Re: residue() confusion

[Ben Abbott]

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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>>
>>   
> 
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