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In electronic communication systems, there is a concept called center frequency. A bandpass filter has upper cutoff and lower cutoff frequencies. Center frequency supposed to be in the middle of these.

Why is the center frequency of a band-pass filter is given by the geometric average of the two cutoff frequencies instead of arithmetic average?

edit: found a very thorough explanation: http://www.insula.com.au/physics/1221/L15.html

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Why is the center frequency of a band-pass filter is given by the geometric average of the two cutoff frequencies instead of arithmetic average?

Because its the ratios that are relevant, not the increments.

For example, if you have a bandpass filter from 2 kHz to 20 kHz, it covers a 10:1 range. The center is then half way between these in ratio terms, which is the (square root of 10) = 3.16. This puts the center frequency at (2 kHz)*3.16 = 6.32 kHz. The room between the center and both ends is the same:

  (20 kHz)/(6.32 kHz) = 3.2
  (6.32 kHz)/(2 kHz) = 3.2

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  • \$\begingroup\$ I opened a relevant question: electronics.stackexchange.com/questions/235272/… \$\endgroup\$ – user16307 May 19 '16 at 22:48
  • \$\begingroup\$ If I had known you were going to accept a answer to this other question so quickly (while I was typing my answer), I wouldn't have wasted my time with it. I'll keep this in mind for your future question. \$\endgroup\$ – Olin Lathrop May 20 '16 at 11:11
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Well a simple answer is that you can't make a 100 Hz centre frequency with a 3 dB bandwidth of 200 Hz because you crash into DC. You have to treat it logarithmically. Having said that a lot of bandpass filters are very "tight" and numerically there is little difference between cetre frequency being bang in the middle or \$\sqrt{f_1.f_2}\$

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In the following, I try to describe the way for deriving the wanted formula (geometric mean value).

  • Start with the classical 2nd-order bandpass function (involving the parameter pole quality factor Qp and pole frequency wp).

  • Replace the variable w by wc (3dB cut-off)and - at the same time - set the magnitude of the transfer function to A=Amax/SQRT(2).

  • As a result, you have a quadratic equation for wc which can be solved.

  • The result is an equation for the two cut-off frequencies of the form:

wc1=-X+SQRT(X²+wp²) and wc2=+X+SQRT(X2+wp²) with X=wp/2Qp.

  • It is easay to show that wc1*wc2=wp²

  • Note that wp=wo (pole frequency is identical to midband frequency).

EDIT 1: I forgot to mention that the maximum gain Amax at center frequency is the ratio of the numerator and the midterm (jw) of the denominator (because the most left and the most right terms cancel each other at w=wp).

EDIT 2: Here comes a more desriptive explanation of the fact that the distance of both cut-off frequencies (wc1, wc2) to the center frequency wo is different:

The bandpass transfer function is zero for (a) infinite frequencies as well as (b) for w=0. It is clear that the "way" from the center frequncy wo to infinite frequencies is much larger (infinite) if comnpared with the distance to w=0.

That means: The magnitude decrease in the direction to larger frequencies is "smoother" than in the direction to w=0. For this reason, the frequency difference between upper cut-off frequency wc2 and the center frequency wo (wc2-wo) is larger than the difference (wo-wc1).

For this reason (no symmetry to both sides of wo) the center frequency wo is NOT the arithmetic mean value of both cut-off frequcies. (The transfer function magnitude plot vs. frequency looks symmetrical only in case of a logarithmic frequency scale):

wo=wp=SQRT(wc1*wc2)

log(wo)=[log(wc1)+log(wc2)]/2

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What's important is that filters are generally represented using a logarithmic frequency scale (think of e.g. Bode plots). Let's see what the arithmetic mean becomes on a log scale:

$$\frac{\log( f_a ) + \log( f_b )} { 2 } = \frac{ \log( f_a.f_b )}{2} = \log \sqrt{ f_a . f_b }$$

There you go: arithmetic average of logarithms is equivalent to logarithm of geometric average.

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  • \$\begingroup\$ Here you have shown that an "arithmetic average of logarithms is equivalent to logarithm of geometric average". However, the questionwas WHY we have the geometric average for linear scaling. This is the reason for the observed "arithmetc average" for log scaling - but it is not really an explanation. \$\endgroup\$ – LvW Jul 27 '16 at 10:39
  • \$\begingroup\$ I am showing that, on a Bode gain plot (i.e. log-log) the center frequency is visually equidistant from both cutoff frequencies, showing why we would think about arithmetic average, and I explain why the geometric average is used instead. Questioning the use of the log scale is another problem imo. \$\endgroup\$ – Florian Castellane Jul 27 '16 at 11:15

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