Topic: Control Systems
Given $f(t)=L^{-1}\left[\frac{3s+1}{s^{3}+4s^{2}+(k-3)s}\right]$. If $\lim_{t\rightarrow\infty}f(t)=1$ then the value of k is
- A. 1
- B. 2
- C. 3
- D. 4
Answer: D. 4
Explanation: (Note: The term $(k-3)$ in the denominator is presumed to be $(k-3)s$ to yield a valid non-infinite steady state based on FVT).
Using the Final Value Theorem: $\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$.
$sF(s) = s \left[ \frac{3s+1}{s(s²+4s+(k-3))} \right] = \frac{3s+1}{s²+4s+(k-3)}$.
Taking the limit as $s \to 0$, we get $\frac{1}{k-3}$.
Given $\lim_{t \to \infty} f(t) = 1$, we have $\frac{1}{k-3} = 1 \Rightarrow k-3 = 1 \Rightarrow k=4$.