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Geometric Mean vs. Arithmetic Mean in Bandpass Filters

  1. Nov 14, 2013 #1

    JJBladester

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    Gold Member

    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

    [tex]f_{0}=\sqrt{f_{1}f_{2}}[/tex]

    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.
     
  2. jcsd
  3. Nov 17, 2013 #2

    NascentOxygen

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    Staff: Mentor

    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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