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Re: residue() confusion


From: Doug Stewart
Subject: Re: residue() confusion
Date: Sat, 22 Sep 2007 22:56:47 -0500
User-agent: Thunderbird 1.5.0.13 (Windows/20070809)

Not yet I was letting some users try it first to make sure that there are no more problems before we commit it.
So give it a try and report back to the list.
Doug

Ben Abbott wrote:
Thanks Doug,

Is this fix included in 2.9.14?



Doug Stewart wrote:
Here is my results for this question

num = [1 0 1];
den = [1 0 18 0 81]; [a,p,k,e] = residue(num,den)

a =

  1.0e+06  *

   5.927582766769742 + 2.314767228467131i
   5.927582730209724 - 2.314767214190160i
  -5.927582717669299 - 2.314767374340102i
  -5.927582681109279 + 2.314767360063129i

p =

   0.000000016264485 + 2.999999993648592i
   0.000000016264485 - 2.999999993648592i
  -0.000000016264485 + 3.000000006351409i
  -0.000000016264485 - 3.000000006351409i

k = []
e =

                  1
                  1
                  1
                  1

 >>   [a,p,k,e] = residue2(num,den)
a =

  -0.000000000000000 - 0.092592592592593i
   0.000000000000000 + 0.092592592592593i
   0.222222222810316 + 0.000000001505971i
   0.222222222810316 - 0.000000001505971i

p =

   0.000000016264485 + 2.999999993648592i
   0.000000016264485 - 2.999999993648592i
  -0.000000016264485 + 3.000000006351409i
  -0.000000016264485 - 3.000000006351409i

k = []
e =

                  1
                  1
                  2
                  2



the second one is my "fixed" code and this agrees with Matlab.

You can get my code at:

www.dougs.homeip.net/octave/residue2.m

Doug





Henry F. Mollet wrote:
Your concern is justified. I don't know how to do partial fractions by
hand
when there is multiplicity. Therefore I checked results by hand using s =
linspace (-4i, 4i, 9) as a first check. It appears that Matlab results
are
correct if I take into account multiplicity of [1 2 1 2]. Octave results
appear to be incorrect.
Henry
octave-2.9.14:29> s =
-0 - 4i 0 - 3i 0 - 2i 0 - 1i 0 + 0i 0 + 1i 0 + 2i 0 + 3i 0
+ 4i

Using left hand side of equation:
octave-2.9.14:30> y=(s.^2 + 1)./(s.^4 + 18*s.^2 + 81)
y =
 Columns 1 through 6:
  -0.30612 + 0.00000i       NaN +     NaNi  -0.12000 - 0.00000i   0.00000
-
0.00000i   0.01235 - 0.00000i   0.00000 - 0.00000i
 Columns 7 through 9:
  -0.12000 + 0.00000i       NaN +     NaNi  -0.30612 + 0.00000i

Using right hand side of equation with partial fraction given by Matlab:
octave-2.9.14:31> yMatlab= (0 - 0.0926i)./(s-3i) + (0.2222 -
0.0000i)./(s-3i).^2 + (0 + 0.0926i)./(s+3i) + (0.2222 +
0.0000i)./(s+3i).^2

yMatlab =
 Columns 1 through 6:
  -0.30611 + 0.00000i       NaN +     NaNi  -0.11997 + 0.00000i   0.00001
+
0.00000i   0.01236 + 0.00000i   0.00001 + 0.00000i
 Columns 7 through 9:
  -0.11997 + 0.00000i       NaN +     NaNi  -0.30611 + 0.00000i

Using right hand side of equation with partial fraction given by Octave:
octave-2.9.14:32> yOctave=(-3.0108e+06 - 1.9734e+06i)./(s-3i) +
(3.0108e+06
+ 1.9734e+06i)./(s-3i).^2 + (-3.0108e+06 + 1.9734e+06i)./(3+3i) +
(3.0108e+06 - 1.9734e+06i)./(s+3i).^2

yOctave =
 Columns 1 through 5:
  -2.9632e+06 + 2.3337e+06i         NaN +       NaNi  -2.9095e+06 +
2.1230e+06i  -6.2042e+05 + 4.4801e+05i  -1.8417e+05 - 1.7290e+05i
 Columns 6 through 9:
  -1.2708e+05 - 1.0447e+06i  -1.3307e+06 - 4.0746e+06i         NaN +
NaNi  -5.2185e+06 + 1.9084e+06i

**********************************

on 9/22/07 2:14 PM, Ben Abbott at address@hidden wrote:

I was more concerned about the differences in "a"

I suppose I'll need to do a derivation and check the correct answer.

On Sep 22, 2007, at 5:05 PM, Henry F. Mollet wrote:

The result for e should be [1 2 1 2] (multiplicity for both poles).
Note
that Matlab does not even give e.  My mis-understanding of the
problem was
pointed out by Doug Stewart. Doug posted new code yesterday, which
I've
tried unsuccessfully, but I cannot be sure that I've implemented
residual.m
correctly. The corrected code still produced e = [1 1 1 1] for me.
Henry


on 9/22/07 1:31 PM, Ben Abbott at address@hidden wrote:

As a result of reading through Hodel's
http://www.nabble.com/bug-in-residue.m-tf4475396.html post  I
decided to
check to see how my Octave installation and my Matlab installation
responded
to the example

Using Matlab v7.3
--------------------------
 num = [1 0 1];
 den = [1 0 18 0 81];
 [a,p,k] = residue(num,den)

a =

        0 - 0.0926i
   0.2222 - 0.0000i
        0 + 0.0926i
   0.2222 + 0.0000i


p =

   0.0000 + 3.0000i
   0.0000 + 3.0000i
   0.0000 - 3.0000i
   0.0000 - 3.0000i


k =

     []
--------------------------

Using Octave 2.9.13 (via Fink) on Mac OSX
--------------------------
 num = [1 0 1];
 den = [1 0 18 0 81];
 [a,p,k] = residue(num,den)

a =

  -3.0108e+06 - 1.9734e+06i
  -3.0108e+06 + 1.9734e+06i
  3.0108e+06 + 1.9734e+06i
  3.0108e+06 - 1.9734e+06i

p =

  -0.0000 + 3.0000i
  -0.0000 - 3.0000i
   0.0000 + 3.0000i
   0.0000 - 3.0000i

k = [](0x0)
e =

   1
   1
   1
   1
--------------------------

These are different from both the result that
http://www.nabble.com/bug-in-residue.m-tf4475396.html Hodel
obtained , as
well as different from
http://www.nabble.com/bug-in-residue.m-tf4475396.html Mollet's

Thoughts anyone?

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