I have been trying to understand what 100Hz 12dB per octave means in a low pass filter. What does it really mean in full details? Any material can be of help.
There are three parameters of this filter that are described in the phrase "100 Hz 12 dB per octave low pass filter". I'll cover them in reverse order.
- Low pass filter - This means the filter does not change lower frequencies ("passes" those frequencies through) and blocks higher frequencies. Sometimes these filters are called "high cut filters", but that can get confused with a "high shelf filter" so "low pass filter" or "LPF" is more common.
- 12 dB per octave - No filter can completely cut off all signal immediately for some frequencies and completely allow it for others. All filters have a "slope" or "quality" which indicates how much the affected frequencies are attenuated. For a low pass filter, higher frequencies are attenuated more. For this filter, the rate of attenuation is 12 dB per octave. An octave is a doubling of frequency. So if the frequency 1000 Hz is affected by this low pass filter, then the frequency of 2000 Hz (one octave higher) will be attenuated to be 12 dB below the level at 1000 Hz. 4000 Hz will be 12 dB lower than 2000 Hz, and so on. Usually a low pass or high pass filter will be created with certain electronics. The smallest amount of electronic components that makes a low or high pass filter creates a slope of 6 dB per octave. That's called a "one-pole" or "first order" filter. IF you double the components, you create a 12 dB per octave low or high cut filter and that is called "two pole" or "second order", and so on. The famous Moog synthesizers are well known for using a four pole LPF with a slope of 24 dB per octave.
- 100 Hz - This is the *cutoff frequency**. That means it is the frequency below which the signal is not affected and above which the signal is attenuated at 12 dB per octave. Technically, there is a minor effect at the cutoff frequency, because it is defined as the "3 dB down point", or the frequency where the signal has been attenuated by 3 dB.
So let's look at a specific signal and how this filter would affect it. If we create a white noise source (all frequencies with equal level) at a level of 0 dBu (0.775 Vrms).
Below 100 Hz, all the frequencies will passed through at 0 dBu.
At 100 Hz, the signal will be at -3 dBu.
At 200 Hz, the signal will be at -12 dBu (one octave higher, so 12 dB lower, the -3 is not included here because the "corner is cut off" which creates the -3 at 100 Hz)
At 400 Hz, the signal will be at -24 dBu
At 3,200 Hz, the signal will be at -60 dBu
At 12,800 Hz, the signal will be at -84 dBu
A low-pass filter with 100Hz cutoff frequency and 12dB/oct attenuation will attenuate the signal by roughly 12dB per octave passing 100Hz.
If you want to draw a diagram, it will consist of two straight lines meeting at the 100Hz/0dB point, with the line before being constant at 0dB/octave, and the line afterwards going down with a rate of 12dB/octave. Those straight lines are asymptotic: the further you are from their meeting point, the more correct they are. At 100Hz there is the maximum level of incorrectness since the actual attenuation there is 3dB rather than 0dB.
So: significantly below 100Hz, attenuation is 0dB. At 100Hz, attenuation is 3dB. At 200Hz, attenuation will almost be 12dB. At 400Hz, attenuation will be 24dB. At 3200Hz, attenuation will be 60dB.
Strictly speaking, the cutoff attenuation (where the asymptotes of passband and stopband intersect) being at 3dB is a feature particular to Butterworth filters, but since Butterworth filters are "maximally flat" in pass- and stopband, they are the ones with the most pronounced asymptotes, so it makes some sense using them as reference also for other filter types.
A filter specified as 12dB/octave, however, is pretty certainly a 2nd degree Butterworth filter. Chebyshev and elliptic (Cauer) filters have ripple in pass and/or stopband instead of proper asymptotes and higher slopes, Bessel filters have much more linear phase response and lower slopes.