# Causality and ROC of a stable LTI system

So I am looking at a stable LTI system whose input is $x[n]$ and output is $y[n]$. The equation relating the two is here: $$y[n-1]-\frac{10}{3}y[n]+y[n+1]=x[n]$$

I was able to compute its system function $H(z)$ to be: $$H(z)=\frac{z^{-1}}{z^{-2}-\frac{10}{3}z^{-1}+1}$$

I then performed partial fraction expansion to end up with: $$H(z)=\frac{\frac 38}{\left(1-3z^{-1}\right)} - \frac{\frac 38}{\left(1-\frac 13z^{-1}\right)}$$

What I am struggling with is I can't decide what its ROC should be since the system is specified to be stable which means it has to include the $j\omega$-axis. I also need to determine if the system is causal or not. Since a system can only be causal and stable when all its poles are on the left hand plane, I am not sure.

Any help determining both the ROC and causality of the system would be very helpful.

• You should be talking about "unit circle" instead of $j\omega$-axis, because we're in discrete time. – Matt L. May 2 '17 at 8:10

There are 2 poles: one at $z=3$ (i.e., outside the unit circle) and one at $z=\frac13$ (i.e., inside the unit circle). For such a system there are three possible ROCs; one corresponds to an anti-causal system, another one to a causal system, and a third one to a non-causal system with a two-sided impulse response. But only one of those ROCs includes the unit circle, and that's the ROC corresponding to a stable system.
• @user28308: Including the unit circle means that $|z|=1$ must be part of the ROC. So I guess you got it right. – Matt L. May 2 '17 at 8:43
• @user28308: A causal system always has a ROC specified by $|z|>R$ with some real-valued $R$. Think of the Z-transform as a power series. If for a causal sequence this series converges for some $|z|=c$ then it must also converge for all $|z|>c$ (since the series has only negative powers of $z$). – Matt L. May 2 '17 at 8:53