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I have a question from the book FE Electrical and Computer Review Manual by Michael R. Lindeburg page# DE X-1 with its answer as follows: enter image description here

I tried to solve it as follows: enter image description here

enter image description here

I don't understand why is it when z = -2 the system is unstable?

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    \$\begingroup\$ When using z-plane, saying that poles being on the left side implies stability is incorrect. The whole of the s-plane maps to a circle in the z-plane and anything outside that circle means instability. Anything on the left in the s-plane is stable but we're talking about the z-plane now. \$\endgroup\$ – Andy aka Jan 11 at 10:43
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    \$\begingroup\$ Does this help \$\endgroup\$ – Andy aka Jan 11 at 11:19
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    \$\begingroup\$ @OMAR You should read about region of convergence (ROC). For a discrete system to be stable, its ROC must contain the unit circle. Pleas note the answer is "unstable" because of the implied causality of the filter. One could argue the question should have included the ROC (|z| > 2), because assuming the filter was anticausal (|z| < 1, or perhaps 1 < |z| < 2) it could actually be marginally stable. However your solution attempt was just wrong since you tried to apply a s-plane concept to the z-plane. \$\endgroup\$ – Vicente Cunha Jan 11 at 11:36
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    \$\begingroup\$ @OMAR I've never said "the system must contain the unit circle", that sentence does not make much sense. The ROC, region of convergence, associated with the system must contain the unit circle. One way to broadly interpret this is that "the summation of impulse response must be bounded". Please read about ROC, plenty references online. \$\endgroup\$ – Vicente Cunha Jan 11 at 12:41
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    \$\begingroup\$ @OMAR Sorry I don't have sufficient time to formulate a quality answer, that's why I used comments to point you in the right direction. You could read the "Stability and causality" section of this link for an example of how ROC, causality and stability are related. \$\endgroup\$ – Vicente Cunha Jan 11 at 13:01
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The mistake was here: -

enter image description here

You assumed that the rules that apply on the s-plane also apply to the z-plane. They don't because when you map the left hand side of the s-plane to the z-plane you get a unit circle: -

enter image description here

It's a unit circle (amplitude 1) because the s-plane Nyquist frequency is \$\pi\$ radians per second (0.5 Hz). So, everything inside the rectangle on the left side of the s-plane bounded by the top horizontal magenta/purple line down to the lower horizontal dotted black line is within or on the unit circle in the z-plane.

So, given that you calculated z-plane pole values of -1 and -2, the -2 pole is clearly unstable.

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  • \$\begingroup\$ It's always bothered me that the Z transform isn't directly analogous to the Laplace transform. Why is that anyway? \$\endgroup\$ – Hearth Jan 11 at 15:27
  • \$\begingroup\$ @Hearth tis the math. There are some good youtube vids..... youtube.com/watch?v=acQecd6dmxw \$\endgroup\$ – Andy aka Jan 11 at 15:28
  • \$\begingroup\$ Yes, but someone made the choice to use \$z\$ in the z-transform where you'd have \$e^s\$ in the Laplace transform, and that's never quite sat right with me. \$\endgroup\$ – Hearth Jan 11 at 15:29
  • \$\begingroup\$ @TonyStewartSunnyskyguyEE75 not true. \$\endgroup\$ – Andy aka Jan 11 at 18:12

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