PID控制器设计

题目：应用不同的算法给下面各个模型设计PID控制器，并比较各个控制器下闭环系统的性能

①Ga(s)=1/(s + 1)^3②Gb(s)=1/(s + 1)^5③Gc(s)=(-1.5 + 1)/(s + 1)^3

试分别利用整定公式和PID控制器设计程序设计控制器，并比较控制器的控制效果。如果采用离散PID控制器，试比较一般离散PID控制器与增量式PID控制器下的控制效果。 解：

①Ga(s)=1/(s + 1)^3 工程设计法 数学模型如下图

1>比例单独作用 G = tf(1,[1 3 3 1]); Kp = [1:1:5];

for i = 1:length(Kp)

Gc = feedback(Kp(i)*G,1); step(Gc); hold on end

>> gtext('1 Kp = 1'), >> gtext('2 Kp = 2'), >> gtext('3 Kp = 3'), >> gtext('4 Kp = 4'), >> gtext('5 Kp = 5'),

Step Response1.41.215 Kp = 5Amplitude0.84 Kp = 43 Kp = 32 Kp = 20.61 Kp = 10.40.2005101520Time (seconds)25303540 2>积分单独作用 G = tf(1,[1 3 3 1]); Kp = 1;

>> Ti = [1 :1:5]; for i = 1:length(Ti)

Gc = tf(Kp*[Ti(i) 1],[Ti(i) 0]); Gcc = feedback(G*Gc,1); step(Gcc); hold on end

>> gtext('1 Ti = 1'), >> gtext('2 Ti = 2'), >> gtext('3 Ti = 3'), >> gtext('4 Ti = 4'), >> gtext('5 Ti = 5'),

Step Response1.61.41 Ti = 11.22 Ti = 213 Ti = 3Amplitude0.84 Ti = 45 Ti = 50.60.40.2005101520Time (seconds)2530354045 3>微分单独作用 G = tf(1,[1 3 3 1]); Kp = 1; Ti = 1;

Td = [1:1:5];

for i = 1:length(Td)

Gc = tf(Kp*[Ti*Td(i) Ti 1],[Ti 0]); Gcc = feedback(G*Gc,1); step(Gcc); hold on end

gtext('1 Td = 1'), >> gtext('2 Td = 2'), >> gtext('3 Td = 3'), >> gtext('4 Td = 4'), >> gtext('5 Td = 5'),

Step Response1.41 Td = 11.22 Td = 24 Td = 413 Td = 35 Td = 50.8Amplitude0.60.40.2005101520Time (seconds)25303540 PID共同作用下，经SIMULINK仿真取Kp=2，Ti=0.8，Td=1，单位阶跃响应曲线如下

1.41.210.80.60.40.20012345678910 ②Gb(s)=1/(s + 1)^5

过程系统一阶模型近似 s = tf('s');

G = 1/(s + 1)^3;

[K1,L1,T1,G1] = getfolpd(1,G); [K3,L3,T3,G3] = getfolpd(3,G);% >> step(G) %受控对象—蓝线 >> hold

Current plot held

>> step(G1)%由响应曲线识别一阶模型—绿线 >> step(G3)%基于传递函数的辨识方法—红线

Step Response1.41.21Amplitude0.80.60.40.2005Time (seconds)1015 Ziegle-Nichols经验整定公式 s = tf('s');

>> G = 1/(s + 1)^3;

>> [Kc,b,wc,d] = margin(G); >> Tc = 2*pi/wc; >> Kp = 0.4*Kc; >> Ti = 0.8*Tc; >> [Kp,Ti];

>> G1 = Kp*(1 + tf(1,[Ti,0]));%PI >> Kp = 0.6*Kc; >> Ti = 0.5*Tc;

Step Response1.6PI1.4PID1.21Amplitude0.80.60.40.2005101520Time (seconds)25303540 ②Gb(s)=1/(s + 1)^5

过程系统一阶模型近似

Step Response1.41.210.8Amplitude带时间延迟的一阶模型近似基于传递函数的辨识方法0.6受控对象0.40.200510Time (seconds)152025 Ziegle-Nichols经验整定公式

Step Response1.41.210.8Amplitude0.60.40.20010203040Time (seconds)5060708090 ③Gc(s)=(-1.5 + 1)/(s + 1)^3 过程系统一阶模型近似

Step Response1.210.80.6Amplitude0.40.20-0.2-0.405Time (seconds)1015 Ziegle-Nichols经验整定公式

Step Response10.5Amplitude0-0.50102030Time (seconds)405060

本文来源：https://www.bwwdw.com/article/pttx.html