Interpeting results of residue command

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brooke
brooke am 20 Feb. 2024
Kommentiert: Hassaan am 20 Feb. 2024
I am using the residue command to aid in my partial fraction expansion. I am confused about the result for the poles, I have never seen anything like the 1.0e+0.4* before. How do I interpret these results into a new equation?
What I tried: Y=-0.3024i/(s-e-0.4(-1.25+0.6614i)) + 0.3024i/(s-e-0.4(-1.25-0.6614i))
Here are the results:
r =
0.0000 - 0.3024i
0.0000 + 0.3024i
p =
1.0e+04 *
-1.2500 + 0.6614i
-1.2500 - 0.6614i
k =
[]
  1 Kommentar
Dyuman Joshi
Dyuman Joshi am 20 Feb. 2024
Bearbeitet: Dyuman Joshi am 20 Feb. 2024
"I am confused about the result for the poles ..."
Refer to the documentation of the function and see what the outputs correspond to.
"... I have never seen anything like the 1.0e+0.4* before. "
That is the scientific notation.
"How do I interpret these results into a new equation?"
I am not sure if I understand what you mean here.

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Hassaan
Hassaan am 20 Feb. 2024
Bearbeitet: Hassaan am 20 Feb. 2024
@brooke The residue command in MATLAB is used to perform partial fraction expansion of a rational function, which is typically represented as the ratio of two polynomials. The output of residue consists of three elements: the residues (r), the poles (p), and direct terms (k). Here's how to interpret the results:
  • Residues (r): These are the coefficients of the terms in the partial fraction expansion.
  • Poles (p): These are the values of s that make the denominator of the original rational function zero.
  • Direct Terms (k): These are the coefficients of any terms that are not part of the partial fraction decomposition (usually when the numerator degree is greater than or equal to the denominator).
In your results, the 1.0e+04 is MATLAB's way of formatting numbers using scientific notation. It's saying that the poles are actually -1.2500 * 1.0e+04 + 0.6614i * 1.0e+04 and -1.2500 * 1.0e+04 - 0.6614i * 1.0e+04. This notation is used to represent large or small numbers compactly.
So your poles are actually:
p1=12500+6614i
p2=125006614i
And the residues corresponding to these poles are:
r1=0.00000.3024i
r2=0.0000+0.3024i
Therefore, your partial fraction expansion of Y would be:
Y = (0.0000-0.3024i/s-(-12500+6614i)) + (0.0000+0.3024i/s-(-12500-6614i)​)
When you simplify this, remember that s will be your variable (usually representing complex frequency in Laplace transforms), and you can then use these expressions to further work on inverse Laplace transforms or other analyses.
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  2 Kommentare
brooke
brooke am 20 Feb. 2024
Thank you very much I understand now!
Hassaan
Hassaan am 20 Feb. 2024
@brooke You are welcome.

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