# 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: $$\begin{gather} x[n] = (1/4)^nu[-n-1] \\[5pt] h[n] = (1/2)^nu[n]\\[5pt] ROC(X): |z|<1/4\\[5pt] ROC(H): |z|>1/2\\[10pt] y[n] = \sum\limits_{k=-\infty}^{\infty} (1/4)^ku[-k-1](1/2)^{n-k}u[n-k]\\[5pt] y[n]= (1/2)^n\sum\limits_{k=-\infty}^{-1} (1/2)^ku[n-k]\\[5pt] y[n] = (1/2)^n\sum\limits_{k=1}^{\infty} 2^ku[n+k]\\[5pt] \text{Clearly } y[n] \text{ is not defined} \end{gather}$$ 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.]