Initial Value Theorem of Laplace Transform

The initial value theorem is a useful property of the Laplace transform. The transform is named after French mathematician and physicist Pierre-Simon Laplace, whose work also covered celestial mechanics and probability. His name is one of the 72 engraved on the Eiffel Tower.
The initial value theorem (IVT) and final value theorem are limiting theorems. IVT finds the right-hand value f(0⁺) directly from F(s), so an inverse transform is not needed. The 0+ notation matters when a signal changes at the origin.

Conditions for the Initial Value Theorem

1. The relevant Laplace transforms must exist, and f(t) must have a finite right-hand limit at the origin.
2. As t approaches 0⁺, f(t) must approach a defined finite value.

1. For the causal form used here, f(t) = 0 for t < 0 and contains no impulse or higher-order singularity at the origin.

Statement of the Laplace Initial Value Theorem

If f(t) and F(s) are a Laplace-transform pair,

then the initial value theorem is:

Proof of the Laplace Initial Value Theorem

The Laplace transform of f(t) is:

The transform of its generalized derivative f ‘ (t) is:

Apply integration by parts to the integral term.

Substituting Equation 2 into Equation 1 gives:

The derivative rule includes f(0^–), and accounting consistently for any jump at the origin gives:

The lower limit 0^– includes a possible switching event at the origin. For an ordinary causal function without an impulse there, the result on the right side is the value approached from positive time.

Note:
This page uses the causal form of the Laplace transform, for which f(t) is zero before the origin.
Take the limit as s tends to positive infinity in Equation 3.

The exponential weighting then concentrates the transform near t = 0, which gives f(0+) and proves the theorem under the stated conditions.

Applications of the Initial Value Theorem

The initial value theorem determines f(0+) from F(s) without first calculating the full time-domain function.

Example 1:
Find the initial value of f(t) = 2u(t) + 3 cos(t)u(t).
Solution:

Apply the initial value theorem.

The initial value is 5.
Example 2:
Find the initial value represented by this transform.

Solution:

Apply the initial value theorem.

[as s → ∞ the values of s become more and more insignificant hence the result is obtained by simply taking the ratio of leading co-efficient] More precisely, the highest-power terms determine the limit. In this example, the result is the ratio of the relevant leading coefficients after multiplying F(s) by s.

Example 3:
Find the initial value represented by
Solution:
The ordinary-function form of the initial value theorem does not apply because this transform contains a singular term at the origin. The following two methods show why.
Start with the direct limit.
Method 1:

Note:
For a rational F(s) that represents an ordinary finite initial value, the numerator degree must be lower than the denominator degree. Polynomial terms in F(s) indicate impulses or derivatives of impulses at the origin.
Applying the ordinary-value formula here produces an unbounded result rather than a finite initial value.

[this is not possible in practical circuits ] This result does not represent a finite initial circuit variable.
Alternative method:


Apply the inverse Laplace transform.

The result contains an impulse at t = 0. An impulse is concentrated at the origin, so it does not have an ordinary finite value f(0+) to which this form of IVT can be applied.

In circuit analysis, this theorem provides a quick check on a transformed response. Confirm the theorem’s conditions before treating its limit as a physical initial voltage or current.