# Perspective. ECE 3640 Lecture 11 State-Space Analysis. To learn about state-space analysis for continuous and discrete-time. Objective: systems

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1 ECE 3640 Lecture State-Space Analysis Objective: systems To learn about state-space analysis for continuous and discrete-time Perspective Transfer functions provide only an input/output perspective of what is going on in a system. There may be things going on physically that do not appear in a transfer function, due to cancellations, etc. On the other hand, state-space analysis provides a more complete representation. Furthermore, it can be generalized to time-varying systems, multi- input or output systems, and in some applications leads to very explicit design formulations. There is also much that can be done with nonlinear systems in state variable form. We have seen that we can describe an LTIC system using a single differential equation. In state-space analysis, we deal with systems of equations, but make it so that all equations are first order. Sometimes this requires introducing some extra variables. The variables appearing in these equations (with respect to which we differentiate) are called the state variables. The idea behind the name is this: for a first order differential equation, if we know where we are initially (the initial condition), then this provides all of the information we need to determine where to go. In circuits, it is common to choose the voltage across the capacitors and the current through the inductors as state variables. This provides our first example. Example Circuit. Two state variables. KCL: 0.2ẋ i i 2 x 2 Ohm s: i 2(f x ). i 2 3x. We obtain or 0.2ẋ 2(f x ) 3x x 2 ẋ 25x 5x 2 + 0f Notice that everything is expressed in terms of the state variables x and x 2 and the input f.

2 ECE 3640: Lecture State-Space Analysis 2 Next equation: KVL: since i 4 x 2 we obtain We obtain ()ẋ 2 + 2i 4 x 0. ẋ 2 x 2x 2. ẋ 25x 5x 2 + 0f ẋ 2 x 2x 2. Note: every possible output of the circuit every voltage and current can be expressed in terms of the state variables. Try a few. Express in matrix form. Note: we have. First order differential equations. 2. Each equation is expressed in terms of the state variables and the input. For linear equations, we can put the equations in matrix form. Let x (t) x(t). x 2 (t) Let ẋ denote taking the derivatives individually: ] [ẋ (t) ẋ(t) ẋ 2 (t) In the example above, let A B 0 Then we can write ẋ(t) Ax(t) + Bf(t). Note: For nonlinear systems, we can still put them in state variable form, even when we cannot use a matrix for the representation. Example 2 ẋ x x 3 cos(x + x 2 ) + f 2 ẋ 2 (x + x 2 ) 2 ẋ 3 x 3 tan(x 2 /x ) Example 3 (Important) Consider the 3rd order equation (D 3 + a 2 D 2 + a D + a 0 )y(t) f(t). We will introduce some new variables. Let x y x 2 ẏ x 3 ÿ Then ẋ x 2 ẋ 2 x 3 ẋ 3 f(t) a 2 ẋ 3 a ẋ 2 a 0 ẋ + f

3 ECE 3640: Lecture State-Space Analysis 3 We also have an output equation We can stack this as follows: or ẋ ẋ ẋ 2 ẋ 3 y x a 0 a a 2 y [ 0 ] x ẋ Ax + bf y c T x. x 2 x f In general, a set of state-variable equations can be written Note that this could be ẋ i g i (x, x 2,..., x n, f, f 2,..., f j ), y i h i (x, x 2,..., x n, f, f 2,..., f j ), i, 2,..., n i, 2,..., k Nonlinear Multiple inputs Multiple outputs (But may not be!) This represents a considerable degree of flexibility. A general j-input, k-output linear system with n state variables can be written as ẋ ẋ.. ẋ n Aẋ + Bf y Cx + Df where A is n n, B is n j, C is k n and D is k j. (Write out the matrices.) Have a student work the circuit on the board..

4 ECE 3640: Lecture State-Space Analysis 4 Transfer functions and state equations Given a transfer function, we may want to write a state variable equation for it. This is very straightforward by writing a system realization for the transfer function. From the system realization, we let the state variables be the outputs of the integrators. Example 4 Canonical Realization: H(s) 2s + 0 s 3 + 8s 2 + 9s + 2 State equations: ẋ x 2 Output equation: y 2x 2 + 0x. Matrix form: ẋ ẋ 2 ẋ 3 ẋ 2 x 3 ẋ 3 2x 9x 2 8x 3 + f x + f x 2 x 3 y [ ] x Comment on the form of the matrix: companion matrix. Describe form in general. What is the characteristic equation of the companion matrix? What are the eigenvalues? Observer Form Realization: Write equations, then in matrix form. Companion x 2 x matrix. What is the characteristic equation? Eigenvalues?

5 ECE 3640: Lecture State-Space Analysis 5 Series form Realization: Equations: H(s) 2 s + 5 s + s + 3 s + 4 ẇ w + f ẇ 2 2w 3w 2 ẇ 3 5w 2 + ẇ 2 4w 3 Eliminate ẇ 2 using the 2nd eqn. Obtain ẇ w + 0 f Characteristic eqn? Eigenvalues? Parallel Realization: y [ 0 0 ] w. Equations: H(s) 4/3 s + 2 s /3 s + 4 ż z + f ż 2 3z 2 + f ż 3 4z 3 + f y 4 3 z 2z z 3 ż z + f 0 0 4

6 ECE 3640: Lecture State-Space Analysis 6 y [ ] z Characteristic polynomial? Eigenvalues? Note: There are other ways of writing down the state equations from the transfer function. In fact, there are an infinite number of ways! Laplace transform of state equations When we talk about the Laplace transform of a vector, we will mean to apply the transform element by element. Thus, if x (t) x(t) x 2 (t) then We find then that L[x(t)] L[x (t)] X(s). L[x 2 (t)] L[ẋ(t)] sx(s) x(0). From the state equation ẋ Ax + Bf we obtain and from the output equation, Let us solve for X(s) from the first: (Why the identity?) Watch the order! Let Φ(s) (si A). We have Inverse transform: sx(s) x(0) AX(s) + BF(s) Y(s) CX(s) + DF(s). (si A)X(s) x(0) + BF(s) X(s) (si A) [x(0) + BF(s)]. X(s) Φ(s)[x(0) + Φ(s)BF(s).] x(t) L [Φ(s)x(0)] + L [Φ(s)BF(s)]. Identify zero-input components and zero-state components. Output: Y (s) CΦ(s)x(0) + [CΦ(s)B + D]F (s) Transfer function: H(s) CΦ(s)B + D Example 5 ] [ẋ 0 x + ẋ x 2 (Two inputs!) 0 f f 2 y y 2 0 x f x f 0 2 y 3

7 ECE 3640: Lecture State-Space Analysis 7 (Three outputs!) Identify A,B,C,D. Φ(s) (si A) [ s 2 s + 3 ] s(s + 3) + 2 [ s + 3 ] 2 s Transfer function: H(s) [ s+4 (s+)(s+2) 2(s 2) (s+)(s+2) (s+)(s+2) 2s (s+)(s+2) ] + D s+4 (s+)(s+2) 2(s+3) s+2 2(s 2) (s+)(s+2) (s+)(s+2) s+2 2(s 2 +4s+2) (s+)(s+2) Poles and Eigenvalues Recall: X adj(x) det(x) Without worrying about what the adj is, note that the denominator always has the determinant. Thus (si A) adj(si A) det(si A). So the denominator has poles where the eigenvalues of A are! Time domain solution We begin by defining a new function. transition matrix) we define For a square matrix A (as in the state e A (I + A + a2 2 + A3 3! + ) (Taylor series). This is directly analogous to e a for scalars, except that all arithmetic is done using matrices. This is computed using the exmp function in Matlab, not exp. Note (show this) d dt eat Ae At e At A. The solution to the DE ẋ Ax + Bf is given by x(t) e At x(0) + t Show that it works by substitution. Computing the matrix exponential: One way: Example 6 A (si A) 2 2/3. 36 s + 2 2/3 36 s + 0 e A(t τ) Bf(τ) dτ e At L Φ(s) L [(si A) ] s + 2/3 (s + 2)(s + ) s + 2 (s + 4)(s + 9) s + 2 2/3. 36 s +

8 ECE 3640: Lecture State-Space Analysis 8 Taking inverse Laplace transforms element by element we obtain e At 0.6e 9t +.6e 4t 0.33e 9t 0.33e 4t 7.2e 9t + 7.2e 4t.6e 9t 0.6e 4t Linear transformations For the state equations ẋ Ax + Bf y Cx + Df let us create a new variable w P x for an invertible matrix P. Then x P w, and ẋ P ẇ. Substituting we find or where Similarly, where P ẇ AP w + Bf ẇ P AP w + P Bf Âw + ˆBf Â P AP ˆB P B. y Ĉw + ˆDf Ĉ CP ˆD D. Instead of (A, B, C, D) we have (Â, ˆB, Ĉ, ˆD). Do these represent the same system? H(s) C(sI A) B + D Ĥ(s) Ĉ(sI Â) ˆB + ˆD. (Work through details.) Other observations: eigenvalues? Eigenvectors? 0. A special transformation: diagonalizing A Given Â P AP, suppose that we want to find a transformation matrix P such that Â is diagonal. (This is a convenient form, since it decouples all the modes.) How can we find such a P? Let e i be eigenvectors of A, and λ i be the eigenvalues of A, assumed (for our purposes) to be unique. Form Q e e 2 e n and where Then Identify: Λ Â, P Q. AQ QΛ Λ diag(λ, λ 2,..., λ n ). Λ Q AQ

9 ECE 3640: Lecture State-Space Analysis 9 Controllability and observability Example 7 Cascade representation State variable form: H (s) s s s + A 0 [ b ] c T [ 0 ] d 0 Eigenvalues: ±. Diagonalize: det(si A) (s + )(s ) A [ 0; -]; b [;0]; c [ -2]; [u,v] eig(a); % u has eigenvectors, v eigenvalues Q u; % Check: inv(q)*a*q % should be diagonal! P inv(q) Ahat P*A*inv(P) bhat P*b chat c*inv(p) We find Â 0 0 ˆb.5.8 ĉ T [ 2, 0] Write state equations. Second state variable: not observable. Now the second system: A As before, diagonalize: [ ] 0 2 [ b ] c T [ 0 ] A [- 0; -2 ]; b [;]; c [0 ]; [u,v] eig(a); % u has eigenvectors, v eigenvalues

10 ECE 3640: Lecture State-Space Analysis 0 Q u; % Check: inv(q)*a*q % should be diagonal! P inv(q) Ahat P*A*inv(P) bhat P*b chat c*inv(p) We find Â ˆb.442 ĉ T [, 0.707]. Write state equations. Second state variable not controllable. Note that in both cases, the end-to-end transfer function hides some information there is cancellation there. The transfer function provides a potentially inadequate representation of what is going on. In the general case, let us write ż Λz + ˆBf y Ĉz + ˆDf where Λ is a diagonal matrix all the modes uncoupled. If there is a row of zeros in B, then f has no influence on the corresponding state variable. That variable is said to be uncontrollable. If there is a column of zeros in Ĉ, then the corresponding state variable is said to be unobservable. For many purposes, systems should be both controllable and observable. Discrete-time Most of what can be said for continuous time can also be said for discrete time: Solution: x[k + ] Ax[k] + Bf[k] y[k] Cx[k] + Df[k]. k x[k] A k x[0] + A k j Bf[j]. j0 (Show how this works by recursion), starting from x[] Ax[0] + Bf[0]. Z-transform: zx(z) zx[0] AX(z) + BF (z). X(z) (zi A) [zx[0] + BF (z)] H(z) C(zI A) B + D

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