A systematic way to compute the matrix exponential: 𝑒𝐴𝑡=ℒ︀−1{(𝑠𝐼−𝐴)−1} (LTI only).

Laplace Transform Review

  • One approach to computing 𝑒𝐴𝑡 more systematically: use Laplace!
    • 𝐹(𝑠)=ℒ︀{𝑓}≝∫∞0−𝑒−𝑠𝑡𝑓(𝑡)d𝑡
  • Recall important properties:
    • Linearity: if 𝑎1,𝑎2 are constant,
      • ℒ︀{𝑎1𝑓1+𝑎2𝑓2}=𝑎1𝐹1+𝑎2𝐹2
    • Derivative rule:
      • ℒ︀{̇𝑓}=𝑠𝐹(𝑠)−𝑓(0−)

WARNING

  • Linearity does not hold if 𝑎1,𝑎2 are themselves time-varying

Laplace of Vectors

  • Apply to vectors as well (component by component):
    • ℒ︀{𝐱(𝑡)}=[ℒ︀{𝑥1(𝑡)}ℒ︀{𝑥2(𝑡)}⋮ℒ︀{𝑥𝑛(𝑡)}]=[𝑋1(𝑠)𝑋2(𝑠)⋮𝑋𝑛(𝑠)]=𝐗(𝑠)
  • Derivative:
    • ℒ︀{̇𝐱(𝑡)}=[𝑠𝑋1(𝑠)−𝑥1(0−)𝑠𝑋2(𝑠)−𝑥2(0−)⋮𝑠𝑋𝑛(𝑠)−𝑥𝑛(0−)]=𝑠𝐗(𝑠)−𝐱0=𝑠𝐼𝐗(𝑠)−𝐱0
    • with
      • 𝐱0=[𝑥1(0−)⋮𝑥𝑛(0−)]
  • Linearity holds in the linear algebra sense:
    • ℒ︀{𝐴𝐱(𝑡)}=𝐴𝐗(𝑠)if𝐴is constant

Homogeneous Equation

  • Apply the LT to the homogeneous eq’n ℒ︀{̇𝐱(𝑡)=𝐴𝐱(𝑡)}:
    • 𝑠𝐼𝐗(𝑠)−𝐱0=𝐴𝐗(𝑠)⇒𝐗(𝑠)=[𝑠𝐼−𝐴]−1⏟Φ(𝑠)𝐱0
    • Φ(𝑠): the “resolvent matrix”
  • 𝐗(𝑠)=Φ(𝑠)𝐱0𝐱(𝑡)=ℒ︀−1{Φ(𝑠)𝐱0}=ℒ︀−1{Φ(𝑠)}𝐱0
  • But we know 𝐱(𝑡)=𝑒𝐴𝑡𝐱0

Matrix Exponential via Laplace

  • 𝑒𝐴𝑡=ℒ︀−1{Φ(𝑠)}=ℒ︀−1{(𝑠𝐼−𝐴)−1}

Structure of the Resolvent

  • Note:
    • (𝑠𝐼−𝐴)−1=𝑄(𝑠)𝑟(𝑠)
    • 𝑄(𝑠)=Adj(𝑠𝐼−𝐴): 𝑛×𝑛 matrix of polynomials in 𝑠; the max degree of each polynomial is 𝑛−1
    • 𝑟(𝑠)=det(𝑠𝐼−𝐴): scalar polynomial in 𝑠 of degree 𝑛
  • ⇒ each entry of (𝑠𝐼−𝐴)−1 is a ratio of polynomials in 𝑠
  • ⇒ each ratio has the same denominator
  • ⇒ find 𝑒𝐴𝑡 by inverse transforming each of these rational functions ⇒ use partial fractions

Matrix Exponential by Partial Fractions

  • 𝐴=[0−61−5]
  • (𝑠𝐼−𝐴)=[𝑠6−1𝑠+5]𝑄(𝑠)=[𝑠+5−61𝑠]
  • 𝑟(𝑠)=𝑠(𝑠+5)+6=𝑠2+5𝑠+6=(𝑠+2)(𝑠+3)
  • 𝑒𝐴𝑡=ℒ︀−1[𝑠+5(𝑠+2)(𝑠+3)−6(𝑠+2)(𝑠+3)1(𝑠+2)(𝑠+3)𝑠(𝑠+2)(𝑠+3)]
  • Look at the (1,1) entry:
    • 𝑠+5(𝑠+2)(𝑠+3)=𝑐1𝑠+2+𝑐2𝑠+3⇒ℒ︀−1{⋅}=𝑐1𝑒−2𝑡+𝑐2𝑒−3𝑡
    • 𝑐1=[(𝑠+2)(𝑠+5(𝑠+2)(𝑠+3))]𝑠=−2=31=3𝑐2=[(𝑠+3)(𝑠+5(𝑠+2)(𝑠+3))]𝑠=−3=2−1=−2
    • (1,1) entry of 𝑒𝐴𝑡 here is 3𝑒−2𝑡−2𝑒−3𝑡
  • Repeat for the other entries (exercise!):
    • 𝑒𝐴𝑡=[3𝑒−2𝑡−2𝑒−3𝑡6(𝑒−3𝑡−𝑒−2𝑡)𝑒−2𝑡−𝑒−3𝑡3𝑒−3𝑡−2𝑒−2𝑡]
  • ⇒ every entry of 𝑒𝐴𝑡 is a linear combination of the same 2 scalar exponential functions 𝑒−2𝑡,𝑒−3𝑡
    • b/c each inverse transform has the same denominator (𝑠+2)(𝑠+3)
  • The same 𝑒𝐴𝑡 is found with eigenvectors in Dyadic Expansion

Possible slip in notes

  • Where the full 𝑒𝐴𝑡 is written out, the notes label the matrix 𝐴=[0−51−6]; the result shown is that of 𝐴=[0−61−5] from the start of the example, so the label looks like a slip

Poles and Eigenvalues

  • Conclude: generally each entry of 𝑒𝐴𝑡 will be a linear combination of the scalar exponential functions {𝑒𝑝𝑘𝑡}, where 𝑝𝑘 are the roots of the char. poly
    • 𝑟(𝑠)=det(𝑠𝐼−𝐴)
    • ⇒ these roots are the poles of the system
  • But note
    • 𝑟(𝑠)=0=det(𝑠𝐼−𝐴)
    • also defines the eigenvalues of 𝐴: equivalently {𝑒𝜆𝑘𝑡}, 𝜆𝑘 eigenvalues of 𝐴

Repeated poles

  • We will also get terms like 𝑡𝑙𝑒𝜆𝑘𝑡 if poles/eigenvalues are repeated, like
    • 𝑟(𝑠)=(𝑠+1)(𝑠+2)2
  • ⇒ will treat the repeated pole case separately later (Jordan Normal Form)