La Place Transforms

NTTF

CONTROL SYSTEMS

LAPLACE TRANSFORMS

Control Systems - Laplace transforms 2NTTF

Introduction• A linear real-time control system can be replaced

with a mathematical model for the purpose of analysis.

• The model is in the form of linear differential equations.

• Solutions of these differential equations completely describe the control system characteristics, including the transient response.

• Several techniques, available in calculus for obtaining the solutions of these differential equations, some of which are quite demanding.


Introduction• Laplace transformation is a method that allows the

solutions of linear differential equations to be obtained without much complexity.

• In this section, the concept of Laplace transforms is introduced, and some of the transform properties are discussed.

• A rigorous treatment of transforms is not intended. The emphasis is on using a transform table to perform forward and inverse transformations for determining the response of control systems.


Transformations• The Laplace transform is a method of operational

calculus that takes a function of time (time domain) and converts it into a function of complex variable s (frequency domain, or s domain).

• In some ways, using a Laplace transform is analogous to logarithmic transformation.

• Before the advent of calculators and computers, logarithms were used for performing multiplication, division, and exponential calculations.

• These two processes are fairly comparable.


Logarithmic Process• Logarithm tables are used to convert numerical

terms.

• Arithmetic operations are carried out to reduce the expression to a single numerical term.

• Logarithm tables are used to find antilogarithms, yielding the desired result.


Laplace Transformation Process• Laplace transform tables are used to transform

a differential equation.

• The transformed equation is simplified to isolate the desired variable.

• Inverse transformation is applied to obtain the desired time equation.


Laplace Transform• Laplace transforms are useful in control system

analysis. • The time response of a control system can be

obtained by first applying a Laplace transform and then taking its inverse transform.

• Because the forward transformation leads into a frequency domain, the frequency response of a control system can be obtained directly from the transformed expression.

• The transfer function of a control system is defined in s domain and provides valuable information about stability and performance of a closed-loop system.


Laplace TransformThe Laplace transform of a function of time, f(t), is defined as the integral

This function is defined for every s, which results in convergence of the integral. The variable s is a complex variable (s =σ+ jw).


Laplace Transforms• Forward Laplace Transformation

– The process of converting a time domain function into s domain is known as forward Laplace transformation, or simply forward transformation.

• Inverse Laplace Transformation– The process of converting an s-domain function

back into a time-domain function is called the inverse Laplace transformation, or simply inverse transformation.


Transform Notation• The forward Laplace transformation process is

indicated by the letter L;– for example, L(f(t)) = F(s).

• The inverse Laplace transformation process is indicated by L -1; – for example, f(t) ==L–1 (F(s)).

• Lowercase letters with or without a t in parentheses are used to indicate functions of time; – for example, f(t), x(t), y(t), f, x, and y.


Transform Notation• Uppercase letters with an s in parentheses are

used to indicate transformed functions; – for example, F(s), X(s), and Y(s).

• Uppercase letters without a t or an s in parentheses are used to indicate constants; – for example, F, X, and Y.


Examples

The following are some examples of constant, time-domain, and s-domain terms.

a. Y constantb. x(t) time-domain functionc. v time-domain functiond. X(s) s-domain functione. E constantf. Z(s) s-domain function


Rules of Transformation• A number of rules have been developed to aid

in performing forward and inverse Laplace transformation.


Rule 1• Multiplication (or Division) by a Constant

– When a function is multiplied by a constant, the Laplace transformation is the product of the original transform and the constant.

– Let K be a constant and let L(f(t)) = F(s), then

L[Kf(t)] = KF (s)

• Similarly, if C is a constant and L[f(t)] = F(s), then


Rule 2• Sum (or Difference) of Two Functions

– The transform of a sum of two time functions is equal to the sum of their individual transforms.

– Let L[f1(t)] = F1(s),and L[f2(t)] = F2(s),then

– L[ f1(t) + f2(t)] = L[ f1(t)] + L[ f2(t)]

– = F1(s) + F2(S)


Rule 3• Derivative of a Function

– First Derivative• The transform of the first derivative of a time function

is given as

where F(s) =L[f(t)] and the term f(O) is the value of function f(t) at time t = 0 (also known as the initial value).


Rule 3• Second Derivative

– The transform of the second derivative of a time function is given as

where F(s) = L[f(t)], f(O) is the value of function f(t) at time t =0 (initial value), and df(O)/dt is the value of the first derivative of function f(t) at time t = o.


Note on rules

For a function with zero initial values, the transformation simplifies to


Example• The current i(t) flowing into a 1-µF capacitor

is related to capacitor voltage v through the following equation.

• Determine the Laplace transform of the current. Initially, the capacitor has no voltage across it (v(O) = 0, zero initial condition).


Solution• Take the Laplace transform of both sides of

the equation

• Because the initial value of capacitor voltage v(O) is zero, the final expression is

1(s) = sV(s)


Example• Repeat above example, assuming that the

capacitor was initially (at time t = 0) charged, with the voltage across capacitor being 1.5 V.


SolutionFrom the previous example, the transformed equation is

I(s) = sV(s) – v(O)

Substituting the value of v(O) (initial condition),

I(s) = sV(s) – 1.5


Rule 4• Integral of a Function

– The transform of the first integral of a time function is given as

– where F(s) = L[f(t)], f(O)is the value of function f(t) at time t =0 (initial value), and ∫f(O)dt is the value of integral of function at time t = 0 (initial value). For functions with zero initial values, the transformation simplifies to


Example• Find the Laplace transform of the following.

Assume zero initial conditions.– a. ∫ i(t)dt– b. 10∫ i(t)dt


Solutiona. Here i(t) is the time function. The Laplace transform of i(t) is


Solutionb. Because 10 is a constant, it is transparent to the transformation process (rule 2).


Rule 5 • Initial Value Theorem

– The initial value (t → 0) of a time function f(t) whose transform is F(s) is given by the following limit:

– This theorem is useful in determining the initial value of the function f(t) from F(s) without performing the inverse Laplace transformation.


Example• Current through a series RL circuit, when subjected to an

applied voltage, is given by , the following equation

– Where E= Battery voltage (10V)

– R=series resistance (100Ω)

– L= series inductance (10mH) • Determine the (instantaneous) value of current immediately

after the application of voltage. Assume that before the voltage was applied, no current was flowing in the circuit and that there was no magnetic field present across the coil.


Solution Because the Laplace transform of the current l(s) is given, the initial value of the current can be obtained simply by applying the initial value theorem.

This is the expected result, because inductance offers almost an infinite resistance to current buildup from zero (initial state) value.


Rule 6 • Final Value Theorem

– The final value (t → ∞) of a time function f(t) with transform F(s) is given by the following limit:

– This theorem is useful in determining the final value (steady state) of the function f(t) from F(s) without performing the inverse Laplace transformation.


Example• . Determine the final (steady-state) value of

current in last example


Solution

The final value of current in the circuit (time t → ∞) can be obtained by applying the final value theorem.

La Place Transforms

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