I am trying to understand why a system with a single pole inside the unit circle is stable. For example, take a system with one pole at $z=\frac{1}{2}$. The literature says the system is stable. As a Physics major with not that much intuition for pole-zero plots I first tried to solve the differential equation that this transfer function represents:


Which, I re-write to find

$Y(z)\cdot\big(z-\frac{1}{2}\big)=X(z)\cdot z$

Which I transformed to the time domain giving


For the impulse response $\dot{x}=\dot{\delta}(t)=0$, so the solution to my equation looks like


And this looks quite unstable to me!

Of course, I also tried doing the same by doing a long-tail division of the transfer function, so I got


The inverse transform of this gives

$y_{\delta}(t)=2^{-t} = \big(\frac{1}{2}\big)^t$

This, I agree, is a stable system.

I've tried looking all over but I haven't found an example of how to go from the differential equation to a stable system. I have seen that some other questions had answers that included something like "look at the characteristic equation, when $\lambda < 1$ then the system is stable. I suppose I would disagree there - since $e^{\lambda t}$ would only be stable for negative $\lambda$.

What am I missing?

  • 1
    $\begingroup$ z^-1 is a delay operator s is a differentiation operator $\endgroup$
    – user28715
    Oct 10, 2018 at 16:22

2 Answers 2


What you are missing is that this is about a discrete-time system, because we're talking about poles and zeros in the complex $z$-plane and about poles inside or outside the unit circle. So there is no differential equation, but there is a difference equation:


The corresponding impulse response is


where $u[n]$ is the unit step sequence.

A continuous-time system with a pole at $s=\frac12$ would indeed be unstable, because for stability, all poles of a continuous-time system must be in the left half plane. In discrete time this corresponds to the requirement of all poles being inside the unit circle $|z|=1$.

  • $\begingroup$ Thank you for the clarification. I never realised that discrete-time and continuous-time systems could be so fundamentally different. I guess I always thought you could discretize a continuous system whenever you like and things would be fine. Do you happen to know of a (real-life) example where this difference occurs? $\endgroup$ Oct 10, 2018 at 19:29
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    $\begingroup$ @AntoinePost: You can indeed discretize a continuous-time system, and there are several ways to do that, but due to the nature of discrete-time signals and systems, the imaginary axis of the s-plane maps to the unit circle in the z-plane (think of aliasing), so that's how the condition "all poles in the left half-plane" turns into "all poles inside the unit circle". $\endgroup$
    – Matt L.
    Oct 10, 2018 at 19:41

You're conflating the discrete-time definition of a system with the continuous-time representation of a system.

Your discrete-time

$$Y(z)\cdot\big(z-\frac{1}{2}\big)=X(z)\cdot z$$

does not transform to:


but to: $$ y[n+1] - \frac{1}{2} y[n] = x[n+1] $$ or $$ y[n] = \frac{1}{2} y[n-1] + x[n] $$ which has an impulse response of $$ h[n] = \left(\frac{1}{2}\right)^n u[n] $$ which is definitely stable.

The problem you're seeing is that, for continuous-time systems (those described by differential equations), the stability criterion is that the poles are in the left-half plane (i.e. negative real axis).

  • 2
    $\begingroup$ Beat you to it by 45 seconds ... :) +1 $\endgroup$
    – Matt L.
    Oct 10, 2018 at 16:22
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    $\begingroup$ @MattL. Curse you, Red Baron! As Snoopy says when he's pretending to be an ace pilot. $\endgroup$
    – Peter K.
    Oct 10, 2018 at 16:24
  • $\begingroup$ I would suggest changing the word "conflating" to "confusing", or something less accusatory in tone $\endgroup$
    – Robert L.
    Oct 10, 2018 at 18:18
  • $\begingroup$ @CarlosDanger To me "confusing" is more accusatory. "Conflating" just means "flowing together" (of ideas), which is what I was after... Any other suggestions? Feel free to edit! :-) $\endgroup$
    – Peter K.
    Oct 10, 2018 at 19:50
  • $\begingroup$ @Downvoter? Any reason for the down vote? $\endgroup$
    – Peter K.
    Oct 11, 2018 at 11:55

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