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 $\mathit{x}\mathrm{(\mathit{t})}$ is a time domain function, then its Laplace transform is defined as −

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

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

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

Initial Value Theorem

The initial value theorem of Laplace transform enables us to calculate the initial value of a function $\mathit{x}\mathrm{(\mathit{t})}$[i.e.,$\:\:\mathit{x}\mathrm{(0)}$] 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

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

Then,

$$\lim_{t \rightarrow \mathrm{0}}\mathit{x}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathit{x}\mathrm{(0)}\:\mathrm{=}\:\lim_{s \rightarrow \infty}\mathit{sX}\mathrm{(\mathit{s})}$$

Proof

From the definition of unilateral Laplace transform, we have,

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

Taking differentiation on both sides, we get,

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

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

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

Now, taking $\lim_{s\rightarrow\infty}$on both sides, we have,

$$\lim_{s \rightarrow \infty}\lbrace\:\int_{\mathrm{0}}^{\infty}\mathit{\frac{\mathit{dx}\mathrm{(\mathit{t})}}{\mathit{dt}}\mathit{e^{-st}}}\:\mathit{dt}\rbrace\:\mathrm{=}\:\lim_{s \rightarrow \infty}\lbrace\mathit{sX}\mathrm{(\mathit{s})}-\mathit{x}\mathrm{(0)}\rbrace$$ $$\Rightarrow\mathrm{0}\:\mathrm{=}\:\lim_{s \rightarrow \infty}\mathit{sX}\mathrm{(s)}-\mathit{x}\mathrm{(0)}$$ $$\Rightarrow\mathit{x}\mathrm{(0)}\:\mathrm{=}\:\lim_{s \rightarrow \infty}\mathit{sX}\mathrm{(s)}$$

Therefore, we have,

$$\lim_{t \rightarrow 0}\mathit{x}\mathrm{(\mathit{t})}\:\mathrm{=}\:\mathit{x}\mathrm{(0)}\:\mathrm{=}\:\lim_{s \rightarrow \infty}\mathit{sX}\mathrm{(s)}$$

Numerical Example

First determine $\mathit{x}\mathrm{(\mathit{t})}$ and then verify the initial value theorem of the function given by,

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

Solution

The given function is,

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

Taking inverse Laplace transform of $\mathit{X}\mathrm{(\mathit{s})}$, we have,

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

Therefore, the initial value of the function is,

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

Again, by the initial value theorem, we obtain,

$$\mathit{x}\mathrm{(0)}\:\mathrm{=}\:\lim_{s \rightarrow \infty}\mathit{sX}\mathrm{(\mathit{s})}\:\mathrm{=}\:\lim_{s \rightarrow\infty}\mathit{s}[\frac{\mathrm{1}}{\mathrm{(\mathit{s}+\mathrm{3})}}]$$ $$\Rightarrow\mathit{x}\mathrm{(0)}\:\mathrm{=}\:\lim_{s \rightarrow\infty}[\frac{\mathrm{1}}{\mathrm{(1+\frac{3}{\mathit{s}})}}]\:\mathrm{=}\:\mathrm{1}$$

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