Final answer:
The final value of f(t) is 7.50 when evaluated using the final-value theorem on the function F(s) = 40(s+3)/s(s+4)².
Step-by-step explanation:
To apply the final-value theorem to the Laplace-transformed function F(s) = 40(s+3)/s(s+4)² we assume that the system is stable meaning that all poles of F(s) are in the left half of the complex plane (i.e., they all have negative real parts).
The final-value theorem states that the final value of f(t) as t approaches infinity is equal to the limit of s multiplied by F(s) as s approaches zero:
f(∞) = lims→0(s · F(s))
Plugging in our function, we get:
f(∞) = lims→0(s · 40(s+3) / s(s+4)²) = lims→0(40(s+3) / (s+4)²)
Since the s term in the numerator and the denominator cancel out, we are left with the limit of (s+3) as s approaches zero:
f(∞) = lims→0(40(s+3) / (4)²) = 40(3) / 16 = 120 / 16 = 7.5
Therefore the final value of f(t) expressed using three significant figures, is 7.50.