It is the rate of attenuation above the crossover point.
Crossovers are not brick walls. They attenuate the frequencies above (since we are talking about a lowpass crossver) the crossover point at a certain rate, called the slope. The steeper the slope (the higher the number), the more attenuation it applies.
12db & 24db is the rate (slope) of attentuation. It attenuates the frequencies above the crossover point at a rate of 12db/oct, or 12 decibels per octave. The 24db slope attenuates the frequencies above the crossover point at a rate of 24db/oct, or 24 decibels per octave.
So, if you have the crossover point set at 70hz, then with the 12db/oct slope the signal will be down 12db @ 140hz compared to it's level at 70hz. But for the 24db/oct slope, at 140hz the signal would be attenuated by 24db.
Put it on whichever sounds best. That simple.
- Thread Starter
- #3
which one would u put it as if the crossover is set at 80 hz
nissanrider06
10+ year member
VooDoo
hold on! so many hz in an octive i thought it was 10??
Whichever sounds best.
There's no right or wrong, this or that. It is simply whatever sounds best. Try it both ways, leave it set to whichever sounds best.
One octave lower is half the frequency, one octave higher is double the frequency.
Take my 70hz as an example again. One octave above 70hz is 140hz. One octave below 70hz is 35hz.
One octave above 100hz would be 200hz. One octave below 100hz would be 50hz.
For one octave higher multiply the frequency by two. For one octave lower, multiply the frequency by .5
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