I’ve been working with parametric EQ settings lately; here’s a quick cheat sheet.
We perceive the frequencies of sounds logarithmically. Each doubling of frequency is an octave. Thus, the difference between 40 and 80 Hz sounds the same as the difference between 4000 and 8000 Hz. Even though the latter difference is 10 times greater, it sounds the same to us. This gives a range of audible frequencies between 9 to 10 octaves, which is much wider than the range of frequencies of light that we can see.
Two frequencies 1 octave apart have a frequency ratio of 2:1; one has twice the frequency of the other. A half octave is halfway between them on a logarithmic scale. That is, some ratio R such that f1 * R * R = f2. Since f2 = 2 * f1, R is the square root of 2, or about 1.414. Sanity check: 40 * 1.414 = 56.6, and 56.6 * 1.414 = 80. Thus 56.6 Hz is a half-octave above 40, and a half-octave below 80. Even though 60 Hz is the arithmetic half-way point between 40 and 80 Hz, to our ears 56.6 sounds like the half-way point between them.
More generally, the ratio for the fractional octave 1/N, is 2^(1/N). Above, N=2 so the half-octave ratio is 1.414. If N=3 we have 1/3 octave ratio which is 2^(1/3) = 1.260. Here is a sequence taken to 4 significant figures:
- 1 octave = 2.000
- 3/4 octave = 1.682
- 1/2 octave = 1.414
- 1/3 octave = 1.260
- 1/4 octave = 1.189
- 1/5 octave = 1.149
- 1/6 octave = 1.122
- 1/7 octave = 1.104
- 1/8 octave = 1.091
- 1/9 octave = 1.080
- 1/10 octave = 1.072
- 1/11 octave = 1.065
- 1/12 octave = 1.059
The last is special because in western music there are 12 notes in an octave. With equal temperament tuning, every note has equally spaced frequency ratios. Thus the frequency ratio between any 2 notes is the 12th root of 2, which is 1.059:1. Every note is about 5.9% higher in frequency than the prior note.
Bandwidth with Q
Another way to express the frequency range or bandwidth of a parametric filter is Q. Narrow filters have big Q values, wide filters have small Q values. A filter 2 octaves wide (1 octave on each side of the center frequency) has Q = 2/3 = 0.667.
For a total bandwidth of N octaves (N/2 on each side of center frequency), the formula is:
Q = sqrt(2^N) / (2^N - 1)
Here are some example values. You can check them by plugging into the formula.
- N=2, Q=0.667
- N=1.5, Q=0.920
- N=1, Q=1.414
- N=2/3, Q=2.145
- N=1/2, Q=2.871
Note that these N octave fractions are total width, which is twice the above table which shows octave on each side of the center frequency.
Whatever tool you’re using for this, make sure you know whether it expects total bandwidth around the center frequency, or bandwidth on each side. And make sure you know whether it expects frequency ranges as raw ratios, fractions of an octave, or Q.
The above formula comes straight from any textbook. But these Q factors may give wider ranges than expected, due to an assumption it makes. This assumption is that the range of the filter is where the peak amplitude (at its center) drops to half its value. So the filter is still taking effect at these edges. If you want the filter to taper to zero at the edges, you need to use a bigger Q value to get a narrower filter. Roughly speaking, this means multiply the Q value by 2.0.
For example consider a filter that is -4 dB at 3,000 Hz, 3/4 octave wide on each side. That is a ratio of 1.682:1, so this filter tapers to zero at 3,000 / 1.682 = 1,784 and 3,000 * 1.682 = 5,045 Hz. Total width is 1.5 octaves (5,045 / 1,784 = 2.83 = 2^1.5). The above formula says this is Q=0.92. But that will be a wider filter. It will reduce to half (roughly +2 dB) at 1,784 and 5,045 Hz. If you want it to taper to zero at these edged then use Q = 0.92 * 2.0 = 1.84.
Note: this is an approximate / rough guide.
Suppose you are analyzing frequency response and see a peak between frequencies f1 and f2. You want to apply a parametric EQ at the center point that tapers to zero by f1 and f2.
First, find the logarithmic midpoint. Compute the ratio f2 / f1 and take its square root to get R. Multiple f1 by R, or divide f2 by R and you’ll have the logarithmic midpoint.
For example if f1 is 600 Hz and f2 is 1700 Hz, the ratio is 2.83:1, so R = sqrt(2.83) = 1.683. Double check our work: 600 * 1.683 = 1010 and 1010 * 1.683 = 1699. Close enough.
So 1,010 Hz is the logarithmic midpoint between 600 and 1700 Hz. We center our frequency here and we want it to taper to zero by 600, and 1700. That range is a ratio of 1.683 on each side, which in the above list is 3/4 octave, or Q=0.920. Multiply Q by 2.0 to get Q=1.84 since you want this filter to have no effect (taper to zero) at these 2 endpoint frequencies. So now we know the center frequency and width of our parametric EQ.