Final answer:
The steady-state for the signal with the given transform V(s) is zero, as any transient behavior eventually dies out. The poles of the transform suggest that the signal will contain decaying exponential components over time.
Step-by-step explanation:
To find the steady-states for signals with the given transform V(s) = 20 / (s(2s+5)), we must consider the inverse Laplace transform. The steady-state of a system in the Laplace domain typically corresponds to the behavior of the signal as t → ∞. In this case, as the system reaches steady-state, the exponential terms go to zero, leaving the constant terms. Since the transform V(s) consists of simple polynomial terms in s, and there's a term with s in the denominator, the stead-state value as t → ∞ is zero for this particular function.
Characterising the corresponding signals involves looking at the poles of the transform function. The poles are at s = 0 and s = -2.5. This indicates that there are components of the signal that will decay exponentially over time, with a constant component due to the s = 0 pole that will persist in the absence of other factors. Since the pole at s = 0 determines the final steady-state value, and it's a simple pole, the resulting steady-state value of the time-domain signal is zero. Thus, any initial transient behavior in the signal will eventually die out, leaving a steady-state signal of zero.