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Given BW in octaves to find Q and given Q to find BW in octaves.
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BW = Δf = f2 − f1 = f0/Q f1 = f02/f2 = f2 − BW f2 = f02/f1 = f2 + BW Q = f0/BW
Conversion formula: 'octave bandwidth' N to quality factor Q:
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Conversion formula: Quality factor Q to 'octave bandwidth' N:
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Also known is this longer formula with 4 Qs; see its development at:
Bandwidth in octaves versus Q in bandpass filters − RaneNote 170
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Frequency ratio of an octave:
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Formula to convert quality factor Q to 'bandwidth in octaves' N,
but with logarithmus naturalis:
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And the very short formula to convert quality factor Q
to 'bandwidth in octaves' N, but with sinh-1 :
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Conversion chart or table
'bandwidth in octaves' N to quality factor Q
| BW in octaves |
Filter Q |
BW in octaves |
Filter Q |
BW in octaves |
Filter Q |
BW in octaves |
Filter Q |
|||
| 1/80 | 115.4 | 1 | 1.41 | 4 | 0.267 | 7 | 0.089 | |||
| 1/60 | 86.6 | 1 1/4 | 1.12 | 4 1/4 | 0.242 | 7 1/4 | 0.082 | |||
| 1/50 | 72.1 | 1 1/3 | 1.04 | 4 1/3 | 0.234 | 7 1/3 | 0.079 | |||
| 1/40 | 57.7 | 1 1/2 | 0.92 | 4 1/2 | 0.220 | 7 1/2 | 0.075 | |||
| 1/30 | 43.3 | 1 2/3 | 0.82 | 4 2/3 | 0.207 | 7 2/3 | 0.071 | |||
| 1/25 | 36.1 | 1 3/4 | 0.78 | 4 3/4 | 0.200 | 7 3/4 | 0.068 | |||
| 1/20 | 28.9 | 2 | 0.67 | 5 | 0.182 | 8 | 0.063 | |||
| 1/16 | 23.1 | 2 1/4 | 0.58 | 5 1/4 | 0.166 | 8 1/4 | 0.058 | |||
| 1/12 | 17.3 | 2 1/3 | 0.56 | 5 1/3 | 0.161 | 8 1/3 | 0.056 | |||
| 1/10 | 14.4 | 2 1/2 | 0.51 | 5 1/2 | 0.152 | 8 1/2 | 0.053 | |||
| 1/8 | 11.5 | 2 2/3 | 0.47 | 5 2/3 | 0.143 | 8 2/3 | 0.050 | |||
| 1/6 | 8.65 | 2 3/4 | 0.45 | 5 3/4 | 0.139 | 8 3/4 | 0.048 | |||
| 1/5 | 7.20 | 3 | 0.40 | 6 | 0.127 | 9 | 0.044 | |||
| 1/4 | 5.76 | 3 1/4 | 0.36 | 6 1/4 | 0.116 | 9 1/4 | 0.041 | |||
| 1/3 | 4.32 | 3 1/3 | 0.35 | 6 1/3 | 0.113 | 9 1/3 | 0.039 | |||
| 1/2 | 2.87 | 3 1/2 | 0.33 | 6 1/2 | 0.106 | 9 1/2 | 0.037 | |||
| 2/3 | 2.14 | 3 2/3 | 0.30 | 6 2/3 | 0.100 | 9 2/3 | 0.035 | |||
| 3/4 | 1.90 | 3 3/4 | 0.29 | 6 3/4 | 0.097 | 9 3/4 | 0.034 | |||
| 10 | 0.031 |
Q factor as a function of the bandwidth in octaves N
| Bandwidth in octaves N |
Filter Q factor |
| 3.0 wide | 0.404 low |
| 2.0 | 0.667 |
| 1.5 | 0.920 |
| 1.0 | 1.414 |
| 2/3 | 2.145 |
| 1/2 | 2.871 |
| 1/3 | 4.318 |
| 1/6 | 8.651 |
| 1/12 small | 17.310 high |
| Notice: A low Q factor gives a broad band (wide) bandwidth or a high Q factor gives a narrow band (small) bandwidth. |
| A high filter quality means narrow-band filtering (notch), with a large Q factor. This results in steep filter flanks with a small bandwidth. A low filter quality means broad-band filtering, with a small Q factor. This results in flat filter flanks with a large bandwidth. The larger the Q the more narrow the resonance peak. The smaller the Q the more broad the resonance peak. |
| The Q factor or the bandwidth does not tell the "steepness" in dB/oct. |
| Slope in dB/oct = steepness of the filter flanks ● Only with high pass and low pass filters − not with bell curves ● |
| Note: The Q factor (quality factor) or the bandwidth is not convertable to the "slope" as dB/oct. There are mastering equalizers with false information regarding the filter setting as "Slope in dB/octave" and not Q factor (width), see: Filter slope or steepness (dB/oct) is not bandwidth = Slope in dB/oct or steepness of filter slopes is not the bandwidth. |
| With "quality" is not meant how valuable the signal is. It is meant the quality of the filter. At a filter with flat slopes many frequencies are influenced around the cutoff frequency. The filter has therefore a larger bandwidth. The so-called quality factor is given with a low number specified. If the filter has steep slopes, its bandwidth is smaller. Here a few frequencies below and above its cutoff frequency are affected and the quality factor Q is specified as a high number. |
| Why is the bandwidth and the cutoff frequency found at a level of "−3 dB"? Full width at half maximum (FWHM). That is the point where the energy (power) is fallen to the 1/2 value or 0.5 = 50 percent of the initial energy quantity. There the voltage is fallen to the value of √(1/2) = 1/√2 or 0.71 = 70.1 percent of the initial voltage as field quantity. A 3 dB voltage drop is a decrease of 29,29 % to 70,71 %. |
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