From help-octave-request at bevo dot che dot wisc dot edu Thu Dec  6 18:06:00 2001
Subject: Re: using conv vs. filter 
From: "Stephen W. Juranich" <sjuranic at crow dot ee dot washington dot edu>
To: "E. Joshua Rigler" <Jrigler at colorado dot edu>
Cc: Help-Octave <help-octave at bevo dot che dot wisc dot edu>
Date: Thu, 6 Dec 2001 18:05:02 -0600


> I'm looking at the code for conv.m in the octave libraries.  It does
> some sort of zero-padding, then calls the "filter" function.  The output
> is a vector that is length(a)+lenght)b)+1.

If you're using digital filtering techniques, then convolving two signals 
(call them 'a' and 'b'), then the length of the result of the convolution is 
going to be length(a)+length(b)-1 (not +1).  The only reason this will not be 
true is if you're sampling below the Nyquist rate, at which point you have 
aliasing (generally considered a Bad Thing [tm]).

> Now, I am trying to convolve an filter with both negative, zero, and
> positively lagged coefficients with an input time series.  In order for
> my time-steps to be synchronized between the input time series, and the
> output from conv.m, how would I index the output?  I guess I just don't
> really understand how the conv.m function works in octave.

I assume you're talking about a non-causal filter.  Let's say that your first 
non-zero coefficient in the filter is a[-k] and the input signal starts at 
b[0].  Then the filter is going to look k samples into the future and start 
convolving there.  So you will start getting output at time = -k.

> At the moment, I use a filter that has an equal number of positive and
> negatively lagged coefficients, and the output seems reasonable when I
> do something like:
> 
> output = conv (filter, input);
> output = output (length(filter)/2 : length(output)- \
>                                     length(filter)/2);
> 
> 1) Does what I have seem right?
> 2) What do I do if my filter has fewer/more negatively lagged
> coefficients than positively lagged coefficients?

As I mentioned before you _expect_ the result to be longer than the two 
inputs, so unless you *really* know what you're doing (and you'd have to be 
doing something really funky), you should keep them all.

Having said all that, It's been a couple of years since my last linear systems 
class, so some of the details might have been clouded up in my mind 
(especially that stuff in the second paragraph). :)

HTH


--------------------------------------------------------------------------
Stephen W. Juranich                             sjuranic at ee dot washington dot edu
Electrical Engineering             http://students.washington.edu/sjuranic
University of Washington                http://ssli.ee.washington.edu/ssli



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