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).