Question

Find the initial value, if it exists of the signal with Laplace transform given below:
\[ X(s) = \frac{s + 4}{s^2 + 3s + 5} \]

• A. 1
• B. 4/5
• C. 5/4
• D. \infty

Hint:

To find the initial value of a signal, multiply its Laplace transform by \( s \) and take the limit as \( s \to \infty \).

Answer 1

The Correct Option is A

To find the initial value of a signal from its Laplace transform, we use the formula: \[ x(0) = \lim_{s \to \infty} sX(s) \] Substituting the given \( X(s) = \frac{s + 4}{s^2 + 3s + 5} \): \[ x(0) = \lim_{s \to \infty} \frac{s(s + 4)}{s^2 + 3s + 5} \] As \( s \to \infty \), the highest power of \( s \) in both the numerator and denominator is \( s^2 \), so the expression simplifies to: \[ x(0) = 1 \] Thus, the initial value is 1, and the correct answer is option (1).