Solving Dynamic Equations: The State Transition Matrix

Transcription

1 Overview Solving Dynamic Equations: The State Transition Matrix EGR 326 February 24, 2017 Solutions to coupled dynamic equations Solutions to dynamic circuits from EGR 220 The state transition matrix Discrete time Understanding the A matrix as an operator on the state vector Continuous time Expanding the concept of the scalar exponential equation to a matrix exponential CT: Response of RL Circuit Natural and Forced Response Find the expression for v(t), for all time Solve the dynamic equation for this sytem i s u(t) R L v n (t) = V 0 e t/τ i n (t) = I 0 e t/τ or KVL: KCL: v o = L di L dt = i R R i L + i R = i S so i R = ( i L i S ) i f (t) = I S ( 1 e t/τ ) L di L dt = ( i i L S ) R L x = ( x u) R 1

2 Total Response i(t) = i f (t)+ i n (t) v(t) = v f (t)+ v n (t) State Transition Property How the system evolves from one moment in time to the next Oscilloscope trace of v(t) for 2 nd order system: v(t) = V o e t τ +V s 1 e t ( τ ) v(t) = v( )+[v(0) v( )]e t τ 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) Solutions for Dynamic System Models Describe how the state vector evolves from one state to the next Mathematically Graphically What are the two sources that drive this evolution? 1) 2) 2

3 State Transition Property For the multi-dimensional, state-space systems, we want to know how the state vector evolves: How to visualize the state vector, x = [v c, i L ], as a single entity, moving through state space Demo: Romeo.m One axis is Romeo s love for Juliet The other axis is Juliet s love for Romeo State-space, plot x1 vs. x2, not x vs. t and y vs. t Both the A matrix, and the initial conditions may change from case to case. Square Matrices & Linear Algebra Functions with SQUARE Matrices x = Ax + Bu x[k+1] = Ax[k] + Bu[k] DT: Discrete time Understanding the A matrix as an operator on the state vector Practice examples on handout 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] Generalize this for the state vector: x[k] = Φ[k,τ ]x[τ ] Discrete Time Scalar Equation Discrete time Geometric growth x[k+1] = a x[k] DT can be used to describe growth in human or other populations vegetation consumption of raw materials accumulation of interest on a loan DT can also result from sampling CT processes or systems 3

4 Recap: Discrete Time Equation The variables to define are x the object, concept, population that is changing, that is dynamic k time step, discrete time a constant rate of growth a > 1 for positive growth x[0] initial condition, magnitude, state Solve with forward recursion (HW 2) x[1] = a x[0] x[2] = a x[1] = a (a x[0]) = a 2 x[0] x[k] = a k x[0] Discrete Time Systems Solving DT homogeneous system Basic Equation: x[k+1] = Ax[k] Solve recursively once the initial state values are known x[1] = Ax[0] x[2] = Ax[1] = AAx[0] Continue solving for x[3], x[4] x[k] (next slide) DT State Transition Matrix Group the system matrices to form the state transition matrix F(k, 0) = AA A (multiplying A k times) = A k for constant coefficient systems Such that x[k] = F(k, 0) x[0] = A k x[0] Multiplication of any initial state vector by this matrix tells you the state of the system (the state vector) at time step k DT State Transition Matrix Generalize the state transition matrix to F(k, l) = A[k-1]A[k-2] A[l] = A k-l for constant coefficient systems for transition from state at time t = l to time t = k 4

5 Complete Solution to DT System Solve x[k + 1] = Ax[k] + Bu[k] Objectives Find algebraic expression for the solution Understand the interpretation of the solution Proposed solution x[k] = Φ(k, 0)x[0]+ Φ(k, l +1)Bu[l] Interpretation is analogous to that for CT solution = A k x[0]+ A k l Bu[l] Practice with the DT Solution Since the system is linear, the general solution can be computed via superposition 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 Practice with the DT Solution The natural response + the forced response(s) = the total response Natural Response 6" 7" 8" 9" 1 6.0# 5.0# 4.0# 6" 7" 8" 9" 1 5

6 Forced Responses Total (Complete) Response Using Superposition 6" 7" 8" 9" 1 0# 1# 2# 3# 4# 5# 6# 7# 8# 9# 10# 6" 7" 8" 9" 1 6.0# 5.0# 4.0# Total (Complete) Response (Using Superposition) 6" 7" 8" 9" 1 6.0# 5.0# 4.0# Complete Solution to DT System Solve x[k + 1] = Ax[k] + Bu[k] Objectives Find algebraic expression for the solution Understand the interpretation of the solution Proposed solution x[k] = Φ(k, 0)x[0]+ Φ(k, l +1)Bu[l] Interpretation is analogous to that for CT solution = A k x[0]+ A k l Bu[l] 6

7 Summary Understanding forms of solutions to dynamic system equations (state space equations) Definition and use of the state transition matrix Continuous time examples for next class 7