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I have seen many times the analogy that the High Pass Filter (HPF) is a differentiator, something that explains some interesting facts, like that a non-negative signal might have negative components as in the case of differentiating a positive signal.

1) Is there an equivalent analogy for the LPF? How about the BPF? 1) Is there a mathematical formulation that prove these results?

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    \$\begingroup\$ Short: Low-pass filter is an integrator, (assuming a net DC of 0), band pass is the two strung together. \$\endgroup\$
    – Connor Wolf
    Mar 19, 2015 at 0:34

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The corollary is an integrator for a LPF. If you have a resistor feeding into a capacitor, then you get your input signal from the beginning of time integrated up to now.

Here's the math behind what I describe.

Band pass would be a LPF + a HPF so potentially an integrator feeding into a differentiator.

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Lowpass: Can be considered as an integrating device for signals far above the 3dB corner frequency only!

Highpass: Can be seen as a differentiator for low frequencies only (far below the 3dB frequency).

The reason: Integrating/differentiating requires (ideally) a phase shift of -/+ 90 deg. This is approximately true for the mentioned frequency ranges only.

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  • \$\begingroup\$ Good answer. And just for completeness: band pass filter would be a differentiator for an input signal with frequencies much below the lower cut-off, and an integrator when frequencies are much above the higher cut-off, as long as the phase shift has the correct \$\pm 90º\$ and that the gain increases/decreases with a first order slope in the considered region. \$\endgroup\$
    – Roger C.
    Mar 19, 2015 at 9:31
  • \$\begingroup\$ Also for completeness: A "correct" value of 90 deg phase shift can never be achieved with a first order circuit. We only can approach this goal. \$\endgroup\$
    – LvW
    Mar 19, 2015 at 10:59
  • \$\begingroup\$ In these ranges, the signal will be much attenuated anyway, so then the analogy is 'misleading' in the sense that it is not in the useful operating region of the filter, right? \$\endgroup\$
    – student1
    Mar 19, 2015 at 20:21
  • \$\begingroup\$ No - it is not "misleading", I think. The analogy is correct - however, you are right that the practical usefulness is limited, as long as passive circuits are concerned. But we have - in principle - the same situation with an active integrator, which - in fact - also is a first order lowpass (because the gain at dc is not infinite). And the same rule applies: It integrates far above the corner frequency (which, in this case, can be made much lower than for passive circuits). \$\endgroup\$
    – LvW
    Mar 19, 2015 at 21:04

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