# Initial-value & Final -value Theorems:-

Initial-value Theorem:-

Use:- to find out the initial value of a signal x(t) without using inverse Laplace Transform.

$x(0^{-})=&space;\lim_{t\rightarrow&space;0^{-}}x(t)=\lim_{S\rightarrow&space;\infty&space;}s\&space;X(S)$.

Proof:-

we know that  $L\left&space;\{&space;\frac{dx(t)}{dt}&space;\right&space;\}\leftrightarrow&space;S\&space;X(S)-x(0^{-})$.

$L\left&space;\{&space;\frac{dx(t)}{dt}&space;\right&space;\}=\int_{0^{-}}^{\infty&space;}&space;\frac{dx(t)}{dt}\&space;e^{-St}\&space;dt$.

i.e,       $\int_{0^{-}}^{\infty&space;}&space;\frac{dx(t)}{dt}\&space;e^{-St}\&space;dt=S\&space;X(S)-x(0^{-})$ .

$\lim_{s\rightarrow&space;\infty&space;}\&space;\int_{0^{-}}^{\infty&space;}&space;\frac{dx(t)}{dt}\&space;e^{-St}\&space;dt=&space;\lim_{s\rightarrow&space;\infty&space;}\&space;S\&space;X(S)-&space;\lim_{s\rightarrow&space;\infty&space;}\&space;x(0^{-})$ .

$0=&space;\lim_{s\rightarrow&space;\infty&space;}\&space;S\&space;X(S)-&space;\lim_{s\rightarrow&space;\infty&space;}\&space;x(0^{-})$ .

$\lim_{s\rightarrow&space;\infty&space;}\&space;S\&space;X(S)=&space;\lim_{s\rightarrow&space;\infty&space;}\&space;x(0^{-})$ .

$\&space;x(0^{-})&space;=&space;\lim_{s\rightarrow&space;\infty&space;}\&space;S\&space;X(S)$ .

Hence proved.

final-value Theorem:-

Use:- to find out the final value of a signal x(t) without using inverse Laplace Transform.

$x(\infty&space;)=&space;\lim_{t\rightarrow&space;\infty&space;}x(t)=\lim_{S\rightarrow&space;0&space;}s\&space;X(S)$.

Proof:-

we know that  $L\left&space;\{&space;\frac{dx(t)}{dt}&space;\right&space;\}\leftrightarrow&space;S\&space;X(S)-x(0^{-})$.

$L\left&space;\{&space;\frac{dx(t)}{dt}&space;\right&space;\}=\int_{0^{-}}^{\infty&space;}&space;\frac{dx(t)}{dt}\&space;e^{-St}\&space;dt$.

i.e,       $\int_{0^{-}}^{\infty&space;}&space;\frac{dx(t)}{dt}\&space;e^{-St}\&space;dt=S\&space;X(S)-x(0^{-})$ .

$\lim_{s\rightarrow&space;0&space;}\&space;\int_{0^{-}}^{\infty&space;}&space;\frac{dx(t)}{dt}\&space;e^{-St}\&space;dt=&space;\lim_{s\rightarrow&space;0&space;}\&space;S\&space;X(S)-&space;\lim_{s\rightarrow&space;0&space;}\&space;x(0^{-})$ .

$x(\infty&space;)-x(0^{-})=&space;\lim_{s\rightarrow&space;0&space;}\&space;S\&space;X(S)-&space;\lim_{s\rightarrow&space;0&space;}\&space;x(0^{-})$ .

$x(\infty&space;)-x(0^{-})=&space;\lim_{s\rightarrow&space;0&space;}\&space;S\&space;X(S)-&space;x(0^{-})$

$x(\infty&space;)=&space;\lim_{s\rightarrow&space;0&space;}\&space;S\&space;X(S)$ .

Hence proved.

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## Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems.

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