过程控制_金以慧_第三章课后习题答案

3.7

1.t=[0:5:85];

y=[0.650,0.651,0.652,0.668,0.735,0.817,0.881,0.979,1.075,1.151,1.213,1.239,1.26 2,1.311,1.329,1.338,1.350,1.351];

deltu=5

length(y)

ywd=1.351;

delty=ywd-y(1);%y的变化值

K=delty/deltu

y=(y-y(1))/delty; %标幺化

plot(t,y);

结果如下：K =0.1402

由下图曲线用切线法可以得到:

T=35.85 τ=18.23

2.用动态特性参数法选择PID参数：

ε=K/T=0.1402/35.85=0.0039

δ=0.85*ε*τ=0.85*0.0039*18.23=0.0604

Ti=2.0*τ=2.0*18.23=36.46

Td=0.5*τ=9.115

3.11（a）、

得到:Kc=0.23 Tcr=11.8

δcr=1/Kc=1/0.23=4.35

δ=2.2δcr=2.2*4.35=9.565;

Ti=0.85*Tcr=0.85*11.8=10.03

(b)、

得到:Kc=2.06 Tcr=2.4

δcr=1/Kc=1/2.06=0.485

δ=1.67δcr=1.67*0.485=0.81; Ti=0.5*Tcr=0.5*2.4=1.2

Td=0.125*Tcr=0.3

3.13

1.用边界稳定法求

得到Kc=1.97 Tcr=3.2 δcr=1/Kc=1/1.97=0.508 2.δ=1.67*δcr=0.848

Ti=0.5*Tcr=1.6

TD=0.125*Tcr=0.4

3.α=1.07

F=1+α1+1.07=1.2675

δ=F=1.2675*0.848=1.0748

Ti===1.2623

Td==1.2675*0.4=0.507

3.14

1.用matlab绘图

deltu=0.02;

t=[0,10,20,40,60,80,100,140,180,250,300,350];

h=[0,0.01,0.5,2,5,9,13.5,22,29.5,36,39,39.5];

hwd=39.5;%y的稳定值

delth=hwd-h(1);%y的变化值

K=delth/deltu

h=h/delth; %标幺化并去除延迟t=t-10;

plot(t,h);

结果如下：K = 1975

由下图曲线用切线法可以得到

T =176.97 τ=30.73 (1)、调节器整定参数:

得到Kc=0.0049 Tcr=112 δcr=1/Kc=1/0.0049=204.1 δ=2.2δcr=2.2*204.1=449.02;

Ti=0.85*Tcr=0.85*112=95.2