Lecture10-SignalFlowGraphs

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Page : 1 EE406 Control Systems Lecture 10 : Signal Flow Graphs

UCSI University Faculty of Engineering Kuala Lumpur, Malaysia Department of Mechatronics

Lecture 10

Signal Flow Graphs

Mohd Sulhi bin Azman Lecturer

Department of Mechatronics UCSI University [email protected]

1 August 2011

Contents

• Signal Flow Graphs

• Block diagrams to signal flow graphs conversion • State space matrix to signal flow graphs

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Introduction

• It is an alternative to block diagrams.

• Consists of only braches, which represent systems and nodes, which represent signals.

• A system is represented by a line with an arrow showing the direction of signal flow through the system.

• We write the transfer function which is adjacent to the line.

• A signal is a node, with the signal’s name written adjacent to the node.

Signal Flow Graph

• Signal-flow graph components: a. system;

b. signal;

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Block Diagram To Signal Flow Graph

• Building signal-flow graphs: a. cascaded system nodes

b. cascaded system signal-flow graph;

Block Diagram To Signal Flow Graph

• Building signal-flow graphs: c. parallel system nodes

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Block Diagram To Signal Flow Graph

• Building signal-flow graphs: e. feedback system nodes

f. feedback system signal-flow graph

Example 1

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Solution to Example 1

• Step I : Develop the signal nodes

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Solution to Example 1

• Step III : Simplify

State space matrix and signal flow graphs

• We can convert a state space matrix and represent them in the form of a signal flow graph.

• Before that, we note the following relationship:

1

integrator differentiator

s

s x

= = = =

∫

ɺ

{ }

( ) 1 ( ) ( ) (0) ( )

I s

i dt I s

s s

x sX s x sX s

= =

= − =

∫

ɺ L

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Example 2

• Represent the following state equation in signal flow graph.

1 1 2 3

2 1 2 3

3 1 2 3

1 2 3

2

5

3

2

6

2

5

3

4

7

4

6

9

x

r

x

r

x

r

y

x

=

−

+

= −

−

+

= −

−

+

= −

+

ɺ

• Step I: Place the nodes.

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• Step III : Form dx₁/dt Connect the

derivative of the state variable with the state variably by placing the definiting integral.

Solution to Example 2

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• Step V : Form dx₃/dt.

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Transfer Function and Signal Flow Graph

• Sometimes, the signal flow graph is also known as the phase variable diagram.

• We can actually convert a transfer function and represent it in signal flow graph.

Example 3

• Represent the following transfer function in phase variable form:

( ) 1 ( )

C s

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• Cross multiply and take inverse Laplace transform.

{

}

{

}

1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

s a C s R s s a C s R s sC s aC s R s d

c t ac t r t dt

c t ac t r t r t c t ac t

− − + = + = + = + = = + = − ɺ ɺ L L ( ) ( ) d

c t c t dt = ɺ

Different Form of Signal Flow Graphs

• There are two forms (actually, there are many, but we focus on two) of signal flow graph that is commonly studied and used in control systems:

– Cascade form – Parallel form

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Cascade Form

• Remember that phase variables are variables that are successive derivative of each other.

• The cascade form of state variable looks at the transfer function in its original form, meaning that we do not need to expand the transfer function into its partial form.

Example 4

• Represent the following transfer function in phase variable form, and from the phase variable diagram, obtain a state space matrix.

( ) 24

( ) ( 2)( 3)( 4)

C s

R s = s+ s+ s+

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Solution to Example 4

• The system can be represented in the block diagram such as:

• Remember that the derivative of a state variable will be at the input to each integrator:

• Then, write the state equation at the input to each integrator:

• Thus giving:

1 4 1 0 1 0

x − x

       

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General Conclusion – Cascade Form

• In general, for the following transfer function:

• The cascade form of the state space matrix is:

(

1

)(

2

)(

3

) (

)

( )

λ λ λ λ

n

T s

s s s s

µ = − − − ⋯ − 1 2 1 2 1

λ ₁ ₀

λ ₁ λ ₁ λ n n n x x r x µ −                         = +                  _{ }     x ⋮

ɺ ⋱ ⋱ ⋮ ⋮

⋮ ⋮ Lower echelon (all zero) Upper echelon (all zero)

Parallel Form

• Another alternative to cascade form is the parallel form.

• In parallel form, we expand the transfer function into its partial form.

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Example 5

• Represent the following transfer function in phase variable form, and from the phase variable diagram, obtain a state space matrix.

( ) 24

( ) ( 2)( 3)( 4)

C s

R s = s+ s+ s+

• In parallel form, we expand the transfer function into its partial form.

( ) 24 12 24 12

( ) ( 2)( 3)( 4) 2 3 4

C s

R s = s+ s+ s+ = s+ −s+ +s+

This fraction is in “parallel” (addition) to

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Solution to Example 4

• Next, we draw a signal flow graph:

• The state space matrix is then equals to:

1 1

2 2

3 3

2 0 0 12

0 3 0 24

0 0 4 12

x x

x x r

x x

−

       

 ₌ ₋    _{+ −} 

       

   ₋     

       

ɺ ɺ ɺ

[

]

1

( ) 1 1 1

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General Conclusion – Parallel Form

• In general, for the following transfer function:

• The parallel form of the state space matrix is:

(

1

)(

2

)(

3

) (

)

1 1 2 2

( )

λ λ λ λ λ λ λ

n

n n

k

k k

T s

s s s s s s s

µ = = + + + − − − ⋯ − − − ⋯ − 1 1 2 1 2 1 λ λ λ λ n n n n x k x r x k −                         = +                         x ⋮

ɺ ⋱ ⋮ ⋮

⋮ ⋮ Lower echelon (all zero) Upper echelon (all zero)

Some Known Forms

• We can also represent the equations in the following canonical form, which will be suitable for state space analysis:

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Controller Canonical Form

• This form is obtained by ordering the phase variables in reverse order.

• The controller canonical form contain the

coefficients of the characteristic polynomial in the bottom or top row of the matrix,

respectively.

Example 5

• Represent the following transfer function in controller canonical form:

2

3 2

( ) 7 2

( ) 9 26 24

C s s s R s s s s

+ + =

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• For this transfer function, let us separate the numerator and the denominator:

• Next, we observe that the output expression is:

• Taking the inverse Laplace transform gives:

(

2

)

1

( ) 7 2 ( )

C s = s + s+ X s

(

)

{

}

1 2

1

1 1 1

( ) 7 2 ( )

( ) 7 2

C s s s X s c t x x x

− ₌ ₊ ₊

= +ɺɺ ɺ +

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Solution to Example 5

• Then, let us draw/sketch the signal flow graph:

• Figure (a) shows the phase-variable form.

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• The state space representation is:

[

]

1 1 2 2 3 3 1 2 3

9 26 24 0

1 0 0 0

0 1 0 1

1 7 2

x x

x x r

x x x y x x − − −          ₌     ₊                             = _{ }     ɺ ɺ ɺ

Observer Canonical Form

• It is so named because it is extensively used in the design of observers, in state space analysis.

• It is a representation that yields a left companion system matrix.

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Example 6

• Represent the following transfer function in controller canonical form:

2

3 2

( ) 7 2

( ) 9 26 24

C s s s R s s s s

+ + =

+ + +

• Divide the numerator with the highest power of “s”, in this case, this is s3_.

• Next, we cross multiply this system:

2

3 3 3 2 3

3 2

2 3

3 3 3 3

7 2 7 2

1 ( )

9 26 24

( ) ₁

s s

C s _s _s _s _s _s

s s s

R s

s s s s s s s

+ + + +

= =

+ + +

7 2 9 26 24

   

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• Next, we combine terms of like power of integration to give:

• Now, let us draw the signal flow graph. State with three integrations, as shown below:

[

]

2

[

]

3

[

]

1 1 1

( ) ( ) 9 ( ) 7 ( ) 26 ( ) 2 ( ) 24 ( )

C s R s C s R s C s R s C s

s s s

= − + − + −

First integrator Second

integrator Third

integrator

Solution to Example 6

• The first term tells us about the first integration (1/s). The output C(s) is formed by integrating [R(s)-9C(s)]. Now place this term in the first integrator.

• The second term (1/s2_{) tells us that the terms}

[R(s)-26C(s)] must be integrated twice. Now place this term in the second integrator.

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[R(s)-Page : 47 EE406 Control Systems Lecture 10 : Signal Flow Graphs

Solution to Example 6

• Write the state variables at the outputs of the integrator:

• And the output is:

1 1 2

2 1 3

3 1

9

26 7

24 2

x x x r

x x r

= − + + = − + + = − +

ɺ ɺ ɺ

1

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• The state space matrix is then:

[

]

1 1 2 2 3 3 1 2 3

9 1 0 1

26 0 1 7

24 0 0 2

1 0 0

x x

x x r

x x x y x x −          _{= −}     ₊           ₋                  = _{ }     ɺ ɺ ɺ

The left “companion” matrix

Next Step

• Textbook reference : Chapter 5.

• Homework 9 has been posted on the course website. Attempt them. You do not have to submit Homework 9 as it will not be graded.