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Is there an explicit formulation for n-cascaded RC filters?

The resistor and capacitor values are not necessarily same. Assume that:
\$ R_i \neq R_j \$ and \$ C_i \neq C_j \$ for \$ i \neq j\$ and \$ 1 \leq i,j \leq n \$.

I built a 5th degree filter with \$R_i=1k\Omega\$, \$C_i=1\mu F\$ and \$V_{in}=1\sin(2\pi \times 1kHz) \; V\$. The output signal was nearly smooth for a \$1k\Omega\$ lead. And rise of the output signal was visible by eye when the input was a step signal.

I tried to find transfer response of this 5th order filter, but after the 3rd order, the size of A4 paper became insufficient for writing the expressions.

  • \$\begingroup\$ I think your image may be missing \$\endgroup\$ – Andy aka Nov 11 '13 at 8:19
  • \$\begingroup\$ It is showing up to me... Anyway: there is no analytical expression for an n-stage RC filter. I tried to crack this problem a couple of months ago and it just gets very unwieldy, even when you can generalize R1=R2=...=Rn. The best way to go is just to regard each stage as a shaped attenuator with finite input and output impedance and multiply them together numerically. \$\endgroup\$ – user36129 Nov 11 '13 at 8:25
  • \$\begingroup\$ I can see it now!! \$\endgroup\$ – Andy aka Nov 11 '13 at 9:33

There isn't an explicit formula for n cascaded RC low pass filters (as you have shown) because as you move from left to right, R2 and C2 load the output of R1 and C1 thus changing R1&C1's response and because R3 and C3 are doing the same to R2 and C2 this in turn reflects on the loading of R1 and C1.

Using 2 port networks and matrix maths does help you solve more easily for several sections but the formulas do tend to get long after a few sections and solving for the cut-off frequency and Q is very tiresome.

The best solution is to use a simulator like LTSpice.

Given that you have an op-amp in your circuit acting as a buffer it wouldn't be unreasonable to use this to help. This sallen-key calculator is very good and gives you the response and formulas for a 2nd order low pass filter: -

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Cascading 2 of these will probably get you a better response than cascading five passive RC filters and the beauty of this method is that you can cascade them without interaction between components because the op-amp acts as a buffer.

  • \$\begingroup\$ I've sometimes wondered if the subdivision of filters into 2-pole sections primarily for ease of figuring out what a filter is doing? I would expect that adding an extra pole using a passive RC combos won't quite yield a full "pole"'s worth of performance, but if the impedance of the second RC is large relative to the first, the performance should be comparable, and I would think that a 9-pole filter using 3 op amps should be able to yield cutoff-frequency performance comparable to an 8-pole filter using four, though I'm not sure about phase performance. \$\endgroup\$ – supercat Nov 11 '13 at 17:22
  • \$\begingroup\$ @supercat I did the very same last week; three cascaded LCR low pass filters with approx 3 times the impedance for subsequent stages. You can only push it so far though! \$\endgroup\$ – Andy aka Nov 11 '13 at 18:10
  • \$\begingroup\$ Certainly I wouldn't expect a 16-pole cascaded RC filter to work very well, but I would think that getting an extra pole or two on each stage of a filter should be entirely practical since in some filter topologies a low-impedance op-amp output will be fed into a high-impedance input so it would seem one could replace an RC stage with either two stages that have a 100:1 impedance variation between them, or three stages with a 10:1 impedance separation. Even just getting three poles per op amp instead of four would seem a pretty big win. \$\endgroup\$ – supercat Nov 11 '13 at 18:23
  • \$\begingroup\$ It's easier with LCRs to do that because losses are smaller.... just small value resistors to control Q. RC it's harder or at least seems to be. \$\endgroup\$ – Andy aka Nov 11 '13 at 18:27

The behavior of any series-parallel combination of resistors, capacitors, and inductors may be modeled at any given frequency using complex arithmetic by regarding each capacitor and inductor as though it were a resistor whose "resistance" [actually impedance] is an imaginary number; a resistor in series with a capacitor thus behaves like a "resistor" with a complex impedance whose real part comes from the resistor and whose imaginary part comes from the capacitor. The normal formula for parallel resistors (1/(1/R1+1/R2)) works just as well for impedances as for simple resistances. Using complex maths, one could compute the voltage output of a cascaded RC filter in terms of the input using a somewhat-reasonable-looking formula. Unfortunately, those forumulae become very messy if one wants to use real numbers to compute the real (in-phase) and imaginary (out-of-phase) components of the final signal, since division by a complex number generally requires normalizing it, so something like (a+bj)/(c+dj) becomes ((ac-bd)+(bc+ad)j)/(c^2-d^2). Nesting such expressions very deeply causes them to become unwieldy very quickly.

Still, if you have a package which can easily handle math with complex numbers, you should be able to come up with a complex transfer function and then plot it. Alternatively, if you have a circuit simulator with a Bode plotter function (which stimulates a circuit with various frequencies and plots the response) you could use that to plot the behavior of your circuit numerically.


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