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What is the merit of fractional bandwidth?

$$f_{fractional}=\frac{f_{higher}-f_{lower}}{f_{mid}}$$

The same absolute bandwidth will have a smaller fractional bandwidth, if the mid frequency is higher, ok. It is reciprocal to the Q-factor, ok. Still, I could not find a good explanation why I would use the fractional bandwidth to describe for example a filter instead of absolute.

The only source I found is this page, look for paragraph How to judge fractional and relative bandwidth?

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  • \$\begingroup\$ Fractional bandwidth is related to Q. Increasing fmid increases Q - sharper, harder to implement. \$\endgroup\$ – Russell McMahon Jun 4 '14 at 16:54
  • \$\begingroup\$ Ok, for example: 1 MHz bandwidth around 9.5 MHz will have 0.11 fractional bandwidth and thus Q is much sharper than 1 MHz bandwidth around 1.5 Mhz with 0.666 fractional bandwidth. This basically means the higher the operation frequency of a device, the harder to implement because the filter response will be sharper unless one increases the absolute bandwidth? So the fractional bandwidth gives the extra information of implementability, while losing the direct information about frequency operation range?! \$\endgroup\$ – Irenaius Jun 5 '14 at 11:10
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In some cases it works out to be more convenient. Some things (like human perception of pitch, and filter responses) respond to frequency logarithmically. That is, higher frequency octaves have more bandwidth than low frequency octaves.

It's the same reason low-pass or high-pass filter topologies have their roll-off described in dB/octave rather than dB/Hz. For example, we know that all first-order filters have a roll-off of 6dB/octave. This holds if we are designing a filter for 1kHz or 1GHz.

The same sort of issue applies to band-pass and band-reject filters. If we took a band-pass filter 100Hz wide at 1kHz, then scaled all the component values to move it up to 1GHz, then its bandwidth would be 100MHz. However, it would have the same fractional bandwidth.

Thus, it's convenient to specify the bandwidth of a particular topology as a fractional bandwidth because it does not change as the passband is moved by scaling all the component values.

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  • \$\begingroup\$ Just a small comment: There is no 1st order Butterworth response. All first order responses are identical. Butterworth starts only at 2nd order. \$\endgroup\$ – LvW Jun 4 '14 at 13:09
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"Still, I could not find a good explanation why I would use the fractional bandwidth to describe for example a filter instead of absolute. "

Well I think, using the fractional bandwidth instead of the absolute bandwidth makes sense, because it is not sufficient to state that the (absolute) bandwidth of a bandpass filter is 1kHz (for example). Thus, to get a feeling about the technical requirements to realize such a filter it is important to know if it is centered at 5 kHz or 5Mhz. However, there is no principle difference if we are using the absolute bandwidth to describe the sharpness (selectivity) of the bandpass filter or the quality factor Q which is the inverse value. But I think, in most cases (filter tables or filter design programs) we are using the Q value (which also appears in most transfer function forms). Remember: For a bandpass, the Q value (selectivity) is identical to the pole Q which is defined based on the pole location in the s-plane. Thus we have the same principle to describe transfer functions for lowpass, highpass and bandpass based on the pole location.

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