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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?

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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)?

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): -

enter image description here

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.

Is the same pattern followed when we connect another identical stage to increase it to 6th order?

Nope - it isn't butterworth any more.

Is it true that this does not hold true for passive filters e.g RC filter due to "loading of following stages". Why?

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.

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  • \$\begingroup\$ Andy, This explanation is very clear. I take this top mean that all the tools connecting together multiple 2nd order stages are taking the "easier approach" i.e shortcut with drawbacks. I have noticed that I only find 2nd order sallen key filters very rarely 3rd order. I guess this has to do with single stage higher order filter being more sensitive to component tolerance. \$\endgroup\$
    – quantum231
    Mar 9, 2016 at 10:18
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    \$\begingroup\$ @quantum231 you can find stuff on the internet for multi stage butterworth filters and they are relatively easy to design but, the more stages you have there is a greater problem with component tolerances. Q for a 2nd order filter is easy - the pole is at 45 deg and, Q = \$\dfrac{1}{2\cdot cos(45 deg)}\$ = 0.7071. With a 4th order filter the poles are split either side of 45 degrees as per the diagram in my answer, hence Q1 is 0.5412 and Q2 is 1.3066. \$\endgroup\$
    – Andy aka
    Mar 9, 2016 at 10:25
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    \$\begingroup\$ It's also important to have the low Q stage first in the chain because this minimizes signal overshoot right at the start and therefore the op-amp output isn't liable to hit the rails thus making life easier for the high Q stage at the end. Very important this is! \$\endgroup\$
    – Andy aka
    Mar 9, 2016 at 10:27
  • \$\begingroup\$ Ok. I understand that a multistage filter will be required for multiple (more than 2) poles. However, all stages are different and locate poles differently on the circle in the complex plane ! They all work together as a single filter. However, the component values for each stage are chosen to take care of loading of each stage on the previous stage and loading can cause the actual transfer function to become different? \$\endgroup\$
    – quantum231
    Mar 9, 2016 at 13:07
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    \$\begingroup\$ Loading effects are minimal in active filters - the op-amp output will remain like a perfect voltage source for all reasonable load currents (up to 5mA or greater on many op-amps). The next stage might have an input impedance of 10 kohm and this won't be a measurable problem. As for a 3 pole butterworth, it's a less generic solution because the loading of the 2nd order butterworth on the 1st order RC low pass is not convenient and also the 1st order filter has an output impedance that modifies the 2nd order response. I haven't looked in detail as to whether they have taken this into account. \$\endgroup\$
    – Andy aka
    Mar 9, 2016 at 13:32
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A passive filter is designed to work with specific terminations. The termination resistance is every bit as important at controlling the transfer function as the components within the filter. For instance an LC filter may require a 50\$\Omega\$ load at each end. An RC filter might require to be driven from a short circuit, and be loaded with an open circuit.

If you cascade a second passive section directly, you change the loading the first section drives into, and the second is driven from, changing the transfer function that each section produces.

If you place a buffer between the sections so that each section still sees the correct terminal impedance, so a 50 ohm isolation amplifier for LC filters, or a high input impedance buffer for a RC filters, then the individual sections retain their original transfer functions, and the resultant is the product of the original functions.

The transfer function of an active filter need not depend on the terminations. A Sallen Key filter has a zero impedance output, and is designed to be driven from a zero impedance input. Its transfer function does depend on being driven from a zero impedance, if that changes the transfer function will change.

However, when we cascade these filters directly, the zero impedance output of the first stage correctly drives the second stage input. Their transfer functions are automatically the product of the individual sections.

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