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Z transform - simple question

  1. Oct 15, 2006 #1
    hi guys.
    Just want to clarify something with the z transform and convergence.

    Do you guys aggree with this.

    If an expression does not converge you CAN STILL find the z transform (using power series) for the system HOWEVER this will not be the closed form z transform.

    just wondering - because in any z transform table you are given a region of convergence - but this is for closed form z transforms.
    I suppose all systems are stable if they converge.

  2. jcsd
  3. Oct 15, 2006 #2
    sorry - i think my last statement may be a bit to generalized:"I suppose all systems are stable if they converge."

    you can get a system that converges but is not necessarily stable. E.g.
    z/(z-a) where region of convergence is defined by, mod |z| > |a|

    but a can be bigger then unity (which wil make the system unstable) but the system will still converge as long as z > a. Do you guys aggree?
    makes sense beacuse i just read somewhere that for a system to be stable , the region of convergence must include the unit circle.
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