Geometric Mean vs. Arithmetic Mean in Bandpass Filters

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The geometric mean is preferred for defining the center frequency of a bandpass filter due to its geometric-mean symmetry, which arises from the reciprocal relationship of transformed lowpass filter points. When analyzing the frequency response on a logarithmic scale, the geometric mean accurately reflects the peak response, unlike the arithmetic mean, which would suggest a peak at a midpoint that does not correspond to the actual response. For example, with corner frequencies of 1Hz and 100Hz, the geometric mean yields a center frequency of 10Hz, aligning with the peak response, while the arithmetic mean suggests 50Hz, which is incorrect. This distinction highlights the importance of using the geometric mean in the context of bandpass filters. Understanding these concepts is crucial for accurate filter design and analysis.
JJBladester
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Why is the geometric mean used to define the center frequency of a bandpass filter instead of the arithmetic mean?

I read in this book that

1. All the lowpass elements yield LC pairs that resonate at ω = 1.
2. Any point of the lowpass response is transformed into a pair of points of the bandpass filter. The frequencies of the pair of points are reciprocals. This means that, after frequency scaling, we can write

f_{0}=\sqrt{f_{1}f_{2}}

where f1 and f2 are the scaled frequncies of the transforms of a single LP point, and f0 was scaled from ω = 1. This effect tells us that the bandpass filter has geometric-mean symmetry.

I get that you need to use the geometric mean when multiplying (scaling) elements but I'm still having a hard time seeing why we can't just use the arithmetic mean in the case of defining the bandpass filter center frequency.
 
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I'll try a qualitative justification. If you plot the separate responses (dB vs. frequency) on a logarithmic frequency scale (the x-axis), the centre of the response does indeed seem to be about midway. But I said this is when plotted on a logarithmic scale, so using the arithmetic mean would miss the mark by a long way.

To throw in some figures. Suppose we have one corner frequency at 1Hz, and another at 100Hz. On a log paper plot, midway between these values corresponds to 10Hz and that's about where the response peaks. The response certainly does not peak near 50Hz.
 
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