is Q on a parametric equalizer octave dependant, or frequency dependant? Q = center frequency / bandwidth ex Q=.4 @1000 would the bandwidth be 800-1200 (this would be frequency dependant) or would the bandwidth be aprox 850-1250 octave dependant (since the octave logarithmic....) on visual scales that are logarithmic you always see it symetrical which would lead me to believe that it would be something like 850-1250 yet... i'm lacking documentation and what i've seen taught is the other.....
I've always seen/heard Q referred to as a db per octave parameter. The example of Q=.4 is arbitrary in my world.......I have EQ's with 6.5 being the highest Q value, others with 24 as the highest, but the slopes are the same at the extremes.
A simpler analogy to your question would be..... It is like brassiere cup sizes! An "A" cup, is a skinny titty. An "E" cup will cover your face! It's simple! 42B Ms. Remy Ann David
they don't answer my question.... let me rephrase my question concerns the bandwidth of Q factors in equalizers in this diagram is the x axis logarithmic or liniar? i would assume it's logarithmic since we are dealing with Hz (octave perportions) yet inall thel theoretical calcualtions i've seen it liniar
So redbort, what exactly are you asking?? "Q" simply refers to the width at a certain frequency. A narrow Q, is simply a notch filter, 1/3 to 1/8 octave wide. A wide Q, generally refers to an octave width or more. A parametric equalizer can be swept between frequencies and the Q can also be adjusted (George Massenburg loves these and made the parametric equalizer very popular but then he knows what he's doing with them). Generally equalizers are linear within their boost and/or cut when speaking of a bell shaped curve. Reciprocal equalizers have the same bell shape, relative to boost and cut. Not all equalizers are reciprocal. Those particular equalizers may have a narrow Q, for boost and a wide Q, for cut and vice versa and so are not reciprocal, i.e. you cannot undo what you did. As a green engineer, you should probably stay away from parametric type equalizers, since you do not quite understand the concept. Narrow Q equalizers are generally not very musical sounding and are used for more corrective purposes. A wider Q or program oriented equalizer is generally much more musical sounding and therefore for general music applications makes more sense. If you have a narrow Q equalizer, you'll probably find yourself boosting and/or cutting with much more amplitude boost and/or cut just to hear the effect. A wider Q equalizer will require less boost and/or cut in amplitude to hear a more enormous difference. Most console program equalizers generally reside within a 1 Q, or 1 octave spread. This was very p-Q-liar Ms. Remy Q David
Remy's right, don't let the semantics do you in. Q is a tool, doesn't matter in the real world if the slope is linear or logarithmic, just learn how to use it. My cathartic moment came over the"J factor" when learning about phase angles re: impedance. I couldn't get my brain wrapped around the square root of -1, even though it was the only value that would generate an answer. My professor told me: "Someone a lot smarter than everyone in this room found it, and it works. Think of it like a wrench, the right tool for the job". I started getting better math grades after that.
i just wanted to know how the bell of the EQ will affect frequencies above and below the center frequencies will it effect an equal amount of the octave above and below the center freqency (logarithmic) or will the effect weigh stronger in the frequencies below the center frequency (liniar) this has nothing to do with how wide the bell is, nor how much dB boost, just the details of the quality it's not semantics because i'm trying to understand exactly what my ears are hearing while i'm training them in white noise boost cut exercises with parametric EQ's..... this is my last try, i promise is it the red line or the black line? red or black both cases center freqency is 200 the bandwidth is 190
Redbort, I would assume that it is octave dependant. Otherwise a Q that effects lets say one octave, spanning from 40 to 80 hz, if swept up, would effect for example only 4000 to 40040 hz, rather than 4000 to 8000hz. I don't have any facts on this, just using logic.
ahhh, you're completely missing the point obviously it will not be 4000-4040 because the center frequency is changing and Q is a ratio between the center freqency and the bandwidth... i know throwing insults won't help me get my answer but are people here mathematically incompetent? how about this for a question for the Q spanning an octave 40 hz - 80 hz what will the center freqency be? 60? (liniar) or lower?
OY........ You are missing the point because your point is pointless. You're trying to be analytical with white noise as opposed to music. An equalizer set to 40 hertz that is a one octave equalizer will symmetrically affect all frequencies below and above 40 hertz, within a one octave range. I think you will find that 20 and 80 hertz are affected equally, with the peak of maximum gain at 40 hertz. Now what is this supposed to accomplish for you? It won't make you a better engineer. Only your ears can do that for you. But that may not be your objective here??? I'm not sure what is? Why are you fixating on this? What is it supposed to accomplish? In what ways do you think this should benefit you in the long run? It can be helpful to play with an equalizer, with white noise and/or pink noise, to familiarize yourself with the sound of the audible spectrum. To evaluate the difference in "Q". But not much else. Your diagrams mean nothing. They're really not applicable to anything you are asking. Sorry, this is all I can help you with. Ms. Remy Ann David
"ahhh, you're completely missing the point obviously it will not be 4000-4040 because the center frequency is changing and Q is a ratio between the center freqency and the bandwidth... " I think you missed the point. I used the 4000-4040 as an example to prove that Q isn't linear, which if it were, a Q that effected 40-80 hz swept up would only affect 4000-4040 hz which it obviously does not. 40 to 80 hz is an octave as is 4000 to 8000hz. Q is logarithmic just like audio, look at a frequency chart. In answere to your question/test, bandwidth of 40-80hz, center would be 56.5 hz. This really is pointless though, pun intended.
thank you there's an answer i understand my only point was that you often see it written in books with examples explaining Q that the center freqnecy would be 60, which we all know and agree to be false thank you again i know i fixate, the details are always a hangup, even more when Idon't understand problem fixed, thanks again RO!
The term Q is short for "Quality" and really relates to the reactance of a capacitor or inductor, the higher the Q the better the "quality" of these reactive components. Somehow it found it's way into audio terms. But Q is really just a way to express bandwidth, and the Q can be found by the formula Q=f/BW where f = the centre frequency and BW = the -3dB points either side of the centre frequency. So, if you have a filter centred at 1000Hz, and the -3dB points are 2000Hz and 500Hz (a difference of 1500Hz), the Q will = 0.6666. (1000/1500). In your picture, most audio EQ's will look like the redline and be symetrical about the centre frequency (however, the frequency scale is still log), however, some are not and I think the Prismsound mastering EQ has unsymetrical slopes, like your black line, in which case the Q description becomes inaccurate. In the above example, the 500Hz point is actually 1 octave below the centre frequency of 1000Hz and the 2000Hz point 1 octave above. Hope this helps with your understanding. Tim
yeah it does tim it's nice to know that i may sometimes come across an EQ has unsymetrical slopes this way i don't take it for grated every time