Vai's Uni Notes

General Solution of the State Equation

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The forced response of ̇𝐱=𝐴(𝑡)𝐱+𝐵(𝑡)𝐮, 𝐲=𝐶(𝑡)𝐱+𝐷(𝑡)𝐮, built from the fundamental matrix and the state transition matrix.

Is the Conserved Quantity Still Conserved?

  • For the homogeneous response, 𝛂=Ψ(𝑡0)−1𝐱0=Ψ(𝑡)−1𝐱(𝑡) for all 𝑡≥𝑡0, and we showed Ψ(𝑡) is in fact invertible
    • i.e. dd𝑡[𝛂]=𝟎=dd𝑡[Ψ−1(𝑡)𝐱(𝑡)] when 𝐮(𝑡)=𝟎
  • For the non-homogeneous response (𝐮(𝑡)≠𝟎), look at
    • dd𝑡[Ψ−1(𝑡)𝐱(𝑡)]
  • When 𝐮(𝑡)≠𝟎:
    • dd𝑡[Ψ−1𝐱]=[dd𝑡Ψ−1]𝐱+Ψ−1̇𝐱=[dd𝑡Ψ−1]𝐱+Ψ−1[𝐴𝐱+𝐵𝐮]
  • Using the derivative of the inverse, dd𝑡Ψ−1=−Ψ−1𝐴:
    • dd𝑡[Ψ−1𝐱]=(−Ψ−1𝐴)𝐱+Ψ−1[𝐴𝐱+𝐵𝐮]⇒dd𝑡[Ψ−1𝐱]=Ψ−1𝐵𝐮

Derivation

  • Integrate both sides:
    • ∫𝑡𝑡0dd𝜏[Ψ−1(𝜏)𝐱(𝜏)]d𝜏=∫𝑡𝑡0Ψ−1(𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏[Ψ−1(𝜏)𝐱(𝜏)]𝑡𝜏=𝑡0=∫𝑡𝑡0(⋅)d𝜏
  • So
    • Ψ−1(𝑡)𝐱(𝑡)−Ψ−1(𝑡0)𝐱(𝑡0)⏟𝐱0=∫𝑡𝑡0Ψ−1(𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏
  • so that
    • 𝐱(𝑡)=Ψ(𝑡)Ψ−1(𝑡0)⏟Φ(𝑡,𝑡0)𝐱0+∫𝑡𝑡0Ψ(𝑡)Ψ−1(𝜏)⏟Φ(𝑡,𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏

General Solution of the State Equation

  • 𝐱(𝑡)=Φ(𝑡,𝑡0)𝐱0⏟𝐱zi(𝑡)+∫𝑡𝑡0Φ(𝑡,𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏⏟𝐱zs(𝑡)
  • 𝐱zi(𝑡): “zero input” response (due only to 𝐱0)
  • 𝐱zs(𝑡): “zero (initial) state” response (due only to 𝐮(𝑡))

Output Response

  • Similar decomposition for the output:
    • 𝐲(𝑡)=𝐶(𝑡)𝐱(𝑡)+𝐷(𝑡)𝐮(𝑡)=𝐶(𝑡)𝐱zi(𝑡)⏟𝐲zi(𝑡)+𝐶(𝑡)𝐱zs(𝑡)+𝐷(𝑡)𝐮(𝑡)⏟𝐲zs(𝑡)
  • 𝐲zi(𝑡)=𝐶(𝑡)Φ(𝑡,𝑡0)𝐱0𝐲zs(𝑡)=𝐷(𝑡)𝐮(𝑡)+∫𝑡𝑡0𝐶(𝑡)Φ(𝑡,𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏

Impulse Response Matrix

  • Equivalently
    • 𝐲zs(𝑡)=∫𝑡𝑡0𝐺(𝑡,𝜏)𝐮(𝜏)d𝜏
    • 𝐺(𝑡,𝜏): 𝑝×𝑚 “impulse response matrix”

Impulse Response Matrix

  • 𝐺(𝑡,𝜏)=𝐶(𝑡)Φ(𝑡,𝜏)𝐵(𝜏)+𝐷(𝜏)𝛿(𝑡−𝜏)
  • where 𝛿(𝑡−𝜏) is the “delta function” (unit impulse), which has the property that for any function 𝑓(𝑡),
    • ∫𝑓(𝜏)𝛿(𝑡−𝜏)d𝜏=𝑓(𝑡)(sifting property)
  • For LTI systems this becomes 𝐺(𝑡)=𝐶𝑒𝐴𝑡𝐵+𝐷𝛿(𝑡), the inverse Laplace transform of the transfer function matrix

Interpretation: Time-Varying Coefficients

  • Rewrite the general state solution as
    • 𝐱(𝑡)=Ψ(𝑡)[Ψ−1(𝑡0)𝐱0+∫𝑡𝑡0Ψ−1(𝜏)𝐵(𝜏)𝐮(𝜏)d𝜏]⏟𝛂(𝑡)
  • or
    • 𝐱(𝑡)=Ψ(𝑡)𝛂(𝑡)=∑𝑛𝑘=1𝛼𝑘(𝑡)𝛙𝑘(𝑡)
    • (the “conserved” 𝛂 now evolves, driven by the input)
  • Note
    • ̇𝛂(𝑡)=Ψ−1(𝑡)𝐵(𝑡)𝐮(𝑡)
  • Check:
    • dd𝑡𝐱(𝑡)=dd𝑡[Ψ(𝑡)𝛂(𝑡)]=̇Ψ(𝑡)𝛂(𝑡)+Ψ(𝑡)̇𝛂(𝑡)=𝐴(𝑡)Ψ(𝑡)𝛂(𝑡)+Ψ(𝑡)Ψ−1(𝑡)𝐵(𝑡)𝐮(𝑡)̇𝐱(𝑡)=𝐴(𝑡)𝐱(𝑡)+𝐵(𝑡)𝐮(𝑡)✓

LTI Systems

  • For LTI systems:
    • Φ(𝑡,𝜏)=𝑒(𝑡−𝜏)𝐴=∑∞𝑘=0(𝑡−𝜏)𝑘𝑘!𝐴𝑘
  • Note: since the STM for LTI depends only on the elapsed time, 𝑡−𝜏, we can w.l.o.g. assume 𝑡0=0. So

LTI Solution

  • 𝐱(𝑡)=𝑒𝐴𝑡𝐱0+∫𝑡0𝑒𝐴(𝑡−𝜏)𝐵𝐮(𝜏)d𝜏
  • [Note: 𝑒𝐴(𝑡−𝜏) is calculated by replacing "𝑡" in the expansion of 𝑒𝐴𝑡 with "𝑡−𝜏", i.e. scalars 𝑒𝑎𝑡→𝑒𝑎(𝑡−𝜏)]

Graph View

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