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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:

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

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

$\dot{y}-\frac{1}{2}y=\dot{x}$

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

$y_{\delta}(t)=c_1e^{\frac{1}{2}t}$

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

$H(z)=\sum\limits_{n=0}^{\infty}2^{-n}z^{-n}$

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?

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    $\begingroup$ z^-1 is a delay operator s is a differentiation operator $\endgroup$ – Stanley Pawlukiewicz Oct 10 '18 at 16:22
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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:

$$y[n]=\frac12y[n-1]+x[n]\tag{1}$$

The corresponding impulse response is

$$h[n]=\left(\frac12\right)^nu[n]\tag{2}$$

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 $|z|=1$.

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  • $\begingroup$ 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? $\endgroup$ – Antoine Post Oct 10 '18 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 '18 at 19:41
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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:

$$\dot{y}-\frac{1}{2}y=\dot{x}$$

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).

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    $\begingroup$ Beat you to it by 45 seconds ... :) +1 $\endgroup$ – Matt L. Oct 10 '18 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 '18 at 16:24
  • $\begingroup$ I would suggest changing the word "conflating" to "confusing", or something less accusatory in tone $\endgroup$ – Robert L. Oct 10 '18 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 '18 at 19:50
  • $\begingroup$ @Downvoter? Any reason for the down vote? $\endgroup$ – Peter K. Oct 11 '18 at 11:55

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