# CHAPTER 2 THE BASIC OF CONTROL THEORY

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CHAPTER 2THE BASIC OF CONTROLTHEORY

1. Laplace Transform Review.

Laplace Transform Review. Laplace Transform is defined as, L [ f (t )] F (s) f (t )e st dt0 Where s s jw is a complex variable. By knowing f(t) we can find thefunction F(s) which is called Laplace transform of f(t). Inverse Laplace1 s j stL [ F ( s)] f (t ) F(s)eds s j 2 j 1The inverse Laplace transform allows us to find f(t) given F(s).

1) The Laplace Transform cont. Transform table:f(t)F(s)1.δ(t)12.u(t)1sStep functiont u(t)1s2Ramp function3.4.tn u(t)5.e-atu(t)6.sin wt u(t)7.cos wt u(t)n!s n 11s aws2 w 2ss2 w 2Impulse functionf(t)8. Ae-atcos wt u(t)9.Be-atsinwt u(t)F(s)A s a s a 2 w 2Bw s a 2 w 2

1) The Laplace Transform cont. TransformProperties

Exercise 1: Laplace Transform.Find the Laplace transform off(t) Ae-atu(t)Solution: F (s) L [ f (t )] f (t )e st dt0 Ae e dt A e ( s a )t dt at st0 .A ( s a )t es aA s a0 t 0

1) The Laplace Transform cont. Example: Find the Laplace Transform for the following.i.Unit function:ii.Ramp function:iii.Step function:f (t ) 1f (t ) tf (t ) Ae at

1) The Laplace Transform cont. Transform Theoremi.Differentiation Theoremdf (t )L{} sF ( s) f (0)dtd 2 f (t )2 (0)L{} sF(s) sf(0) fdt 2ii.Integration Theorem: t F (s)L f ( )d s 0 iii.Initial Value Theorem:f (0) lim sF ( s)t iv.Final Value Theorem:lim f (t ) lim sF ( s)t s 0

1) The Laplace Transform cont. The inverse Laplace Transform can be obtained using:s jw1 stf (t ) F(s)eds 2 j s jw Partial fraction method can be used to find the inverse Laplace Transformof a complicated function.We can convert the function to a sum of simpler terms for which we knowthe inverse Laplace Transform.F ( s) F1 ( s) F2 ( s) Fn ( s)f (t ) L 1 F1 (s) L 1 F2 (s) L 1 Fn (s) f1 (t ) f 2 (t ) f n (t )

1) The Laplace Transform cont. We will consider three cases and show that F(s) can be expandedinto partial fraction:i.Case 1:Roots of denominator A(s) are real and distinct.ii.Case 2:Roots of denominator A(s) are real and repeated.iii.Case 3:Roots of denominator A(s) are complex conjugate.

1) The Laplace Transform cont. Case 1: Roots of denominator A(s) are real and distinct.Example 1:2F( s ) ( s 1 )( s 2 )Solution:ABF ( s) s 1 s 222 s 1 s 2f (t ) 2e t 2e 2tIt is found that:A 2 and B -2

Example 2:s 3Y ( s) ( s 1)(s 2)Problem: Find the Inverse Laplace Transform for the following.Solution:s 3( s 1)(s 2)ABY (s) s 1 s 2s 3A (s 1)(s 1)(s 2)Y (s) s 1 2 B (s 2)21 t 2 ty(t) 2e es 1 s 2 .s 3(s 1)(s 2)12s 2 1

1) The Laplace Transform cont.Case 2: Roots of denominator F(s) are real and repeated.Example 1:2F ( s) ( s 1)(s 2) 2Solution:ABCIt is found that:F ( s) A 2, B -2 and C -2s 1 s 2 ( s 2) 2222 s 1 s 2 ( s 2) 2 f (t ) 2e t 2e 2t 2te 2t

Example 2: Find the Inverse Laplace Transform ofX s 3s 4( s 1)(s 2) 2Solution:Step 1: Use the partial fraction expansion of X(s) to writeABCX s (s 1) (s 2) ( s 2) 2Solving the A, B and C by the method of residues(3s 4)A( s 2) 2B( s 1)(s 2)C ( s 1) 222( s 1)(s 2)( s 1)(s 2)( s 2) ( s 1) ( s 2) 2 ( s 1)14

Cont’d Example(3s 4) A( s 2) 2 B ( s 2)(s 1) C ( s 1) A( s 2 4 s 4) B ( s 2 3s 2) C ( s 1) ( A B ) s 2 ( 4 A 3B C ) s ( 4 A 2 B C )so, comparecoefficient ,A B 0 (1)4 A 3B C 3 ( 2)4 A 2 B C 4 (3)(3) ( 2); B 1B 1.From(1)A B 0A 1SubstituteB andA, int o( 2)4(1) 3( 1) C 3C 2.15

A 1, B -1 and C 2112X s (s 1) (s 2) (s 2) 2Step 2: Construct the Inverse Laplace transform from the above partial-fractionterm above.- The pole of the 1st term is at s -1, so- The pole of the 2nd term is at s -2, so1e u (t ) ( s 1) tLuLue 2t u (t ) 1( s 2)-The double pole of the 3rd term is at s -1, so12te u (t ) ( s 2) 2 tLuStep 3: Combining the terms. .x(t ) e t u (t ) e 2t u (t ) 2te 2t u (t ).16

1) The Laplace Transform cont. Case 3: Roots of denominator F(s) are complex conjugate.Example:3F ( s) s(s 2 2s 5)Solution:ABs CIt is found that:F ( s) 2A 3/5, B -3/5s s 2s 5and C -6/53 5 3 s 2 2 s 5 s 2s 5 3 5 3 ( s 1) (1 2)(2) s 5 ( s 1) 2 22 3 3 t1f (t ) e (cos 2t sin 2t )5 52

Exercise 2: Laplace Transform Function DifferentialEquation.d2ydy 4t 9 2y 6edt 2dty (0 ) 2dy (0 ) 4dtSolution: s Y (s) 2s 4 9 sY (s) 2 2Y (s) s 6 4262s 14Y ( s) 22(s 4)(s 9s 2) s 9s 2 .

2. Block Diagram

2) Block Diagram A block diagram of a system is a practical representation of the functionsperformed by each component and of the flow of signals.Input Cascaded sub-systems:Transfer FunctionG(s)Output

2) Block Diagram cont. Feedback Control System

2) Block Diagram cont. Feedback Control SystemThe negative feedback ofthe control system is givenby:Ea(s) R(s) – H(s)Y(s)Y(s) G(s)Ea(s)Therefore,Y (s) G(s)[R(s) H (s)Y (s)]Y ( s)G( s) R( s ) 1 G ( s ) H ( s )

2) Block Diagram cont. Reduction Rules

2) Block Diagram cont. Reduction Rules

2) Block Diagram cont. Problem 1:

2) Block Diagram cont.Problem 1: -G1 G2 G3G4H1H3U(s)G1- U(s)H2/ G4- G2G3H1/ G2H2/ G4H3 G1G4Y(s)Y(s)

2) Block Diagram cont.Problem 1:G1G2G3G4Y(s) U(s) H1/ G2- H2/ G4- H3 G1U(s)G1G 2 G3 G 4 H H1 G2 G3 G4 1 2 H 3 G1 G2 G4 Y(s)

2) Block Diagram cont. Problem 1:U(s)U(s)G1G2 G3 G4 H H1 G2 G3 G4 1 2 H 3 G1 G2 G4 Y(s)G1G2 G3G41 G3 G4 H 1 G2 G3 H 2 G1G2 G3 G4 H 3 Y(s)

2) Block Diagram cont. Problem 2:

2) Block Diagram cont. Problem 3:Reduce the system to a single transfer function

3. Signal-flow graph

3) Signal Flow Graph A signal flow graph is a graphical representation of therelationships between the variables of a set of linear algebraicequations.The basic element of a signal flow graph is a unidirectional pathsegments called branch.The input and output points or junctions are called nodes.A path is a branch or continuous sequence or branches that can betraversed from one signal node to another signal node.A loop is a closed path that originates and terminates on the samenode, and along the path no node is met twice.Two loops are said to be non-touching if they do not have a samecommon node.

3) Signal Flow Graph cont. Signal flow graph of control systems

3) Signal Flow Graph cont. Signal flow graph of control systems

3) Signal Flow Graph cont. Mason’s Gain Formula for Signal Flow GraphTij Pijk ijkk Where,Pijk ijk: kth path from variable xi to xj: Determinant of the graph: Cofactor of the path PijkΔ 1 - (sum of all different loop gains) (sum of the gain products of all combinations of 2 nontouching loops)- (sum of the gain products of all combinations of 3 nontouching loops) .

3) Signal Flow Graph cont. Example 1: Transfer function of interacting system

3) Signal Flow Graph cont. Example 1: Transfer function of interacting systema) The paths connecting input R(s) to output Y(s) are:P1 G1G2G3G4P2 G5G6G7G8b) There are four individual loops:L1 G2H2L2 G3H3L3 G6H6L4 G7H7

3) Signal Flow Graph cont. Example 1: Transfer function of interacting systemc) Loops L1 and L2 does not touch loops L3 and L4. Therefore, thedeterminant is: 1 ( L1 L2 L3 L4 ) ( L1 L3 L1 L4 L2 L3 L2 L4 )d) The cofactor of the determinant along path 1 is evaluated byremoving the loops that touch path 1 from . Therefore have:L1 L2 0and, 1 1 ( L3 L4 )Similarly, the cofactor for path 2 is: 2 1 ( L1 L2 )

3) Signal Flow Graph cont. Example 1: Transfer function of interacting systeme) Therefore, the transfer function of the system is:Y ( s)P P T ( s) 1 1 2 2R( s ) G1G2G3G4 (1 L3 L4 ) G5G6G7G8 (1 L1 L2 ) 1 L1 L2 L3 L4 L1L3 L1L4 L2 L3 L2 L4

3) Signal Flow Graph cont. Problem 1:Obtain the closed-loop transfer function by use of Mason’s GainFormula

3) Signal Flow Graph cont. Problem 2:Obtain the closed-loop transfer function by use of Mason’s GainFormula

4. Review state space variable

Introduction The basic questions that will be addressed in state-space approachinclude:i.What are state-space models?ii.Why should we use them?iii.How are they related to the transfer function used in classicalcontrol system?iv.How do we develop a space-state model?

4) State-Space ModelA representation of the dynamics of Nth-ordersystem as a first-order equation in an N-vector,which is called the state.Convert the Nth-order differential equation thatgoverns the dynamics of the system into N firstorder differential equation.

4) State-Space Model The state of a system is described by a set of first-orderdifferential equations written in terms of the state variable.

4) State-Space Model In a matrix form, we have: State vector:

4) State Space ModelInput equationOutput equation

Laplace Transform Review. Laplace Transform is defined as, Where s s jwis a complex variable. By knowing f(t) we can find the function F(s)which is called Laplace transform of f(t). Inverse Laplace T

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