Maker Pro
Maker Pro

FIR filter question

J

Jamie Morken

Jan 1, 1970
0
Hi,

When using a lowpass FIR filter, with 100kHz sampling and 1kHz -3dB
cutoff frequency, is there a minimum number of coefficients required for
the filter to operate properly similar to a Nyquist minimum? I was
thinking that there would be 100 or even 200 coefficients required since
the cutoff frequency is 100 times lower than the sampling frequency so
that the low frequency data can remain in the coefficient data long
enough, but I am not sure if this is correct..

cheers,
Jamie
 
M

Mark

Jan 1, 1970
0
Hi,

When using a lowpass FIR filter, with 100kHz sampling and 1kHz -3dB
cutoff frequency, is there a minimum number of coefficients required for
the filter to operate properly similar to a Nyquist minimum?  I was
thinking that there would be 100 or even 200 coefficients required since
the cutoff frequency is 100 times lower than the sampling frequency so
that the low frequency data can remain in the coefficient data long
enough, but I am not sure if this is correct..

cheers,
Jamie

the number of taps you need is related to the ratio of cutoff freq to
sampling freq just the way you say....but it is also related to the
STEPPNESS of the transition band. a steeper cutoff needs more
taps...

so to use the least number of taps you want the Fs as low as possible
and you want the cutoff as gradual as possible...

get a filter design program and play with it

these guys
http://www.mds.com/
used to have a free eval version... QED lite..

looks ike they want $99 for it now..maybe you can find another free
program...experimnet with the number of taps and the type of window...

Mark
 
V

Vladimir Vassilevsky

Jan 1, 1970
0
Jamie said:
Hi,

When using a lowpass FIR filter, with 100kHz sampling and 1kHz -3dB
cutoff frequency, is there a minimum number of coefficients required for
the filter to operate properly similar to a Nyquist minimum? I was
thinking that there would be 100 or even 200 coefficients required since
the cutoff frequency is 100 times lower than the sampling frequency so
that the low frequency data can remain in the coefficient data long
enough, but I am not sure if this is correct..


First, having a FIR filter with 100kHz sampling and 1kHz cutoff doesn't
look like a good idea. For the cases like that, you should use IIR
and/or make the sampling rate lower.

Very roughly, the required length of the FIR LPF can be estimated as a
period of the cutoff frequency. There are more exact formulae for the
FIR length depending on the passband flatness and stopband attenuation.


Vladimir Vassilevsky
DSP and Mixed Signal Design Consultant
http://www.abvolt.com
 
Hi,

When using a lowpass FIR filter, with 100kHz sampling and 1kHz -3dB
cutoff frequency, is there a minimum number of coefficients required for
the filter to operate properly similar to a Nyquist minimum?  I was
thinking that there would be 100 or even 200 coefficients required since
the cutoff frequency is 100 times lower than the sampling frequency so
that the low frequency data can remain in the coefficient data long
enough, but I am not sure if this is correct..

cheers,
Jamie

I have a website that allows you to buils and see simple FIR filters.
It is located here. Perhaps this will help you get some insight.

http://www.fourier-series.com/fourierseries2/convolution.html
 
J

Jamie Morken

Jan 1, 1970
0
Phil said:
To the OP:

Do you need some specific property of a FIR filter, e.g. perfectly flat
group delay (for symmetric filters) or an impulse response that goes to
exactly 0?

If not, you can probably do this with a few-tap IIR design.

Hi,

I've used IIR filters before, but for this project I have an FPGA so am
trying out FIR, also Altera Quartus has a neat built in FIR filter
builder. Someone told me IIR can have a problem with oscillation, not
sure if it is a problem in reality for most projects or just a
theoretical idea from the infinite impulse response.. I guess this is
what you mean by "impulse response goes to exactly zero".

Quartus doesn't have an IIR filter builder, probably because they don't
use enough FPGA resources :)

cheers,
Jamie
 
Top