Help: How to find out the 3db bandwidth of a second order system.

[wangmengde]

For a second order system with the transfer function: T=1/(s²+sω/Q+ω²), how can I get the conclusion that when Q»5, the 3db bandwidth is approximately ω/Q?
Can anybody give me some hints on making this derivation?

 

[LvW]

For a second order system with the transfer function: T=1/(s²+sω/Q+ω²), how can I get the conclusion that when Q»5, the 3db bandwidth is approximately ω/Q?
Can anybody give me some hints on making this derivation?

Who told you this? The frequency w is a running variable - thus, the 3dB-bandwidth cannot contain w.
Perhaps you mean the bandwidth BW of a 2nd order bandpass? In this case, the numerator contains "s" and BW=wp/Q with wp= pole or center frequency.

 

[FvM]

The shown transfer function is of the low-pass type, it's unusual to specify a bandwidth for a low-pass.

 

[barry]

The bandwidth for a LP is from 0 to the -3dB frequency, i.e., the BW IS the frequency at -3db. That's not an unusual specification.

But LVW is right. The cutoff frequency (or BW) is NOT a function of Q.

 

[FvM]

the BW IS the frequency at -3db

The more general terms are pass band and stop band frequency in my view. You may want to name the pass band frequency bandwidth in this case. But what's the "bandwidth" of a 2nd order high-pass?

 

[barry]

infinity.