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.

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

Proof:-

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

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

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

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

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

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

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

Hence proved.

final-value Theorem:-

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

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

Proof:-

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

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

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

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

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

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

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

Hence proved.

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Author: Lakshmi Prasanna Ponnala

Completed M.Tech in Digital Electronics and Communication Systems. View all posts by Lakshmi Prasanna Ponnala

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