Why are band pass filters asymmetrical?

[Ben B]

Here's something that's confusing to me. Say you have a one-octave band pass filter, centered around 1000 Hz. The mathematics of such a filter place the minimum and maximum frequencies at about 707 Hz and 1414 Hz, respectively.

If that is so, the midpoint between those minimum and maximum frequencies is actually around 1060 Hz (not the original 1000 Hz from which the calculations began).

Why?

I'm finding this to be the case for all band pass filters. When I do the calculations, the original center frequency is never the ending center frequency (although it's always close).

What gives here?

-Ben B

[j_j]

Originally Posted by Ben B

Here's something that's confusing to me. Say you have a one-octave band pass filter, centered around 1000 Hz. The mathematics of such a filter place the minimum and maximum frequencies at about 707 Hz and 1414 Hz, respectively.

If that is so, the midpoint between those minimum and maximum frequencies is actually around 1060 Hz (not the original 1000 Hz from which the calculations began).

Why?

I'm finding this to be the case for all band pass filters. When I do the calculations, the original center frequency is never the ending center frequency (although it's always close).

What gives here?

-Ben B

There is a difference between 'z' domain filters (sampled) and 's' (analog) filters in this regard. 's' domain works, effectively, on the log domain, and the "center" is the geometric mean of the lower and upper cutoffs, i.e. sqrt(707 * 1414), which is going to get you back to your 1kHz.

The 'z' domain could do several things, depending on how the filter is designed. If it's a single pole-pair, then it will exhibit the symmetry that you originally expected. This is not the usual way of building 'z' domain filters, and as such, there may be substantial differences that are too complex to explain in a short article.

[hcholm]

When dealing with frequencies, you'll normally multiply, not add. Frequency is perceived by the ear in a logarithmic manner. Frequency intervals in a musical scale are defined by factors to multiply by, not differences to add.

Jumping an octave is the same as doubling the frequency. In this case, the octave is from 707 Hz to 2 * 707 Hz = 1414 Hz. Half an octave (= the tritone interval) is the square root of 2, approx. 1.414. Thus 707 * sqrt(2) ~= 707 * 1.414 ~= 1000. Multiply again to get to 1000 * sqrt(2) = 1414, an octave above 707.

A general formula to find the midpoint is min * sqrt(max/min). Here: 707 * sqrt(1414/707) = 707 * sqrt(2) ~= 1000.

[Ben B]

Awesome!

I suspected something along these lines, akin to multiplying frequency ratios to obtain the ratios of larger intervals, as in Just Intonation.

You're helped me understand this much more clearly. I appreciate it!

-Ben B

[j_j]

Originally Posted by hcholm

When dealing with frequencies, you'll normally multiply, not add. Frequency is perceived by the ear in a logarithmic manner. Frequency intervals in a musical scale are defined by factors to multiply by, not differences to add.

Indeed, but that is not why an analog filter works on a logarithmic frequency scale.

Just to point out that the relationship is due to the physics of the situation, the cause is physics, and the results are both the critical band scale of the ear (which is not even close to logarithmic at low (under 700Hz) frequencies, by the way) and the way analog filters work.

Jumping an octave is the same as doubling the frequency. In this case, the octave is from 707 Hz to 2 * 707 Hz = 1414 Hz. Half an octave (= the tritone interval) is the square root of 2, approx. 1.414. Thus 707 * sqrt(2) ~= 707 * 1.414 ~= 1000. Multiply again to get to 1000 * sqrt(2) = 1414, an octave above 707.

A general formula to find the midpoint is min * sqrt(max/min). Here: 707 * sqrt(1414/707) = 707 * sqrt(2) ~= 1000.

Now on that we agree. The midpoint of any analog filter consisting of one pole pair is the geometric, not arithmetic, mean of the edges.

[Ben B]

Yes, yes, yes. This all makes perfect sense to me now. Thank you so much.

-Ben B