# Final 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 −

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

## Final Value Theorem

The final value theorem of Laplace transform enables us to find the final value of a function$\mathit{x}\mathrm{(\mathit{t})}$[i.e.,$\mathit{x}\mathrm{(\infty)}$] directly from its Laplace transform X(s) without the need for finding the inverse Laplace transform of X(s).

### Statement

The final value theorem of Laplace transform states that, if

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

Then,

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

### Proof

From the definition of the unilateral Laplace transform, we have,

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

Taking differentiation with respect to time on both sides, we get,

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

Now, 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 have,

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

By taking the $\lim_{s \rightarrow 0}$ on both sides of the above expression, we get,

$$\lim_{s \rightarrow 0}[\int_{\mathrm{0}}^{\infty}\mathit{\frac{\mathit{dx}\mathrm{(\mathit{t})}}{\mathit{dt}}\mathit{e^{-st}}}\:\mathit{dt}]\:\mathrm{=}\:\lim_{s \rightarrow \infty}[\mathit{sX}\mathrm{(\mathit{s})}-\mathit{x}\mathrm{(0^-)}]$$ $$\Rightarrow\int_{\mathrm{0}}^{\infty}\mathit{\frac{\mathit{dx}\mathrm{(\mathit{t})}}{\mathit{dt}}}\:\mathit{dt}\:\mathrm{=}\:\lim_{s \rightarrow 0}[\mathit{sX}\mathrm{(\mathit{s})}-\mathit{x}\mathrm{(0^-)}]$$ $$\Rightarrow[\mathit{x}\mathrm{(\mathit{t})}]_0^\infty\:\mathrm{=}\:\lim_{s \rightarrow 0}[\mathit{sX}\mathrm{(s)}-\mathit{x}\mathrm{(0^-)}]$$ $$\Rightarrow\mathit{x}\mathrm{(\infty)}-\mathit{x}\mathrm{(0^-)}\:\mathrm{=}\:\lim_{s \rightarrow 0}[\mathit{sX}\mathrm{(\mathit{s})-\mathit{x}\mathrm{(0^-)}}]$$ $$\Rightarrow\mathit{x}\mathrm{(\infty)}\:\mathrm{=}\:\lim_{s \rightarrow 0}\mathit{sX}\mathrm{(\mathit{s})}$$

Therefore, we have,

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

Note − In order to apply the final value theorem of Laplace transform, we must cancel the common factors, if any, in the numerator and denominator of sX(s).If any poles of sX(s) after cancellation of the common factor lie in the right half of the s-plane, then the final value theorem does not hold.

## Numerical Example

First Determine $\mathit{x}\mathrm{(\mathit{t})}$ and then verify the final value theorem for the function $\mathit{X}\mathrm{(\mathit{s})}\:\mathrm{=}\;\mathrm{1}/\mathrm{(\mathit{s}+3)}$.

### Solution

The given function is,

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

Taking the inverse Laplace transform, we have,

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

Therefore, the final value of the given function is,

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

Now, using the final value theorem, we obtain,

$$\mathit{x}\mathrm{(\infty)}\:\mathrm{=}\lim_{s \rightarrow 0}\mathit{sX}\mathrm{(s)}\:\mathrm{=}\:\lim_{s \rightarrow 0}\mathit{s}\begin{bmatrix} \frac{\mathrm{1}}{\mathrm{(\mathit{s}+{\mathrm{3}})}} \end{bmatrix}$$ $$\Rightarrow\mathit{x}\mathrm{(\infty)}\:\mathrm{=}\:\lim_{s \rightarrow0}\begin{bmatrix}\frac{\mathrm{1}}{\mathrm{(1+\frac{3}{\mathit{s}})}} \end{bmatrix}\:\mathrm{=}\:\mathrm{0}$$

Hence, the final value theorem for the given function is proved.

Updated on: 07-Jan-2022

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