Second Order Systems, Interactive

• Overdamped Case (ζ>1)

  For the overdamped case:

  \[\begin{array}{c} H\left( s \right) = \frac{{\omega _0^2}}{{{s^2} + 2\zeta {\omega _0}s + \omega _0^2}} = \frac{{\omega _0^2}}{{{s^2} + 2\zeta {\omega _0}s + \omega _0^2}} = \frac{{{\alpha _1}{\alpha _2}}}{{\left( {s + {\alpha _1}} \right)\left( {s + {\alpha _1}} \right)}}\\ {\rm{distinct}}\;{\rm{real}}\;{\rm{poles}}\;{\rm{at}}\;s = - {\alpha _{1,2}} = - \zeta {\omega _0} \pm {\omega _0}\sqrt {{\zeta ^2} - 1} \\ {y_\gamma }\left( t \right) = 1 - \frac{{{\alpha _2}{e^{ - {\alpha _1}t}} - {\alpha _1}{e^{ - {\alpha _2}t}}}}{{{\alpha _2} - {\alpha _1}}} \end{array}\]
  • In the time domain there is not oscillation, nor is there any overshoot. As ζ increases, the system gets slower and looks more like a first order response (because of the dominant pole approximation).
  • In the frequency domain (Bode Plot), the response is flat until the frequency reaches α₂ (the lower frequency pole) at which point it starts decreasing at 20 dB per decade until it reaches the second pole at α₁ where it breaks down to -40dB per decade. Note that if α₁>>α₂, that the magnitude of Bode plot is alread quite small at ω=α₁ , so the higher frequency pole matters less (this is the frequency domain argument for making the dominant pole approximation).

• Critically Damped Case (ζ=1)

  For the critically damped case

  \[\begin{array}{c} H\left( s \right) = {\left. {\frac{{\omega _0^2}}{{{s^2} + 2\zeta {\omega _0}s + \omega _0^2}}} \right|_{\zeta = 1}} = \frac{{\omega _0^2}}{{{{\left( {s + {\omega _0}} \right)}^2}}} = \frac{{{\alpha ^2}}}{{{{\left( {s + \alpha } \right)}^2}}}\\ {\rm{double}}\;{\rm{pole}}\;{\rm{at}}\;s = - \alpha = - {\omega _0}\\ {y_\gamma }\left( t \right) = 1 - {e^{ - \alpha t}}\left( {1 + \alpha t} \right) \end{array}\]
  • In the time domain there is not oscillation, nor is there any overshoot..
  • In the frequency domain (Bode Plot), the response is flat until the frequency reaches α at which point it starts decreasing at 40 dB per decade.
  • Though the math needed is a special case, the behavior of the system is very similar to that of the under (or over) damped case with ζ near 1.

• Underdamped Case (ζ<1)

  For the underdamped case:

  \begin{gathered}
  H\left( s \right) = \frac{{\omega _0^2}}{{{s^2} + 2\zeta {\omega _0}s + \omega _0^2}} = \frac{{{\alpha ^2} + \omega _d^2}}{{{{\left( {s + \alpha } \right)}^2} + \omega _d^2}} = \frac{{{\alpha ^2} + \omega _d^2}}{{{s^2} + 2\alpha s + {\alpha ^2} + \omega _d^2}} \\
  {\text{complex}}\;{\text{conjugate}}\;{\text{poles}}\;{\text{at}}\;s = - \alpha \pm j{\omega _d} = - \zeta {\omega _0} \pm j{\omega _0}\sqrt {1 - {\zeta ^2}} \\
  {y_\gamma }\left( t \right) = 1 - \frac{{\sqrt {{\alpha ^2} + \omega _d^2} }}{{{\omega _d}}}{e^{ - \alpha t}}\sin \left( {{\omega _d}t + \operatorname{atan} \left( {\frac{{{\omega _d}}}{\alpha }} \right)} \right) , \quad or\\ {y_\gamma }\left( t \right) = 1 - \frac{1}{{\sqrt {1 - {\zeta ^2}} }}{e^{ - \zeta {\omega _0}t}}\sin \left( {\sqrt {1 - {\zeta ^2}} {\omega _0}t + \operatorname{acos} \left( \zeta \right)} \right) \\
  \end{gathered}
  
  • The real part of the pole locations is -α=-ζω₀. The imaginary part of the pole locations are ±ω_d=ω₀√(1-ζ²). The distance of the pole from the origin is ω₀, and the angle from the horizontal axis is θ=acos(ζ). Refer to pole zero diagram.
  • Time Domain
    • Effect of ω_d and α: There are oscillations at a frequency ω_d, or period T=2π/ω_d (note ${\omega _d} = {\omega _0}\sqrt {1 - {\zeta ^2}}$ so for small values of ζ, ω_d≈ω₀). The decay envelope goes as e^-αt. To see this, try changing α and the frequency of oscillation will not change (the pole moves horizontall so the imaginary part, which determines frequency of osicllation, remains constant. Try changing ω_d, and the pole moves vertically with a constant envelope of decay, but the frequency of oscillation changes.
    • Effect of ω₀ and ζ: As you increase ω₀, the speed of the system increases, but the shape of the step response remains the same - it is simply scaled (expanded or compressed) in time. As you decrease ζ, the system becomes less damped and oscillates more.
  • In the frequency domain (Bode Plot) the magnitude of the Bode plot will have a peak if ζ<1/√2=0.707. The location of the peak is at ${\omega _{peak}} = {\omega _0}\sqrt {1 - 2{\zeta ^2}}$ and the peak value is $\left| {H\left( {j{\omega _{peak}}} \right)} \right| = 1/\left( {2\zeta \sqrt {1 - {\zeta ^2}} } \right)$ (in dB use $20\cdot log_{10}\left( {\left| {H(j{\omega _{peak}})} \right|} \right)$). For small ζ use ω_peak≈ω₀ and $\left| {H(j{\omega _{peak}})} \right| = 1/\left( {2\zeta } \right)$. The peak hieght is determined solely by ζ, and (with the approximation) the peak location is determined solely by ω₀.
  
  • Things to try:
    • Set ζ=0.2, and ω₀=1. Now increase ω₀ and note that the system speeds (the step response changes more quickly; the Bode plot shifts to the right) but the shape (i.e., peak heights...) does not. If you double ω₀, system doubles in speed. This is because wherever ω₀ appears in the step response, it is multpiplied by time, t. So doubling ω₀, makes the system twice as fast. In other words as the poles move away from the orgin (recall that disctance is ω₀) the system gets faster.
    • Set ζ=0.2, and ω₀=1. Now vary ζ. As ζ decreases, the system becomes less stable, i.e., the step response oscillates more and the Bode plot peak increases (if you were to set ζ=0, the oscillations would continue forever, and the Bode peak would go to infinity) As you increase ζ the system settles down, and there are no socillations if you make ζ≥1.
    • Set ζ=0.2, and ω₀=2. Now vary α. Note that the frequency of oscillation is constant (look at where the oscillations cross y_γ(t)=1). This is because the frequency of the decaying sinusoid is given by ω_d.
    • Set ζ=0.2, and ω₀=3. Now vary ω_d. Note that the frequency of oscillation changes, but the decay envelope does not (if you were to draw "envelope" functions, one through the lower peaks and one through the upper peaks, these envelopes don't change). This is because the decay coefficient of the decaying sinusoid is given by α.

References