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Finally, the feedback system is reduced and multiplied by Gi(s) to yield the equivalent transfer function shown in Figure 5.10(c). Collapse sum- ming junctions; b[r]

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242 Chapter 5 Reduction of Multiple Subsystems

R(s)

G(s) R(s)G(s)

R(s)

Vfi

R(s) G(s)

G(s)

R(s)G(s)

R(s)

1

G{s) R(s)

FIGURE 5.8 Block diagram

algebra for pickoff p o i n t s

-equivalent forms for moving a

block a to the left past a

pickoff point; b to the right

past a pickoff point

m

R(s)G(s)

G(s) R(s)G(s) R(s)

R(s)G(s)

R(-i)

G(s)

G(s)

G(s)

R(s)G(s)

R(s)G(s)

R(s)G(s)

along with the forms studied earlier in this section, can be used to reduce a block diagram to a single transfer function In each case of Figures 5.7 and 5.8, the equivalence can be verified by tracing the signals at the input through to the output and recognizing that the output signals are identical For example, in Figure 5.7(a),

signals R(s) and X(s) are multiplied by G(s) before reaching the output Hence, both block diagrams are equivalent, with C(s) = R(s)G(s) ^fX{s)G(s) In Figure 5.7(b),

R(s) is multiplied by G(s) before reaching the output, but X(s) is not Hence, both

block diagrams in Figure 5.7(b) are equivalent, with C(s) = R(s)G(s) =f X(s) For

pickoff points, similar reasoning yields similar results for the block diagrams of Figure 5.8(A) and (b)

Let us now put the whole story together with examples of block diagram reduction

FIGURE 5.9 Block diagram

for Example 5.1

Example 5.1

Block Diagram Reduction via Familiar Forms

PROBLEM: Reduce the block diagram shown in Figure 5.9 to a single transfer function

R(s)

G x (s)

7 \ _ + / v A _ ,

—T

G 2 (s)

H x (s)

H 2 (s)

H 3 (s)

—*~ G 3 (5) cm

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5.2 Block Diagrams 243 SOLUTION: 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)

Second, recognize that the three feedback functions, Hi(s), H 2 (s), and H^(s), are

connected in parallel They are fed from a common signal source, and their outputs are

summed The equivalent function is Hi(s) — Hi{s) + Hs(s) Also recognize that G2(s)

and G${s) are connected in cascade Thus, the equivalent transfer function is the

product, G 3 (s)G2(s) The results of these steps are shown in Figure 5.10(6)

Finally, the feedback system is reduced and multiplied by Gi(s) to yield the

equivalent transfer function shown in Figure 5.10(c)

R(s)

+\ -*»

G 2 (s) -~

H { (s)

H 2 (s)

N^s)

G 3 (5) C(s)

(a)

R(s)

•- B X {A-H&)+H&)

ib)

m G 3 (s)G 2 (s)G } (s)

C(s)

1 + G 2 {s)G 2 (s)[H { {s) - H 2 (s) + H 3 (s)]

C(s)

(c)

FIGURE 5.10 Steps in solving Example 5.1: a Collapse sum-ming junctions; b form equi-valent cascaded system in the forward path and equivalent parallel system in the feedback path; c form equivalent feed-back system and multiply by

cascaded G t (s)

Example 5.2

Block Diagram Reduction by Moving Blocks

PROBLEM: R e d u c e t h e system shown in Figure 5.11 to a single transfer function

R(s) + , 0 ^ 1 ( 5 )

2 (s) w±/6vW

V 7 (J)

V 6 (s)

lids)

fUs)

O&i as)

w H-sis)

FIGURE 5.11 Block diagram for Example 5.2

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244 C h a p t e r 5 R e d u c t i o n of Multiple Subsystems

SOLUTION: In this example we make use of the equivalent forms shown in

Figures 5.7 and 5.8 First, move G 2 (s) to the left past the pickoff point to create

parallel subsystems, and reduce the feedback system consisting of G3(s) and H 3 (s)

This result is shown in Figure 5.12(A)

Second, reduce the parallel pair consisting of VG 2 (s) and unity, and push

Gi(s) to the right past the summing junction, creating parallel subsystems in the feedback These results are shown in Figure 5.12(6)

Bb}+/r>Mft

®m m +/Q, vm *W

vm Hi(s)

V 7 {s)

G 2 {s)

H 2 (s)

V 4 (s)

1

G 2 (s)

+

i®- G 3 (s)

l+G 3 (s)H 3 (s)

C(s)

R

^ +/Ov V ^ +

<gK G l (s)G 2 {5) V A (s)

H 2 (s)

C,(5)

tf,(s)

Gds) + 1

G 3 (s)

1 + G 3 (s)H 3 (s)

C(s)

m

R(s) +

<S<5—- G ] (s)G 2 {s) V 4 (s)

H 2 (s)

GM +#,(*)

\G 2 (s) )\l + G 3 {s)H 3 (s)

C(s)

R(s) G,(i)G 2 (5)

1 + G 2 (s)H 2 (s) + 0,(5)02(5)//,(5)

V 4 (5)

-+ 1 G3(5) \

G 2(5) ){l + G 3 (s)H 3 (s)j

C(5)

R(s) G,(5)G 3 (5)[1 + G2(5)]

[1 + G 2 (5)tf 2 (5) + G,(5)G 2 (5)//,(5)][1 + G 3 (5)// 3 (5)]

C(s)

FIGURE 5.12 Steps in t h e block diagram reduction for E x a m p l e 5.2

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