题目:
G
(
s
)
=
1
s
2
(
s
+
1
)
(
2
s
+
1
)
G(s)=\frac{1}{s^2(s+1)(2s+1)}
G(s)=s2(s+1)(2s+1)1
1. 正常的解题
G
(
s
)
=
1
s
2
(
s
+
1
)
(
2
s
+
1
)
G(s)=\frac{1}{s^2(s+1)(2s+1)}
G(s)=s2(s+1)(2s+1)1
解: 第一步:
G
(
j
ω
)
=
1
(
j
ω
)
2
(
j
ω
+
1
)
(
2
j
ω
+
1
)
=
−
1
ω
∗
1
ω
∗
1
j
ω
+
1
∗
1
2
j
ω
+
1
=
1
ω
2
∗
ω
2
+
1
∗
(
2
ω
)
2
+
1
e
−
π
2
−
π
2
−
a
r
c
t
a
n
ω
−
a
r
c
t
a
n
2
ω
=
1
ω
2
∗
ω
2
+
1
∗
(
2
ω
)
2
+
1
e
−
π
−
a
r
c
t
a
n
ω
−
a
r
c
t
a
n
2
ω
G(jω)=\frac{1}{(jω)^2(jω+1)(2jω+1)}=-\frac{1}{ω}*\frac{1}{ω}*\frac{1}{jω+1}*\frac{1}{2jω+1}=\frac{1}{ω^2*\sqrt{ω^2+1}*\sqrt{(2ω)^2+1}}e^{-\frac{π}{2}-\frac{π}{2}-arctanω-arctan2ω}=\frac{1}{ω^2*\sqrt{ω^2+1}*\sqrt{(2ω)^2+1}}e^{-π-arctanω-arctan2ω}
G(jω)=(jω)2(jω+1)(2jω+1)1=−ω1∗ω1∗jω+11∗2jω+11=ω2∗ω2+1∗(2ω)2+11e−2π−2π−arctanω−arctan2ω=ω2∗ω2+1∗(2ω)2+11e−π−arctanω−arctan2ω
∴
∣
G
(
j
ω
)
∣
=
1
ω
2
∗
ω
2
+
1
∗
(
2
ω
)
2
+
1
\therefore|G(jω)|=\frac{1}{ω^2*\sqrt{ω^2+1}*\sqrt{(2ω)^2+1}}
∴∣G(jω)∣=ω2∗ω2+1∗(2ω)2+11
∠
G
(
j
ω
)
=
−
π
−
a
r
c
t
a
n
ω
−
a
r
c
t
a
n
2
ω
\angle{G(jω)}=-π-arctanω-arctan2ω
∠G(jω)=−π−arctanω−arctan2ω