Modern Control Engineering

50 Chapter 2 / Mathematical Modeling of Control Systems

Solution.From the figure, we obtain

(2–52)

(2–53)

By substituting Equation (2–53) into Equation (2–52), we obtain

(2–54)

By substituting Equation (2–52) into Equation (2–53), we get

(2–55)

Solving Equation (2–54) for C 1 ,we obtain

(2–56)

Solving Equation (2–55) for C 2 gives

(2–57)

Equations (2–56) and (2–57) can be combined in the form of the transfer matrix as follows:

Then the transfer functions C 1 (s)/R 1 (s),C 1 (s)/R 2 (s),C 2 (s)/R 1 (s)andC 2 (s)/R 2 (s)can be obtained

as follows:

Note that Equations (2–56) and (2–57) give responses C 1 andC 2 ,respectively, when both inputs

R 1 andR 2 are present.

Notice that when R 2 (s)=0,the original block diagram can be simplified to those shown in

Figures 2–25(a) and (b). Similarly, when R 1 (s)=0,the original block diagram can be simplified

to those shown in Figures 2–25(c) and (d). From these simplified block diagrams we can also ob-

tainC 1 (s)/R 1 (s),C 2 (s)/R 1 (s),C 1 (s)/R 2 (s),andC 2 (s)/R 2 (s),as shown to the right of each corre-

sponding block diagram.

C 2 (s)

R 1 (s)

=-

G 1 G 2 G 4

1 - G 1 G 2 G 3 G 4

,

C 2 (s)

R 2 (s)

=

G 4

1 - G 1 G 2 G 3 G 4

C 1 (s)

R 1 (s)

=

G 1

1 - G 1 G 2 G 3 G 4

,

C 1 (s)

R 2 (s)

=-

G 1 G 3 G 4

1 - G 1 G 2 G 3 G 4

B

C 1

C 2

R = D

G 1

1 - G 1 G 2 G 3 G 4

-

G 1 G 2 G 4

1 - G 1 G 2 G 3 G 4

-

G 1 G 3 G 4

1 - G 1 G 2 G 3 G 4

G 4

1 - G 1 G 2 G 3 G 4

TB

R 1

R 2

R

C 2 =

- G 1 G 2 G 4 R 1 +G 4 R 2

1 - G 1 G 2 G 3 G 4

C 1 =

G 1 R 1 - G 1 G 3 G 4 R 2

1 - G 1 G 2 G 3 G 4

C 2 =G 4 CR 2 - G 2 G 1 AR 1 - G 3 C 2 BD

C 1 =G 1 CR 1 - G 3 G 4 AR 2 - G 2 C 1 BD

C 2 =G 4 AR 2 - G 2 C 1 B

C 1 =G 1 AR 1 - G 3 C 2 B

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