Vai's Uni Notes

Equilibria and Trim Conditions

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A special (and very common) choice of reference solution for linearization.

Equilibria

  • The 𝐳∗=𝟎 pendulum case in the linearization example is representative of a special category of reference solutions known as equilibria
  • Such solutions are defined by
    • ̇𝐳∗=𝟎⇒𝐳∗=constant
  • The state eq’n with the above restrictions:
    • 𝟎=𝐟(𝟎,𝐳∗,𝐰∗)
  • This is no longer a DE, but rather a system of nonlinear equations which can be solved (analytically or numerically) for 𝐳∗ given 𝐰∗ (or alternately for 𝐰∗ given 𝐳∗)

Trim Conditions

  • In A/C applications these equations are often referred to as the “trim conditions”

Pendulum Equilibria

  • For the pendulum, equilibria satisfy
    • 0=̇𝑧∗1−𝑧∗2⇒𝑧∗2=00=𝑚𝑙2̇𝑧∗2+𝑚𝑔𝑙sin𝑧∗1+𝑏(𝑧∗2)⏟𝑏(0)=0−𝑤∗
  • ⇒𝑚𝑔𝑙sin𝑧∗1=𝑤∗or𝑧∗1=sin−1(𝑤∗𝑚𝑔𝑙)
  • where 𝑤∗ is the “trim” torque applied at the pivot, i.e.
    • 𝑤∗=0⇒𝑧∗1=0𝑤∗=12𝑚𝑔𝑙⇒𝑧∗1=𝜋6(30°)𝑤∗=√22𝑚𝑔𝑙⇒𝑧∗1=𝜋4(45°)𝑤∗=𝑚𝑔𝑙⇒𝑧∗1=𝜋2(90°)etc.
  • For these equilibria, 𝐸 and 𝐺 in the linearization are unchanged, but
    • 𝐹=[0𝑚𝑔𝑙cos𝑧∗1−1𝑏′(0)]
    • The 𝑚𝑔𝑙cos𝑧∗1 entry changes depending on the specific equilibrium used in the linearization!
  • How do these changes affect properties of the corresponding linearized model?

Linearized Dynamics at Different Equilibria

  • Return to the pendulum’s linearized I/O DE (derived here):
    • ̈𝑦(𝑡)+𝛼1(𝑡)̇𝑦(𝑡)+𝛼0𝑦(𝑡)=𝛽0𝑢(𝑡)
  • Consider the equilibrium solutions, for which ̇𝐳∗=𝟎⇒𝐳∗=const. For such solutions
    • 𝛼0(𝑡)=𝑔𝑙cos𝜃∗(const)𝛼1(𝑡)=𝑏′(0)𝑚𝑙2(const)
    • so the dynamics are LTI
  • Suppose for convenience the numerical values
    • 𝑔𝑙=5,𝑏′(0)𝑚𝑙2=2
  • (Equilibrium condition: 𝑚𝑔𝑙sin𝜃∗=𝑤∗)

Case 1: θ* = 0 (w* = 0)

  • Char poly is 𝑠2+2𝑠+5, with poles −1±2𝑗
  • ⇒𝑦ℎ(𝑡)=𝑐1𝑒−𝑡cos(2𝑡)+𝑐2𝑒−𝑡sin(2𝑡)=𝐴𝑒−𝑡cos(2𝑡+𝜑)
  • Stable: 𝑦ℎ(𝑡)→0⇒𝜃(𝑡)→0

Case 2: θ* = π/4

  • 𝛼0=5√2≈3.54 ⇒ poles −1±1.59𝑗 (frequency changed)
  • ⇒𝑦ℎ(𝑡)=𝐴𝑒−𝑡cos(1.59𝑡+𝜑)
  • 𝑦ℎ(𝑡)→0: stable. But 𝜃(𝑡)↛0; instead 𝜃(𝑡)→𝜋4
    • Sketch: 𝜃(𝑡) starts at 𝜃(𝑡0) and oscillates (decaying) about the dashed line 𝜋4

Case 3: θ* = 80°

  • 𝛼0=𝑔𝑙cos80°≈0.868
  • Char poly 𝑠2+2𝑠+0.868 has poles −1.36, −0.64
  • ⇒𝑦ℎ(𝑡)=𝑐1𝑒−1.36𝑡+𝑐2𝑒−0.64𝑡
  • Stable ⇒𝑦ℎ(𝑡)→0
    • Sketch: 𝜃(𝑡) rises from 𝜃(𝑡0) to 80° with a purely exponential transient now (no oscillation)

Case 4: θ* = 90° (w* = mgl)

  • 𝛼0=𝑔𝑙cos90°=0
  • Char poly 𝑠2+2𝑠 ⇒ poles 0,−2
  • ⇒𝑦ℎ(𝑡)=𝑐1+𝑐2𝑒−2𝑡
  • 𝑦ℎ(𝑡)↛0 here: Not stable

Case 1 Revisited: θ* = 180°

  • Equilibrium condition 𝑚𝑔𝑙sin𝜃∗=𝑤∗
    • 𝑤∗=0⇒𝜃∗=0 or 𝜃∗=±180°
  • For 𝜃∗=180°: 𝛼0=𝑔𝑙cos𝜃∗=−5
  • Char poly is 𝑠2+2𝑠−5, with poles −3.45, 1.45
  • 𝑦ℎ(𝑡)=𝑐1𝑒−3.45𝑡+𝑐2𝑒1.45𝑡
  • Note 𝑦ℎ(𝑡)→∞ here (unstable)
  • So the same physical system, linearized about different equilibria, yields LTI models with very different behavior (oscillatory, purely exponential, marginal, unstable)

Graph View

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