# Region of convergence of transfer function

I posted this question Mathematics SE and got no answer so I have posted it here.

I learned in my signal processing class that an LTI system can be defined using a linear constant coefficient differential equation. Whenever we have 'initial rest' condition, the LTI system is causal. Initial rest condition is basically the output $$y(t)$$ is $$0$$ till $$x(t)$$ becomes non-zero for the first time.

$$\sum_{i=0}^N a_i y^{(i)} = \sum_{i=0}^M b_i x^{(i)} \hspace{1cm} initial \hspace{0.25cm} rest \hspace{0.25cm} condition$$

We get the Laplace transform of its transfer function as

$$H(s) = \frac{\displaystyle\sum_{i=0}^M b_i s^{i}}{\displaystyle\sum_{i=0}^N a_i s^{i}}$$

and more importantly, its ROC is the half plane right of the rightmost pole (condition for causality).
My question is how can I modify the initial condition to get any ROC I desire? For example, what should be the initial condition if I want my ROC to be left half plane, left of the leftmost pole? What if I want my ROC to be a vertical strip between two adjacent poles having real parts as $$\alpha$$ and $$\beta$$, ROC $$= \{s| \alpha < \textrm{Re}\{s\} < \beta\}$$?
I have no idea how to approach this problem.

• As a hint consider why it is called the “region of convergence”: try to solve the Laplace Transform for a specific s that is not in the ROC and convince yourself that the answer is unbounded. Next consider the case of an anti-causal system (0 for positive time) and repeat that experiment. This should give you further insight to your question. – Dan Boschen Apr 21 at 13:59