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You are testing a new device to determine its characteristics. You find the magnitude of its frequency response shows a peak at 500 Hz, and the measured overshoot of the step response is 40% of the steady state level. Find the system’s characteristics and confirm your answers by generate plots of its frequency and step response using either Simulink or Pspice.

I'm trying to figure out how to find the natural frequency with the given info.

$\frac{500\,\text{Hz}}{140\,\%}\approx 357.145$; I'm assuming my $r_o = 357.145$, where we settle at.

Given percentage overshoot (40%) I can solve for our damping ratio $\zeta = 0.279$. I have formulas to solve for Natural Frequency ($\omega_n$) but not enough known variables to solve for it. Does anyone know how to solve for $\omega_n$ here?

In our case, $\beta = 1.288$, if that helps.

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  • $\begingroup$ What is $r_0$? The final value of the step response? What is $\beta$? $\endgroup$ – Matt L. Feb 7 '16 at 20:25
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Using this formula for the percentage overshoot, the damping ratio is given by

$$\zeta=\sqrt{\frac{\log(0.4)^2}{\pi^2+\log(0.4)^2}}=0.28$$

which is more or less the same value that you obtained. Consequently, you have an underdamped system. The peak frequency $\omega_{max}$ of the frequency response is related to the damping ratio and to the natural frequency in the following way (link):

$$\omega_{max}=\omega_n\sqrt{1-2\zeta^2}$$

With $\zeta=0.28$ and $f_{max}=500\text{ Hz}$ this results in a natural frequency $f_n=544.51\text{ Hz}$. Apart from a scaling factor, a second order system is completely characterized by $\zeta$ and $\omega_n=2\pi f_n$.

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  • $\begingroup$ Thanks for the help. I did run across the wn=2 pi fn. Thanks for replying $\endgroup$ – Dslethal Feb 8 '16 at 20:56
  • $\begingroup$ @Dslethal: If the answer was helpful you can accept it by clicking on the green check mark to its left. $\endgroup$ – Matt L. Feb 8 '16 at 20:56

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