Second Order Transfer Function Question regarding Overshoot

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SUMMARY

The discussion centers on the application of the overshoot formula for a closed-loop second-order transfer function represented as $$\frac{10-s}{0.3s^2+3.1s+(1+24K_{C})}$$ when subjected to a step input. The formula $$\frac{A}{B}=e^{\frac{-\pi \zeta}{\sqrt{1-\zeta^2}}}$$ is questioned for its validity in this context. Participants suggest splitting the transfer function into a low-pass and a bandpass component, each having established overshoot expressions in relevant textbooks. The numerical value of $$K_C$$ is crucial for determining the final response.

PREREQUISITES
  • Understanding of second-order transfer functions
  • Familiarity with overshoot calculations in control systems
  • Knowledge of damping ratio ($\zeta$) and its impact on system response
  • Experience with simulation tools for step response analysis
NEXT STEPS
  • Research the impact of varying $$K_C$$ on the system's overshoot and stability
  • Learn about low-pass and bandpass filter characteristics in control systems
  • Explore simulation tools like the one mentioned (http://sim.okawa-denshi.jp/en/detatukeisan.htm) for visualizing step responses
  • Study the Nyquist plot and its relevance in analyzing system stability
USEFUL FOR

Control engineers, systems analysts, and students studying control theory who are interested in understanding overshoot behavior in second-order systems.

Tom Hardy
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Homework Statement


If I have a closed loop second order transfer function such as:

$$\frac{10-s}{0.3s^2+3.1s+(1+24K_{C})}$$

Can I still use this formula for overshoot (when a step input is applied) ?:
$$\frac{A}{B}=e^{\frac{-\pi \zeta}{\sqrt{1-\zeta^2}}}$$
Where B is the step input size

I don't think you can but I'm not sure, can someone confirm?
 
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Tom Hardy said:

Homework Statement


If I have a closed loop second order transfer function such as:

$$\frac{10-s}{0.3s^2+3.1s+(1+24K_{C})}$$

Can I still use this formula for overshoot (when a step input is applied) ?:
$$\frac{A}{B}=e^{\frac{-\pi \zeta}{\sqrt{1-\zeta^2}}}$$
Where B is the step input size

I don't think you can but I'm not sure, can someone confirm?
https://en.wikipedia.org/wiki/Overshoot_(signal)
 
Split your xfr function into two parts: one is a low-pass 2nd order, the other a bandpass 2nd order. For both, there are expressions for overshoot in textbooks etc. Your answer depends on the numerical value of KC.

You then just add the time responses of the two separated xfr functions.
@nascent, thanks for the link! wow, even does nyquist plot!
 

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