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band pass filter - butterworth.vi

I want to use a band-pass filter with a high cut off freq of 0.01 Hz and low cut off freq of 1 Hz. My sampling rate is 20 Hz.

Why does this not work?

It works if I switch the cutt of freq values, but I need it to work this way.

I use other applications that do this perfectly, I feel like I am missing something...

 

 

Thanks!!!!Smiley Happy

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Message 1 of 4
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It works just fine.  If you want all the data suppressed to ~1E-18, that is.

 

The detailed help file for the Buterworth Filter VI states:

 

"The values for high cutoff freq: fh and low cutoff freq: fl must observe the following relationship:

0 < f1 < f2 < 0.5fs

where f1 is low cutoff freq: flf2 is high cutoff freq: fh, and fs is sampling freq: fs."

 

I suspect your problem is terminology. You want the high pass part of the filter to cut off at 0.01 Hz, meaning that it will pass everything above that frequency.  But that is what LabVIEW calls the low cutoff frequency for a bandpass filter.

 

Lynn

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Message 2 of 4
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Hey, i have a problem with butterworth bandpass filter too..
i am generating Sine waveform, which i can control the frequency, and connected it to butterworth filter with fh 25 hz and fl 3 hz. It does reduce the amplitude, but in the wrong way, i think.. I dont know if i missed something. I just want to testing the butterworth filter in labview, to match the butterworth theory. If you would kindly give me some light, i will appreciate it. 

 


Regards..

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Message 3 of 4
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In the sine generator you specify a sampling rate of 1000 Samples per second, but for the filter (which does not work on waveforms, but on arrays) you specify 100 Samples per second. That should of course be the same value.

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