Butterworth Filter

2nd order means 2 poles but what does 'dual pass' mean?

EMG: ElectroMyeloGraph?

The EMG data has to be filtered, yes.

Going for a "2nd order dual pass butterworth filter with the cut-off frequency of 40 Hz" tells us that you're not paying attention to what's important here.

The guy who wrote the paper you're trying to copy used a certain filter. Fine. What are you trying to accomplish with this filter? EMG data changes relatively slowly, AFAIK, compared to sample rates of some modern-day ADCs.

If the particular textbook/examination papers for that particular filter are what you're looking for, then please go make love to the books on your own. If it's a flagrant test question, then go find out if your d!ck reaches your a55.

Otherwise, filtering for EMG could make a really fun topic for sci.electronics.design ! I'd bet there's some spikes below 1Hz, but wildly varying, which would make the tracking of them all that much more fun!

I'd use a cap and a resistor, and fudge around with the values until I got a usable output.

Anyone?

Cheers! Rich

Maybe dual pass means you have to put the data through the filter twice, once forwards, and once backwards.

--Mac

Hey Rich, no one asked you for your a55 hole opinion.

probably means 2 pole. there are some "sallen and key" websites that will help

martin

"dual pass" could possibly mean a notch filter. A notch has two passbands and a bandlimited stopband. The requirement is probably to notch the power line frequency.

Yeah, this would indicate a digital filter. But when an analog filter is meant, maybe it is a cascaded Butterworth, which forms a Linkwitz-Riley filter of 4th order. The word "dual pass" is meaningful only inthe digital domain.

--
ciao Ban
Bordighera, Italy

Maybe not, but I do note that you've singled this out of several other options.

So, I'm guessing that it really _was_ a test question, am I right? (or homework - same difference)

And I really, really wish that googlegroupies would clue up to copying and pasting context, if they can't find that other google option, which supposedly automatically quotes context when you post a followup.

Thanks, Rich

----- Original Message ----- From: Newsgroups: sci.electronics.design Sent: Thursday, June 23, 2005 12:20 AM Subject: Butterworth Filter

Hello Dan,

I bet it means run the signal twice through this filter. So it's finally a 4th degree filtering. The trick is to run it backward through the filter in the second pass. This removes nonlinear phase and delay. It can be only done with with digital signal processing.

"The raw X-, Y- and Z-coordinates of each IRED were digitally filtered by a dual pass through a

2nd-order Butterworth filter with a cut-off frequency of 20 Hz (equivalent to a 4th-order filter with no phase lag and a cut-off of ~16 Hz)."

Best regards, Helmut

Hi Helmut,

can you clarify "run it backward" please?

do you mean take N samples, n = 0...(N-1)

and feed through the filter, giving y0...yN-1

Then starting with the last output YN-1 and working backwards to the first output Y0, feed them through the same filter to give yy0...yyN-1 ?

So the second pass through the filter is in negative time, hence the phase lags cancel.

Cheers Terry

Hello Terry,

yes I have exactly tried that. The only thing we additionally need is reversing the order of the samples after the second run through the filter. y(N)=y(0), Y(N-1)=y(1), .... I had a hard time with Scilab to check this, because I had no experience with it before. The result has been indeed a filtered signal with zero delay.

The example below runs a stored data sequence two times through a 4th degree lowpass filter. The second pass is done with the sequence reversed(last sample first). The result is a zero delay dual pass 4th order Butterworth filter.

Best regards, Helmut

Scilab example

-------------- ! The for-loops reverse the data sequence. ! fg=0.02*fs fs=10kHz

t=(0:1e-4:0.1);

sigbase=sin(2*%pi*t*50)+0.5*sin(2*%pi*t*150);

signoise=sigbase+0.5*sin(2*pi*t*1100);

[hz]=iir(4,'lp','butt', [0.02 0], [0 0]);

y1=rtitr(hz(2),hz(3),signoise);

y2for=rtitr(hz(2),hz(3),y1);

for j=1:1001, y1rev(1,j)=y1(1002-j);end ;

y20rev=rtitr(hz(2),hz(3),y1rev);

for j=1:1001, y2rev(1,j)=y20rev(1002-j);end ;

xbasc plot2d(t,signoise,style=3); plot2d(t,sigbase,style=1); plot2d(t,y2for,style=4); plot2d(t,y2rev,style=5);

Hello, sorry, I forgot the % in front of pi in one place.

"Helmut Sennewald" schrieb im Newsbeitrag news:d9kbvb$5lr$01$ snipped-for-privacy@news.t-online.com...

signoise=sigbase+0.5*sin(2*%pi*t*1100);

Hi Helmut,

thanks for that. I've archived it, its a neat idea, and has got me thinking.

Cheers Terry