Why does the Residue function returns complex coefficients of a rational function....??
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Hi
Even though I ensured that all my poles ,residues are complex conjugate pairs and the direct term is real, the residue function returns complex coefficients. May I know the reason why this is happening in matlab.
For eg:
Poles=[ -0.297252868065168 - 11.6108351815607i -0.297252868065168 + 11.6108351815607i -19.9444674513931 - 8.76592887488931i -19.9444674513931 + 8.76592887488931i -8.52965151869893 - 23.6423888144931i -8.52965151869893 + 23.6423888144931i];
Res=[ -0.219160922557423 + 0.314530473001705i -0.219160922557423 - 0.314530473001705i 3.62282522955231 + 177.563841204173i 3.62282522955231 - 177.563841204173i -4.53489652493515 - 28.5303076418931i -4.53489652493515 + 28.5303076418931i];
Direct_term=0.00210679058580560;
[b a]=residue(Res,Poles,Direct_term);
Why am I getting complex co-eff in 'b' ?? Please let me know asap....
Thanks Venu
回答（1 个）
the cyclist
2011-10-15
0 个投票
Looks to me that the imaginary parts are tiny, and are just computer round-off error.
4 个评论
venu
2011-10-15
Walter Roberson
2011-10-15
It would help if you could show the actual outputs you get. People do not always have MATLAB handy to run tests with at the time they are reading questions.
venu
2011-10-15
the cyclist
2011-10-15
venu, when I run that code, I get that the largest imaginary part is 15 orders of magnitude smaller than the largest real part. I still believe this is roundoff error.
If you need better that this, then maybe you need to solve this analytically, and not numerically. I don't know for sure, but maybe the Symbolic Math Toolbox handles this.
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