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Second Order Transfer Function Question regarding Overshoot

  • #1
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1

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?
 

Answers and Replies

  • #2
NascentOxygen
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  • #3
donpacino
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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?
https://en.wikipedia.org/wiki/Overshoot_(signal)
 
  • #4
rude man
Homework Helper
Insights Author
Gold Member
7,634
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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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