# Order of the filter

I can't seem to figure out the order of the filter. With the knowledge i have, "The order of the filter depends on how much change there is in the (20x)dB/decade in the amplitude response."

If it was a addition of 20dB/decade i would understand.But apparantely, i see 30dB/decace. I have no idea what is the order of filter now!

The Circuit from where i got this response is:

From what i know, this is a band pass filter formed by (L1 and C2).

This resembles old CryBaby Wah-pedal. It had a sweepable band boost filter or more precisely a sweepable high-pass filter with some resonance boost. This is an active filter where the result is formed by a feedback loop that can be varied by turning a pot. This is not a bandpass filter that consists L1 and C2.

In pure math the order is the total number of reactive components (=inductors and capacitors in the signal and feedback paths. If 2 reactive components of the same type happen to be purely in series or parallel, they should be counted only as one.

In practice the most remarkable effect (here the wah) can be caused by a subcircuit. The others affect remarkably only at the ends of the frequency range. For example C1 only cuts some bass and makes a gap for DC.

The measures XXX desibels per octave or decade are not good for this. They are developed for easy comparisons between the steepnesses or selectivities between frequency selective filters. This filter is an equalizer, it's not for killing some frequencies.

The formula 20 db/decade is a kind of approximation, valid at higher frequencies (w>>1).

In a typical first order low pass filter, the gain is $$10.log(1+\omega ^{2}) \approx 10.log(\omega ^{2}) = 20.log(\omega)$$

This approximation is accurate, provided that the frequency is large. And for large frequencies, from 1K to 10K, it looks like the roll-off is -20 db/decade. More over, for lesser frequencies, the internal capacitance of the BJT plays some role, but as the frequencies increases, the gain added by these capacitance decreases .

• I rather think that for frequencies below the resonance peak the capacitive properties internally to the BJT plays NO role at all. These effects come into play for very large frequecies only (transit frequency of the BJT). – LvW Nov 30 '16 at 10:08

Yes - the whole circuit has a bandpass characteristic. However, the frequency response of the circuit - in particular, the rising part of the transfer curve for frequencies below the maximum - is determined both by the L-C resonant circuit as well as all coupling capacitors and the associated time constants within the circuit. Hence, it is not surprising that the transfer curve is not identical to a simple 2nd-order L-C bandpass. It is rather a "mixture" between bandpass and high-pass elements. This interpretation is supported by the fact that the falling characteristic (for large frequencies) approaches (nearly) the expected 20dB/dec drop.