LEC 1 - Laplace transform and block diagrams PDF

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Laplace and Z transform model based dynamic systems and control, Laplace and Z transform model based dynamic systems and control, Laplace and Z transform model based dynamic systems and control...

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Control Systems Design Lecture 1

UFMFW7-15-3

Tool 1 (descriptive) --- Laplace transform and block diagrams

1 Introduction The full introduction of the Laplace transform can be found from following website http://en.wikipedia.org/wiki/Laplace_transform 1.1 What is the Laplace transform? Mathematically Laplace transform is defined as follows 

L f (t )  F (s )  f (t ) exp( st )dt (1.1) 0

Where f(t) is a function of time t such as f ( t ) t t 0 , s is a complex variable, L is the Laplace transform operator indicating that the quantity that it prefixes is to be transformed by the Laplace integral 

f (t ) exp(  st )dt

, F(s) is the Laplace transform of f(t).

0

The reverse process of finding the time function f(t) from the Laplace transform F(s) is called the inverse Laplace transform, which mathematically expressed L 1  F ( s )  f (t ) 

1 2j

c j

F ( s) exp(st )ds

(t  0)

(1.2)

c j

1.2 Why need the Laplace transform? The Laplace transform is an operational method that can be used advantageously for solving linear differential equations which represent most of linear dynamic systems. By use of Laplace transform, one can convert many common functions, such as sinusoidal functions, damped sinusoidal functions, exponential functions, into algebraic functions of a complex variable s. Operations such as differentiation and integration can be replaced by algebraic operations in the complex plane. Then the solution of a differential equation can be obtained by taking inverse Laplace transform. This process can be illustrated by following schematic diagram Analytical approach

DE (Differential Equation)

Solution

Inverse Laplace transform

Laplace transform Algebraic equation

Fig. 1 The procedures to obtain solution of a differential equation An advantage of using the Laplace transform is that it allows the effective utilisation of graphical techniques for predicting the system performance without actually solving the system differential

1

equations. Another advantage is that, when one solves the differential equation, both the transient components and steady state component of the solution can be obtained simultaneously. 2 Properties 2.1 Linearity (superposition principle)

L[ Af ( t)]  AL[ f ( t)]  AF( s)

L 1 [ AF( s)]  AL 1 [ F( s)]  Af ( t)

L[ f1 (t )  f 2 (t )]  F1 ( s ) F2 (s )

L 1 [ F1 (s ) F2 (s )]  f 1 (t )  f 2 (t )

where A is a constant. 2.2 Real differentiation theorem

d f (t )] sF( s)  f ( 0) dt n d n k f ( t)]  s n F( s)   s n k f (0) dtn  k k 1 L[

L[

dn dt n

(2.2)

where F(s) = L[f(t)] and f(0) is the initial value of f(t) at t = 0. 2.3 Real integration theorem

L[ f ( t) dt] 

F( s) f  1 ( 0)  s s

where F(s) = L[f(t)] and f

1

(2.3)

( 0 )  f ( t ) dt , valued at t = 0.

2.4 Final value theorem

lim f (t )  lim sF( s ) (2.4) t  s 0 2.5 Initial value theorem

lim f ( t)  lim sF( s) (2.5) t 0 s  which is the counterpart of the final value theorem. 2.6 Unit impulse function (Dirac delta function)

 0 t t 0  (t  t 0 )   t  t 0 

 ( t 

(2.6)

t0 ) dt 1



whose Laplace transform is a unit constant. The properties of Laplace transform are summarised in Appendix. 3 A general formula for obtaining the inverse Laplace transform of (1.2)

2

(2.1)

This residual theorem based technique is presented as

 d m 1  1 f (t )  L 1  F (s )  Re s F (s )e st   ( s  si ) m F( s) e st m 1 m 1 !     ds 









 s si

(3.1)

where si is F(s)'s pole and m is the order of the pole s = si . 4 Solve linear differential equations by Laplace transform In this section we are concerned with the application of the Laplace transform method in solving linear, time invariant, differential equations. Consider a general expression of the linear time invariant differential equation

(n) ( n 1) (m ) (m  1) . . a 0 y  a1 y an  1 y  an y  b0 x  b1 x bm  1 x  bm x

n m

(4.1)

where y is the dependent variable (known as the output in control engineering) and x is the independent variable (known as the input correspondingly). ai and bi are the associated constant (time invariant) coefficients and n is the order of the equation. The Laplace transform method yields the complete solution (complementary solution and particular solution) of linear, time invariant, differential equations. Classical methods for finding the complete solution of a differential equation require the evaluation of the integration constants from the initial conditions. In the case of the Laplace transform method, however, this requirement is unnecessary because the initial conditions are automatically included in the Laplace transform of the differential equation. If all initial conditions are zero, then the Laplace transform of the differential equation is obtained simply by replacing d/dt with s, d2/dt2 with s2, and so on. In solving linear, time invariant, differential equations by the Laplace transform method, two steps are needed. 1. By taking the Laplace transform of each term in the given differential equation, convert the differential equation into an algebraic equation in s and obtain the expression for the Laplace transform of the dependent variable by rearranging the algebraic equation. 2. The time solution of the differential equation is obtained by finding the inverse Laplace transform of the dependent variable. Example: Find the solution x(t) of the differential equation x  3 x  2 x 1

x( 0) 0,

x (0 ) 0

(4.2)

where a and b are constants. 5 Block diagrams A control system may consist of a number of components. To show the functions performed by each component, in control engineering, we commonly use a diagram called block diagram . This section explains what a block diagram is, presents a method for obtaining block diagrams for physical systems so simplify such diagrams. 5.1 Block diagram of an open loop system

3

Input R(s)

Output Y(s) Transfer function Go(s)

Fig. 2 An open loop system description Example two: Depict the block diagram of the transfer function obtained in E2 Lecture 1. Input P(s)

Output X(s)

1 m b 2 s  s

Fig. 3 Block diagram of the transfer function in E2 Lecture 1 5.2 Block diagram of a closed loop system Summing point

Input R(s)

Branch point Output Y(s)

E(s)

+

Go(s) -

Fig. 4 A closed loop system description Explanation to feedforwad path, feedback path, loop The closed loop transfer function is derived as follows

Y ( s) Go ( s) E( s) (5.1) E( s)  R( s)  Y( s) Therefore

Y ( s)  G( s) E( s)  Go ( s)[ R( s)  Y( s)]

(5.2)

which gives the closed transfer function

Gc ( s) 

G ( s) Y( s)  o R( s) 1  Go ( s)

(5.3)

The open loop transfer function Go (s) can be worked out when the closed loop transfer function Go (s) is known

4

Go (s ) 

Gc ( s ) 1  G c (s )

(5.4)

Example three: Work out the closed loop transfer function in Fig. 4.5 R(s)

+

G1 ( s )

Y(s)

+

-

G2 ( s )

G3 ( s )

-

Fig. 5 A closed loop system Exercise 1 E1 Using Laplace transform method to check if the solution x(t) of the differential equation

.. . x  2 x  5 x 3

x( 0)  0

. x( 0)  0 (E 1)

is coincident with that obtained in the exercise of lecture one. Using Matlab program invLT2.m confirms the solution. E2 By applying the final value theorem, find the final value of f(t) whose Laplace transform is given by

F (s ) 

10 s s  1

(E 2)

Verify this result by taking the inverse Laplace transform of F(s) and taking

t .

E3 1) Work out closed loop transfer function Gc ( s) from following block diagram, 2) derive the closed loop system output time response to a unit step input, 3) let R(s) be a step input, using Matlab/Simulink plot the output response and input curves against time within one graph paper. 4) Compare if the simplified block diagram and the original block diagram generate the same output response. R(s)

+

G1 ( s )

Y(s)

+

-

G2 ( s ) -

Fig. E1 A closed loop system where

G1 ( s) 0 .5

G2 ( s) 

1 s

1 G3 ( s)  2 (E 3) s  s 1

Appendix 1) Table of selected Laplace transforms

5

G3 ( s )

2) Properties Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s):

6

7...