Initial Value Theorem of Laplace Transform

---

---

Laplace Transform

The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain or s-domain.

Mathematically, if x(t) is a time domain function, then its Laplace transform is defined as −

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \int_{-\infty}^{\infty}\: x(t) e^{-st} \: dt\:\:\dots\:(1)}$$

Equation (1) gives the bilateral Laplace transform of the function x(t). But for the causal signals, the unilateral Laplace transform is applied, which is defined as −

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \int_{0}^{\infty}\: x(t) e^{-st} \: dt\:\:\dots\:(2)}$$

Initial Value Theorem

The initial value theorem of Laplace transform enables us to calculate the initial value of a function x(t) [i.e., x(0)] directly from its Laplace transform X(s) without the need for finding the inverse Laplace transform of X(s).

Statement

The initial value theorem of Laplace transform states that, if

$$\mathrm{x(t) \:\overset{LT}\longleftrightarrow\: X(s)}$$

Then,

$$\mathrm{\lim_{t \:\to\: 0} \:x(t) \:=\: x(0) \:=\: \lim_{s \:\to\: \infty}\: s X(s)}$$

Proof

From the definition of unilateral Laplace transform, we have,

$$\mathrm{L[x(t)] \:=\: X(s) \:=\: \int_{0}^{\infty}\: x(t) e^{-st} \: dt}$$

Taking differentiation on both sides, we get,

$$\mathrm{L\left[ \frac{dx(t)}{dt} \right] \:=\: \int_{0}^{\infty}\: \frac{dx(t)}{dt}\: e^{-st} \: dt}$$

By the time differentiation property $\mathrm{\left[i.e..,\:\frac{dx(t)}{dt}\: \overset{LT}\longleftrightarrow\: s X(s) \:-\: x(0^-)\right]}$ of Laplace transform, we get,

$$\mathrm{L\left[ \frac{dx(t)}{dt} \right]\:=\:\int_{0}^{\infty}\:\frac{dx(t)}{dt}\:e^{-st}\:dt\:=\:s X(s)\:-\:x(0^-)}$$

Now, taking $\mathrm{\lim_{s\:\to\:\infty}}$ on both sides, we have,

$$\mathrm{\lim_{s\:\to\: \infty} \left\{ \int_{0}^{\infty} \:\frac{dx(t)}{dt}\: e^{-st} \: dt \right\} \:=\: \lim_{s\:\to \: \infty} \left\{ s X(s) \:-\: x(0) \right\}}$$

$$\mathrm{\Rightarrow\: 0 \:=\: \lim_{s \:\to\: \infty} s X(s) \:-\: x(0)}$$

$$\mathrm{\Rightarrow \:x(0) \:=\: \lim_{s\: \to\: \infty} s X(s)}$$

Therefore, we have,

$$\mathrm{\lim_{t\:\to\:0} x(t) \:=\: x(0) \:=\: \lim_{s\: \to\: \infty}\: s X(s)}$$

Numerical Example

First determine x(t) and then verify the initial value theorem of the function given by,

$$\mathrm{X(s) \:=\: \frac{1}{s\:+\:3}}$$

Solution

The given function is,

$$\mathrm{X(s) \:=\: \frac{1}{s\:+\:3}}$$

Taking inverse Laplace transform of X(s), we have,

$$\mathrm{x(t) \:=\:L^{-1} \left[ X(s) \right]\:=\: L^{-1} \left[ \frac{1}{s\:+\:3} \right]}$$

$$\mathrm{\Rightarrow\: x(t) \:=\: e^{-3t}}$$

Therefore, the initial value of the function is,

$$\mathrm{x(0) \:=\: \left[ x(t) \right]_{t\:=\:0}}$$

$$\mathrm{\Rightarrow\: x(0) \:=\: \left[ e^{-3t} \right]_{t\:=\:0} \:=\: e^0 \:=\: 1}$$

Again, by the initial value theorem, we obtain,

$$\mathrm{x(0) \:=\: \lim_{s\:\to\:\infty}\: s X(s)\:=\:\lim_{s\:\to\:\infty}\:s \left[ \frac{1}{s\:+\:3} \right]}$$

$$\mathrm{\Rightarrow\: x(0) \:=\: \lim_{s\:\to\:\infty} \left[ \frac{s}{s\:+\:3} \right] \:=\: 1}$$

Hence, the initial value theorem is verified for the given function.