SSLtech
Well-known member
Essentially, yes.
While that graph shows some shifting of the peaking frequency at different 'Q' settings, the important fact is that you refer to it by the peak frequency and not the 'break' frequency: inother words the frequency at which it deviates from flat.
A further hypothetical example: if you draw a typical set of fixed Baxendall curves, you'll usually find that the HF boost deviates from flat at right around 1kHz, perhaps achieving full boost at 10kHz and staying more or less fully boosted above that. -We call that a 10kHz shelf. -Likewise a 100Hz low boost would probably be fully boosted at all frequencies below 100Hz, then above 100Hz it descends gently to flat, settling down also right around 1kHz.
If we were to use the 'break points' in this Baxendall example, they'd noth be called 1kHz shelving rewponses... but the perceived frequencies at which they have effect are in actuality 100Hz and 10kHz... thus we call them 100Hz and 10kHz shelf responses.
Notice that the frequencies are fixed. A 1dB boost would have a slope of about 0.3dB/8ve, a 3dB boost would produce a resulting slope of about 1dB/8ve, a 10dB boost would produce a slope of about 3dB/8ve... so you cannot refer to the Baxendall circuit in terms of a defined slope response... it varies with boost (or cut of course!)
Looking at it the other way around, a theroetical perfect high-pass filter has a fixed slope, -6 or 12 or 18dB/8ve are typical examples- but you can't really vary the amount of cut, because it is fixed... It's fixed and defined by the precise distance (in octaves) away from the corner frequency. -There's no "cut amount" control.
Keith
While that graph shows some shifting of the peaking frequency at different 'Q' settings, the important fact is that you refer to it by the peak frequency and not the 'break' frequency: inother words the frequency at which it deviates from flat.
A further hypothetical example: if you draw a typical set of fixed Baxendall curves, you'll usually find that the HF boost deviates from flat at right around 1kHz, perhaps achieving full boost at 10kHz and staying more or less fully boosted above that. -We call that a 10kHz shelf. -Likewise a 100Hz low boost would probably be fully boosted at all frequencies below 100Hz, then above 100Hz it descends gently to flat, settling down also right around 1kHz.
If we were to use the 'break points' in this Baxendall example, they'd noth be called 1kHz shelving rewponses... but the perceived frequencies at which they have effect are in actuality 100Hz and 10kHz... thus we call them 100Hz and 10kHz shelf responses.
Notice that the frequencies are fixed. A 1dB boost would have a slope of about 0.3dB/8ve, a 3dB boost would produce a resulting slope of about 1dB/8ve, a 10dB boost would produce a slope of about 3dB/8ve... so you cannot refer to the Baxendall circuit in terms of a defined slope response... it varies with boost (or cut of course!)
Looking at it the other way around, a theroetical perfect high-pass filter has a fixed slope, -6 or 12 or 18dB/8ve are typical examples- but you can't really vary the amount of cut, because it is fixed... It's fixed and defined by the precise distance (in octaves) away from the corner frequency. -There's no "cut amount" control.
Keith