Question

The Laplace transform of a signal $X(t)$ is $$X(s) = \frac{4s + 1}{s^2 + 6s + 3}.$$ The initial value $X(0)$ is:

• A. $0$
• B. $4$
• C. $1/6$
• D. $4/3$

Hint:

For the Laplace transform $X(s)$, the Initial Value Theorem states: $$X(0) = \lim\limits_{s \to \infty} s X(s).$$

Answer 1

The Correct Option is D

Step 1: Use the initial value theorem. $$\lim\limits_{t \to 0} X(t) = \lim\limits_{s \to \infty} s X(s).$$ Step 2: Compute limit. $$\lim\limits_{s \to \infty} s \cdot \frac{4s + 1}{s^2 + 6s + 3}.$$ Dividing numerator and denominator by $s$: $$\lim\limits_{s \to \infty} \frac{4s^2 + s}{s^2 + 6s + 3} = \lim\limits_{s \to \infty} \frac{4 + \frac{1}{s}}{1 + \frac{6}{s} + \frac{3}{s^2}}.$$ Step 3: Evaluating the limit. $$\lim\limits_{s \to \infty} \frac{4}{1} = 4/3.$$ Step 4: Selecting the correct option. Since $X(0) = 4/3$, the correct answer is (D).