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In control system design it is necessary, to analyse the performance and stability
of a proposed system before it is built or implemented. Many analysis
techniques use transformed variables to facilitate mathematical treatment of the problem. In
the analysis of continuous time dynamical systems, this generally involves the use of Laplace Transforms
A table of laplace transforms is available to transform real time domain variables to
laplace transforms. The necessary operations are carried out and the laplace transforms obtained
in terms of s are then inverted from the s domain to the time (t) domain. This tranformation
from the s to the t domain is called the inverse transform...
Laplace Transform Operations.
Table showing selection of Laplace Transforms.
Derivation of table values
examples on how the table values have been derived are provided below.
From above the transform for a unit step i.e f(1) is easily obtained by setting a = 0 ( e 0.t) = 1)
From above the transform for cos ωt and sin ωt is obtained by setting a = jω
Laplace transform example
An example of using Laplace transforms is provided below
The manipulation of the laplace tranform equation into a form to enable a convenient inverse transform often involves use of partial fractions...
Partial Fraction Expansion...Notes are also provided for partial fractions on webpage Partial Fractions
The splitting up of a ratio of polynomials is often necessary to produce simpler ratios
from which inverse Laplace transforms are more conveniently obtained. The most favoured
procedure for converting using hands-on (as opposed to using computers) is the "Heaviside
cover up" procedure..
Partial Fraction Expansion process using the Heaviside cover up method
The Laplace operations generally result in a ratio
This must be proper in that the order of the denominator D(s) must be higher than the numerator N(s).
If the function is not proper then the numerator N(s) must be divided by the denominator using the long division method.
a 1,a 1 etc are the roots of D(s).
To obtain a 1 simply multiply both sides of the equation by (s - a 1 ) letting s = a 1 This results in all terms on the RHS becoming zero apart from a 1...
G(s).( s-a 1 )| s = a 1 = a 1
The LHS is multiplied be (s - a 1) thus cancelling out (s - a 1)in the denominator. ...
and all instances of s are then replaced by a 1
When the denominator has repeated roots the breakdown into partial fractions is treated differently as shown below...
The factor b 0 is obtained in exactly the same way as above..
The factors b 2 to b r is obtained by progressive differentiation of G(s)(s-b) and dividing the result by the factorial of the level of differentiation (if d/ds 2 then divide by 2!, if , d/ds 3 then divide by 3!(6))...
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Last Updated 28/01/2013