# Why is the ROC of Laplace transform independent of imaginary part of s?

An integral is defined as converging if it yields a finite value upon application of limits of integration. It is divergent otherwise.

Now sticking to the mathematical notation of Laplace transform, we have for a causal function $$x(t) = u(t)$$: $$X(s) = \int_0^\infty x(t)e^{-st}dt = \int_0^\infty e^{-st}dt$$

$$X(s) = -\frac{1}{s}e^{-st}\Biggr|_{0}^{\infty} = \frac{1}{s} = \frac{1}{(\sigma + jw)}$$

By multiplying numerator and denominator with $$(\sigma - jw)$$, we get: $$X(s) = \frac{(\sigma - jw)}{(\sigma^2 +\omega^2)}$$

For laplace transform not to exist, the denominator must become 0. Hence in this contrived example, both $$\sigma$$ and $$w$$ must be 0.

Conversely, if $$\sigma = 0$$ and $$w \neq 0$$, $$X(s)$$ exists as $$\frac{-j}{\omega}$$ with a magnitude of $$\frac{1}{\omega}$$.

Unless my basics are messed up, why in the literature is $$jw$$ disregarded as influencing $$X(s)$$? In other words, why is the ROC dependent only on $$\sigma$$?

I think your misunderstanding is that you manipulate an expression, which is only valid for $$\text{Re}\{s\}>0$$, in order to show for which values of $$s$$ it might be valid.

Note that the integral

$$\mathcal{L}\{u(t)\}=\int_0^{\infty}e^{-st}dt$$

only converges for $$\text{Re}\{s\}>0$$, so the result $$1/s$$ does not make any sense for other values of $$s$$.

Only the real part of $$s$$ in the term $$e^{-st}$$ inside the integral can provide the necessary damping such that the integrand decays sufficiently fast for the integral to converge. That's why the region of convergence only depends on the real part of $$s$$.

• Ok. In the end, the laplace transform is fourier transform of a signal multiplied by exp(-sigma). So sigma should be chosen such that the integration of the original signal multiplied by exp(-sigma) yields a finite value. In this example, if sigma was negative, then the integrand is an ever increasing signal leading to divergence.
– Raj
Apr 20, 2019 at 3:16