Laplace’s Transform: Re-visit�
Presented by
Dr. Asim Halder
Dept. of Applied Electronics and Instrumentation Engineering
Haldia Institute of Technology
Basic Mathematics for Linear Control Systems
Pre-requisite
Pre-requisite for Linear Control Theory
simultaneously.
Pre-requisite
Definition of the Laplace Transform
(1.1)
Derivation of Laplace Transform of Standard Functions
The Laplace transform of the unit step-function u(t) is-
Pre-requisite
Since u(t) has the value ‘1’ over the limit of integration, so-
if
u(t)
0
t
1
The exponent ‘α’ is a positive real number.
(1.2)
if
(1.3)
1
t
0
Pre-requisite
Here,‘ω’ is a positive real number.
(1.4)
(1.5)
if
(1.6)
Pre-requisite
(1.7)
(1.8)
if
(1.9)
(1.10)
Pre-requisite
(1.11)
(1.12)
is
(1.13)
(1.14)
Pre-requisite
transformable then-
(1.15)
(1.16)
where is the value of the limit of the derivative of f(t) as the
origin t = 0 is approached from the right side.
Pre-requisite
(1.17)
One point to be noted here that the transform includes the initial conditions, where as in the classical method of solution, initial conditions are introduced separately to evaluate the coefficient of the solution of the differential equation. When all initial conditions are zero, the Laplace transform of the derivative of f(t) is simply .
transformable and the value of its transform is
(1.18)
The term is the constant of integration and is equal to the value of the integral as the origin is approached from the positive right side.
Pre-requisite
The transform of double integral is
(1.19)
The transform of the order integral is
(1.20)
If f(t) and Df(t) are Laplace transformable, if the Laplace transform of f(t) is F(s), and if the limit of f(t) as exists then
(1.21)
The theorem states that behavior of f(t) in the neighborhood of is related to the behavior of sF(s) in the neighborhood of s = 0. If sF(s) has
poles (values of ‘s’ for which becomes infinite) on the imaginary axis (excluding the origin) or in the right-half ‘s’ plane, there is no finite final value of f(t) and the theorem can not be used.
If f(t) is sinusoidal the theorem is invalid, since has poles at
and does not exist.
Pre-requisite
However, for poles of sF(s) at the origin, s = 0, this theorem gives the final value of
This correctly describes the behavior of f(t) as .
If the function f(t) and its 1st derivative are Laplace transformable, if the Laplace transform of f(t) is F(s), and if exists then
(1.22)
The theorem states that the behavior of f(t) in the neighborhood of t = 0 is related to the behavior of sF(s) in the neighborhood of . There are no limitations on the locations of the poles of sF(s).
If the Laplace transform f(t) is F(s) and if has a limit as
then
(1.23)
It states that the division by the variable in the real domain results in integration w. r. t. ‘s’ in s-domain.
Thank you