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I have modeled a system showing the following properties.

  1. It has some negative eigenvalues.
  2. All poles are placed in a left-half plane.
  3. Some zeros are right-half plane.

In my knowledge, eigenvalues demonstrate the location of poles as they are roots of the characteristic equations, and poles are denominators of the transfer function. When I plot the eigenvalues of my model, two of them are negative (complex pairs), While two zeros are located in the right half plane while all the poles lie in left half plane.

Is there any relation of eigenvalues to zeros of a system?

I am using MATLAB for finding eigenvalues and using the pole-zero plot (pzplot) function to analyze the system.

winny
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Dan
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    You've essentially said yourself that the eigenvalues are the poles of the system. Knowledge of the poles does not imply knowledge of the zeros. The eigenvalues alone do not provide a constraint on the zeros. – LetterSized Aug 20 '22 at 18:56
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    Here's one way to think about it. In a state space representation of a system, you might have something like x_dot = Ax +Bu. As we know, the eigenvalues of A are the roots of the characteristic polynomial, which are the poles of the system. The poles determine the form of the system transient response (i.e. the types of exponentials that form the homogenous solution). The zeros play a role when you are considering the forced response, which requires you to look at the B matrix, in addition to the A matrix. The A matrix alone cannot tell you what the zeros are. – LetterSized Aug 20 '22 at 19:01
  • One can also calculate the zeros from a state space model by solving a slightly different eigenvalue problem, [see tzero](https://www.mathworks.com/help/control/ref/lti.tzero.html#bs9klmx-7). – fibonatic Aug 20 '22 at 20:22

1 Answers1

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As you indicated the eigenvalues (poles) are calculated from the denominator of the transfer function.

The zeros are calculated from the numerator, so the eigenvalues cannot influence the zeros. So the answer to your question is no.

RussellH
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