Advanced Control Theory
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1 State Space Solution and Realization
2 Outline State space solution 2
3 Solution of state-space equations x t = Ax t + Bu t First, recall results for scalar equation: x t = a x t + b u t Laplace transforming: sx s x 0 = a X s + b U s (s a)x s = x(0) + b U s Solve for transformed variable: X s = x 0 s a + b s a U s 3
4 Solution of state-space equations Recall Laplace transform L 1 1 s a = eat L 1 F 1 s F 2 (s) = 0 t f 1 t τ f_2 τ dτ Taking inverse Laplace transform of x t = e at x(0) + zero input solution 0 t e a t τ bu τ dτ zero state solution 4
5 Solution of state-space equations Vector-matrix equation: x t = Ax t + Bu t Laplace transforming: sx s x 0 = A X s + B U s (si A)X s = x(0) + BU s X s = si A 1 x(0) + si A 1 BU s Introducing matrix exponential by analogy with scalar exponential function e at = at + 1 2! at ! at ! at 4 + e At = I At + 1 2! At ! At ! At 4 + 5
6 Ref) Matrix exponential e At matrix exponential If A is an (n n) square matrix, then e At is also an (n n) matrix From the following (verify!) 1 s a = 1 s + a s 2 + a2 s 3 + a3 s 4 + si A 1 = I s + A s 2 + A2 s 3 + A3 s 4 + Laplace transform of e At can be found L 1 1 s a = at + 1 2! at ! at ! at 4 + = e at L 1 si A 1 = I At + 1 2! At ! At ! At 4 + = e At 6
7 Solution of state-space equations Note that: L e At = si A 1 From the previous solution in Laplace domain X s = si A 1 x(0) + si A 1 BU s Taking inverse Laplace transform x t = e At x(0) + zero input solution 0 t e A t τ bu τ dτ zero state solution 7
8 State transition matrix State transition matrix: It describes the transition of the states from initial conditions x(0) to those at time t, when there is no input: x t = Φ t x(0) State transition matrix is used to find the solution of linear time varying system x t = A(t)x t + B(t)u t The matrix exponential e At is the state transition matrix for LTI system For LTI, Φ t satisfies: Φ 0 = e A 0 = I, Φ 1 t = e At = Φ t Other properties of Φ(t): Φ t 1 + t 2 = Φ t 2 Φ(t 1 ) 8
9 Computation of state transition matrix In principle, we can compute Φ t = e At = I At + 1 2! At ! At 3 + as a series, terminating when no significant change observed Very simple example: double-integrator plant control orientation of satellite with thrusters J States: θ = F d = u, Φ s U s = 1 Js 2 S-S: 9
10 Computation of state transition matrix Here: Hence: Then, solution to homogeneous equation x t = Ax(t) with initial conditions x 0 = θ 0 ω 0 T is In practice, more sophisticated numerical algorithms are employed to compute the state transition matrix 10
11 Ex. State transition matrix via Laplace transform State transition matrix can be obtained from Laplace transform solution To illustrate method, consider example plant with transfer function Y(s) U(s) = 1 s 2 + 3s + 2 = 1 (s + 1)(s + 2) A state-space realization is x 1 x 2 = x 1 x u For this system y = 1 0 x 1 x 2 si A = s = s s 11
12 Ex. State transition matrix via Laplace transform Hence Φ s = si A 1 = = s (s + 1)(s + 2) 2 (s + 1)(s + 2) s 1 2 s + 3 s 2 + 3s (s + 1)(s + 2) s + 3 (s + 1)(s + 2) = 1 s s s s s s s s + 1 Thus Φ t = L 1 si A 1 = e t + 2e 2t e t 2e 2t 2e t + 2e 2t 2e t e 2t 12
13 Ex. State transition matrix via Laplace transform Calculate with MATLAB (Symbolic math) >>A = [-3 1; -2 0]; >>syms t >>Phi = expm(a*t) Φ t = L 1 si A 1 = e t + 2e 2t e t 2e 2t 2e t + 2e 2t 2e t e 2t Phi = [ 2*exp(-2*t)-exp(-t), exp(-t)-exp(-2*t)] [ -2*exp(-t)+2*exp(-2*t), -exp(-2*t)+2*exp(-t)] can also compute numerical value with MATLAB fn >>Phi1=subs(Phi,t,1) Phi1 =
14 Ex. State transition matrix via Laplace transform Total system response to a unit step input u(t) = 1(t) [U(s) = 1/s] is applied to the system x 1 x 2 = x 1 x u Total response is: 14
15 Ex. State transition matrix via Laplace transform >>A=[-3 1;-2 0]; B=[0;1]; C=[1 0]; D=0; >>G = ss(a, B, C, D); >>ltiview(g) 15
16 Ex. State transition matrix via Laplace transform Zero input response: syms s, syms x10 x20 real xzi = Phi*[x10; x20] xzi = [ (2*exp(-2*t)-exp(-t))*x10+(exp(-t)-exp(-2*t))*x20] [ (-2*exp(-t)+2*exp(-2*t))*x10+(-exp(-2*t)+2*exp(-t))*x20] Zero state U=1/s; Response: xzs = ilaplace(inv(s*eye(2)-a)*b*u) xzs = [ 1/2-exp(-t)+1/2*exp(-2*t)] [ 3/2+1/2*exp(-2*t)-2*exp(-t)] 16
17 Outline Simulation diagram State space realization Controllable canonical form Observable canonical form Modal canonical form 17
18 Simulation diagrams Consider the problem of constructing an analog system to produce a given input-output behavior, using only integrators, gain elements and summing elements (as in an analog computer ) For example, input-output behavior defined by: Then: 18
19 Simulation diagrams Define states as outputs of integrators Now we can write the state-space equations: A state-space realization A matrix is in upper companion form 19
20 Simulation Diagram vs. Block Diagram Simulation diagram, in which: signals are in the time domain; e.g. u(t) the only system elements are integrators, gains and summers Block diagram, in which: the signals are in the Laplace domain; e.g. U(s) the system elements may be arbitrary transfer functions 20
21 Realization Realization: For a given LTI system, a state space model which is constructed from a given transfer function G s is called a realization of G. We were able to realize an ODE with no derivatives on the RHS (i.e., no zeros in the TF) with a simple chain of integrators (previous example) Consider SISO 3 rd order model y + a 2 y + a 1 y + a 0 y = b 2 u + b 1 u + b 0 u where a i, b i are arbitrary real numbers G s = Y(s) U(s) = b 2s 2 + b 1 s + b 0 s 3 + a 2 s 2 + a 1 s + a 0 21
22 Realization: Controllable canonical form How to deal with y + a 2 y + a 1 y + a 0 y = b 2 u + b 1 u + b 0 u i.e. G s = Y(s) = b 2s 2 +b 1 s+b 0 = B(s) U(s) s 3 +a 2 s 2 +a 1 s+a 0 A(s) Introduce a partial state ξ as an auxiliary variable, such that: ξ + a 2 ξ + a 1 ξ + a 0 ξ = u ; this can be realized with a chain of integrators as before Ξ(s) U(s) = 1 A(s) We can then construct the output thus: y = b 2 ξ + b 1 ξ + b 0 ξ; Y s = B s A s U(s) = B s Ξ(s) 22
23 Realization: Controllable canonical form Y(s) U(s) = b 2s 2 + b 1 s + b 0 s 3 + a 2 s 2 + a 1 s + a 0 ξ = u a 2 ξ a 1 ξ a 0 ξ b 2 b 1 y = b 2 ξ + b 1 ξ + b 0 ξ b 0 a 2 a 1 a 0 x 1 x 2 x 3 = a 2 a 1 a x 1 x 2 x u y = b 2 b 1 b 0 x 2 x 3 x 1 23
24 Realization: Observable canonical form Y(s) U(s) = b 2s 2 + b 1 s + b 0 s 3 + a 2 s 2 + a 1 s + a 0 To get another canonical form, rewrite the transfer function (s 3 + a 2 s 2 + a 1 s + a 0 )Y s = b 2 s 2 + b 1 s + b 0 U(s) Divide throughout by s 3 to obtain (1 + a 2 s + a 1 s 2 + a 0 s 3)Y s = b 0 s 3 + b 1 s 2 + b 0 s U(s) Rearrange term then, Y = b 0U a 0 Y s 3 + b 1U a 1 Y s 2 + b 2U a 2 Y s 24
25 Realization: Observable canonical form Y(s) U(s) = b 2s 2 + b 1 s + b 0 s 3 + a 2 s 2 + a 1 s + a 0 b 0 b 1 b 2 a 0 a 1 a 2 x 1 x 2 x 3 = a a a x 1 x 2 x 3 + b 2 b 1 b 0 u y = x 1 x 2 x 3 25
26 Relationships between CCF and OCF Observable canonical form Controllable canonical form 26
27 Realization: Modal canonical form If the poles of transfer function are all distinct, we can do the partial fraction expansion: Each term in the summation can be represented: 27
28 Realization: Modal canonical form Hence, parallel or diagonal realization: The modal states are decoupled Output is a linear combination of the system modes 28
29 Realization: Natural modes We saw previously that by performing a partial fraction expansion of the transfer function of a dynamic system we could obtain a parallel state-space realization for which the A matrix is diagonal so that the state-space equations are decoupled This modal canonical form is very fundamental: the 'natural modes' make independent contributions to the system output each mode has a characteristic 'natural frequency' (eigenvalue) and 'mode shape' (eigenvector) 29
30 Natural motions For 'natural' (unforced) motions (i.e., u 0): We know from before that the unforced response is a combination of simple exponential motions of the form The system is said to be moving in a natural mode when all the states have the same form of motion: The vector p (the mode shape) describes the relative amplitude of the common motion for each state 30
31 Eigenvalue problem If then But That is, We seek a solution for the eigenvalue (or mode shape) p. The condition for a non-trivial solution (i.e., p 0) is that The roots of the characteristic equation are: the "eigenvalues" λ i (i = 1, 2,..., n) of A the "natural frequencies" of the system the "poles" of the transfer function Y(s)/U(s) 31
32 Eigenvalue problem The eigenvalues are the values of λ for which the assumed motion x(t) = pe λt is possible without any external input u For each eigenvalue λ i, we can find the initial conditions x(0) = p which will excite this natural mode, as the solution of the auxiliary equation: The solution p (i) is: the "eigenvector" corresponding to the eigenvalue λ i the "mode shape" of the i-th mode Note that p (i) may be found easily as a vector proportional to any nonzero column of the adjugate of the auxiliary matrix F =[λ i I A] 32
33 Example Auxiliary matrix: Characteristic equation: Eigenvalues: 33
34 First mode: λ 1 = 2 Auxiliary equation: Solution: Note that the eigenvector, or mode shape, may be normalized in various ways set the first or last element to 1, as above set the vector "magnitude" to 1 (unit norm vector): 34
35 Rapid calculation of mode shape Auxiliary matrix: Adjoint: The columns of the adjoint of F are proportional to each other, and to the eigenvector It is thus only necessary to compute one column to find the mode shape 35
36 Motion in first natural mode The first mode may be excited by setting up initial conditions corresponding to the first mode shape: Then 36
37 Motion in first natural mode 37
38 MATLAB calculation of responses % Get controller canonical form >>[A, B, C, D] = tf2ss(1, [1 5 6]); >>G = ss(a, B, C, D); % Check >>G.a ans = % initial condition >>x0 = [-2 1]'; % compute response >>[y, t, x] = initial(g, x0); % plot response >>plot(t, x), grid >>legend('x_1', 'x_2') 38
39 Double check with transition matrix solution % Set up for symbolic computation >>syms t >>Phi = expm(a*t) % transition matrix >>Phi = [ 3*exp(-3*t)-2*exp(-2*t), -6*exp(-2*t)+6*exp(-3*t)] [ exp(-2*t)-exp(-3*t), -2*exp(-3*t)+3*exp(-2*t)] >>x0 = [-2 1]'; % initial condition = mode shape % Compute zero-input response >>X = Phi*x0 X = [ -2*exp(-2*t)] pure 1st mode [ exp(-2*t)] 39
40 Second mode: λ 2 = 3 Auxiliary equation: 40
41 Motion in second natural mode 41
42 Response to arbitrary initial conditions Example: Displace 1 unit and release from rest Construct initial condition as superposition of mode shapes: Then: Response is the sum of contributions from the independent natural modes 42
43 Response to initial displacement 43
44 Modal contributions to output y 44
45 Outline Similarity transform 45
46 Transformations between state variables We have seen that there is no unique set of state variables to realize a given input-output behaviour. That is, there are many s-s model for a given input-output model!! Given one realization, We can form another by a (nonsingular) linear transformation of variables: T = transformation matrix 46
47 Transformations between state variables Then: Hence where This is called a similarity transformation Two systems are called equivalent with each other 47
48 Transfer functions from state-space equations Until now, we have seen that there are many different state space realizations of a given transfer function The reverse process is also of interest; that is, SS TF Consider a linear time invariant state dynamics Start by taking the Laplace Transform of these equations 48
49 Transfer functions from state-space equations which gives And By definition G(s) = C(sI - A) -1 B + D is called the Transfer Function of the system. And C(sI - A) -1 x(0-) is the initial condition response. It is part of the response, but not part of the transfer function. 49
50 Transfer functions between equivalent models Let s get back to two equivalent models where Consider the two transfer functions: 50
51 Transfer functions between equivalent models The transfer function is not changed by putting the state-space model through a similarity transformation 51
52 Evaluation of transfer function matrix To evaluate transfer function from s-s model, we should calculate G(s) = C(sI - A) -1 B + D For a SISO system, G(s) may be evaluated using See Franklin et al. App C.6 For a SIMO system, G(s) may be evaluated using: The MATLAB function ss2tf uses this formulation 52
53 Example Choose Then, state-space equations are: Transfer function: s 5 6 det 1 s 0 1 s 5 det s s 2 1 5s 6 1 ( s 2)( s 3) 53
54 Transformation to modal state variables Let P [p (1) p (2) p (n) ] And modal matrix We have: assume eigenvalues are distinct Hence: That is: The modal matrix P is the similarity transformation matrix between the original state vector x and the modal state vector x d : 54
55 Ex Modal matrix: Similarity transformation: modal canonical form 55
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