I wonder if lop~ object's argument should be the filter's cut-off frequency.

I tested "lop~ 400" with an 1kHz OSC~ input, but then the signal sounds like there is no attenuation at all.

However, in Csound, such a test with lpf18 or biquad opcode (tuned to a one-pole) will attenuate the 1kHz sine accordingly.

So have I misunderstood the argument? I noticed that with lop~ and hip~, the argument in the help is referred to as "**roll-off** frequency" while their right inlets were termed "**cut-off** frequency". Are they different?

Please advise.

]]>I wonder if lop~ object's argument should be the filter's cut-off frequency.

I tested "lop~ 400" with an 1kHz OSC~ input, but then the signal sounds like there is no attenuation at all.

However, in Csound, such a test with lpf18 or biquad opcode (tuned to a one-pole) will attenuate the 1kHz sine accordingly.

So have I misunderstood the argument? I noticed that with lop~ and hip~, the argument in the help is referred to as "**roll-off** frequency" while their right inlets were termed "**cut-off** frequency". Are they different?

Please advise.

]]>anyhoo

http://lists.puredata.info/pipermail/pd-list/2014-05/106928.html

seems like a normalized filter that has a single pole that goes from 1 to 0 on the real axis as the cutoff frequency goes from 0 to 1 radians(so it is definitely an approximation). Every frequency higher than that cutoff is clipped, and the frequencies in that range will have a warped cutoff (half-power) frequency (& frequency response).

from miller's book: http://msp.ucsd.edu/techniques/v0.11/book-html/node140.html

just for fun let's see what the frequency response should be where you measured it:

at 44100 sample rate, a cutoff frequency of 400 would be translated into:

400*2*pi/44100 = .05699

1 - .05699 = .94301 is the pole

so the transfer function will be:

.05699/(1 - .94301(z^-1))

the radian value of 1000 @ samplerate of 44100 is (1000/44100) * 2pi = .142476 radians

the cartesian co-ordinates in the complex plane of .94301(z^-1):

real part: cos(.142476)*.94301 = .933455
imaginary part: -sin(.142476)*.94301 = -.133902

then the distance formula:

sqrt((1 - .933455)^2 + (0 + .133902)^2) = .14952585

so the gain will be: .05699/.14952585 = .38113811

which is the peak amplitude after the lop~ when i measured .. actually quite a bit of attenuation

]]>lop-fix.zip ]]>

lop-fix.zip

Thank you very much for the detailed explanation.

I should've measured the signal output before posting.