chapter 5. reduction of multiple subsystems 1st

Chapter 5.Reduction of Multiple Subsystems_1st

Table of Contents

• Introduction

• Block Diagrams

• Analysis and Design of Feedback Systems

• Signal-Flow Graphs

•Mason’s Rule

Introduction

• Multiple subsystems

• Represented by the interconnection of many subsystems

• Block diagrams and signal-flow graphs representation

Figure 5.1 a. Transfer function ;b. Equivalent block diagram showing phase variables;

Note y(t) = c(t);

Block Diagrams

• Example: Configuration of multiple subsystems

• Subsystem is represented as a block with an input, an output, and a transfer function. Many systems are composed of multiple subsystems, as in Figure 5.2.

Figure 5.2 The space shuttle consists of multiple subsystems. Can you identify those that are control systems or parts of control systems?

Block Diagrams

• Components of a block diagram

Figure 5.3 Components of a block diagram for a linear, time-invariant system

Three basics forms: cascade, parallel, and feedback forms.

Figure 5.4 a. Cascaded subsystems ;b. Equivalent transfer function

• Cascaded Form§ Intermediate signal values are shown at the output of each subsystem.

§ Each signal is derived from the product of the input times the transfer function.

• Parallel Form

• Parallel subsystems have a common input and an output formed by the algebraic sum of the outputs from all of the subsystems.

(a)

Figure 5.5 a. Parallel subsystems ;b. Equivalent transfer function

• Feedback Form

• Basis for our study of control systems engineering

Figure 5.6 a. Feedback control system ;b. Simplified model ;c. Equivalent transfer function

• The typical feedback system, described in detail in Chapter 1, is shown in Figure 5.6(a); a simplified model is shown in Figure 5.6(b).

Directing our attention to the simplified model,

But since ,

• Substituting Eq. (5.2) and solving for the transfer function , we obtain the equivalent, or closed-loop, transfer function shown in Figure 5.6(c),

The product, , in Eq. (5.3) is called the open-loop transfer function, or loop gain.

( ) ( ) ( ) ( )E s R s C s H s= m (5.1)

( ) ( ) ( )C s E s G s=

( )( )( )

C sE s

G s= (5.2)

( ) ( ) ( )/ eC s R s G s=

( )( )

( ) ( )1e

G sG s

G s H s=

±(5.3)

( ) ( )G s H s

Moving blocks

• Moving Block to create familiar Forms

• These equivalences, along with the forms studied earlier in this section, can be used to reduce a block diagram to a single transfer function.

Figure 5.7 Block diagram algebra for summing junctions-equivalent forms for moving a blocka. To the left past a summing junction ;b. To right past a summing junction

Moving blocks

Figure 5.8 Block diagram algebra for pickoff points-equivalent forms for moving a blocka. To the left past a pickoff point ;b. To the right past a pickoff point

Moving blocks

• Ex 5.1 Block diagram reduction via familiar forms• Reduce the block diagram shown in Figure 5.9 to a single transfer

function.

Figure 5.9 Block diagram for Example 5.1

Moving blocks

• Sol ) We solve the problem by following the steps in Figure 5.10. First, the three summing junctions can be collapsed into a single summing junction, as shown in Figure 5.10(a).

Figure 5.10 Steps in solving Example 5.1:a. Collapse summing junctions

Moving blocks

• Second, recognize that the three feedback functions, , are connected in parallel. They are fed from a common signal source, and their outputs are summed. The equivalent function is . Also recognize that are connected in cascade. Thus, the equivalent transfer function is the product, . The results of these steps are shown in Figure 5.10(b).

Figure 5.10 Steps in solving Example 5.1:b. Form equivalent cascaded system in the forward path and

equivalent parallel system in the feedback path

( ) ( ) ( )1 2 3, , andH s H s H s

( ) ( ) ( )1 2 3H s H s H s- +( ) ( )2 3and GG s s

( ) ( )3 2G s G s

Moving blocks

• Finally, the feedback system is reduced and multiplied by to yield the equivalent transfer function shown in Figure 5.10(c).

Figure 5.10 Steps in solving Example 5.1:c. Form equivalent feedback system and multiply by cascaded ( )1G s

( )1G s

Reduction of blocks

• Ex 5.2 Block diagram reduction by moving blocks• Reduce the system shown in figure 5.11 to a single transfer function.

Figure 5.11 Block diagram for Example 5.2

Reduction of blocks

• Sol ) In this example we make use of the equivalent forms shown in Figures 5.7 and 5.8 First, move to the left past the pickoff point to create parallel subsystems, and reduce the feedback system consisting of

and . This result is shown in Figure 5.12(a).

Figure 5.12 Steps in the block diagram reduction for Example 5.2

( )2G s

( )3G s ( )3H s

Reduction of blocks

• Second, reduce the parallel pair consisting of and unity, and push to the right past the summing junction, creating parallel subsystems

in the feedback. These results are shown in Figure 5.12(b)

Figure 5.12 Steps in the block diagram reduction for Example 5.2

( )21/G s

( )1G s

Reduction of blocks

• Third, collapse the summing junctions, add the two feedback elements together, and combine the last two cascaded blocks. Figure 5.12(c) shows these results.

Figure 5.12 Steps in the block diagram reduction for Example 5.2

Reduction of blocks

• Fourth, use the feedback formula to obtain Figure 5.12(d).

• Finally, multiply the two cascaded blocks and obtain the final result, shown in Figure 5.12(e).

Figure 5.12 Steps in the block diagram reduction for Example 5.2

Analysis and Design of Feedback Systems

• Secondary feedback system analysis

• Percent overshoot, settling time, peak time, and rise time can then be found from the equivalent transfer function.

Figure 5.13 Second-order feedback control system

2

( ) ( )( )

1 ( ) ( ) 1( )

( )

K

G s s s aT s

KG s H ss s a

K K

s s a K s as K

+= =

+ ++

= =+ + + +

(5.4)

• Real pole : over damped

• The real pole overlap : critically damped

• Complex number pole : under damped

• K value increases• Increase in the imaginary part of pole

• Fixed settling time

• Peak time reduction

• Percent overshoot increases

2

1,2

4

2 2

a a Ks

-= - ± (5.5)

2

04

aK

æ ö< <ç ÷

è ø

1,22

as = -

2

4

aK

æ ö=ç ÷

è ø(5.6)

2

1,2

4

2 2

a K as j

-= - ±

2

4

aK

æ ö>ç ÷

è ø(5.7)

Analysis and Design of Feedback Systems

• Ex 5.3 Finding transient response• For the system shown in Figure 5.14, find the peak time, percent

overshoot, and settling time.

• Sol ) The close-loop transfer function found from Eq. (5.3) is

• From Eq. (4.18),

Figure 5.14 Feedback system for Example 5.3

( ) 2

25

5 25T s

s s=

+ +(5.8)

25 5nw = = (5.9)

Analysis and Design of Feedback Systems

• From Eq.(4.21),

• Substituting Eq. (5.9) into (5.10) and solving for yields

• Using the values for and along with Eqs. (4.34), (4.38), and (4.42), we find, respectively,

2 5nzw = (5.10)

z

0.5z = (5.11)

znw

20.726second

1p

n

Tp

w z= =

-(5.12)

2/ 1% 100 16.303OS e zp z- -= ´ = (5.13)

41.6secondss

n

Tzw

= = (5.14)

Analysis and Design of Feedback Systems

• Ex 5.4 Gain design for transient response• Design the value of gain, K, for the feedback control system of Figure

5.15 so that the system will respond with a 10% overshoot.

• Sol ) The closed-loop transfer function of the system is

Figure 5.15 Feedback system for Example 5.4

( ) 2 5

KT s

s s K=

+ +(5.15)

Analysis and Design of Feedback Systems

• From Eq. (5.15),

and

Thus,

• Since percent overshoot is a function only of , Eq.(5.18) shows that percent overshoot is a function of k.

A 10% overshoot implies that . Substituting this value for the damping ratio into Eq. (5.18) and solving for K yields

2 5nzw = (5.16)

n Kw = (5.17)

5

2 Kz = (5.18)

z

0.591z =

17.9K = (5.19)

Analysis and Design of Feedback Systems

Signal-Flow Graphs

• Signal flow graphs

• Signal-flow graph consists only of branches, which represent systems and nodes, which represent signals

• Each signal is the sum of signals flowing into it.

Figure 5.16 Signal-flow graph components;a. system;b. signal;c. interconnection of systems and signals

Signal-Flow Graphs

• Ex 5.5 Converting common block diagrams to signal-low graphs• Convert the cascaded, parallel, and feedback forms of the block diagrams

shown in Figures 5.4(a), 5.5(a), and 5.6(b), respectively, into signal-flow graphs.

5.4(a)

5.5(a) 5.6(b)

Signal-Flow Graphs

• Sol ) In each case we start by drawing the signal nodes for that system. Next we interconnect the signal nodes with system branches. The signal nodes for the cascaded, parallel, and feedback forms are shown in Figure 5.17(a), (c), and (e), respectively.

Figure 5.17 Building signal-flow graphs:a. cascaded system nodes(from Figure 5.4(a));c. Parallel system nodes (form Figure 5.5(a));e. feedback system nodes (from Figure 5.6(b))

Signal-Flow Graphs

• The interconnection of the nodes with branches that represent the subsystems is shown in Figure 5.17(b), (d), and (f) for the cascaded, parallel, and feedback forms, respectively.

Figure 5.17 Building signal-flow graphs:b. cascaded system signal-flow graph (from Figure 5.4(a));d. parallel system signal-flow graph (form Figure 5.5(a));f. feedback system signal-flow graph (from Figure 5.6(b))

Signal-Flow Graphs

• Ex 5.6 Converting a block diagram to a signal-flow graph• Convert the block diagram of Figure 5.11 to a signal-flow graph.

• Sol ) Begin by drawing the signal nodes, as shown in Figure 5.18(a). Next, interconnect the nodes, showing the direction of signal flow and identifying each transfer function.

Figure 5.18 Signal-flow graph development :a. signal nodes

Signal-Flow Graphs

• The result is shown in Figure 5.18(b). Notice that the negative signs at the summing junctions of the block diagram are represented by the negative transfer Functions of the signal-flow graph.

Figure 5.18 Signal-flow graph development :b. signal-flow graph

Signal-Flow Graphs

• Finally, if desired, simplify the signal-flow graph to the one shown in Figure 5.18(c) by eliminating signals that have a single flow in and a single flow out, such as .

Figure 5.18 Signal-flow graph development :c. simplified signal-flow graph

( ) ( ) ( ) ( )2 6 7 8, , , and VV s V s V s s

Mason’s Rule

• Signal-flow graphs to single transfer functions that relate the output of a system to it's input

• Mason’s formula components

• Loop gain

• Forward – path gain

• Non-touching loops

• Non-touching – loop gain

Mason’s Rule

• Loop gain

• Ends at the same node

• Without passing through any other node more than once

Figure 5.19 Signal-flow graph for demonstrating Mason’s rule

( ) ( )2 1G s H s (5.20a)

( ) ( )4 2G s H s (5.20b)

( ) ( ) ( )4 5 3G s G s H s (5.20c)

( ) ( ) ( )4 6 3G s G s H s (5.20d)

Mason’s Rule

• Forward – path gain

• Gains found by traversing a path from the input node to the output node of the signal-flow graph in the direction of signal flow.

Figure 5.19 Signal-flow graph for demonstrating Mason’s rule

( ) ( ) ( ) ( ) ( ) ( )1 2 3 4 5 7G s G s G s G s G s G s

( ) ( ) ( ) ( ) ( ) ( )1 2 3 4 6 7G s G s G s G s G s G s

(5.21a)

(5.21b)

Mason’s Rule

v Non-touching loops

• Loops that do not have any nodes in common.

• In figure 5.19, loop dose not touch loops and .

Figure 5.19 Signal-flow graph for demonstrating Mason’s rule

( ) ( )2 1G s H s ( ) ( ) ( ) ( ) ( )4 2 4 5 3,G s H s G s G s H s( ) ( ) ( )4 6 3G s G s H s

Mason’s Rule

• Non-touching – loop gain

• The product of loop gains from non-touching loops taken two, three, four, or more at a time.

Figure 5.19 Signal-flow graph for demonstrating Mason’s rule

( ) ( ) ( ) ( )2 1 4 2G s H s G s H sé ù é ùë û ë û

( ) ( ) ( ) ( ) ( )2 1 4 5 3G s H s G s G s H sé ù é ùë û ë û

( ) ( ) ( ) ( ) ( )2 1 4 6 3G s H s G s G s H sé ù é ùë û ë û

(5.22a)

(5.22b)

(5.22c)

Mason’s Rule

v MASON’S RULE

§ The transfer function, , of a system represented by a signal-flow graph is

where

( ) ( )/C s R s

( )( )( )

k kkTC s

G sR s

D= =

D

å(5.23)

number of forward paths

the th forward-path gain

1 loopgains + nontouching loop gain taken two at a time

nontouching loop gains taken three at a time

+ nontouching loop gains taken four at a tim

k

k

T k

=

=

D = -å å -

-å -

å - e

loop gain terms in that touch the th forward path.

In other words, is formed by eliminating from those loop gain that touch the th forward path.

k

k

k

k

-

D = D -å D

D D

L

Mason’s Rule

• Ex 5.7 Transfer function via Mason’s rule• Find the transfer function, , for the signal-flow graph in Figure

5.20.( ) ( )/C s R s

Figure 5.20 Signal-flow graph for Example 5.7

Mason’s Rule

• Sol) First, identify the forward-path gains. In this example there is only one:

Second, identify the loop gains. There are four, as follows:

Third, identify the non-touching loops taken two at a time. From Eqs. (5.25) and Figure 5.20, we can see that loop 1 dose not touch 2, loop 1 does not touch loop3, and loop 2 does not touch loop 3. Notice that loop 1, 2, and 3 all touch loop 4. Thus, the combinations of non-touching loops taken two at a time are at follows:

( ) ( ) ( ) ( ) ( )1 2 3 4 5G s G s G s G s G s (5.24)

( ) ( )2 1G s H s

( ) ( )4 2G s H s

( ) ( )7 4G s H s

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 4 5 6 7 8G s G s G s G s G s G s G s

(5.25a)

(5.25b)

(5.25c)

(5.25d)

Mason’s Rule

• Thus, the combinations of non-touching loops taken two at a time are at follows:

Loop 1 and loop 2:

Loop 1 and loop 3:

Loop 2 and loop 3:

• Finally, the non-touching loops taken three at a time are as follows:

Loops 1, 2, and 3:

Loops 1, 2, and its definitions, we form and Hence,

(5.26a)( ) ( ) ( ) ( )2 1 4 2G s H s G s H s

( ) ( ) ( ) ( )2 1 7 4G s H s G s H s (5.26b)

( ) ( ) ( ) ( )4 2 7 4G s H s G s H s (5.26c)

D kD

( ) ( ) ( ) ( ) ( ) ( )2 1 4 2 7 4G s H s G s H s G s H s (5.27)

Mason’s Rule

• Loops 1, 2, and its definitions, we form and Hence,

• We form by eliminating from the loop gains that touch the k thforward path:

• Expressions (5.24), (5.28), and (5.29) are now substituted into Eq.(5.23), yielding the transfer function:

(5.28)

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

2 1 4 2 7 4

2 3 4 5 6 7 8

2 1 4 2 2 1 7 4

4 2 7 4

2 1 4 2 7 4

1 [

]

[

]

G s H s G s H s G s H s

G s G s G s G s G s G s G s

G s H s G s H s G s H s G s H s

G s H s G s H s

G s H s G s H s G s H s

D = - + +

+

+ +

+

- é ùë û

D kD

kD D

( ) ( )1 7 41 G s H sD = - (5.29)

( )( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 3 4 5 7 41 1

1G s G s G s G s G s G s H sTG s

-é ù é ùD ë û ë û= =D D

(5.30)