Initial and 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: vikramarka

Completed M.Tech in Digital Electronics and Communication Systems and currently working as a faculty.