One pole, one zero LP/HP¶
- Author or source: ti.dniwni@tretsim
- Created: 2002-08-26 23:34:48
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 | void SetLPF(float fCut, float fSampling)
{
float w = 2.0 * fSampling;
float Norm;
fCut *= 2.0F * PI;
Norm = 1.0 / (fCut + w);
b1 = (w - fCut) * Norm;
a0 = a1 = fCut * Norm;
}
void SetHPF(float fCut, float fSampling)
{
float w = 2.0 * fSampling;
float Norm;
fCut *= 2.0F * PI;
Norm = 1.0 / (fCut + w);
a0 = w * Norm;
a1 = -a0;
b1 = (w - fCut) * Norm;
}
Where
out[n] = in[n]*a0 + in[n-1]*a1 + out[n-1]*b1;
|
Comments¶
- Date: 2006-01-15 01:12:26
- By: ten.tramepyh@renietsretep
what is n? lol...sorry but i mean this seriously! ;)
- Date: 2006-01-16 14:17:35
- By: ku.oc.snosrapsd@psdcisum
n is the index of sample being considered.
out[] is an array of samples being output, and in[] is the input array. you would construct a loop such that:
[Pseudocode]
loop n{0..numsamples-1}
out[n] = in[n]*a0 + in[n-1]*a1 + out[n-1]*b1;
end loop;
[/Pseudocode]
You will need some cleverness so that [n-1] doesn't cause an index error when n=0, but I'll leave that to you :)
- Date: 2006-01-18 09:28:30
- By: ten.tramepyh@renietsretep
whoops - sorry, of course n = number... stupid me ;)
interesting code, i will see if can adapt that to delphi, shouldn´t be a big deal :)
i assume i dont need to place either setHPF or LPF into the samples loop, just the block itself?
- Date: 2006-01-23 10:57:26
- By: ku.oc.snosrapsd@psdcisum
absolutey - set the coefficients outside of the loop. There is the case of changes being made whilst the loop is running, depends what platform/host you are writing for.
I'm a delphi code as well. Feel free to use my posted address if you need to :) DSP
- Date: 2006-07-21 09:07:12
- By: moc.oohay@keelanej
Shouldn't that be float w = 2*PI*fSampling; ???
In which case we can simplify:
void SetLPF(float fCut, float fSampling)
{
a0 = fCut/(fSampling+fCut);
a1 = a0;
b1 = (fSampling-fCut)/(fSampling+fCut);
}
void SetHPF(float fCut, float fSampling)
{
a0 = fSampling/(fSampling+fCut);
a1 = -a0;
b1 = (fSampling-fCut)/(fSampling+fCut);
}
You can keep the norm = 1/(fSampling+fCut) if you like.
- Date: 2020-05-23
- By: JoergBitzer
The equation of the original contributor is correct. It is a first order Butterworth-Filter
H(s') = 1/(1+s') and then denormalized s' = s/wcut and transformed by the bilinear transform
s = 2f_s (z-1)/(z+1). Only the tan-prewarp is missing for fcut/wcut.