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

So natural frequency will be 958.9616 rad/sec

Step-by-step explanation:

We have given amplitude response
`\pm 5` %

Damped frequency f =80 Hz

So
`\omega _d=2\pi\ f=2* 3.14* 80=502.4rad/sec`

So
`e^{(-\pi \zeta )/(√(1-\zeta ^2))}=0.05`

We know that
`\zeta =cos\Theta`

So
`√(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√(1-\zeta ^2)`

`\omega _n=(\omega _d)/(√(1-\zeta ^2))=(502.4)/(√(1-0.690^2))=958.9616rad/sec`

So natural frequency will be 958.9616 rad/sec