
(a) Dividing through by the lead coefficient term R
1
, C
1
, R
2
, and C
2
and introducing new constants
a
1
, a
2
, b
0
, b
1
, and b
2
give
€
v
0
þ a
1
_
v
0
þ a
0
v
0
¼ b
2
€
v
i
þ b
1
_
v
i
þ b
0
v
i
(3:99)
where
a
0
¼
1
R
1
C
1
R
2
C
2
, a
1
¼
R
1
C
1
þ R
1
C
2
þ R
2
C
2
R
1
C
1
R
2
C
2
(3:100)
b
0
¼
1
R
1
C
1
R
2
C
2
, b
1
¼
R
1
C
1
þ R
2
C
2
R
1
C
1
R
2
C
2
, b
2
¼ 1(3:101)
Constructing the simulation diagram for the system starts with the following two equations, which
are equivalent to Equation 3.99 (see Section 2.4):
€
z þ a
1
_
z þ a
0
z ¼ v
i
(3:102)
v
0
¼ b
0
z þ b
1
_
z þ b
2
€
z (3:103)
Solving for
€
z in Equation 3.102 and substituting the result in Equation 3.103 yield
v
0
¼ b
0
z þ b
1
_
z þ b
2
[v
i
a
0
z a
1
_
z](3:104)
¼ (b
0
a
0
b
2
)z þ (b
1
a
1
b
2
)
_
z þ b
2
v
i
(3:105)
The simulation ...