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Old 26th January 2010, 03:19 AM   #1
dooper is offline dooper  Canada
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Default Passive crossover/filter question

I searched the site and many other places with no luck. Just one simple question...

A second-order LC filter (high- or low-pass) has a specified cutoff or resonant frequency (f3). At this frequency the output of the connected driver is down 3 dB as compared to its output without the filter connected.

Let's say that f3=500 Hz and we're dealing with a low-pass filter. In theory, what would be the amount of attenuation at 1000 Hz? Would it be 12 dB or 15 dB?

While we're at it, what would be the theoretical attenuation at 2 kHz? 24 dB?
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Old 26th January 2010, 03:57 AM   #2
DcibeL is offline DcibeL  Canada
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1st order filters have a 20 dB/decade (or 6 dB/octave) slope.
2nd order filters have a 40 dB/decade (or 12 dB/octave) slope.
3rd order filters have a 60 dB/decade (or 18 dB/octave) slope.
etc...

An octave is x2 the frequency
A decade is x10 the frequency.
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Old 26th January 2010, 04:00 AM   #3
infinia is offline infinia  United States
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slopes of filters are determined by the order except close to the cutoff. For figuring attenuation at least an octave or greater away the cutoff freq is calculated as the reference ie at O dB. so for your example the answer is 12 dB. we use an asymptote for filter slopes. http://upload.wikimedia.org/wikipedi..._High-Pass.PNG

1st 6 dB/octave
2nd 12 dB/octave
3rd 18 dB/oct
4th 24 dB/o
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