Vai's Uni Notes

State Transition Matrix

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The unique matrix that maps the state at one time to the state at another for ̇𝐱=𝐴(𝑡)𝐱.

Definition

  • From the fundamental matrix, every homogeneous solution is 𝐱(𝑡)=Ψ(𝑡)𝛂 ∀𝑡≥𝑡0, where 𝛂=Ψ(𝑡0)−1𝐱0. Thus
    • 𝐱(𝑡)=Ψ(𝑡)Ψ−1(𝑡0)⏟Φ(𝑡,𝑡0)𝐱0∀𝑡≥𝑡0

State Transition Matrix

  • 𝐱(𝑡)=Φ(𝑡,𝑡0)𝐱0Φ(𝑡,𝜏)=Ψ(𝑡)Ψ(𝜏)−1,Ψfundamental
  • Φ is the “state transition matrix” (STM) for the system: unique to the system (any choice of fundamental matrix gives the same Φ)
  • With the STM, the general solution of the state eq’n is
    • 𝐱(𝑡)=Φ(𝑡,𝑡0)𝐱0+∫𝑡𝑡0Φ(𝑡,𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏
  • ⇒ if we know Ψ(𝑡) (or Φ(𝑡,𝜏)) we know everything about the system. What can we say about these generally?
    • Φ(𝑡,𝜏) is an 𝑛×𝑛 matrix, defined by 𝑛2 functions of 𝜏,𝑡
    • Usually we will not know these functions exactly, and must argue using only generic properties of any STM

Properties of Any STM

  1. Φ(𝑡,𝑡)=Φ(𝜏,𝜏)=𝐼 for any 𝑡,𝜏
  2. Φ(𝑡,𝜏) is nonsingular for any 𝑡,𝜏 ⇒detΦ(𝑡,𝜏)≠0
    • detΦ(𝑡,𝜏)=det[Ψ(𝑡)Ψ−1(𝜏)]=detΨ(𝑡)detΨ(𝜏)−1=detΨ(𝑡)[detΨ(𝜏)]−1
    • But we know detΨ(𝑡)≠0 (Lemma 2), and the same proof shows detΨ(𝑡)≠∞ at any finite 𝑡
    • ⇒detΦ(𝑡,𝜏)≠0 ✓
  3. Inverse:
    • Φ(𝑡,𝜏)−1=Φ(𝜏,𝑡)
    • Pf:
    • Φ(𝑡,𝜏)−1=[Ψ(𝑡)Ψ−1(𝜏)]−1=(Ψ−1(𝜏))−1Ψ−1(𝑡)=Ψ(𝜏)Ψ−1(𝑡)=Φ(𝜏,𝑡)✓
  4. Derivatives:
    • a)𝜕𝜕𝑡Φ(𝑡,𝜏)=𝐴(𝑡)Φ(𝑡,𝜏)b)𝜕𝜕𝜏Φ(𝑡,𝜏)=−Φ(𝑡,𝜏)𝐴(𝜏)
    • Pf a):
    • 𝜕𝜕𝑡Φ(𝑡,𝜏)=𝜕𝜕𝑡[Ψ(𝑡)Ψ−1(𝜏)]=[𝜕𝜕𝑡Ψ(𝑡)]Ψ−1(𝜏)=𝐴(𝑡)Ψ(𝑡)Ψ−1(𝜏)=𝐴(𝑡)Φ(𝑡,𝜏)✓
    • Pf b) (using the derivative of the inverse, dd𝑡Ψ−1=−Ψ−1𝐴):
    • 𝜕𝜕𝜏Φ(𝑡,𝜏)=Ψ(𝑡)[𝜕𝜕𝜏Ψ−1(𝜏)]=Ψ(𝑡)[−Ψ−1(𝜏)𝐴(𝜏)]=−Φ(𝑡,𝜏)𝐴(𝜏)✓
  5. For any 𝑡0,𝑡1,𝑡2:
    • Φ(𝑡2,𝑡0)=Φ(𝑡2,𝑡1)Φ(𝑡1,𝑡0)
    • the “composition” or “semi-group” property
    • Pf:
    • Φ(𝑡2,𝑡0)=Ψ(𝑡2)Ψ(𝑡0)−1Φ(𝑡1,𝑡0)=Ψ(𝑡1)Ψ(𝑡0)−1Φ(𝑡2,𝑡1)=Ψ(𝑡2)Ψ−1(𝑡1)
    • Φ(𝑡2,𝑡1)Φ(𝑡1,𝑡0)=[Ψ(𝑡2)Ψ(𝑡1)−1][Ψ(𝑡1)Ψ(𝑡0)−1]=Ψ(𝑡2)Ψ(𝑡0)−1=Φ(𝑡2,𝑡0)✓

Composition, Intuitively

  • Consider the homogeneous solution 𝐱(𝑡)=Φ(𝑡,𝑡0)𝐱0 (with 𝐱0=𝐱(𝑡0)), 𝐱∈ℝ2:
  • State trajectory in the x1-x2 plane: Phi(t1,t0) maps x0 to x1, Phi(t2,t1) maps x1 to x2, and Phi(t2,t0) maps x0 directly to x2 (from Prof. Sanner's notes)
    • 𝐱1=𝐱(𝑡1)=Φ(𝑡1,𝑡0)𝐱0𝐱2=𝐱(𝑡2)=Φ(𝑡2,𝑡0)𝐱0=Φ(𝑡2,𝑡1)𝐱1
  • Note also this implies
    • Φ(𝑡,𝜏)=Φ(𝑡,0)Φ(0,𝜏)=Φ(𝑡,0)Φ(𝜏,0)−1

Constructing the STM

  • Can we actually constructively determine the STM for an arbitrary 𝐴(𝑡)? i.e. find the 𝑛2 functions of 𝑡,𝜏
  • Generally no, but we can try:
    1. Find a family of fundamental solutions
      • remember ̇𝐱=𝐴𝐱 ⇔
      • ̇𝑥1=𝑎11𝑥1+𝑎12𝑥2+…+𝑎1𝑛𝑥𝑛⋮̇𝑥𝑛=𝑎𝑛1𝑥1+…+𝑎𝑛𝑛𝑥𝑛
    2. Numerically solve for Φ(𝑡,𝑡0) on a grid of 𝑡>𝑡0 points
    3. Brute force series expansion (may possibly simplify): the “Peano-Baker” series

Peano-Baker Series

  • Φ(𝑡,𝜏)=𝐼+∫𝑡𝜏𝐴(𝜎)d𝜎+∫𝑡𝜏𝐴(𝜎1)∫𝜎1𝜏𝐴(𝜎2)d𝜎2d𝜎1+∫𝑡𝜏𝐴(𝜎1)∫𝜎1𝜏𝐴(𝜎2)∫𝜎2𝜏𝐴(𝜎3)d𝜎3d𝜎2d𝜎1+…(infinite # terms)
  • IF
    • 𝐴(𝑡)∫𝑡𝜏𝐴(𝜎)d𝜎=[∫𝑡𝜏𝐴(𝜎)d𝜎]𝐴(𝑡)
    • (i.e. 𝐴(𝑡) commutes with its integral) for every 𝑡,𝜏
  • THEN
    • Φ(𝑡,𝜏)=∑∞𝑘=01𝑘!𝑀(𝑡,𝜏)𝑘where𝑀(𝑡,𝜏)=∫𝑡𝜏𝐴(𝜎)d𝜎
  • Recall the scalar exponential function satisfies
    • 𝑒𝑚=∑∞𝑘=0𝑚𝑘𝑘!
  • By analogy the above power series defines the matrix exponential function. Common to write (when the commutativity condition is satisfied!):
    • Φ(𝑡,𝜏)=exp[∫𝑡𝜏𝐴(𝜎)d𝜎]
    • where by “exp” is meant the formal power series above
    • This looks comfortingly familiar, but we must be careful: see Matrix Exponential

Constant A

  • As a final possible simplification, if 𝐴 is constant, then the commutativity condition is definitely satisfied, and
    • 𝑀(𝑡,𝜏)=∫𝑡𝜏𝐴d𝜎=𝐴∫𝑡𝜏d𝜎=(𝑡−𝜏)𝐴
  • ⇒Φ(𝑡,𝜏)=exp[(𝑡−𝜏)𝐴]=𝑒(𝑡−𝜏)𝐴
  • (Note: can also write 𝑒𝐴(𝑡−𝜏) when 𝐴 is understood to be constant)

Other Results Involving the STM

  • Under a change of state coordinates 𝐱=𝑃(𝑡)𝐳: Φ𝑥(𝑡,𝜏)=𝑃(𝑡)Φ𝑧(𝑡,𝜏)𝑃−1(𝜏) (proof)
  • For periodic 𝐴(𝑡), the STM factors into a periodic part and an LTI part: Floquet Theory

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