Reading the LTI response ̇𝐱=𝐴𝐱+𝐵𝐮, 𝐲=𝐶𝐱+𝐷𝐮 mode by mode, using the right eigenvectors 𝐯𝑘, left eigenvectors 𝐰𝑘 and eigenvalues 𝜆𝑘 of 𝐴 (distinct eigenvalues; see Dyadic Expansion).

Homogeneous Response

  • Example application #2 of the dyadic expansion: solving the homogeneous eq’n
    • ̇𝐱(𝑡)=𝐴𝐱(𝑡),𝐱(0)=𝐱0
  • We know 𝐱(𝑡)=𝑒𝐴𝑡𝐱0. Use the dyadic expansion:
    • 𝐱(𝑡)=[∑𝑛𝑘=1𝑒𝜆𝑘𝑡𝐯𝑘𝐰𝖳𝑘]𝐱0
    • (the 𝐰𝖳𝑘𝐱0 are dot products!)

Modal Form of the Homogeneous Response

  • 𝐱(𝑡)=∑𝑛𝑘=1(𝐰𝖳𝑘𝐱0)⏟weights𝑒𝜆𝑘𝑡𝐯𝑘⏟modes
  • 𝑒𝜆𝑘𝑡𝐯𝑘: the “modes” of the state eq’n
  • 𝐰𝖳𝑘𝐱0: amount of the 𝑘th mode present in the response (constant) = the extent the 𝑘th mode is “excited” by the ICs
    • 𝐰𝖳𝑘𝐱0=𝐰𝑘⋅𝐱0: dot product of the left eigenvector with the ICs
  • 𝐯𝑘: right eigenvectors of 𝐴; 𝐰𝑘: left eigenvectors of 𝐴; 𝜆𝑘: eigenvalues of 𝐴; 𝐱0=𝐱(0): initial conditions

Initial Condition Along an Eigenvector

  • Suppose, for example,
    • 𝐱0=𝛽𝐯𝑗for some𝑗=1,…,𝑛
    • (i.e. the IC lies along one of the right eigenvectors)
    • Sketch: in the 𝑥1-𝑥2 plane with eigenvectors 𝐯1,𝐯2, the IC 𝐱0=𝛽𝐯1 sits on the 𝐯1 line and 𝐱(𝑡) moves straight along that line
  • Then, by bi-orthonormality:
    • 𝐱(𝑡)=∑𝑛𝑘=1𝛽(𝐰𝖳𝑘𝐯𝑗)⏟=0if𝑘≠𝑗,=1if𝑘=𝑗𝑒𝜆𝑘𝑡𝐯𝑘
    • ⇒𝐱(𝑡)=𝛽𝑒𝜆𝑗𝑡𝐯𝑗for all𝑡
  • Only the 𝑗th mode is present in the response! (No other modes contribute to 𝐱(𝑡) for such an IC)

Forced Response

  • Forced solutions:
    • ̇𝐱(𝑡)=𝐴𝐱(𝑡)+𝐵𝐮(𝑡)
  • Column partition 𝐵=[𝐛1,𝐛2,…,𝐛𝑚]:
    • 𝐵𝐮(𝑡)=∑𝑚𝑗=1𝐛𝑗𝑢𝑗(𝑡)⇒̇𝐱(𝑡)=𝐴𝐱(𝑡)+∑𝑚𝑗=1𝐛𝑗𝑢𝑗(𝑡)
    • 𝐛𝑗 describes how the 𝑗th input affects the rate of change of 𝐱(𝑡)
    • ⇒ see this more explicitly in the solutions for 𝐱(𝑡):
  • 𝐱(𝑡)=𝑒𝐴𝑡𝐱0+∫𝑡0𝑒𝐴(𝑡−𝜏)𝐵𝐮(𝜏)d𝜏=𝑒𝐴𝑡𝐱0+∑𝑚𝑗=1∫𝑡0𝑒𝐴(𝑡−𝜏)𝐛𝑗𝑢𝑗(𝜏)d𝜏=∑𝑛𝑘=1(𝐰𝖳𝑘𝐱0)𝑒𝜆𝑘𝑡𝐯𝑘+∑𝑚𝑗=1∑𝑛𝑘=1∫𝑡0(𝑒𝜆𝑘(𝑡−𝜏)𝐯𝑘𝐰𝖳𝑘)𝐛𝑗𝑢𝑗(𝜏)d𝜏
    • (𝐰𝖳𝑘𝐛𝑗: dot product again!)
  • Hence
    • 𝐱(𝑡)=∑𝑛𝑘=1𝑒𝜆𝑘𝑡𝐯𝑘{⋅}
    • where
    • {⋅}=𝐰𝖳𝑘𝐱0+∑𝑚𝑗=1(𝐰𝖳𝑘𝐛𝑗)∫𝑡0𝑒−𝜆𝑘𝜏𝑢𝑗(𝜏)d𝜏
    • Each term of the sum: amount of the 𝑘th mode due to input 𝑗
    • 𝐰𝖳𝑘𝐛𝑗: “strength” of the impact of the 𝑗th input on the 𝑘th mode
  • 𝐰𝑘⟂𝐛𝑗 ⇔ the 𝑗th input has no impact on the 𝑘th mode

Output Response

  • 𝐲(𝑡)=𝐶𝐱(𝑡)+𝐷𝐮(𝑡)=𝐶[∑𝑛𝑘=1𝑒𝜆𝑘𝑡𝐯𝑘{⋅}]+𝐷𝐮(𝑡)
  • Row partition:
    • 𝐶=[𝐜𝖳1⋮𝐜𝖳𝑝]𝐷=[𝐝𝖳1⋮𝐝𝖳𝑝]
  • Then for 𝑖=1,…,𝑝:
    • 𝑦𝑖(𝑡)=∑𝑛𝑘=1𝑒𝜆𝑘𝑡(𝐜𝖳𝑖𝐯𝑘){⋅}+𝐝𝖳𝑖𝐮(𝑡)
    • 𝐜𝖳𝑖𝐯𝑘: how much of the 𝑘th mode is “visible” in the 𝑖th output
  • If 𝐜𝑖⟂𝐯𝑘 ⇒ the 𝑘th mode is not visible in the 𝑖th output
  • Let 𝑧𝑘(𝑡)=𝑒𝜆𝑘𝑡{⋅} in the above eq’ns. Then
    1. Expansion:
      • 𝐱(𝑡)=∑𝑛𝑘=1𝑧𝑘(𝑡)𝐯𝑘
      • 𝑧𝑘(𝑡): coordinates of 𝐱(𝑡) in the basis given by {𝐯𝑘}𝑛𝑘=1
    2. and
      • ̇𝑧𝑘(𝑡)=dd𝑡𝑧𝑘(𝑡)=𝜆𝑘𝑒𝜆𝑘𝑡{⋅}⏟=𝑧𝑘(𝑡)+𝑒𝜆𝑘𝑡dd𝑡{⋅}⏟=∑𝑚𝑗=1(𝐰𝖳𝑘𝐛𝑗)𝑢𝑗(𝑡)
  • so that

Decoupled Modal Dynamics

  • ̇𝑧𝑘(𝑡)=𝜆𝑘𝑧𝑘(𝑡)+∑𝑚𝑗=1(𝐰𝖳𝑘𝐛𝑗)𝑢𝑗(𝑡)
  • Decoupled!!
  • Laplace:
    • 𝑍𝑘(𝑠)=1𝑠−𝜆𝑘∑𝑚𝑗=1(𝐰𝖳𝑘𝐛𝑗)𝑈𝑗(𝑠)
    • and
    • 𝑌𝑖(𝑠)=∑𝑛𝑘=1(𝐜𝖳𝑖𝐯𝑘)𝑍𝑘(𝑠)+∑𝑚𝑗=1𝑑𝑖𝑗𝑈𝑗(𝑠)
    • ⇒ the TFM in partial-fraction form
  • 𝐱(𝑡)=∑𝑛𝑘=1𝑧𝑘(𝑡)𝐯𝑘
    • expresses the solution in a new basis, given by the eigenvectors of 𝐴
  • This is sometimes called the “modal basis” and 𝑧𝑘(𝑡) the “modal coordinates”:
  • State trajectory x(t) from x0 in the x1-x2 plane, with its components x1(t), x2(t) along the axes and its modal coordinates z1(t), z2(t) along the eigenvectors v1, v2 (from Prof. Sanner's notes)
  • 𝐱(𝑡)=∑𝑛𝑘=1𝑧𝑘(𝑡)𝐯𝑘=𝑉𝐳(𝑡)𝑉=[𝐯1,…,𝐯𝑛],𝐳(𝑡)=[𝑧1(𝑡)⋮𝑧𝑛(𝑡)]
  • Then
    • 𝐴𝐱(𝑡)+𝐵𝐮(𝑡)=̇𝐱(𝑡)=𝑉̇𝐳(𝑡)(𝐴𝐱(𝑡)=𝐴𝑉𝐳(𝑡))
    • ⇒̇𝐳(𝑡)=𝑉−1𝐴𝑉⏟𝐴𝑚𝐳(𝑡)+𝑉−1𝐵⏟𝐵𝑚𝐮(𝑡)
  • Output: 𝐲=𝐶𝐱+𝐷𝐮=𝐶(𝑉𝐳)+𝐷𝐮

Modal Form of System Dynamics

  • ̇𝐳=𝐴𝑚𝐳+𝐵𝑚𝐮𝐲=𝐶𝑚𝐳+𝐷𝑚𝐮
  • 𝐴𝑚=𝑉−1𝐴𝑉=Λ=diag{𝜆𝑘}(modal𝐴)𝐵𝑚=𝑉−1𝐵(modal 𝐵)𝐶𝑚=𝐶𝑉𝐷𝑚=𝐷
  • Block diagram: 𝐮(𝑡)→ [system, 𝐱 or 𝐳] →𝐲(𝑡)
    • The internal details are different, but 𝐮 and 𝐲 are the same in both models ⇒ same input/output behavior
  • There is nothing special about the eigenvector matrix 𝑉 here: see State Coordinate Transformations

Complex Eigenvalues: Block Modal Form

  • The modal coordinate transform results in the decoupled ̇𝑧𝑘=𝜆𝑘𝑧𝑘+∑𝑚𝑗=1(𝐰𝖳𝑘𝐛𝑗)𝑢𝑗
  • We might want to modify this if 𝜆𝑘 is complex, since then the solutions 𝑧𝑘(𝑡) are also complex and hard to interpret physically

Block Modal Form

  • Suppose
    • 𝜆1,2=𝜎1±𝑗𝜔1𝜆3,4=𝜎3±𝑗𝜔3𝜆5,…,𝜆𝑛real, non-repeated
  • There exists a nonsingular 𝑃 such that
    • 𝐴𝑝=[𝜎1−𝜔1𝜔1𝜎1𝜎3−𝜔3𝜔3𝜎3𝜆5⋱𝜆𝑛]
    • the “block modal form” (zeros off the blocks)
    • Pure diagonalization would instead give diag{𝜎1+𝑗𝜔1,𝜎1−𝑗𝜔1,𝜎3+𝑗𝜔3,𝜎3−𝑗𝜔3,𝜆5,…}
  • 𝑃=[Re{𝐯1},Im{𝐯1},Re{𝐯3},Im{𝐯3},𝐯5,…,𝐯𝑛]
    • 𝜆1=𝜎1+𝑗𝜔1⇒𝐯1
    • 𝜆2=𝜎1−𝑗𝜔1⇒𝐯2; 𝜆2=𝜆1⇒𝐯2=𝐯1 (so only 𝐯1, 𝐯3 are needed)
  • Then
    • 𝑒𝐴𝑝𝑡=[𝑒𝜎1𝑡cos𝜔1𝑡−𝑒𝜎1𝑡sin𝜔1𝑡𝑒𝜎1𝑡sin𝜔1𝑡𝑒𝜎1𝑡cos𝜔1𝑡𝑒𝜎3𝑡cos𝜔3𝑡−𝑒𝜎3𝑡sin𝜔3𝑡𝑒𝜎3𝑡sin𝜔3𝑡𝑒𝜎3𝑡cos𝜔3𝑡𝑒𝜆5𝑡⋱]
    • (using the rotation-like and block-diagonal forms)
  • and
    • 𝐱(𝑡)=𝑃𝐳(𝑡)=𝑃𝑒𝐴𝑝𝑡𝐳0=𝑧1(𝑡)Re{𝐯1}+𝑧2(𝑡)Im{𝐯1}⏟1st complex mode+𝑧3(𝑡)Re{𝐯3}+𝑧4(𝑡)Im{𝐯3}⏟2nd complex mode+𝑧5(𝑡)𝐯5+…
    • Oscillations for the 1st complex mode occur in a plane spanned by {Re{𝐯1},Im{𝐯1}}
    • Oscillations for the 2nd complex mode occur in the plane spanned by Re{𝐯3},Im{𝐯3}

Possible slip in notes

  • The notes write the second eigenvalue as 𝜆2=𝜎2−𝑗𝜔2 before noting 𝜆2=𝜆1; written here as 𝜎1−𝑗𝜔1

Exciting a Single Oscillatory Mode

  • Suppose
    • 𝐳0=𝐳(0)=[𝑎𝑏0⋮0]
  • Then
    • 𝐳(𝑡)=𝑒𝐴𝑝𝑡𝐳0=𝑎[𝑒𝜎1𝑡cos𝜔1𝑡−𝑒𝜎1𝑡sin𝜔1𝑡0⋮0]⏟𝑧1(𝑡)entry+𝑏[𝑒𝜎1𝑡sin𝜔1𝑡𝑒𝜎1𝑡cos𝜔1𝑡0⋮0]⏟𝑧2(𝑡)entry
  • ⇒𝐱(𝑡)=𝑒𝜎1𝑡(𝑎cos𝜔1𝑡+𝑏sin𝜔1𝑡)Re{𝐯1}+𝑒𝜎1𝑡(𝑏cos𝜔1𝑡−𝑎sin𝜔1𝑡)Im{𝐯1}
    • ⇒ only the 1st mode is present, and creates changes in 𝐱(𝑡) in only 2 directions: Re{𝐯1},Im{𝐯1}
  • Note that
    • 𝐱0=𝑃𝐳0=𝑎Re{𝐯1}+𝑏Im{𝐯1}
  • Thus, if 𝐱0∈span{Re{𝐯1},Im{𝐯1}}, then
    1. The solution contains only the 1st mode (oscillations with frequency 𝜔1)
    2. 𝐱(𝑡)∈span{Re{𝐯1},Im{𝐯1}} for all 𝑡>0
    • i.e. the modal oscillations occur in a planar subset of the original state space
  • Similarly 𝐱0∈span{Re{𝐯3},Im{𝐯3}} would uniquely excite the 2nd oscillatory mode, and the resulting 𝐱(𝑡) would be confined to this planar subset of state space

Polar Visualization

  • An alternate visualization: let the complex eigenvector 𝐯 be expressed in polar form:
    • 𝐯=[𝑟1𝑒𝑗𝜑1𝑟2𝑒𝑗𝜑2⋮𝑟𝑛𝑒𝑗𝜑𝑛]
  • For a mode with such an eigenvector, the corresponding oscillations in each component of 𝐱(𝑡) will have relative amplitude 𝑟𝑖 and relative phase 𝜑𝑖
    • i.e. the ratio of the amplitude of oscillations in 𝑥1(𝑡) compared to the amplitude of oscillations in 𝑥3(𝑡) would be 𝑟1/𝑟3
    • and the difference in phase between oscillations in 𝑥1(𝑡) and 𝑥3(𝑡) would be 𝜑1−𝜑3
  • Another reason to find an alternate 𝑃 for LTI: repeated (real) eigenvalues → Jordan Normal Form