Example Problems and Solutions 51

+–

R 1 C 1

R 1

C 1

1 – G 1 G 2 G 3 G 4

G 1

G 1

G 3 G 4 – G 2

+–

R 1 C 2

G 3

G 1 –G 2 G 4

=

+–

R 2 C 2

R 2

C 2

1 – G 1 G 2 G 3 G 4

G 4

G 4

G 2 G 1 – G 3

=

R 1

C 2

1 – G 1 G 2 G 3 G 4

• 
  

  G 1 G 2 G 4

  
  

+–

R 2 C 1

G 2

G 4 – G 3 G (^1) R
2
C 1
1 – G 1 G 2 G 3 G 4

• 
  

  G 1 G 3 G 4

  
  

(a)

(b)

(c)

(d)

Figure 2–25
Simplified block
diagrams and
corresponding
closed-loop transfer
functions.

A–2–6. Show that for the differential equation system

(2–58)

state and output equations can be given, respectively, by

(2–59)

and

(2–60)

where state variables are defined by

x 3 =y$-b 0 u$-b 1 u# -b 2 u=x# 2 - b 2 u

x 2 =y# -b 0 u# -b 1 u=x# 1 - b 1 u

x 1 =y-b 0 u

y=[1 0 0]C

x 1

x 2

x 3

S +b 0 u

C

x# 1

x# 2

x# 3

S =C

0

0

• a 3

1

0

• a 2

0

1

• a 1

SC

x 1

x 2

x 3

S + C

b 1

b 2

b 3

Su

y%+a 1 y$+a 2 y# +a 3 y=b 0 u%+b 1 u$+b 2 u#+b 3 u