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Firstly, some basic theory….

The transfer function of a simple low pass filter is:-

Low Pass Filter

This transfer function has one pole. The gain goes from 1 to almost zero as frequency increases and the phase goes from 0 to -90 deg. giving -20dB roll off at -90 deg. This transfer function is referred to as “a simple lag”. The gain magnitude Bode plot is:-

Low Pass Magnitude Response

My question is - Is there any practical RC circuitry which will cancel out this pole? That is to say circuitry which has a cancelling zero of 1+Ts as its transfer function. This transfer function is referred to as “a simple lead”. The zero would have a gain increase of 20dB/decade at +90 degrees starting at the same w = 1/T frequency as the pole which it is cancelling. The overall result of cascading the low pass pole with the added circuitry would be 0dB gain and 0 deg. phase shift at the output over the full bandwidth. The transfer function of this pole cancelling circuitry would have a gain Bode plot of:-

Pole Canceller Pass Magnitude Response

Intuitively I would expect a simple RC high pass filter to do the job of cancelling out the pole of the simple RC low pass filter but this is not the case. A simple RC high pass filter has the transfer function:-

High Pass Transfer Function

As can be seen, this transfer function has a zero, s in its numerator but there is also a pole in the denominator. The high pass filter gain Bode plot looks like this:-

High Pass Magnitude Response

As is apparent from the Bode plot, the high pass filter does not cancel out the low pass filter. In fact if the two filters were to be cascaded, the overall gain peaks at a maximum of ½ that of the individual filters at w = 1/T before falling away and the phase goes from +90 deg. to -90 deg. as frequency increases. (An attenuated bandpass filter).

So in a practical sense, is there any RC circuitry which will perfectly cancel out the phase and magnitude of a pole?

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  • \$\begingroup\$ In reality, "practical" and "perfect(ly)" always contradict. \$\endgroup\$ – Huisman Dec 14 '19 at 14:23
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    \$\begingroup\$ 1/ Not an RC circuit as your output is increasing which is not possible with only passive components 2/ Not perfectly as all circuits, even active ones, will roll off if the frequency gets high enough. \$\endgroup\$ – Oldfart Dec 14 '19 at 14:23
  • \$\begingroup\$ Nothing cancels perfectly. Usually, when trying to do this, you end up with a dipole, and a consequent long-tail on the step response. \$\endgroup\$ – Chu Dec 14 '19 at 14:42
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    \$\begingroup\$ Why would you practically want to cancel a RC filter? Why not pick R=0 \$ \Omega \$ and leave out the cap in the first place? \$\endgroup\$ – Huisman Dec 14 '19 at 14:51
  • \$\begingroup\$ Why bother taking an asprin when you shouldn't have got the headache in the first place? \$\endgroup\$ – James Dec 14 '19 at 15:12
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No. You may be able to get close, but not an exact cancellation. That's why you need to be careful trying to cancel a 2nd pole (the one that causes the gain to roll off at -40db/decade) or the pole in a second order filter because if you miss, you could still end up with 0 deg phase margin and a possible oscillator.

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Intuitively I would expect a simple RC high pass filter to do the job of cancelling out the pole of the simple RC low pass filter but this is not the case.

Indeed, because now you have no path for DC to pass through the circuit. You need a resistor as well:

enter image description here

Taken from this manuscript on pole zero cancellation in photodetectors: https://ieeexplore.ieee.org/document/6154091

So in a practical sense, is there any RC circuitry which will perfectly cancel out the phase and magnitude of a pole?

It depends on what you mean by perfect. Real circuits have noise, so you will never unattenuate a strongly attenuated circuit perfectly. Furthermore, real circuits have additional parasitic capacitances, so your filter is only addressing the lowest order pole. In practice however this can work quite well. For photosensors, getting 5 or even 10x higher bandwidth is certainly possible if you are motivated enough to very precisely tune the capacitor and resistor values.

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    \$\begingroup\$ It's becoming apparent that I shouldn't have used that word "perfect(ly)" \$\endgroup\$ – James Dec 14 '19 at 15:40
  • \$\begingroup\$ Well, you know how anal engineers are, and there are a lot of them on this forum ;- \$\endgroup\$ – SteveSh Dec 15 '19 at 18:52

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