Geometric Mean vs. Arithmetic Mean in Bandpass Filters

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SUMMARY

The geometric mean is utilized to define the center frequency of a bandpass filter due to its inherent geometric-mean symmetry, which arises from the transformation of lowpass filter elements. Specifically, the relationship f0 = √(f1f2) illustrates that the frequencies of the transformed points are reciprocals. In contrast, using the arithmetic mean would inaccurately represent the center frequency, especially when plotted on a logarithmic scale, as demonstrated by the example of corner frequencies at 1Hz and 100Hz, where the peak response occurs at 10Hz rather than 50Hz.

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  • Understanding of bandpass filter design principles
  • Familiarity with logarithmic frequency scales
  • Knowledge of geometric and arithmetic means
  • Basic concepts of lowpass and highpass filter transformations
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  • Explore the impact of logarithmic scaling on frequency response plots
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Electrical engineers, signal processing specialists, and audio engineers seeking to optimize bandpass filter designs and understand the mathematical foundations of frequency response 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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