2018南京航天航空大学920自动控制原理第三题
设
P
I
PI
PI控制器为
G
c
(
s
)
=
K
p
(
T
i
s
+
1
)
T
i
s
G_c(s)=\frac{K_p(T_is+1)}{T_is}
Gc(s)=TisKp(Tis+1)
系统的开环传递函数
G
(
s
)
=
10
K
p
(
T
i
s
+
1
)
T
i
s
(
s
+
1
)
(
2
s
+
1
)
G(s)=\frac{10K_p(T_is+1)}{T_is(s+1)(2s+1)}
G(s)=Tis(s+1)(2s+1)10Kp(Tis+1)
系统的闭环传递函数
Φ
(
s
)
=
10
K
p
(
T
i
s
+
1
)
T
i
s
(
s
+
1
)
(
2
s
+
1
)
+
10
b
K
p
(
T
i
s
+
1
)
\Phi(s) = \frac{10K_p(T_is+1)}{T_is(s+1)(2s+1)+10bK_p(T_is+1)}
Φ(s)=Tis(s+1)(2s+1)+10bKp(Tis+1)10Kp(Tis+1)
扰动传递函数
Φ
N
(
s
)
=
10
T
i
s
(
s
+
1
)
+
10
T
i
s
G
N
(
s
)
T
i
s
(
s
+
1
)
(
2
s
+
1
)
+
10
b
K
p
(
T
i
s
+
1
)
\Phi_N(s)=\frac{10T_is(s+1)+10T_isG_N(s)}{T_is(s+1)(2s+1)+10bK_p(T_is+1)}
ΦN(s)=Tis(s+1)(2s+1)+10bKp(Tis+1)10Tis(s+1)+10TisGN(s)
1.
输出变换
c
(
t
)
=
2
−
4
3
e
−
0.5
t
s
i
n
(
3
2
t
+
π
3
)
c(t)=2-\frac{4}{\sqrt3}e^{-0.5t}sin(\frac{\sqrt3}{2}t+\frac{\pi}{3})
c(t)=2−34e−0.5tsin(23t+3π)
设系统闭环传递函数为
Φ
(
s
)
=
K
s
2
+
2
ζ
ω
n
s
+
ω
n
2
\Phi(s)=\frac{K}{s^2+2\zeta\omega_ns+\omega_n^2}
Φ(s)=s2+2ζωns+ωn2K
由输出表达式可知，该系统式二阶系统，且输出稳态值为输入的两倍，则可得到以下方程
{
K
=
2
ω
n
2
1
−
ζ
2
=
3
2
−
ζ
ω
n
=
−
0.5
ω
n
1
−
ζ
2
=
3
2
\begin{cases} K=2\omega_n^2\\ \sqrt{1-\zeta^2}=\frac{\sqrt3}{2} \\ -\zeta\omega_n=-0.5 \\ \omega_n \sqrt{1-\zeta^2}=\frac{\sqrt3}{2} \end{cases}
⎩⎪⎪⎪⎨⎪⎪⎪⎧K=2ωn21−ζ2=23−ζωn=−0.5ωn1−ζ2=23
解得
{
ζ
=
0.5
ω
n
=
1
\begin{cases} \zeta= 0.5\\ \omega_n=1 \end{cases}
{ζ=0.5ωn=1
系统的闭环传递函数为
Φ
(
s
)
=
2
s
2
+
s
+
1
\Phi(s)=\frac{2}{s^2+s+1}
Φ(s)=s2+s+12
与原式对比，①
T
i
=
1
T_i=1
Ti=1
Φ
(
s
)
=
5
K
p
/
T
i
s
2
+
0.5
s
+
5
b
K
p
/
T
i
\Phi(s)=\frac{5K_p/T_i}{s^2+0.5s+5bK_p/T_i}
Φ(s)=s2+0.5s+5bKp/Ti5Kp/Ti
不成立
②
T
i
=
2
T_i=2
Ti=2
{
b
=
0.5
K
p
=
0.4
\begin{cases} b=0.5 \\ K_p=0.4 \end{cases}
{b=0.5Kp=0.4
2.
由稳态输出表达式可知，系统的扰动误差为零，则
G
N
(
s
)
=
−
(
s
+
1
)
G_N(s)=-(s+1)
GN(s)=−(s+1)
输出幅值是输入的两倍，且相位差为
90
°
90°
90°，即
{
A
(
1
)
=
K
(
ω
n
2
−
1
)
2
+
4
ζ
2
ω
n
2
=
2
φ
(
1
)
=
−
arctan
2
ζ
ω
n
ω
n
2
−
1
=
−
90
°
\begin{cases} A(1)=\frac{K}{\sqrt{(\omega_n^2-1)^2+4\zeta^2\omega_n^2}}=2 \\ \varphi(1)=-\arctan\frac{2\zeta\omega_n}{\omega_n^2-1}=-90° \end{cases}
⎩⎨⎧A(1)=(ωn2−1)2+4ζ2ωn2K=2φ(1)=−arctanωn2−12ζωn=−90°
解得
{
K
=
2
ζ
=
0.2
ω
n
=
1
\begin{cases} K=2 \\ \zeta=0.2 \\ \omega_n=1 \end{cases}
⎩⎪⎨⎪⎧K=2ζ=0.2ωn=1
综上
{
G
N
(
s
)
=
−
(
s
+
1
)
G
c
(
s
)
=
0.4
(
2
s
+
1
)
2
s
b
=
0.5
\begin{cases} G_N(s)=-(s+1) \\ G_c(s)=\frac{0.4(2s+1)}{2s} \\ b=0.5 \end{cases}
⎩⎪⎨⎪⎧GN(s)=−(s+1)Gc(s)=2s0.4(2s+1)b=0.5
本文地址：https://blog.csdn.net/qq_41062383/article/details/109006905