Solving Dynamic Equations: The State Transition Matrix II

Transcription

1 Reading the Text Solving Dynamic Equations: The State Transition Matrix II EGR 326 February 27, 2017 Just a reminder to read the text Read through longer passages, to see what is connected to class topics. Be aware of proofs Practice with the examples Focus on the concepts Class time alone is not enough to learn the course material! Overview The state transition matrix A function of the system matrix, A, that determines how the state vector transitions, or evolves, over time. Continuous time Expand the concept of the scalar exponential equation to a matrix exponential Discrete time (previous class) Understand the A matrix as an operator on the state vector Solutions for Dynamic System Models 1) Analytical Closed-form expression (equation) For example, x(t) = X 0 e αt Plot the behavior of x(t) 2) Numerical Use a computer to simulate how x(t) behaves as time progresses Plot the behavior of x(t) 1

2 Recap: DT State Transition Property Assume that x[k] is any solution to x[k +1] = Ax[k] For individual, scalar variables, we know: x[k] = a k x[0] Complete Solution to DT System Solve x[k + 1] = Ax[k] + Bu[k] Solution x[k] = Φ(k, 0)x[0]+ k 1 k 1 l=0 = A k x[0]+ A k l Bu[l] l=0 Φ(k,l +1)Bu[l] Recap of the DT Solution For linear system models, the general solution can be computed via superposition à The total response from several inputs is equal to The sum of their individual responses, Plus an initial condition term Sketch the graphical solution for the system, from [k = 0 to k = 10] x[k+1] = 0.75 x[k] + u[k] x[0] = 3 Input: u[1] = 4, u[3] = 4, u[5] = 2 5" 4" 3" 2" 1" 0" Total (Complete) Response Using Superposition Init%Cond%&%Input%% 0" 1" 2" 3" 4" 5" 6" 7" 8" 9" 10" à 6.0# 5.0# 4.0# 3.0# 2.0# 1.0# 0.0# 0# 1# 2# 3# 4# 5# 6# 7# 8# 9# 10# 2

3 State Transition Property Visualize state-space For the multi-dimensional, state-space systems, we want to know how the state vector evolves in state-space How to visualize the state vector, e.g. x = [v c, i L ], as a single entity, moving through state-space Natural Response Solution For a scalar expression, we know: x(t) = e αt x(0)!!! =!1 +!! + 1 2!!!! + 1 6!!!! +! Write this out for x(t) = e 3t (5) and find x(1), x(2) For the dynamic matrix equations, we find (pp 52-53):!(!) =!! +!" + 1 2!!!! + 1 6!!!! +! observe that this looks a lot like the series for the exponential, e αt Natural Response Solution Define the matrix exponential as e At I + At + A2 t 2 2! + A3 t 3 3! Ak t k +... k! The homogeneous solution is thus seen to be x(t) = e At x(0) Solving with the Matrix Exponential Natural Response Example: Given a system matrix of! =! Such that! =! ! !!! x 0 = 1 $ # & " 1 % Task: Find solution expressions for x(t) 3

4 State Transition Matrix Define a matrix, Φ(t, ) That will advance the state vector From its initial position at time To its position at time t x(t) = Φ(t, )x( ) (Note, this is the natural response only, so far) State Transition Matrix Define a matrix, Φ(t, τ) that will advance the state vector from its position at time τ to its position at time t: Φ(t,τ ) e At e Aτ If τ is selected to be time = 0, then: Φ(t) = e At = e A(t τ ) x(t) = e A(t τ ) x(τ ) = e At x(0) Complete Solution for CT Systems For a basic state-space model, with initial conditions and a forcing function (or input source) x(t) = Ax(t)+ Bu(t) The complete solution must include the natural response and the forced response x(t) = x n (t)+ x f (t) Complete Solution for CT Systems Using the matrix exponential for the state transition matrix, the solution is x(t) = Φ(t, )x( )+ Φ(t,τ )Bu(τ ) dτ x(t) = e A ( t ) x( )+ e A( t τ ) Bu(τ ) dτ t t 4

5 * Interpret the General Solution * x(t) = e A ( t ) t x( )+ e A( t τ ) Bu(τ ) dτ Left hand side is a vector à the state vector First term, on the right hand side is a matrix times a vector = a vector This is the response to initial conditions Natural response The integral: the integrand is also a vector for each value of τ. The terms are then summed (continuously), and so this is a vector This is the system response to inputs Forced response General Solution Format The Claim of Linear Dynamic System Theory is that: ALL Dynamic Systems behave according to combinations of Exponential functions Geometric sequences All Systems in the universe are understood to behave according to our two solution expressions (CT & DT) General Solution Format Continuous time solution is an exponential expression Discrete time solution is a geometric expression For both CT and DT, our solutions are The sum of the natural response plus the forced response There are two terms in the integral equation and in the summation equation Summary Solutions to scalar dynamic equations are CT: exponential DT: geometric series Solutions to dynamic system equations use the state transition matrix Matrix exponential of system matrix Powers of the system matrix Solutions are the sum of the natural and forced response 5