Question

A second order system is to be subjected to inputs below 80Hz and is to operate with an amplitude response of ±5 percent. Calculate the minimum value of natural frequency ωn, on to accomplish this goal.

Answer 1

Answer:

So natural frequency will be 958.9616 rad/sec    

Explanation:

We have given amplitude response  $\pm 5$ %

Damped frequency f =80 Hz

So  $\omega _d=2\pi\ f=2\times 3.14\times 80=502.4rad/sec$

So $e^{\frac{-\pi \zeta }{\sqrt{1-\zeta ^2}}}=0.05$

We know that $\zeta =cos\Theta$

So $\sqrt{1-\zeta ^2}=sin\Theta$

So $e^{-\pi cot\Theta }=0.05$

${-\pi cot\Theta }=-3$

$cot\Theta =0.954$

$\Theta =46.34$

$cos\Theta =\zeta =0.690$

Now we know that $\omega _d=\omega _n\sqrt{1-\zeta ^2}$

$\omega _n=\frac{\omega _d}{\sqrt{1-\zeta ^2}}=\frac{502.4}{\sqrt{1-0.690^2}}=958.9616rad/sec$

So natural frequency will be 958.9616 rad/sec