Re: I have modified firls.m to include all types of FIRs plus HT and diff

[je suis]

> Your code errors out on all three, so there's a bit of a Matlab
> compatibility issue.

Thank you for the answers. I realize now that, as I went along with
deciphering the code and the mathematics behind the original file, I
changed the way I wanted the function to work since I was deciphering
it with wxMaxima, and then retranslating it to Octave. So if it's a
mess, don't hold out on the blames.

> ok, Matlab help for firls has the following as test inputs to firls:

Considering these lines, the way this new firls.m should be used is this:

> % Example of a length 31 lowpass filter.
>     h=firls(30,[0 .1 .2 .5]*2,[1 1 0 0]);

h=firls(30,[0 .2 .4 1],[1 1 0 0],[1 1])
h =
   1.27199776427391e-03
   1.75973248875545e-03
  -4.74050123698976e-04
  -5.07655778150969e-03
  -7.30622779100661e-03
  -1.39834985893641e-03
   1.18182305510126e-02
   2.08600394665460e-02
   1.08509633977440e-02
  -2.01948738097687e-02
  -5.03421398292692e-02
  -4.26075899832328e-02
   2.73002961309274e-02
   1.44919701151268e-01
   2.57048286517311e-01
   3.03233271924369e-01
   2.57048286517311e-01
   1.44919701151268e-01
   2.73002961309274e-02
  -4.26075899832328e-02
  -5.03421398292692e-02
  -2.01948738097687e-02
   1.08509633977440e-02
   2.08600394665460e-02
   1.18182305510126e-02
  -1.39834985893641e-03
  -7.30622779100661e-03
  -5.07655778150969e-03
  -4.74050123698976e-04
   1.75973248875545e-03
   1.27199776427391e-03

> % Example of a length 45 lowpass differentiator.
>     h=firls(44,[0 .3 .4 1],[0 .2 0 0],'differentiator');
h=firls(44,[0.3 0.4],[0.2 0],[1 10])
h =
   0.0001935680319982
   0.0005837361716208
   0.0006318704675258
  -0.0000649881094611
  -0.0010909168811554
  -0.0012746155567639
   0.0000830256283033
   0.0019781831198726
   0.0022112619647656
  -0.0002511958142674
  -0.0034193660983087
  -0.0035798846980063
   0.0006828426031918
   0.0058134841226126
   0.0057701601029457
  -0.0015617139368161
  -0.0103256404335874
  -0.0104086300033443
   0.0030055573421858
   0.0228147964389532
   0.0338124265801903
   0.0250410954472427
   0.0000000000000000
  -0.0250410954472427
  -0.0338124265801903
  -0.0228147964389532
  -0.0030055573421858
   0.0104086300033443
   0.0103256404335874
   0.0015617139368161
  -0.0057701601029457
  -0.0058134841226126
  -0.0006828426031918
   0.0035798846980063
   0.0034193660983087
   0.0002511958142674
  -0.0022112619647656
  -0.0019781831198726
  -0.0000830256283033
   0.0012746155567639
   0.0010909168811554
   0.0000649881094611
  -0.0006318704675258
  -0.0005837361716208
  -0.0001935680319982

I think this one differs because of the unimplemented 1/f^2 weighting.
I used [1 10].

> % Example of a length 26 type 4 highpass filter.
>     h=firls(25,[0 .4 .5 1],[0 0 1 1],'h');
h=firls(25,[0 0.4 0.5 1],[0 0 1 1],[1 1],1)
h =
   0.00480437391569623
   0.01050613053429931
  -0.00340207115372934
  -0.01833526416773617
  -0.00228782077798131
   0.02741907886546163
   0.01471652010061388
  -0.03658489209828520
  -0.03855957915035270
   0.04447014544831961
   0.08964774885148744
  -0.04980513866503054
  -0.31257721229656388
   0.55169152299314106
  -0.31257721229656388
  -0.04980513866503054
   0.08964774885148744
   0.04447014544831961
  -0.03855957915035270
  -0.03658489209828520
   0.01471652010061388
   0.02741907886546163
  -0.00228782077798131
  -0.01833526416773617
  -0.00340207115372934
   0.01050613053429931
   0.00480437391569623

This one is different, I'm wrong somewhere in there, unless there is
something hidden under the hood?

At any rate, I will modify the script to match Matlab's choice of
input, even if, honestly, and without infatuation, the way I did it
seems more practical. But compatibility is compatibility. I'll work on
it tomorrow, now it's late here. Should I also modify the output to be
row vector?

Vlad