ruffrecords wrote: ↑Tue Sep 09, 2025 3:46 pm
Gold wrote: ↑Mon Sep 08, 2025 4:37 pm
ruffrecords wrote: ↑Mon Sep 08, 2025 3:46 pm
There is no official agreed upon definition of a centre frequency for a shelving EQ. There are essentially two turnover frequencies; one where the the boost/cut begins and one where it ends. Personally I now use low Q bell curves instead of shelves because they allow the control of the out of audio band response whereas shelves do not.
I look at a shelf like a 'half bell'. For a low shelf the center frequency would be where the boost begins. At what would be the center frequency of a bell. the upper frequency. The lower frequency, the equivalent to the skirt of the bell would be where the boost ends.
How these are labeled is up for debate. For low shelves it makes sense to me to label the EQ point according to the 'center frequency' of the shelf. For high shelves it makes more sense to me to label according to where the 'skirt of the bell' starts. But for technical purposes I'd label it the same as a low shelf. where the 'center frequency' defines the label. It's done this way on a Sontec. They call the high shelf 10K when that's where the boost flattens out or is the 'center frequency'. But from a sound standpoint I'd call it a 1K high shelf because that's about where the bell skirt is. Where the skirt is depends on the Q of the shelf.
What's the formula for figuring this out?
Unfortunately, the formula you require depends critically on what you define as "where the boost begins". Is this 0.001dB of boost or 0.1dB of boost or 0.5dB of boost? A similar problem exists if you try to define things in terms of where the boost ends. I think you will find it far more beneficial to look at the actual curves obtained using a simulation. This will give you a much greater insight into haw the EQ works.
"Standard" RC or LC network formulae for this EQ are equally flawed because they give the theoretical frequency of 3dB boos or cut. Unfortunately, for an EQ with a maximum boost of 6dB, the 3dB point is very inaccurate when given by the standard formulae simply because it does not account for the 6dB boost limit.
As I mentioned in an earlier post, the intention is to make both arms of the resistive ladder equal to 600 ohms. With that in mind I have attached a simple LTspice schematic to illustrate a high shelf boost using a 100nF capacitor.
simpleshelf.png
Perhaps you could tell me where you think the boost begins and ends?
Cheers
ian
From the circuit in the LTSpice simulation you showed I derived exact formulas if anyone is interested.
(IF THE IMAGES DO NOT LOAD, HIT REFRESH IN YOUR BROWSER)
The transfer function is given by
- tr_func.png (8.96 KiB) Viewed 6442 times
So there is a zero at
- zero.png (3.34 KiB) Viewed 6442 times
and a pole at
- pole.png (4.45 KiB) Viewed 6442 times
The frecuency at which a certain attenuation ('Att') in dB occurs is given by
For the special case where both resistors are equal to a value R, the equation reduces to
- sc_1.png (9.93 KiB) Viewed 6442 times
Furthermore, if both resistors are equal and ATT = 3, which is the 3dB frequency, the equation becomes
- sc_2.png (7.11 KiB) Viewed 6442 times
So for Ian's circuit the 3dB frequency is at 3.75 kHz.
Please note that whatever equation you use the value of ATT must lie on the interval
- int.png (8.31 KiB) Viewed 6409 times
which for the case of equal resistors is between 0 and 6 dB.
So lets say that both resistors are equal to 600 ohms and we want to make the shelf Attenuate 1 dB at 1 kHz, then we would need a cap of
Just for fun: Lets make a shelf with a maximum attenuation of 12 dB and a -3dB frequency at 1 kHz.
The maximum attenuation is given by
- att_max.png (5.25 KiB) Viewed 6442 times
Solving for R1
- R1andR2.png (5.53 KiB) Viewed 6442 times
If we choose R2 = 1K, then R1 is roughly 3K.
Now we can substitute these values in the big equation at the top with f0=1000 and Att=3 and solve for C
The simulator agrees, there is a small difference due to rounding but good enough for rock n' roll
