Question

Given X(S)= (s+5)/s²(s+2)(s+4)

Apply Initial Value Theorem (IVT) and find x(0+)

Answer 1

Applying the Initial Value Theorem, the value of x(0+) is 0.

Step-by-step explanation:

Applying Initial Value Theorem (IVT) to find x(0+)

Initial Value Theorem (IVT):

For a Laplace transform X(s) of a function x(t) that satisfies certain conditions (including being zero for t > 0 and having no impulses or higher-order singularities at the origin), the following holds:

lim_{s -> ∞} sX(s) = x(0+)

Given:

X(s) = (s+5)/s²(s+2)(s+4)

Steps:

1. Apply the limit rule of the Laplace transform:

lim_{s -> ∞} sX(s) = lim_{s -> ∞} s[(s+5)/s²(s+2)(s+4)]

2. Simplify the expression:

lim_{s -> ∞} sX(s) = lim_{s -> ∞} [(s^2 + 5s)/s^3(s+2)(s+4)]

3. Apply the fact that any term with a lower power than the dominant term (s^3) will become zero in the limit:

lim_{s -> ∞} sX(s) = lim_{s -> ∞} [s^2/s^3] = 0

4. According to the IVT, this limit equals x(0+):

x(0+) = 0

Therefore, x(0+) = 0.