Stability of system with poles inside unit circle - conflict with differential equation

Question

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?

Answer 1

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 $|z|=1$.

Answer 2

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).