2018哈尔滨工业大学801控制原理第七题
(1)
线性部分，系统的开环传递函数为
G
(
s
)
=
10
(
10
27
s
+
1
)
s
2
G(s)=\frac{10(\frac{10}{27}s+1)}{s^2}
G(s)=s210(2710s+1)
频率特性
G
(
j
ω
)
=
10
(
10
27
j
ω
+
1
)
−
ω
2
G(j\omega)=\frac{10(\frac{10}{27}j\omega+1)}{-\omega^2}
G(jω)=−ω210(2710jω+1)
当
ω
=
0
+
\omega=0^+
ω=0+时，
∣
G
(
j
ω
)
∣
=
∞
|G(j\omega)|=\infin
∣G(jω)∣=∞，
∠
G
(
j
ω
)
=
−
180
°
\angle G(j\omega)=-180°
∠G(jω)=−180°；当
ω
=
∞
\omega=\infin
ω=∞时，
∣
G
(
j
ω
)
∣
→
0
|G(j\omega)|\to 0
∣G(jω)∣→0，
∠
G
(
j
ω
)
=
−
90
°
\angle G(j\omega)=-90°
∠G(jω)=−90°；与实轴无交点；
非线性部分，由齿隙特性的描述函数图可知，当 A = 0.1 A=0.1 A=0.1时， N ( A ) = 0 N(A)=0 N(A)=0， − 1 N ( A ) → − ∞ -\frac{1}{N(A)}\to-\infin −N(A)1→−∞， ∠ N ( A ) = − 90 ° \angle N(A)=-90° ∠N(A)=−90°；当 A → ∞ A\to\infin A→∞， N ( A ) = 1 N(A)=1 N(A)=1， − 1 N ( A ) = − 1 -\frac{1}{N(A)}=-1 −N(A)1=−1， ∠ N ( A ) = 0 ° \angle \N(A)=0° ∠N(A)=0°，图略。
− 1 N ( A ) -\frac{1}{N(A)} −N(A)1曲线由不稳定区域穿越到稳定区域，故 G ( j ω ) G(j\omega) G(jω)曲线和 − 1 N ( A ) -\frac{1}{N(A)} −N(A)1曲线的交点为自持振荡点，会出现自持振荡。
(2)
根据
∣
G
(
j
ω
)
∣
=
∣
1
N
(
A
)
∣
,
∠
G
(
j
ω
)
=
−
π
−
∠
N
(
A
)
|G(j\omega)|=\Bigg|\frac{1}{N(A)}\Bigg|, \quad \angle G(j\omega)=-\pi-\angle N(A)
∣G(jω)∣=∣∣∣∣∣N(A)1∣∣∣∣∣,∠G(jω)=−π−∠N(A)
取
∠
N
(
A
)
=
−
45
°
\angle N(A)=-45°
∠N(A)=−45°，读取图中对应的参数
∣
N
(
A
)
∣
=
0.66
|N(A)|=0.66
∣N(A)∣=0.66
ω
=
2.7
r
a
d
/
s
∣
G
(
j
ω
)
∣
=
5.75574939
∣
1
N
(
A
)
∣
=
0.1737393226
\omega=2.7rad/s \\ |G(j\omega)|=5.75574939 \\ \Bigg|\frac{1}{N(A)}\Bigg|=0.1737393226
ω=2.7rad/s∣G(jω)∣=5.75574939∣∣∣∣∣N(A)1∣∣∣∣∣=0.1737393226
不符合；
取
∠
N
(
A
)
=
−
50
°
\angle N(A)=-50°
∠N(A)=−50°，读取图中对应的参数
∣
N
(
A
)
∣
=
0.31
|N(A)|=0.31
∣N(A)∣=0.31
ω
=
3.2177347
r
a
d
/
s
∣
G
(
j
ω
)
∣
=
3.536640707
∣
1
N
(
A
)
∣
=
0.2827541961
\omega=3.2177347rad/s \\ \quad |G(j\omega)|=3.536640707 \\ \Bigg|\frac{1}{N(A)}\Bigg|=0.2827541961
ω=3.2177347rad/s∣G(jω)∣=3.536640707∣∣∣∣∣N(A)1∣∣∣∣∣=0.2827541961
取
∠
N
(
A
)
=
−
50.5
°
\angle N(A)=-50.5°
∠N(A)=−50.5°，读取图中对应的参数
∣
N
(
A
)
∣
=
0.3
|N(A)|=0.3
∣N(A)∣=0.3
ω
=
3.455842407
r
a
d
/
s
∣
G
(
j
ω
)
∣
=
3.31941772
∣
1
N
(
A
)
∣
=
1
0.69
=
0.3012576555
\omega=3.455842407rad/s \\ \quad |G(j\omega)|=3.31941772 \\ \quad \Bigg|\frac{1}{N(A)}\Bigg|=\frac{1}{0.69}=0.3012576555
ω=3.455842407rad/s∣G(jω)∣=3.31941772∣∣∣∣∣N(A)1∣∣∣∣∣=0.691=0.3012576555
G
(
j
ω
)
G(j\omega)
G(jω)曲线和
−
1
N
(
A
)
-\frac{1}{N(A)}
−N(A)1曲线相交，读图可得
a
A
=
0.68
\frac{a}{A}=0.68
Aa=0.68
解得
A
=
a
0.68
=
0.1470588235
A=\frac{a}{0.68}=0.1470588235
A=0.68a=0.1470588235