Does connecting 2 stages of Sallen Key topology Butterworth characteristic filter only create a 4th order Butterworth filter with same cutoff frequency (-3dB point) but double the roll-off (-40dB per decade instead of -20dB per decade)?
Is the same pattern followed when we connect another identical stage to increase it to 6th order?
Is it true that this does not hold true for passive filters e.g RC filter due to "loading of following stages". Why?
I want to understand that if we can just multiply transfer function of each stage in active filter, why does same not apply to passive filters?
Best Answer
No it doesn't create a 4th order butterworth in the classical sense - it will still be flat in the pass band but have a less precise roll-off area compared to a classical butterworth 4th order filter.
A multi-order butterworth filter has poles equally distributed around a circle in the pole zero diagram. The diameter of the circle is the natural resonant frequency for each stage (common to all): -
If you cascaded two identical 2nd order stages you'd end up with double poles at 45 degrees and the overall Q factor would be 0.5. If you look at any classical butterworth filter design and you multiply all the individual Q factors for each stage, the overall Q factor is 0.7071 - this doesn't happen when you cascade two individual 2nd order butterworth filters because 0.7071 x 0.7071 = 0.5.
An 8th order I recently designed has Q factors of 0.509795579, 0.601344886, 0.899976223 and 2.562915448. Multiply them all together and you get 0.707107072 which is near enough the reciprocal of the sq root of 2.
Nope - it isn't butterworth any more.
You are missing the point - it neither holds true for active or passive filters. However, it's worse for passive filters due to loading effects.