Disjoint ROCs of input and system function I am trying to understand how the output behaves when the input and the system function do not have a common region of convergence (ROC) for an LSI system.
Consider an LSI system with $x[n]$, $h[n]$, $y[n]$ as input, impulse response, output signal respectively:
We know that
\begin{gather}
Y(z) = X(z)\cdot H(z)\\[5pt]
\text{with } \ ROC(Y) \supseteq ROC(X) \ \cap \ ROC(H)
\end{gather}
So what if $ROC(X) \ \cap \ ROC(H)=\phi$, what would be the output? Is it even properly defined?
I tried with some examples considering $x[n]$ and $h[n]$ with disjoint ROCs, the output is not defined,
here is one example:
\begin{gather}
x[n] = (1/4)^nu[-n-1] \\[5pt]
h[n] = (1/2)^nu[n]\\[5pt]
ROC(X): |z|&lt;1/4\\[5pt]
ROC(H): |z|&gt;1/2\\[10pt]
y[n] = \sum\limits_{k=-\infty}^{\infty} (1/4)^ku[-k-1](1/2)^{n-k}u[n-k]\\[5pt]
y[n]= (1/2)^n\sum\limits_{k=-\infty}^{-1} (1/2)^ku[n-k]\\[5pt]
y[n] = (1/2)^n\sum\limits_{k=1}^{\infty} 2^ku[n+k]\\[5pt]
\text{Clearly } y[n] \text{ is not defined}
\end{gather}
My question is whether this is true in every case that the output would not be defined or there are some special $x[n]$ and $h[n]$ with disjoint ROCs where $y[n]$ is properly defined.
If the ROCs don't overlap, the corresponding convolution sum doesn't converge, hence the output is undefined. Note that in your question it is assumed that the $\mathcal{Z}$-transforms of the input as well as of the system's impulse response both exist.
You can construct examples for which neither the $\mathcal{Z}$-transform of the input nor of the output exist, yet the input and output signals are perfectly defined. E.g., the input sequence $x[n]=A\cos(\omega_0n)$ has no $\mathcal{Z}$-transform, neither has the corresponding output sequence $y[n]=B\cos(\omega_0n+\phi)$, where $B$ and $\phi$ depend on the system's frequency response. [In this example it is assumed that the ROC of the system's transfer function $H(z)$ includes the unit circle, i.e. the system is stable.]