What is the initial value theorem in Laplace domain?

What is the initial value theorem in Laplace domain?

• A. lim_{t→0+} x(t) = lim_{s→∞} s X(s)
• B. lim_{t→0+} x(t) = lim_{s→0} X(s)
• C. lim_{t→0+} x(t) = lim_{s→∞} X(s)/s
• D. lim_{t→0+} x(t) = lim_{s→0} s X(s)

Answer: (A)

The initial value is captured by looking at how the Laplace transform behaves when s is very large. Since X(s) = ∫0^∞ x(t) e^{-st} dt, a large s makes the exponential decay extremely fast, so the integral is dominated by the value of x just after t = 0, namely x(0+). If x(t) is well-behaved at 0 (piecewise continuous and of exponential order), we can approximate X(s) ≈ x(0+)/s for large s. Multiplying by s then gives s X(s) → x(0+), so the initial value theorem states lim_{t→0+} x(t) = lim_{s→∞} s X(s). This is the correct relationship for the start of the signal. The other limits don’t isolate the initial value: for example, taking s to zero relates to the final value under the final-value interpretation, and dividing by s or taking s to infinity of X(s) does not recover x(0+).

The initial value is captured by looking at how the Laplace transform behaves when s is very large. Since X(s) = ∫0^∞ x(t) e^{-st} dt, a large s makes the exponential decay extremely fast, so the integral is dominated by the value of x just after t = 0, namely x(0+). If x(t) is well-behaved at 0 (piecewise continuous and of exponential order), we can approximate X(s) ≈ x(0+)/s for large s. Multiplying by s then gives s X(s) → x(0+), so the initial value theorem states lim_{t→0+} x(t) = lim_{s→∞} s X(s). This is the correct relationship for the start of the signal. The other limits don’t isolate the initial value: for example, taking s to zero relates to the final value under the final-value interpretation, and dividing by s or taking s to infinity of X(s) does not recover x(0+).