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An \$n^{th}\$ order circuit has \$n\$ reactive components. So a circuit with 2 capacitors and 2 inductors (total of 4 reactive components) should be considered a 4th order circuit.

However, I have encountered countless resources which refer to the following filter as second order even though, it clearly has 4 reactive components: Filter Source: https://electronicbase.net/band-pass-filter-calculator/#passive-bandpass-filter-2nd-order

Why is such a circuit referred to as a second order? Isn't it a 4th order filter with a roll off of 80 dB/dec?

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It is the denominator of the corresponding transfer function that determines the order of the filter.

  • A simple RLC combination (lowpass, highpas, bandpass) has a second-order denominator. Hence, it is a second-oder filter - even when the rising/falling slope of the magnitude approaches 20dB/Dec only (bandpass case).

  • Therefore, the shown bandpass circuit (2L and 2C) - together with the required termination resistors - will give a 4th-order bandpass.

  • Remember: A 1st-order lowpass can be transformed (lowpass-bandpass transformation) into a second-order bandpass. Hence, this gives the lowest order bandpass function; there will be no 1st-order bandpass.

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An \$n^{th}\$ order circuit has \$n\$ reactive components. So a circuit with 2 capacitors and 2 inductors (total of 4 reactive components) should be considered a 4th order circuit.

It's not that simple. You need to check the contribution of each reactive component to the frequency response.

Connecting 10 parallel-connected capacitors across the input terminals doesn't make the filter 10th order. Because those capacitors don't have almost any contribution to the frequency response unless the input source is a current source.

As for the circuit in your question, as tobalt stated in their answer, the boundaries of the band-pass filter are plotted with a slope of ±40dB/decade (or ±12dB per octave). So the filter is a 2nd order BP filter.

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The other two answers already do the job, but I'll add that there is a little bit of semantics involved: a 2nd order filter, and a 2nd order bandpass filter. The first refers, in general, to any filter, and it represents the number of states (reactive elements or delays). The second one refers specifically to a bandpass filter, so it shows its origin as being the lowpass prototype (as you correctly mention in the comments). And, to make things even more confusing, if you were to refer to a bandpass filter in terms of the 1st description, you could say that a 2nd order bandpass is a 4th order filter -- which is true.

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  • \$\begingroup\$ "...a 2nd-order bandpass is a 4th order filter" ? That doesn't just sound contradictory to me - it is contradictory. A 2nd-order bandpass (lowest possible order) has two poles and one zero., which is responsible for the first order slopes (20 dB/dec). Why do you think that the transfer function would be of 4th order? \$\endgroup\$
    – LvW
    Nov 27 '21 at 9:07
  • \$\begingroup\$ Quote (C. Lindquist: Active network Design with Signal Filtering,): "The order of a filter is a measure of flter complexity. A filter is said to be of order k where k is the number of finite poles (n) or zeros (m) which ever is greater. Mathematically, k=max(n,m) and in most situations where n>m, the filter order is k=n." \$\endgroup\$
    – LvW
    Nov 27 '21 at 9:36
  • \$\begingroup\$ @LvW That's why I tried to disambiguate the language that is used: "a 2nd order filter, and a 2nd order bandpass filter" are two different things. The 2nd order bandpass refers to the lowpass prototype -> bandpass perspective, to the fact that it has 40 dB/dec slopes, just like the lowpass prototype. So from this perspective it's only a 2nd order, but it needs to be specified that it is a 2nd order bandpass. I agree it's confusing, because it appeals to grammar, rather than engineering. In your answer you're talking about the general case, where the transfer function dictates the title ... \$\endgroup\$ Nov 27 '21 at 10:13
  • \$\begingroup\$ ...but this refers to the attenuation. That's why in OP's picture they're saying, explicitely, "a 2nd order bandpass filter", not just a 2nd order filter. I agree it's confusing, but it does make sense from this perspective. \$\endgroup\$ Nov 27 '21 at 10:15
  • \$\begingroup\$ I must admit that I cannot see any confusing situation. A 2nd order lowpass has two poles and a second-order bandpass also has two poles. Where is the problem? I think, in this context the attenuation (slope of the rising and/or falling magnitude) is not a reliable indication of the filter order (remember: A 2nd order allpass has (ideally) a constant magnitude.) It works only the other way round: A 2nd-order lowpass with -40dB/dec, a 2nd oder highpass with +40 dB/dec and a 2nd-oder bandpass with +-20dB/dec. Just to say "a 2nd oder filter" is not enough to describe the filter type. \$\endgroup\$
    – LvW
    Nov 27 '21 at 11:02
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There is no 1st order filter (I mean RC or LR) with a bandpass response. A single reactive element only makes half a bandpass.

Therefore, for a 1st order bandpass you need already 2 reactive elements and so forth.

The shown filter would rolloff at 40 db/dec to either side, so it is considered a 2nd order filter.

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  • \$\begingroup\$ Would it be safe to say that the order of a bandpass filter is based on the order of its low-pass prototype? \$\endgroup\$
    – qrk
    Nov 26 '21 at 9:34
  • \$\begingroup\$ Yes - I think so. It is "based on" but not identical! When the order of the corresponding lowpass is "n", then the lowpass-to-bandpass transformation results in a bandpass of the order "2n". Therefore, the order of a bandpass is always even. \$\endgroup\$
    – LvW
    Nov 28 '21 at 9:38

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