Understanding \$ j \omega \$ axis in transfer function
We start with
$$H(s)$$
in the frequency domain, and since $$s =\sigma+j\omega$$
We can replace s with
$$H(\sigma+j\omega)$$
But, in steady state the real part (sigma) is zero, so we get,
$$H(j\omega)$$
To clarify the &quot;transient&quot; (sigma) vs. &quot;steady-state&quot; (omega) , here is damped cosine,
$$h(t)=e^{-0.8t}cos(5t)$$

Re-writing with only exponentials (using euler),
$$h(t)=e^{-0.8t}(\frac{e^{j5t}+e^{-j5t}}{2})$$
Here you can see the exponential with the real value (-0.8) is the transient influence, and those with the imaginary (j5 and -j5) are steady-state.  In this case the steady-state settles out to zero, so i'll give another example that does not.
Let,
$$f(t)=e^{-0.8t}cos(5t)-cos(5t)$$

Re-writing with all exponentials,
$$f(t)=e^{-0.8t}(\frac{e^{j5t}+e^{-j5t}}{2})-(\frac{e^{j5t}+e^{-j5t}}{2})$$
For completeness, here are the Laplace transforms of the above two time functions h(t) and f(t),
$$\mathcal{L}[h(t)]=H(s)=\frac{s+0.8}{s^2+1.6s+25.64}$$
and,
$$\mathcal{L}[f(t)]=F(s)=\frac{0.8s^2+0.64s-20}{(s^2+25)(s^2+1.6s+25.64)}$$
and now they are in the s-domain and you can analyze them with frequency domain tools/methods.