Re: [SI-LIST] : Passivity of a Linear System

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From: J. Eric Bracken ([email protected])
Date: Thu Dec 02 1999 - 14:32:45 PST

The way to test it is to compute the transfer function from the poles
and zeros:

                (s-z1)*(s-z2)*...*(s-zm)
        H(s) = ---------------------------
                (s-p1)*(s-p2)*...*(s-pn)

Assuming that H(s) represents the input impedance or input admittance
of the linear system, then passivity is equivalent to being "positive
real":

  Re[ H(s) ] >= 0 when Re[s] >= 0

There are other conditions (complex poles and zeros must be in conjugate
pairs, duh), but this is usually sufficient.

So, to check passivity you just need to plot Re[ H(j*omega) ] for
omega from 0 to infinity on a very fine grid and make sure it never
goes negative.

As for causality, you can't tell from the poles and zeros. Any given
set of poles and zeros can be interpreted in a multitude of ways, some
of which give you causal, anticausal, or two-sided signals. You have
to know the region of convergence for the Laplace transform. See
Oppenheim, Wilsky and Young's book on Signals and Systems (Prentice
Hall) for more information.

--Eric

>>>>> "CSchuster" == Schuster, Christian writes:

    CSchuster> Dear SI-List-Listeners,

    CSchuster> I know that the following question is somewhat beyond
    CSchuster> the scope of the usual list topics but nevertheless I
    CSchuster> hope to get some feedback from you ...

    CSchuster> Question: Given the poles and zeros of a linear system,
    CSchuster> how can I decide if its passive?

    CSchuster> Or weaker: How can I decide if its causal?

    CSchuster> Any input is very much appreciated.

    CSchuster> Best regards,

    CSchuster> Ch. Schuster 

--
J. Eric Bracken, Ph.D.                          Tel: 1.412.261.3200 x135
Manager, Signal Integrity R&D                   Fax: 1.412.471.9427
Ansoft Corp., Four Station Square, Suite 200    [email protected]
Pittsburgh, PA USA 15219-1119                   http://www.ansoft.com   

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