[sdiy] step response of second order filter

[Magnus Danielson]

Harry,

On 01/10/2012 04:30 PM, Harry Bissell wrote:
> Hello all
>
> in a previous thread i asked about the step response of a second order filter (SVF). The consensus was that the
> step response (overshoot, undershoot, settle) was an inevitable consequence of the (2nd order) filter itself.
>
> Does the "q" of such a filter make any difference.  ie at a value of .707 I'd expect one overshoot, one undershoot, then settling.
> Higher resonance values extend this.
>
> What about if the Q was less than .707 ? would the filter then be underdamped ?

The Q has definite impact on the step response. The higher Q, the longer 
ringing, as you can expect. Similarly, very low Q will result in a 
modest overshot, if any, and then settle. The slope will be controlled 
by the bandwidth.

So,
The overdamped (low Q/high damping factor) will be a rather eventless thing.
The critically damped (Q=0.707/d=1.414) will be a modest overshot and 
then settle,
The underdamped (high Q/low damping factor) will be a long ringing 
oscillation superimposed ontop of the rising slope.

The ringing will be of the peak frequency.

All this is well-established in control theory. Where the position of 
poles, zeros and its effects on impulse response, step response and 
frequency response is well established.

I find myself watching scopes and making comments on step response 
behaviours on second degree systems quite regularly, so I feel it is a 
relevant knowledge for me in the real world.

Cheers,
Magnus