 1 ζ 2 dt [ 1 + critically damped ζ = 1 e At = e ωnt ωn t t ωnt 2 1 ω n t ] over-damped ζ > 1 e At = e ζωnt cosh ω dt + ζ sinh ω 1 1 ζ 2 dt ω d sinh ω d t ω d sinh ω ζ2 dt cosh ω d t ζ sinh ω 1 ζ2 dt 1

9 Linear Time Invariant Systems 9 8 Eigenvalues and Diagonalization Consider the dynamics matrix A of a linear time invariant, (LTI) system. If the input, u(t) is zero, then ẋ = Ax. Assuming a solution of the form x(t) = xe λt, and substituting this solution into ẋ = Ax, results in: or xλe λt = A xe λt xλ = A x, (23) which is a standard eigenvalue problem, in which x is the eigenvector and λ is the eigenvalue. If A is a n n matrix, then there are n eigenvalues, λ 1,, λ n and n associated eigenvectors, x 1,, x n. For the dynamics matrix given in equation (7), there are two eigenvalues. λ 1,2 = c 2m ± c 2 4m k (24) m = ζω n ± ω n ζ2 1 (25) The dynamics matrix contains all the information required to determine the natural frequencies and damping ratios of the system. The n eigenvectors can be assembled, column-by-column into a matrix, X = [ x 1 x 2 x n ]. Pre-multiplying the eigen-problem by X 1, X 1 A X = X 1 XΛ = diag(λi ) = λ 1... λ n This is called a diagonalization of the dynamics matrix A. Now, consider the linear transformation of coordinates, x(t) = Xq(t), q(t) = X 1 x(t). Substituting this change of coordinates into equation (3), X q(t) = A Xq(t) + Bu(t),

10 1 CEE 629. System Identification Duke University Fall 217 H.P. Gavin and pre-multiplying by X 1 q = X 1 A Xq + X 1 Bu, or q = λ 1... λ n q + X 1 Bu, q() = X 1 x o q = Āq + Bu, q() = X 1 x o (26) y = C Xq + Du y = Cq + Du (27) The n differential equations q i = λ i q i + B i u are uncoupled. The state q i (t) is independent of all the other states q j (t), j i. Note that if Λ is complex, so are X, Ā, B, and C. Consider one of the un-coupled equations from equation (26), for the unforced case u = q i (t) = λ i q i (t), q i () = 1. This equation has a solution q i (t) = e λ it, where λ i is, in general, a complex value, and q i + q i is real-valued. λ i = σ i ± iω i, q i (t) + q i (t) = 1 2 eλ it eλ i t = 1 2 eσ it (cos ω i t + i sin ω i t) eσ it (cos ω i t i sin ω i t) = e σ it cos ω i t,

11 Linear Time Invariant Systems 11 9 Jordan Forms If a square matrix A R n n has distinct eigenvalues, then it can be reduced to a diagonal matrix through a similarity transformation X 1 A X = Λ = λ λ n where X R n n is the matrix of linearly independent eigenvectors of A. (28) If A has repeated eigenvalues, the it can be reduced to a diagonal matrix only if all n eigenvectors are linearly independent. If A has repeated eigenvalues and two or more of the eigenvectors associated with the repeated eigenvalues are not linearly independent, then it is not similar to a diagonal matrix. It is, however, similar to a simpler matrix called the Jordan form of A, X 1 A X = J J n (29) where the square submatrices in J i = λ i 1 λ i 1 λ i λ i 1 λ i (3) are called Jordan blocks. In a Jordan block, repeated eigenvalues are on the diagonal and 1 s are just above the diagonal. Consider an eigenvalue λ i of matrix A R n n with multiplicity k. There are k eigenvectors, x i, associated with the eigenvalue λ i. If all pairs of these

12 12 CEE 629. System Identification Duke University Fall 217 H.P. Gavin k eigenvectors are linearly dependent, (there exist real values α i such that x 1 = α i x i i = 2,..., k), then this set of linearly dependent eigenvectors span a one-dimensional subspace, and a single Jordan block of dimension k k will be associated with the eigenvalue λ i. On the other hand, if the eigenvalue λ i has b i linearly independent eigenvectors, then there will be b i separate Jordan blocks associated with λ i. The similarity transformation into Jordan form X 1 A X = J may be written A X = X J, in which the columns of X are [ x1 x 2 x n ]. So A[ x 1 x 2 x n ] = [ x 1 x 2 x n ]J (31) Writing the columns of X associated with a single Jordan block of λi, A[u 1 u 2 u 3 u k ] = [u 1 u 2 u 3 u k ] λ i 1 λ i 1 λ i λ i 1 λ i k k [Au 1 Au 2 Au 3 Au k ] = [u 1 λ i u 1 + u 2 λ i u 2 + u 3 λ i u k 1 + u k λ i ], and associating columns of the left and right hand side of this equation, Au 1 = u 1 λ i [A λ i I]u 1 = Au 2 = u 1 + u 2 λ i [A λ i I]u 2 = u 1 Au 3 = u 2 + u 3 λ i [A λ i I]u 3 = u 2 [A λ i I] 2 u 3 = u 1.. Au k = u k 1 + u k λ i [A λ i I]u k = u k 1 [A λ i I] k 1 u k = u 1 The size of the Jordan block is k k where u k is not linearly independent of u k+1. An equivalent criterion is rank[a λ i I] k = rank[a λ i I] k+1. (32) So, to find the vectors of the transformation matrix X in a particular Jordan block of λ i, start first by finding k such that the above equation is satisfied.

13 Linear Time Invariant Systems 13 The next step is to find a vector in the null space of [A λ i I] k. start by setting u 1 equal to one of the eigenvectors of λ i. Then iterate on u n = [A λ i I] 1 u n 1 (33) until u n and u n 1 are not linearly independent. If there is more than one Jordan block associated with the repeated eigenvalue λ i, the transformation vectors associated with the remaining Jordan blocks may be found by restarting the iterations with the other eigenvectors associated with λ i. If A R n n and k is the multiplicity of the eigenvalue λ i, then the size of the largest Jordan block of eigenvalue λ i is k such that rank(a λ i I) k = rank(a λ i I) k+1. (34) For a similar derivation of this see pages of C-T Chen, Introduction to Linear Systems Theory, Hold, Rinehart and Winston, 197. If  Rn n has the Jordan block form λ 1 1 λ 1 1 λ λ 1 1 λ 1 then eât = n n e λ 1t t e λ 1t t 2 e λ1t /2! t 3 e λ1t /3! t n 1 e λ1t /(n 1)! e λ 1t t e λ 1t t 2 e λ1t /2! t n 2 e λ1t /(n 2)! e λ 1t t e λ 1t t n 3 e λ1t /(n 3)! e λ 1t t e λ 1t e λ 1t n n (35) (36)

14 14 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 1 Transfer Function and Frequency Response Function Taking the Laplace transform of equation (3), (and considering the particular part of the solution, i.e., ignoring the effects of initial conditions) gives s x(s) = Ax(s) + Bu(s), (37) y(s) = Cx(s) + Du(s), (38) which can be written as y(s) in terms of u(s) as follows, y(s) = H(s) u(s) = [C[sI A] 1 B + D] u(s) (39) This is the transfer function of the linear dynamic system. The same may be carried out on any transformed equation of state, Ā = T 1 AT, B = T 1 B, C = CT (e.g., equations (26) and (27)), resulting in y(s) = H(s) u(s) = [ C[sI Ā] 1 B + D] u(s). Recall that the matrix Λ = diag(λ i ), so, H(λ i ) =. Equation (39) may be used to determine the complex-valued frequency response function of any dynamic system, by evaluating the transfer function along the line s = iω. y(ω) = H(ω) u(ω) = [C[iωI A] 1 B + D] u(ω) (4) Assuming that the inputs u(t) are sinusoidal with frequency ω and unit amplitude, the magnitude of the frequency response function, H(ω) = Re[H(ω)] 2 + Im[H(ω)] 2 gives the amplitude of the responses y(t). The phase of H(ω), H(ω) = arctan Im[H(ω)] Re[H(ω)] gives the phase angle between the input u(t) and the output y(t). So if u(t) = cos(ωt), then y(t) = H(ω) cos(ωt + H(ω)). A graph of H(ω) and H(ω) is called a Bode plot. In Matlab, [mag,pha] = bode(a,b,c,d); generates a Bode plot for a system defined by matrices A, B, C, and D.

15 Linear Time Invariant Systems Singular Value Spectra In systems with multiple inputs and multiple outputs (MIMO), the strength of the output depends upon the relative amplitudes and phases of the inputs. It is possible that the effect of two non-zero inputs properly scaled in amplitude and phase, could have a relatively small effect on the responses, while a different combination if inputs could have a very strong affect on the system response. Further, for harmonic inputs (and outputs), the scaling and phasing of the inputs to achieve a strong or weak response is dependent upon the frequency. The singular value decomposition of the frequency response function matrix y(ω) = H(ω) (m r) u(ω) = U H (ω) Σ H (ω) V T H (ω) u(ω) = k σ Hk (ω)[u Hk (ω)v T Hk(ω)] u(ω) = k σ Hk (ω)u Hk (ω) (v T Hk(ω) u(ω)) (41) provides the means to assess how inputs can be scaled for maximal or minimal effect. Here, columns of U H (ω) and V H (ω) are the left and right singular vectors of H(ω). For complex-valued frequency responses, singular vectors are complex-valued. (The singular values are always real (σ H1 σ H2... )). For inputs u(ω) proportional to v H1 (ω), y(ω) 2 is maximized and is proportional to u H1 (ω). If r < m and σ Hr > then inputs proportional to V Hr (ω) have the weakest coupling to the outputs. On the other hand, if r > m there is some linear combination of inputs that (at a particular frequency ω) will result in no output at all. Right singular vectors spanning this kernal of H(ω), [v H(m+1) (ω),..., v Hr (ω)], is an ortogonal basis for these inputs Plots of σ H1 (ω) and σ Hr (ω) indicate the frequency-dependence of the largest amplification and the smallest amplification for any linear combination (and phasing) of inputs, in the context of steady-state harmonic response.

16 16 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 12 Zeros and Poles of MIMO LTI Systems In the Laplace domain, the dynamics equation is sx(s) = Ax(s) + Bu(s), or [ si A, B ] If there is a value of s such that n (n+r) x(s) u(s) rank([si A, B]) < n, =. then there is a non-zero u(s) for which x(s) can not be uniquely determined (x(s) could be zero in one or more components). Similarly, the output response in the Laplace domain is sx(s) = Ax(s), y(s) = Cx(s), or si A x(s) =. C y(s) If there is a value of s such that rank (n+m) n si A C < n then the non-trivial null space of this matrix, x(s), corresponds to y(s) =. In other words, there is a subspace of the state-space that does not couple to the output. These rank conditions for controllability and observability are called the Popov-Belevitch-Hautus (PBH) tests. Note that for any s λ i (the eigenvalues of A), rank(si A) = n, and that rank(λ i I A) < n. If the columns of B do not span the N (λ i I A), then u(s) does not couple to the i-th mode of the system, and λ i is an input-decoupling zero. If a system has one or more input decoupling zeros, then there are inputs that can not affect a subspace of the state space, and the system is called uncontrollable. Similarly, if the rows of C do not span N (λ i I A T ), then y(s) does not couple to the i-th mode of the system, and λ i is an output-decoupling zero. If a system has one or more output decoupling zeros, then there are states that do not affect the output, and the system is called unobservable.

17 Linear Time Invariant Systems 17 The matrix-valued MIMO transfer function may be represented as a system of two sets of linear equations in the Laplace domain x(s) = (si A) 1 Bu(s) y(s) = Cx(s) + Du(s), (42) which is equivalent to si A B C D x(s) u(s) = y(s). (43) This is called the Rosenbrock System Matrix (RSM) formulation. For a system with a zero output, A B C D + s I x(s) u(s) = (44) which is a generalized eigenvalue probelem when D is square. Note that in this eigenvalue problem, the matrix multiplying the eigenvalue s is not invertible, and requires a numerical method such as the QZ decomposition. 1 Eigenvectors corresponding to finite values of s satisfying this generalized eigenvalue problem, define the magnitudes and phases of inputs u(s) (and of the corresponding states x(s)), such that the output y(s) is zero. In other words, at values of s for which the rank of the Rosenbrock System Matrix is less than (n + min(rankb, rankc)), there exists a set of non-zero inputs such that the output is zero. Values of s satisfying the generalized eigenvalue probem (44) are called invariant zeros. 2 Eigenalues of A that are not also zeros are called poles. 1 C.B. Moler and G.W. Stewart, An Algorithm for Generalized Matrix Eigenvalue Problems, SIAM J. Numer. Anal., 1(2) (1973), H.H. Rosenbrock, The zeros of a system, Int l J. Control, 18(2) (1973):

18 18 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 13 Asymptotic Stability of LTI Systems The dynamics matrix fully specifies the stability properties of LTI systems. The following statements are equivalent: The LTI system is asymptotically stable. All the eigenvalues of the dynamics matrix have a negative real part. There exists a matrix R > such that the matrix P > solves: = A T P + P A + R. (45) The third statement is a result of Lyapunov stability theory. Define the energy in a system as V (x(t)) = x(t) T P x(t) where P >. The equation V (x(t)) = x(t) T P x(t) is called a Lyapunov function. The rate of change of this energy is V (x(t)) = x(t) T P ẋ(t) + ẋ(t) T P x(t). Substituting the free response dynamics, ẋ(t) = Ax(t), into V, V (x(t)) = x(t) T A T P x(t) + x(t) T P Ax(t) = x(t) T [A T P + P A]x(t) So if A T P + P A is negative definite, V (x(t)) decreases monotonically, for any non-zero value of the state vector. The condition > A T P +P A is equivalent to = A T P + P A + R for a matrix R >. The equation A T P + P A + R = is called a Lyapunov equation. The following statements are equivalent: A is asymptotically stable all eigenvalues of A have real parts that are negative R > s.t. P > satisfies the Lyapunov equation A T P + P A + R =. R > s.t. the integral P = e ATt Re At dt converges and satisfies the Lyapunov equation A T P + P A + R =.

19 Linear Time Invariant Systems Observability and Controllability If the solution P to the left Lyapunov equation = A T P + P A + C T C (46) is positive definite, then the system defined by A and C is called observable, meaning that the initial state can be inferred from the time series of free responses, y(t) = Cx(t) The matrix of free output responses from independent initial conditions on every state, x() = I n is Y (t) = Ce At. This matrix of independent free responses has m rows and n columns. The covariance of Y T (t) is called observability gramian and solves the left Lyapunov equation, above. P = Y T (t)y (t) dt = If the solution Q to the right Lyapunov equation (eatt C T )(Ce At ) dt (47) = AQ + QA T + BB T (48) is positive definite, then the system defined by A and B is called controllable, meaning that the controls u acting on the system ẋ = Ax + Bu can return the system to x = from any initial state x() in finite time. The matrix of state response sequence from independent impulses on each input, u(t) = I r δ(t) is X(t) = e At B. This matrix of independent state impulse responses has n rows and r columns. The covariance of X(t) is called the controllability gramian. and solves the right Lyapunov equation, above. Q = X(t)X T (t) dt = (eat B)(B T e ATt ) dt (49) These Lyapunov equations and gramians are useful in determining the norm of LTI systems.

20 2 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 15 H 2 norms of Continuous-Time LTI systems Consider the MIMO system transfer function of a dynamic system, y(s) = H(s)u(s). The H 2 norm provides a way to measure the size of H(s). 3 There are three ways to view, motivate, define, or interpret the H 2 norm The Frobeneus Norm The H 2 norm of MIMO systems is defined in terms of the Frobeneus norm. The Frobeneus norm of a matrix is the square-root of the sum of the squares of all the terms in the matrix. A F = n,m 1/2 A 2 ij i,j = [A 2 1,1 + A 2 1,2 + + A 2 1,n + A 2 2,1 + A 2 2,2 + + A 2 2,n +... A 2 m,1 + A 2 m,2 + + A 2 m,n] 1/2 If a matrix A is real (A R n m ) then the Frobeneus norm of A is A F = [ tr AA T] 1/2 If a matrix A is complex (A C n m ) then the Frobeneus norm of A is A F = [tr AA ] 1/2 The Frobeneus norm of A is the same as the Frobeneus norm of A T. A F = A T F = [tr AA T ] 1/2 = [tr A T A] 1/2 3 The objective function for linear quadratic control synthesis is the H 2 norm of the closed-loop transfer function.

21 Linear Time Invariant Systems The H 2 norm of a system in terms of unit impulse response The H 2 norm of an LTI system is defined in terms of the Frobeneus norm of its matrix of unit impulse response functions H(t). H 2 2 = = H(t) 2 F dt (5) tr [H T (t)h(t)] dt where P is the observability gramian, = tr (BT e ATt C T )(Ce At B) dt = tr B T (eatt C T )(Ce At ) dt B = tr B T P B (51) P = (eatt C T )(Ce At ) dt 15.3 The H 2 norm of a system in terms of unit white noise response The H 2 norm of an LTI system may also be defined in terms of the covariance of system responses y(t) to standard white noise. H 2 2 = lim E [y T (t)y(t)] t (52) = lim E tr [y T (t)y(t)] t ( ) (53) = lim E tr [y(t)y T (t)] t = lim E y(t)y T (t) 2 t F = lim E tr [Cx(t)x T (t)c T ] t = lim tr C E [x(t)x T (t)]c T t = lim tr CQ(t)C T t where Q(t) = E [x(t)x T (t)] is the (non-negative definite) state covariance matrix. Defining Q to be the limit of Q(t) as t approaches infinity, An r dimensional unit white noise process u(t), H 2 2 = CQC T (54) u(t) = w 1 (t). w m (t)

22 22 CEE 629. System Identification Duke University Fall 217 H.P. Gavin has scalar components w i (t) having the following properties: E [w i (t)] = t... expected value E [w i (t 1 )w i (t 2 )] = δ(t 1 t 2 ) t 1, t 2... auto correlation E [w i (t 1 )w j (t 2 )] = i, j, t 1, t 2 and i j... cross correlation These properties, combined, result in facts that E [u(t 1 )u T (t 2 )] = δ(t 1 t 2 )I r and that the power spectral density of unit white noise is 1 for all frequencies. Now, examining the time-evolution of the state covariance matrix, we will see that the state covariance is the controllability gramian, in the limit as t. Q(t) = d dt E [x(t)xt (t)] = E [ẋ(t)x T (t) + x(t)ẋ T (t)] = E [(Ax + Bu)x T (t) + x(t)(ax + Bu) T ] = E [Axx T + xx T A T ] + E [Bux T + xu T B T ] = AQ(t) + Q(t)A T + [ ( E Bu(t) e At x o + t e A(t τ) Bu(τ) dτ ) T + = AQ(t) + Q(t)A T + [ t E B u(t)u T (τ)b T e AT (t τ) dτ + = AQ(t) + Q(t)A T + B t δ(t τ)b T e AT (t τ) dτ + t = AQ(t) + Q(t)A T BBT BBT t ( t ) e At x o + e A(t τ) Bu(τ) dτ e A(t τ) Bu(τ)u T (t) dτb T ] e A(t τ) Bδ(t τ) dτb T u T B T ] = AQ(t) + Q(t)A T + BB T (55) the solution to this differential equation is Q(t) = e At Q()e ATt + where the initial state covariance matrix Q(), is t Q() = E [x()x T ()] eaτ BB T e ATτ dτ

23 Linear Time Invariant Systems 23 For asymptotically stable systems, e At approaches zero as t approaches infinity, and the transient response e At Q()e ATt also approaches zero. Hence lim Q(t) = t t eaτ BB T e ATτ dτ which is the definition of the controllability gramian. Equivalently, for asymptotically stable systems, lim t and Q is the controllability gramian. Q(t) = = AQ + QA T + BB T Using the property of the Frobeneus norm that A F = A T F, the two interpretations of the H 2 norm of of an LTI system presented in the preceding pages (the interpretation in terms of white noise and the interpretation in terms of unit impulse responses) can be shown to be equivalent. where H 2 2 = tr B T P B = tr CQC T, = A T P + P A + C T C and = AQ + QA T + BB T Proof 1: H 2 2 = H(t) 2 F dt = tr (CeAt B) T (Ce At B) dt = tr B T e ATt C T Ce At dt B = tr B T P B = tr C e At BB T e ATt dt C T = tr CQC T Proof 2: tr(b T P B) = tr(bb T P ) = tr(( AQ QA T )P ) = tr( AQP QA T P ) = tr( QP A QA T P ) = tr(qc T C) = tr(c T QC)

24 24 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 15.4 The H 2 norm of a system in terms of frequency response The third interpretation of the H 2 norm is in the frequency domain. This interpretation is an expression of Parseval s Theorem, H(t) 2 F dt = 1 2π H(iω) 2 F dω (56) To prove that this identity is true, we need a couple of facts. First we can see that if A is asymptotically stable, then e At dt = A 1. Proof: e At dt = A 1 Ae At dt = A 1 ( ) d dt eat dt = A 1 e At] = A 1 Second we show that the Laplace transform of e At is (si A) 1 L{e At } = e At e st dt = e (A si)t dt = (si A) 1 Third we show that the Laplace transform of H(t) is C(sI A) 1 B L{H(t)} = L{Ce At B} = CL{e At }B = C(sI A) 1 B ( ( lim ) e At I) t Fourth, we show that if Q satisfies = AQ + QA T + BB T, then Q is given by Q = 1 2π (iωi A) 1 BB T (iωi A) dω for the special case in which A has been diagonalized, and BB T = I n. Proof: Q = 1 2π = 1 2π = 1 2π = 1 2π (iωi A) 1 BB T (iωi A) dω [( iωi A)(iωI A)] 1 dω [ω2 I + A 2 ] 1 dω diag = 1 2π π( A) 1 = 1 2 A 1, 1 ω 2 + A 2 ii dω

25 Linear Time Invariant Systems 25 and plugging into AQ + QA T + BB T gives, A ( 1 ) 2 A 1 + ( 1 ) 2 A 1 A + I = Now, examining the left hand side of the Parseval equality, tr H(t)H T (t)dt = tr And examining the right hand side of this equality, 1 2π tr H(ω)H (ω) dω Ce At BB T e ATt C T dt = tr CQC T. = 1 2π tr C(ωI A) 1 BB T (ωi A) C T dω = 1 2π tr C (iωi A) 1 BB T (iωi A) dω C T = tr CQC T In summary, the H 2 norm of a system can be equivalently interpreted as the integral of the sum of the squares of the unit impulse responses the covariance of the outputs responding to uncorrelated unit white noise the sum of the areas under the magnitude-squared frequency response functions These norms can be calculated by solving Lyapunov equations for the controllability gramian and the observability gramian. Note, finally, that the H 2 norm of a continuous-time LTI system with D does not exist. Continuous-time LTI systems with D = are called strictly proper.

26 26 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 16 Transforming Continuous Time Systems to Discrete-Time Systems (ZOH) Consider the evolution of the state response from a time t to a time t + t for inputs u(τ) that equal u(t) for t τ t+ t. This is a zero-order hold (ZOH) on u(t) over the interval [t, t + t]. The initial condition to this evolution is x(t), and we wish to find the states x(t + t). Shifting time to be at time t in equation (2) and defining x(k t) x(k), x((k + 1) t) x(k + 1), and, u(τ) u(k t) u(k). for τ t, gives x(k + 1) = e A t x(k) + t e A( t τ) dτ Bu(k) = [ e A t] x(k) + [ A 1 ( e A t I ) B ] u(k), = [A d ] x(k) + [B d ] u(k) (57) where A d and B d are the discrete-time dynamics matrix and the discrete-time input matrix, A d = e A t and B d = A 1 (A d I)B. Note that B d exists even though A may not be invertible. Consider the diagonalization A = X 1 Λ X. The inverse of A may be expressed A 1 = X 1 Λ 1 X, where X X 1 = I. So, A 1 (e At I) = X 1 Λ 1 X X 1 (e Λt I) X = X 1 Λ 1 (e Λt I) X, which contains diagonal terms (e λ i 1)/λ i, and e λ i lim 1 = 1. λ i λ i To compute B d without inverting A, note the following: e A t I = A t + t2 2 A 1 (e A t I) = t + t2 B d = A 1 (e A t I)B = tb + t2 2 and, if M = t3 t4 AA + AAA AAAA + 2 A + t3 6 A n n AA + t4 24 AAA + t3 t4 AB + AAB AAAB +, B n r r n r r, (58)

27 Linear Time Invariant Systems 27 then, e M t so = I + tm + t2 t3 t4 MM + MMM MMMM + = I + t A B + t2 AA AB + t3 AAA AAB = I + ta + t2 2 t2 AA + tb + 2 e M t = A d B d AB + t3 6 AAB +,, (59) and the discrete-time dynamics matrix A d and input matrix B d may be computed using a single matrix exponential computation. In Matlab: [n,r] = size(b); M = [ A B ; zeros(r,n+r) ]; emdt = expm(m*dt); Ad = emdt(1:n,1:n); Bd = emdt(1:n,n+1:n+r); With the matrices A d and B d, the transient response may be computed digitally using equation (57). x(:,1) = x; for p=1:points-1 x(:,p+1) = Ad * x(:,p) + Bd * u(:,p); end

28 28 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 17 Transforming Continuous Time Systems to Discrete-Time Systems (FOH) Using the same time-shifting as in the previous ZOH derivation, but now specifying that the input changes linearly over a time increment, t, u(τ) = u(k) + 1 t (u(k + 1) u(k)) τ u(k) + u (k) τ for τ t, (6) This is a first order hold (FOH) on the inputs over [t, t + t]. x(k t) x(k), u(k t) = u(k) gives Defining x(k + 1) = e A t x(k) + = e A t x(k) + t t e A( t τ) B u(τ) dτ e A( t τ) dτ B u(k) + t e A( t τ) B u (k) τ dτ = e A t x(k) + A 1 ( e A t I ) B u(k) + A 2 ( e A t I A t ) B u (k) = [ e A t] x(k) + [ A 1 ( e A t I ) B ] u(k) + [ A 2 ( e A t I A t ) B ] u (k), = [A d ] x(k) + [B d ] u(k) + [B d] u (k) (61) where A d, B d, and B d are the discrete-time dynamics matrix and the discretetime input matrices. The input matrices B d and B d exist even though A may not be invertible. To compute B d without inverting A, note the following: e A t I = A t + t2 2 A 1 (e A t I) = t + t2 B d = A 1 (e A t I)B = tb + t2 2 and, t3 t4 AA + AAA AAAA + 2 A + t3 6 AA + t4 24 AAA + t3 t4 AB + AAB AAAB +, e A t I A t = A 2 (e A t I A t) = B d = A 2 (e A t I A t)b = t2 t3 t4 AA + AAA AAAA + t2 2 + t3 6 A + t4 24 AA + t2 2 B + t3 t4 AB AAB +,

29 Linear Time Invariant Systems 29 If then, M = A n n B n r n r r n r r I r r r n r r r r (n+2r) (n+2r), (62) e M t = I + tm + t2 2 so, = I + t = A B I I + ta + t2 2 t3 t4 MM + MMM MMMM + + t2 2 AA AB B AA + tb + t2 2 + t3 6 AAA AAB AB + AB + t3 6 AAB + t 2 2 B + t3 6 AB + e M t = A d B d B d,, (63) and the discrete-time dynamics matrix A d and input matrices B d and B d may be computed using a single matrix exponential computation. The discrete-time system with ramp inputs between sample points is then x(k + 1) = A d x(k) + [B d B d/( t)] u(k) + [B d/( t)] u(k + 1) = A d x(k) + B d u(k) B d1 u(k + 1) (64) y(k + 1) = C x(k + 1) + D u(k + 1) (65) In this system, y(k+1) depends on x(k+1), which in turn depends on u(k+1). So even if D =, u(k+1) feeds through to y(k+1). This discrete-time system can be expressed in a new state, x(k) = x(k) B d1 u(k), giving x(k + 1) = A d x(k) + B d u(k) y(k) = C x(k) + D u(k) (66) where B d = B d + A d B d1 and D = D + CB d1.

30 3 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 18 Discrete-Time Linear Time Invariant Systems The transformation from continuous-time to discrete-time models provides a convenient method for digital simulation of transient system responses. In addition, the dynamics of discrete-time systems may be investigated analytically. This section summarizes analytical solutions to discrete-time LTI systems described by x(k + 1) = A x(k) + B u(k) y(k) = C x(k) + D u(k) (67) Discrete-time models imply that the frequencies of harmonic components of dynamic variables are limited to the Nyquist interval, ω [ π/( t), +π/( t)]. Harmonic components with frequencies outside this range are aliased into the Nyquist interval, when they are sampled. Frequency components of analog signals outside the Nyquist range should be filtered-out with so-called (lowpass) anti-alias filters prior to sampling. Hereinafter, the discrete-time dynamics matrix and input matrix are denoted A and B. Expressions in this section involving the continuous-time dynamics and input matrices will have a subscript c, as in A c, B c, and λ c. 19 State Response Sequence Applying the dynamics equation of (67) recursively from a known initial state x() and with a known input sequence, x(1) = Ax() + Bu() x(2) = A 2 x() + ABu() + Bu(1) x(3) = A 3 x() + A 2 Bu() + ABu(1) + Bu(2). x(k) = A k x() + A k 1 Bu() + + ABu(k 2) + Bu(k 1) (68) or, starting at any particular time step, j > and advancing by k time steps, x(j + k) = A k x(j) + k A i 1 Bu(k + j i) (69) i=1 y(j + k) = CA k x(j) + k CA i 1 Bu(j + k i) + Du(j + k) (7) i=1

31 Linear Time Invariant Systems 31 The first terms in (69) and (7) are the free state responses of the state and the output (the homogeneous solution) to the difference equations; the second terms are the forced responses (the particular solution). 2 Eigenvalues and Diagonalization in Discrete-Time The discrete-time system may be diagonalized with the eigenvector matrix of the discrete-time dynamics matrix. X 1 A X = Λ = λ 1... where Λ is the diagonal matrix of discrete-time eigenvalues and the eigenvalue problem is A X = XΛ. Note that the eigenvalues of the discrete-time dynamics matrix are related to the eigenvalues of the continuous-time dynamics matrix through the scalar exponential exp[λ ci t] = λ i. The eigenvectors X of the continuous-time dynamics A are the same as those of the discrete-time dynamics A. The diagonalized system (in modal coordinates) is λ n q(k + 1) = Λ q(k) + X 1 B u(k) y(k) = C X q(k) + D u(k) (71) and the modal response sequence from time step k to time step j is q(j + k) = Λ k q(j) + k Λ i 1 X 1 B u(j + k i) (72) i=1 The stability of a mode of a discrete-time system is determined from the magnitude of the eigenvalue. Considering the free responses of (71) or (72), if λ i > 1 then q i (k) will grow exponentially with k.

32 32 CEE 629. System Identification Duke University Fall 217 H.P. Gavin mode i is stable: λ i < 1 Re(λ ci ) < mode i is unstable: λ i > 1 Re(λ ci ) > A system is stable if and only if all of its modes are stable. Note that the elements of the modal sequence vector q(k+j) may be evaluated individually, since Λ j is diagonal. For systems with under-damped dynamics the continuous-time and discrete-time eigenvalues are complex-valued. The eigenvalues and modal coordinates appear in complex-conjugate pairs. The sum of the complex conjugate modal coordinates q(k) + q (k) is real valued and equals twice the real part of q(k). 21 Discrete-time convolution and Unit Impulse Response Sequence (Markov Parameters) Applying equations (67) and (68) to derive the output sequence y(k) from x() = and u(k) = k <, y() = Du() y(1) = CBu() + Du(1) y(2) = CABu() + CBu(1) + Du(2) y(3) = CA 2 Bu() + CABu(1) + CBu(2) + Du(3). y(k) = k i=1 CA i 1 B u(k i) + Du(k) This is a discrete-time convolution, and it may be re-expressed as y(k) = k 1 i= (73) Y (i) u(k i) (74) where the kernel terms of this convolution are called Markov parameters: Y () = D, Y (k) = CA k 1 B k > (75) The first Markov parameter, Y (), is the feed-through matrix, D. The output evolving from a zero initial state and a unit impulse input (u() = I r, u(k) = for k ) is the sequence of Markov parameters.

33 Linear Time Invariant Systems 33 The scalar-valued sequence of the (p, q) terms of the sequence of matrix-valued Markov parameters, [Y ()] p,q, [Y (1)] p,q, [Y (2)] p,q, is the output element p from a unit impulse at input element q. For asymptotically stable (AS) discrete-time LTI systems, free responses decay asymptotically to zero. So, in principle, the sequence of Markov parameters is infinitely long. But, note that for x() = and u(k) = k <, i= Y (i) u(k i) = k Y (i) u(k i) (76) i= A forced response sequence evolving from x() = with u(k) = k <, is linear in the Markov parameters. [ y() y(1) y(2) y(3) y(j) ] m (j+1) = [ D CB CAB CA 2 B CA j 1 B ] m r(j+1) u() u(1) u(2) u(3) u(j) u() u(1) u(2) u(j 1) u() u(1) u(j 2) u() u(j 3).... u() r(j+1) (j+1) (77)

34 34 CEE 629. System Identification Duke University Fall 217 H.P. Gavin The forced response sequence of a system evolving from x() = with an (assumed) finite sequence of K + 1 Markov parameters is or y(k) = K Y (i) u(k i) i= [ y(k) y(k + 1) y(k + 2) y(k + 3) y(n) ] m (N K+1) = [ D CB CAB CA 2 B CA K 1 B ] m r(k+1) u(k) u(k + 1) u(k + 2) u(k + 3) u(n) u(k 1) u(k) u(k + 1) u(k + 2) u(n 1) u(k 2) u(k 1) u(k) u(k + 1) u(n 2) u(k 3) u(k 2) u(k 1) u(k) u(n 3)..... u() u(1) u(2) u(3) u(n K) (78) r(n K+1) (N K+1) Given input/output time series u(k) and y(k), (k =...N) and an assumption for the model order K, equation (78) provides a linear model for the sequence of K + 1 Markov parameters. In this estimation problem, there are m (N K+1) equations and m r(k+1) unknown model coefficients. The estimation of Markov parameters is a time-domain approach to Wiener filtering and is the first step in the Eigensystem Realization Algorithm (ERA) for identification of state-space models. If we know that the MIMO system to be strictily proper (D = (m r) ) then the model may be expressed without the feedthrough matrix (Y ()) by removing D from the set of Markov parameters and removing the first r rows from the matrix of input data.

35 Linear Time Invariant Systems Frequency Response in Discrete-Time Consider the steady-state harmonic response of a discrete-time LTI system to harmonic inputs. The real-valued harmonic input, state, and output are represented as the sum of complex conjugates at a single frequency ω. u(t) = u(ω)e iωt + u (ω)e iωt (79) x(t) = x(ω)e iωt + x (ω)e iωt (8) y(t) = y(ω)e iωt + y (ω)e iωt (81) At discrete points in time t k = k t, the states at t k+1 are x(k + 1) = x(ω)e iω(k+1) t + x (ω)e iω(k+1) t (82) Substituting equations (79) - (82) into the discrete-time state-space equations, (67), and recognizing that the complex conjugate parts of the response (x(ω)e iωk t and x (ω)e iωk t ) are linearly independent, we obtain x(ω)e iω(k+1) t = Ax(ω)e iωk t + Bu(ω)e iωk t (83) y(ω)e iωk t = Cx(ω)e iωk t + Du(ω)e iωk t (84) Now, factoring-out the e iωk t from each term, x(ω) e iω t = Ax(ω) + Bu(ω) (85) y(ω) = Cx(ω) + Du(ω) (86) Solving for the outputs in terms of the inputs gives the frequency response function in terms of a state-space LTI model in the discrete-time domain, y(ω) = [ C[e iω t I A] 1 B + D ] u(ω) (87) H(z) = C[zI A] 1 B + D (88) where z = e iω t. Equation (88) is analogous to the transfer function in terms of continuous-time state-space models, equation (39).

36 36 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 23 Laplace transform and z-transform The Laplace transform of a continuous-time function y(t) is given by y(s) = L[y(t)] = y(t)e st dt, s C (89) This one-sided Laplace transform applies to causal functions for which y(t) = t <. Now, if we sample y(t) at discrete points in time spaced with interval t, y(k t) = y(k) = y(t)δ(t k t), and take the Laplace transform of the sampled signal, y(s) = k= Recall the property of the delta function, so, y(t)δ(t k t) e st dt. (9) f(t)δ(t τ) dt = f(τ) ( τ > ), (91) y(s) = y(k t)e sk t dt = y(k) ( e s t) k. (92) k= Defining z = e s t, we arrive at the discrete-time version of the Laplace transform... the z-transform: k= y(z) = y(k)z k = Z[y(k)]. (93) k= Now, we can apply the z-transform to the discrete-time convolution y(z) = Z k i= Y (i)u(k i) = z k k= k i= Y (i)u(k i) (94) If u(k) = k <, then k i= Y (i)u(k i) = i= Y (i)u(k i), so y(z) = = z k k= i= i= Y (i)z i Y (i)u(k i) (95) k= z (k i) u(k i) (96) = H(z) u(z) (97)

37 Linear Time Invariant Systems 37 So, convolution in the discrete-time domain is equivalent to multiplication in the z-domain, and the frequency-response is related to the Markov parameters via H(z) = Y (i)z i (98) i= The discrete-time state-space equations (67), the sequence of Markov parameters (75), and the frequency response function (88) are equivalent descriptions for discrete-time linear time-invariant systems. Now consider the steady-state response to a sinusoidal input sequence u(k) = cos(ωk t) = 1 2 y(k) = i= = i= = Y (i)u(k i) Y (i) cos(ω(k i) t) i= Y (i) 1 2 ( e iωk t + e iωk t) ( e iω(k i) t + e iω(k i) t) = 1 2 eiωk t Y (i)z i e iωk t Y (i)z i i= = 1 2 eiωk t H(z) e iωk t H (z) = H(z) cos(ωk t + φ) where the frequency-dependent phase angle of the sinusoidal response is given by tan φ(z) = Im(H(z)) Re(H(z)) A graph of H(ω) and H(ω) is called a Bode plot. i=

38 38 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 24 Lyapunov Equations for Discrete-Time Systems In discrete-time, the free response is x(k + 1) = Ax(k), x() = x o (99) Defining a Lyapunov function as the energy in the system at time k as, V (k) = x T (k)p x(k), P > (1) If the energy in free response decays monotonically, that is, if V (k+1) V (k) < x(k) and then the system is asymptotically stable. V (k + 1) V (k) = x T (k + 1)P x(k + 1) x T (k)p x(k) = x T (k)a T P Ax(k) x T (k)p x(k) = x T (k) [A T P A P ] x(k) (11) So, [A T P A P ] < V (k + 1) V (k) < x(k), and the system x(k + 1) = Ax(k) is asymptotically stable (AS) if and only if there exist positive definite matrices P and Z that solve the left discrete-time Lyapunov equation, A T P A P + Z = (12) The following statements are equivalent: A is asymptotically stable all eigenvalues λ i of A have magnitudes less than 1 Z > s.t. P > is a solution to A T P A P + Z = the series k= A kt ZA k converges. The series P = k= A kt ZA k solves A T P A P + Z =. k= P A kt ZA k Z + A kt ZA k k=1 = Z + A T P A = Z + A T A kt ZA k A k= = Z + A T A T(k 1) ZA k 1 A k=1 = Z + A kt ZA k k=1 (since A = I).

39 Linear Time Invariant Systems Controllability of Discrete-Time Systems A discrete-time system with n states, r inputs, and dynamics matrix A is said to be controllable by inputs coming through input matrix B if there exists a sequence of inputs, [u(), u(1),..., u(n 2), u(n 1), u(n)] that can bring the initial state x() to any arbitrary state x(n) within n steps. Let s consider the sequence of states arising from an input sequence u(k),... (k =,..., n 1) starting from an initial state x() =. x(1) = Bu() x(2) = Ax(1) + Bu(1) = ABu() + Bu(1) x(3) = Ax(2) + Bu(2) = A 2 Bu() + ABu(1) + Bu(2).. x(n) = A (n 1) Bu() + + A 2 Bu(n 3) + ABu(n 2) + Bu(n 1) x(n) = [ B, AB, A 2 B,, A (n 2) B, A (n 1) B ] u(n 1) u(n 2) u(n 3). u(1) u() Think of this last equation as a simple matrix-vector multiplication. (13) x(n) = C n u n (14) The matrix C n has n rows and nr columns and is called the controllability matrix. The column-vector u n is the sequence of inputs, all stacked up on top of each other, into one long vector. If the n rows of the controllability matrix C n are linearly-independent, then any final state x(n) can be reached through an appropriate selection of the control input sequence, u n. If the rank of C n equals n then there is at least one sequence of inputs u n that can take the state from x() = to any state x(n) within n steps. So, if rank(c n ) = n then the pair (A, B) is controllable.

40 4 CEE 629. System Identification Duke University Fall 217 H.P. Gavin If r > 1 then the system is under-determined and the input sequence u n to arrive at state x(n) (in n steps), is computed using the right Moore-Penrose pseudo inverse, which minimizes u n 2 2. The covariance of an infinitely long sequence of state responses to independent impulses (u() = I r ) is the controllability gramian, C C T = Q = A k BB T A kt (15) k= The element Q (i,j) is the covariance of the the inner product of the i-th state unit impulse responses with the j-th state unit impulse response, with unit impulses at each input. In other words, if x iq (k) is the response of the i-th state to a unit imuplse at input q, then Q (i,j) = r k= q=1 x iq (k)x jq (k). (16) The controllability gramian solves the right Lyapunov equation = AQA T Q + BB T (17) These expressions are analogous to equations (48) and (49) in continuoustime. The following statements are equivalent: The pair (A, B) is controllable The rank of C is n. The rank of Q is n. A matrix Q > solves the right Lyapunov equation = AQA T Q + BB T The singular-value decomposition of the controllability matrix equation x(n) = σ Ci u Ci (vci T u n ) (18) i shows that the sequence of inputs proportional to v C1 couples most strongly to the states. Input sequences that lie entirely in the kernel of C n have no effect on the state. Likewise, state vectors proportional to the eigenvector of Q with the largest eigenvalue are most strongly affected by inputs u n And state vectors proportional to an eigenvector of Q with an eigenvalue of zero (if there is one) can not be attained by the control input Bu(k).

41 Linear Time Invariant Systems Observability of Discrete-Time Systems A system with n states, m outputs, and dynamics matrix A is said to be observable by outputs coming out through output matrix C if there exists a sequence of outputs [y(), y(1),..., y(n 3), y(n 2), y(n 1)] from which the initial state x() can be uniquely determined. Let s consider the sequence of outputs arising from a sequence of states, in free response from some initial condition x(). (The initial condition is not equal to zero.) y() = Cx() y(1) = Cx(1) = CAx() y(2) = Cx(2) = CAx(1) = CA 2 x().. y(n 1) = Cx(n 1) = CA (n 1) x() y() y(1) y(2). y(n 1) = C CA CA 2. CA (n 1) x() (19) Think of this last equation as a simple vector-matrix multiplication. y n = O n x() (11) The matrix O n has nm rows and n columns and is called the observability matrix. The column-vector y n is the sequence of outputs, all stacked up on top of each other, into a long vector. If the n columns of the observability matrix, O n, are linearly-independent, then any initial state x() can be determined from the associated sequence of n free-responses. If O n has rank n then the set of all vectors y n can be transformed into a set of vectors x that fill an n-dimensional vector space, via the matrix inverse (or pseudo-inverse) of O n.

42 42 CEE 629. System Identification Duke University Fall 217 H.P. Gavin If m > 1 then the matrix equation is over determined, and the initial state x() associated with the free response outputs y n is computed with the left Moore-Penrose pseudo inverse, which minimizes y n O n x() 2 2. The covariance of an infinitely long sequence of output responses to n independent unity initial conditions (x() = I n ) is the observability gramian, O O T = P = A kt C T CA k (111) The element P (i, j) is the covariance of the inner product of the two free output responses one evolving from x i () = 1 and the other from x j () = 1 (with x k () =, k i, j). In other words, if y pi (k) is the free response of the p-th output to the initial state x i () = 1, x j () =, i j, then P (i,j) = m k= p=1 k= y pi (k)y pj (k). (112) The observability gramian solves the left Lyapunov equation = A T P A P + C T C (113) These expressions are analogous to equations (46) and (47) in continuoustime. The following statements are equivalent: The pair (A, C) is observable. The rank of O is n. The rank of P is n. A matrix P > solves the left Lyapunov equation = A T P A P + C T C The singular-value decomposition of the observability matrix equation y(n) = σ Oi u Oi (voi T x()) (114) i shows that an initial state x() proportional to v O1 couples most strongly to the outputs. Initial states that lie entirely in the kernel of O n (if it exists) have no output through y = Cx. Likewise, initial states proportional to the eigenvector of Q with the largest associated eigenvalue contribute most to the output covariance. And initial states proportional to an eigenvector of Q with an eigenvalue of zero (if there are any) do not affect the output covariance.

43 Linear Time Invariant Systems H 2 norms of discrete-time LTI systems The H 2 norm of asymptotically-stable discrete-time models with D = is expressed in terms of the controllability gramian and observability gramian satisfying the right and left discrete-time Lyapunov equations (17) and (113) and equations (51) and (54). Because the frequency content of signals in discrete-time is limited to the Nyquist interval, the H 2 norm in terms of the frequency response of discretetime systems involves integration around the unit circle, z = 1 H 2 2 = 1 +π 2π tr [ H(e iω t ) T H(e iω t ) ] dω (115) π where H(e iω t ) = H(z) as defined for discrete-time systems in equation (88). In terms of unit impulse responses (Markov parameters), the H 2 norm of a discrete time system is H 2 2 = Y (k) 2 F (116) k= where Y () = D and Y (k) = CA k 1 B for k >. Note that in continuous-time systems, lim ω [H(ω)] = D. Referring to the frequency-domain interpretation of the H 2 norm, equation (56), we see that for systems with D, the integral of H(ω) F over < ω < is not finite, and so the H 2 norm can not be defined in this case. In the time domain, the unit impulse u i (t) = I r δ(t) has responses H(t) = Ce At B + Dδ(t). In this case the integral of H(t) F over < t < involves the integral of δ 2 (t), which is not finite, and so the H 2 norm can not be defined in this case either. The facts that H 2 norms of exactly proper continuous-time systems are not finite, and that they are for discrete-time systems is a consequence of the facts that Dirac delta is defined only in terms of convolution integrals, and that u() = 1 in discrete time is a perfectly reasonable statement. So, a normequivalency of discrete-time systems derived from continuous-time systems necessarily implies that continuous-time inputs have no power outside of the Nyquist interval. An low-pass filtered Dirac-delta is a sinc function,

44 44 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 28.m-functions abcddim.m hpgavin/abcddim.m 1 function [n, r, m] = abcddim (A, B, C, D) 2 % Usage : [ n, r, m] = abcddim (A, B, C, D) 3 % 4 % Check f o r c o m p a t i b i l i t y o f t h e dimensions o f t h e matrices d e f i n i n g 5 % t h e l i n e a r system (A, B, C, D). 6 % 7 % Returns n = number o f system s t a t e s, 8 % r = number o f system inputs, 9 % m = number o f system o u t p u t s. 1 % 11 % Returns n = r = m = 1 i f t h e system i s not c o m p a t i b l e % A. S. Hodel <scotte@eng. auburn. edu> i f (nargin = 4) 16 error ( usage : abcddim (A, B, C, D) ); 17 end n = -1; r = -1; m = -1; 2 21 [an, am] = size (A); 22 i f ( an = am), error ( abcddim : A is not square ); end [bn, br] = size (B); 25 i f (bn = an) 26 error ( sprintf ( abcddim : A and B are not compatible, A :(% dx%d) B :(% dx%d),am,an,bn,br )) 27 end [cm, cn] = size (C); 3 i f (cn = an) 31 error ( sprintf ( abcddim : A and C are not compatible, A :(% dx%d) C :(% dx%d),am,an,cm,cn )) 32 end [dm, dr] = size (D); 35 i f (cm = dm) 36 error ( sprintf ( abcddim : C and D are not compatible, C :(% dx%d) D :(% dx%d),cm,cn,dm,dr )) 37 end 38 i f (br = dr) 39 error ( sprintf ( abcddim : B and D are not compatible, B :(% dx%d) D :(% dx%d),bn,br,dm,dr )) 4 end n = an; 43 r = br; 44 m = cm; % abcddim.m

45 Linear Time Invariant Systems 45 lsym.m hpgavin/lsym.m 1 function y = lsym (A,B,C,D,u,t, x, ntrp ) 2 % y = lsym ( A, B, C, D, u, t, x, ntrp ) 3 % t r a n s i e n t response o f a continuous time l i n e a r system to a r b i t r a r y i n p u t s. 4 % dx/ dt = Ax + Bu 5 % y = Cx + Du 6 % 7 % A : dynamics matrix ( n by n) 8 % B : i n p u t matrix (n by r ) 9 % C : output matrix (m by n) 1 % D : f e e d t h r o u g h matrix (m by r ) 11 % u : matrix o f sampled i n p u t s ( r by p ) 12 % t : v e c t o r o f uniformly spaced p o i n t s in time (1 by p ) 13 % x : v e c t o r o f s t a t e s at t h e f i r s t p o i n t in timen by 1 14 % ntrp : zoh zero order hold, foh f i r s t order h o l d ( d e f a u l t ) 15 % y : matrix o f t h e system o u t p u t s (m by p ) i f (nargin < 8), ntrp = foh ; end [n,r,m] = abcddim (A,B,C,D); % matrix dimensions and c o m p a t a b i l i t y check 2 21 points = size (u,2); % number o f data p o i n t s dt = t (2) - t (1); % uniform time s t e p v a l u e % continuous time to d i s c r t e time c onversion i f strcmp(lower( ntrp ), foh ) % f i r s t order h o l d on i n p u t s 27 M = [ A B zeros (n, r) ; zeros (r, n+r) eye( r) ; zeros (r, n +2* r) ]; 28 else % zero order h o l d on i n p u t s 29 M = [ A B ; zeros (r, m+r) ]; 3 end 31 emdt = expm(m*dt ); % matrix e x p o n e n t i a l 32 Ad = emdt (1:n,1: n); % d i s c r e t e time dynamics matrix 33 Bd = emdt (1:n,n +1: n+r); % d i s c r e t e time i n p u t matrix 34 i f strcmp(lower( ntrp ), foh ) 35 Bd_ = emdt (1:n,n+r +1: n +2* r); % d i s c r e t e time i n p u t matrix 36 Bd = Bd - Bd_ / dt; % d i s c r e t e time i n p u t matrix f o r time p 37 Bd1 = Bd_ / dt; % d i s c r e t e time i n p u t matrix f o r time p+1 38 else 39 Bd = Bd; 4 Bd1 = zeros (n, r); 41 end % i n i t i a l c o n d i t i o n s are zero u n l e s s s p e c i f i e d 44 i f ( nargin < 7 ), x = zeros (n,1); else, x = x; end y = zeros (m, points ); % memory a l l o c a t i o n f o r t h e output y (:,1) = C * x + D * u (:,1); % output f o r t h e i n i t i a l c o n d i t i o n 49 5 for p = 2: points x = Ad * x + Bd * u(:,p -1) + Bd1 * u(:,p); y(:,p) = C * x + D * u(:,p); end i f max(max(abs(imag(y )))) < 1e -12, y = real (y); end % s t r i p out round o f f 59 6 % LSYM

46 46 CEE 629. System Identification Duke University Fall 217 H.P. Gavin dlsym.m hpgavin/dlsym.m 1 function y = dlsym (A,B,C,D,u,t, x) 2 % y = dlsym (A,B,C,D, u, t, x ) 3 % s i m u l a t e s t h e response o f a d i s c r e t e time l i n e a r system to a r b i t r a r y i n p u t s 4 % 5 % x ( k+1) = A x ( k ) + B u ( k ) 6 % y ( k ) = C x ( k ) + D u ( k ) 7 % 8 % A : n by n dynamics matrix 9 % B : n by m i n p u t matrix 1 % C : l by n output matrix 11 % D : l by m f e e d t h r o u g h matrix 12 % u : l by p matrix o f sampled i n p u t s 13 % t : 1 by p v e c t o r o f uniformly spaced p o i n t s in time 14 % x : n by 1 v e c t o r o f i n i t i a l s t a t e s, d e f a u l t s to zero 15 % y : m by p matrix o f t h e system o u t p u t s 16 % [n,m, l] = abcddim (A,B,C, D); 19 2 points = size (u,2); % number o f data p o i n t s y = zeros (l, points ); % memory a l l o c a t i o n f o r t h e output i f ( nargin < 6 ) 25 t = [1: points ]; 26 end i f ( nargin == 7 ) 29 x = x; % i n i t i a l c o n d i t i o n s f o r t h e s t a t e 3 else 31 x = zeros (n,1); % i n i t i a l c o n d i t i o n s are zero 32 end y (:,1) = C * x + D * u (:,1); for p = 2: points x = A * x + B * u(:,p -1); 39 4 y(:,p) = C * x + D * u(:,p); end end % dlsym.m

47 Linear Time Invariant Systems 47 damp.m hpgavin/damp.m 1 function [ wnout, z] = damp (a, delta_t ) 2 % DAMP Natural f r e q u e n c y and damping f a c t o r f o r continuous systems. 3 % [Wn, Z ] = DAMP(A) r e t u r n s v e c t o r s Wn and Z c o n t a i n i n g t h e 4 % n a t u r a l f r e q u e n c i e s and damping f a c t o r s o f A. The v a r i a b l e A 5 % can be in one o f s e v e r a l formats : 6 % 7 % 1) I f A i s square, i t i s assumed to be t h e s t a t e space 8 % A matrix. 9 % 2) I f A i s a row vector, i t i s assumed to be a v e c t o r o f 1 % t h e polynomial c o e f f i c i e n t s from a t r a n s f e r f u n c t i o n. 11 % 3) I f A i s a column vector, i t i s assumed to contain 12 % r o o t l o c a t i o n s. 13 % 14 % When invoked w i t h o u t l e f t hand arguments DAMP p r i n t s t h e 15 % e i g e n v a l u e s with t h e i r n a t u r a l f r e q u e n c y and damping f a c t o r in a 16 % t a b u l a r format on t h e screen [m, n] = size ( a); 19 2 i f (n <1 m <1), wnout =; z =; return; end i f (m == n) 23 r = eig (a); 24 e l s e i f (m == 1) 25 r = ( roots (a )); 26 e l s e i f (n == 1) 27 r = a; 28 else 29 error ( Must be a vector or a square matrix. ); 3 end i f ( nargin == 2 ), r = log (r)/ delta_t ; end % d i s c r e t e time system for k = 1: n 35 wn(k) = abs(r(k )); 36 z(k) = - ( real (r(k)) - 2*eps) / (wn(k) + 2*eps); 37 end [wns, idx ] = sort (abs(wn )); % s o r t by i n c r e a s i n g n a t u r a l f r e q u e n c y 4 wn = wn( idx ); 41 z = z( idx ); 42 r = r( idx ); 43 wd = wn.* sqrt ( abs ( z.ˆ2-1 ) ); i f nargout == % Display r e s u l t s on t h e screen. 46 f p r i n t f ( \n ); 47 f p r i n t f ( Natural Damped \n ); 48 f p r i n t f ( Frequency Frequency Eigenvalue \n ); 49 f p r i n t f ( ( cyc / sec ) Damping ( cyc / sec ) real imag \n ); 5 f p r i n t f ( \n ); for idx = 1:1: n 53 f p r i n t f ( %1.5 f %1.5 f %1.5 f %1.5 f %1.5 f \n, wn(idx )/(2* pi ), z( idx ), wd(idx )/(2* p 54 end return % Suppress output 57 end 58 wnout = wn; 59 6 % DAMP

48 48 CEE 629. System Identification Duke University Fall 217 H.P. Gavin mimobode.m hpgavin/mimobode.m 1 function [mag,pha,g] = mimobode (A,B,C,D,w,dt, figno,ax,leg, tol ) 2 % [ mag, pha,g] = mimobbode (A,B,C,D, dt, w, f i g n o, ax, l e g, t o l ) 3 % p l o t s t h e magnitude and phase o f t h e 4 % steady s t a t e harmonic reponse o f a MIMO d i s c r e t e time l i n e a r dynamic system. 5 % where : 6 % A, B, C,D are t h e standard s t a t e space matrices o f a dynamic system 7 % dt i s t h e sample p e r i o d 8 % w i s a v e c t o r o f f r e q u e n c i e s ( d e f a u l t : w = l o g s p a c e ( 2,2,2) 2 p i 9 % ax i s e i t h e r x, y, n, or b to i n d i d a t e wich ax s h o u l d be log s c a l e d... 1 % x, y, n e i t h e r, or both. The d e f a u l t i s both. The y a x i s f o r t h e phase p l o t 11 % i s always l i n e a r l y s c a l e d. 12 % mag and pha are t h e magnitude and phase o f t h e f r e q u e n c y response f c t n matrix 13 % mag and pha have dimension ( l e n g t h (w) x m x r ) % Henri Gavin, Dept. C i v i l Engineering, Duke U n i v e r s i t y, h e n r i. gavin@duke. edu % Krajnik, Eduard, A simple and r e l i a b l e phase unwrapping algorithm, 18 % h t t p ://www. mathnet. or. kr / mathnet / p a p e r f i l e /Czech/Eduard/ phase. ps 19 2 i f (nargin < 1) tol = 1e -16; end % d e f a u l t rcond l e v e l 21 i f (nargin < 9) leg = []; end 22 i f (nargin < 8) ax = n ; end % d e f a u l t p l o t f o r m a t t i n g 23 i f (nargin < 7) figno = 1; end % d e f a u l t p l o t f o r m a t t i n g 24 i f (nargin < 6) dt = []; end % d e f a u l t to continuous time 25 i f (nargin < 5) w = logspace ( -2,2,2)*2* pi ; end % d e f a u l t f r e q u e n c y a x i s [n,r,m] = abcddim (A,B,C,D); % check f o r c o m p a t i b i l e dimensions 28 nw = length( w); 29 3 %warning o f f I = sqrt ( -1.); 33 In = eye( n); % continuous time or d i s c r e t e time i f ( length( dt) == ) sz = I* w; else sz = exp( I* w* dt ); end G = zeros (nw,m,r); % a l l o c a t e memory f o r t h e f r e q u e n c y response f u n c t i o n, g 39 mag = NaN( nw,m, r); 4 pha = NaN( nw,m, r); for ii =1: nw % compute t h e f r e q u e n c y response f u n c t i o n, G 43 % G( i i, :, : ) = C ( (w( i i ) I In A) \ B) + D; % ( si A) i s i l l c o n d i t i o n e d 44 [u,s, v] = svd( sz( ii )* In - A ); % SVD o f ( si A) 45 idx = max( find ( diag( s) > s (1,1)* tol ) ); 46 char_eq_inv = v (:,1: idx ) * inv(s (1: idx,1: idx )) * u (:,1: idx ) ; 47 G( ii,:,:) = C * char_eq_inv * B + D; 48 end 49 5 mag = abs( G); 51 pha (1,:,:) = -atan2(imag(g (1,:,:)), real (G (1,:,:))); % ve s i g n! 52 pha1 = -atan2(imag(g (1,:,:)), real (G (1,:,:))); % ve s i g n! 53 pha1 = reshape( ones( nw -1,1,1)* pha1, [ nw -1,m, r] ); 54 pha (2: nw,:,:) = pha1 - cumtrapz ( angle(g (2: nw,:,:)./ G (1: nw -1,:,:))); i f length( dt) == 1 % remove out of Nyquest range v a l u e s in DT systems 57 w_out = find (w> pi /dt ); % f r e q u e n c i e s o u t s i d e o f t h e Nyquist range 58 mag ( w_out ) = NaN; 59 pha ( w_out ) = NaN; 6 end

49 Linear Time Invariant Systems 49 dliap.m hpgavin/dliap.m 1 function P = dliap (A, X) 2 % f u n c t i o n P = d l i a p (A,X) 3 % S o l v e s t h e Liapunov e q u a t i o n A P A P + X = f o r P by transforming 4 % t h e A & X matrices to complex Schur form, computes t h e s o l u t i o n o f 5 % t h e r e s u l t i n g t r i a n g u l a r system, and transforms t h i s s o l u t i o n back. 6 % A and X are square matrices. 7 8 % h t t p ://www. mathworks. com/ m a t l a b c e n t r a l / newsreader / v i e w t h r e a d / % From : daly32569@my d e j a. com 1 % Date : 12 Apr, : 2 : % Downloaded : % Transform t h e matrix A to complex Schur form 14 % A = U T U... T i s upper t r i a n g u l a r, U U = I 15 [U,T] = schur( complex (A )); % f o r c e complex schur form s i n c e A i s o f t e n r e a l % Now... P (U T U ) P (U T U ) = X... which means % U P U (T U ) P (U T) = U X U 19 % Let Q = U P U y i e l d s, Q T Q T = U X U = Y 2 21 % S o l v e f o r Q = U P U by transforming X to Y = U X U 22 % Therefore, s o l v e : Q T Q T = Y... f o r Q 23 % Save memory by using P f o r Q. 24 dim = size (A,1); 25 Y = U * X * U; 26 T1 = T; 27 T2 = T ; 28 P = Y; % I n i t i a l i z e P... t h a t i s, i n i t i a l i z e Q 29 for col = dim : -1:1, 3 for row = dim : -1:1, 31 P(row, col ) = P(row, col ) + T1(row, row +1: dim )*( P( row +1: dim, col +1: dim )* T2(col +1: dim, col )); 32 P(row, col ) = P(row, col ) + T1(row, row )*( P(row, col +1: dim )* T2(col +1: dim, col )); 33 P(row, col ) = P(row, col ) + T2(col, col )*( T1(row, row +1: dim )*P( row +1: dim, col )); 34 P(row, col ) = P(row, col ) / (1 - T1(row, row )* T2(col, col )); 35 end 36 end 37 % U P U U T1 P T1 U X 38 % Convert Q to P by P = U Q U. 39 P = U* P*U ; 4 % A P A P + X % check t h a t t h i s i s zero, or c l o s e to i t.

50 5 CEE 629. System Identification Duke University Fall 217 H.P. Gavin invzero.m hpgavin/invzero.m 1 function zz = invzero (A,B,C,D, tol ) 2 % z e r o s = i n v z e r o (A,B,C,D, t o l ) 3 % i n v a r i a n t and d e c o u p l i n g z e r o s o f a continuous time LTI system 4 % t h e c o n d i t i o n f o r an i n v a r i a n t zero i s t h a t t h e p e n c i l [ zi A, B; C D ] i s 5 % rank d e f i c i e n t. For z e r o s t h a t are not p o l e s ( i. e., f o r minimal r e a l i z a i t o n s ) 6 % i n v a r i a n t zeros, z, make H( z ) rank d e f i c i e n t. 7 % method : use QZ decomposition to f i n d t h e g e n e r a l e i g e n v a l u e s o f t h e 8 % Rosenbrock system matrix, padded with zero or randn to make i t square. 9 % t o l : t o l e r a n c e v a l u e f o r rank determination, d e f a u l t = 1e [nn,rr,mm] = abcddim (A,B,C,D); i f nargin < 5, tol = 1e -6; end % make t h e system square by padding with z e r o s or randn, as needed ( c l u g e?) 16 re = mm - rr; 17 me = rr - mm; 18 rm = max(mm,rr ); 19 2 zi = []; 21 for iter = 1: Bx = B; Cx = C; Dx = D; 24 i f iter == 1 % zero padding f o r d e c o u p l i n g z e r o s 25 i f mm > rr, Bx = [ B, zeros ( nn, re) ]; Dx = [ D, zeros ( mm, re) ]; end 26 i f mm < rr, Cx = [ C ; zeros ( me, nn) ]; Dx = [ D ; zeros ( me, rr) ]; end 27 else % randn padding f o r i n v a r i a n t z e r o s 28 i f mm > rr, Bx = [ B, randn( nn, re) ]; Dx = [ D, randn( mm, re) ]; end 29 i f mm < rr, Cx = [ C ; randn( me, nn) ]; Dx = [ D ; randn( me, rr) ]; end 3 end abcd = [ -A, - Bx ; Cx, Dx ]; % Rosenbrock System Matrix 33 ii = [ eye(nn), zeros (nn,rm) ; zeros (rm,nn+rm) ]; 34 i f strcmp(version, ) 35 [pp,qq,abcd,ii] = balance (abcd, ii ); % o c t a v e 36 else 37 [ abcd, ii] = balance ( abcd, ii ); % matlab 38 end 39 zz = -qz ( abcd, ii ); 4 i f iter == 1 41 z1 = zz; 42 else 43 zi = [ zi, zz ]; 44 end 45 end zz = [ z1 ; intersecttol ( zi, tol ) ]; zz = zz ( i s i n f (zz )); % d i s c a r d any z e r o s at i n f i n i t y 5 51 idxr = find (abs(imag(zz )) < 1e -1 ); zz( idxr ) = real (zz( idxr )); % r e a l z e r o s 52 idxc = find (abs(imag( zz )) > 1e -1 ); % complex z e r o s zz = [ zz( idxr ) ; zz( idxc ) ; conj( zz( idxc )) ]; % complex conj p a i r s nz = length( zz ); % Are both t h e p e n c i l o f t h e Rosenbrock System Matrix and 59 % t h e t r a n s f e r f u n c t i o n matrix rank d e f i c i e n t?? 6 % confirm t h a t a l l t h e z e r o s are i n v a r i a n t z e r o s

51 Linear Time Invariant Systems Numerical Example: a spring-mass-damper oscillator For the single degree of freedom oscillator of Section 2, let s say m = 2 ton, c = 1.4 N/mm/s, k = 6.8 N/mm, d o = 5.5 mm and v o = 2.1 mm/s. Note that these units are consistent. (1 N)=(1 kg)(1 m/s 2 )=(1 ton)(1 mm/s 2 ) For these values, the linear time invariant system description of equations (7) and (8) become ẋ(t) = y(t) = x(t) + x(t) u(t), x() = (117) u(t) (118) 1. What are the natural frequencies and damping ratios of this system? >> A = [ 1 ; ] % the dynamics matrix A = >> eig(a) % eigenvalues of the dynamics matrix ans = i i >> wn = abs(eig(a)) % absolute values of the eig s of A are omega_n wn = >> z = -real(eig(a))./ wn % ratio of real eig(a) to omega_n is damping ratio z = >> wd = imag(eig(a)) % imaginary parts of the eig s of A are omega_d wd =

52 52 CEE 629. System Identification Duke University Fall 217 H.P. Gavin So, the natural frequency is 1.84 rad/s (the same as k/m), the damping ratios is 19% (the same as c/ 4mk), and the damped natural frequency is 1.81 rad/s (the same as ω n ζ2 1 ). These calculations can be done in one step with the m-file damp.m >> damp(a) Natural Damped Frequency Frequency Eigenvalue (cyc/sec) Damping (cyc/sec) real imag What is the discrete-time system realization for t =.1 s? >> dt =.1; % time step, s >> Ad = expm(a*dt); Bd = A\(Ad-eye(2))*B; % continuous-to-discrete-time Ad = Bd = e e-3 >> damp(ad,dt) % check the dynamics of discrete-time system Natural Damped Frequency Frequency Eigenvalue (cyc/sec) Damping (cyc/sec) real imag The following analyses are carried out for the continuous-time system model and can also be equivalently carried out for the discrete-time system model.

53 Linear Time Invariant Systems What is the free response of this system to the specified initial conditions x o? >> dt =.1; % time step value, sec >> n = 1; % number of time steps >> t = [:n-1]*dt; % time values, starting at t= >> xo = [ 5.5 ; 2.1]; % initial state (mm, mm/s) >> C = [ ; ]; % output matrix >> y = zeros(2,n); % initialize outputs >> for k=1:n > y(:,k) = C*expm(A*t(k))*xo; > end >> plot(t,y) >> legend( foundation force, N, mass acceleration, mm/sˆ2 ) >> xlabel( time, s ) >> ylabel( outputs, y_1 and y_2 ) The free response is plotted in figure 1. 4 foundation force, N mass acceleration, mm/s 2 3 outputs, y 1 and y time, s Figure 1. Free response of the linear time invariant system given in equations (117) and (118).

54 54 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 4. What is the response of this system to a unit impulse, u(t) = δ(t)? >> B = [ ;.5 ]; % input matrix >> h = zeros(2,n); % initialize unit impulse responses >> for k=1:n > h(:,k) = C*expm(A*t(k))*B; > end >> plot(t,h) >> legend( foundation force, N, mass acceleration, mm/sˆ2 ) >> xlabel( time, s ) >> ylabel( unit impulse responses, h_1(t) and y_2(t) ) The unit impulse response is plotted in figure 2. Note that at h() = CB, which is not necessarily zero foundation force, N mass acceleration, mm/s 2 unit impuse responses, h 1 (t) and h 2 (t) time, s Figure 2. Unit impulse response of the linear time invariant system given in equations (117) and (118).

55 Linear Time Invariant Systems What is the response of this system to external forcing u(t) = 5 cos(πt)? >> D = [ ;.5 ]; % feedthrough matrix >> u = 5 * cos(pi*t); % external forcing >> y = lsym(a,b,c,d,u,t,xo); % use the "lsym" command >> plot(t,y) >> legend( foundation force, N, mass acceleration, mm/sˆ2 ) >> xlabel( time, s ) >> ylabel( forced harmonic responses, y_1(t) and y_2(t) ) The forced harmonic response is plotted in figure input forcing responses, y 1 (t) and y 2 (t) time, s -4-6 foundation force, N mass acceleration, mm/s time, s Figure 3. Forced response of the linear time invariant system given in equations (117) and (118).

56 56 CEE 629. System Identification Duke University Fall 217 H.P. Gavin 6. What is the frequency response function from u(t) to y(t) for this system? >> w = 2*pi*logspace(-1,,1); % frequency axis data >> bode(a,b,c,d,1,w); The magnitude and phase of the steady-state forced harmonic response is plotted in figure 4. Note how the magnitude and phase of the frequency response shown in figure 4 can be used to predict the steady state response of the system to a forcing of u = 1 cos(πt). (The forcing frequency is.5 Hertz). magnitude phase (degrees) frequency (Hertz) 1 Figure 4. Frequency response of the linear time invariant system given in equations (117) and (118). The Laplace-domain transfer functions of the system are plotted in figure 5. This is a fairly simple example. Nevertheless, by simply changing the definitions of the system matrices, A, B, C, and D, and of the input forcing u(t), any linear time invariant system may be analyzed using the same sets of Matlab commands.

57 57 Linear Time Invariant Systems Figure 5. Laplace-domain transfer function of the linear time invariant system given in equations (117) and (118). Poles are marked with x on the Laplace plane (s = σ + iω). The frequency response is shown as the red curve along the iω axis, at σ =. The real parts of the poles are all negative, meaning that the system is asymptotically stable.