【机械】基于simulink模拟给定系统的超前补偿器，比较补偿和未补偿系统对阶跃和斜坡输入的响应

1 内容介绍

Problem: Separately excited DC motor with a voltage of 200 V, armature current of 10.5 A, speed

of 2000 rpm, armature resistance of 0.5 ohm, armature inductance of 0.2 H, field resistance of 400

ohm, field inductance of 200 H, and moment of inertia of 3 kgm2. Load torque is assumed  

proportional to speed as 𝑇𝐿 = 𝑏𝜔 where 𝑏 = 0.0408 Nmsec/rad. Design a lead compensator for

DC motor have a phase margin of 450.

Solution: Solving this numerical we get the values of back emf (194.7 V), power (2044.87 W),

speed in rad/sec (209.43 rad/sec), field current (0.5 A), mutual inductance (1.859 H), and

electromagnetic torque (9.364 Nm).

2 部分代码

%% Program to design the Lead compensator for a given system 

% Consider a numerical from Advanced Control Theory, A Nagoor Kani

% Numerical 1.6: Design a lead compensator for a unity feedback system with

% open loop transfer function G(s)= K/(s(s+1)(s+5)) to satisfy the

% following specifications (i) velocity error constant K_v=50, (ii) Phase

% margin =20 degrees. 

% Name of the Student:

% Roll number:

close all

clear all

clc

%% Uncompensated control system transfer function (assuming H(s)=1)

% Calculate the value of gain K from e_ss and use it. 

num=[250];

den=[1 6 5 0];

G=tf(num,den);

%% Bode plot of the uncompensated system 

figure(1)

bode(G), grid on            % To check PM and GM of the uncompensated system 

title('Bode plot of uncompensated system')

[Gm,Pm,Wcg,Wcp] = margin(G);

%% Lead compensator Design

Pmd=20;                       % Desired Phase Margin (PM)

Phi_m=Pmd-Pm+30;               % Maximum phase lead angle (degree)

% Check different values for the safety factor to get the desired PM

Phi_mr=Phi_m*(pi/180);        % Maximum phase lead angle (radian)

% Determine the transfer function of the lead compensator 

alpha=(1+sin(Phi_mr))/(1-sin(Phi_mr));

Mc=-10*log10(alpha);          % Magnitude to be compensated in db

% Locate the frequency in Figure(1) for Mc

wm=28.5;

p=wm*sqrt(alpha);             % Pole of lead compensator

z=p/alpha;                    % Zero of lead compensator

gain=alpha;

numc=[1 z];

denc=[1 p];

Gc=tf(numc,denc);

% Total forward transfer function of the compensated system

Gt=gain*Gc*G;

%% Comparison of compensated and uncompensated bode plots

figure(2)

bode(G,'--r', Gt,'-'), grid on

legend('Uncompensated system', 'Compensated system')

title('Comparison of compensated and uncompensated bode plots')

%% Since H(s)=1, the feedback transfer function 

Gc1u=feedback(G,1);         % Closed loop TF of uncompensated system

Gclc=feedback(Gt,1);        % Closed loop TF of compensated system

% Comparison of compensated and uncompensated step responses

figure(3)

subplot(2,1,1)

step(Gc1u, '--r'); grid on  

title('Uncompensated system step response')

subplot(2,1,2)

step(Gclc, '-'); grid on  

title('Compensated system step response')

%% Comparison of compensated and uncompensated ramp responses

t=0:0.02:4;

figure(4)

[y1,z1]=step(250, [1 6 5 250 0],t);   % Take num and den coefficients of Gclu with a pole at origin

[y2,z2]=step([2.185e06 6.661e05], [1 2670 1.599e04 2.198e06 6.661e05 0],t);

                                % Take num and den coefficients of Gclc with a pole at the origin

subplot(2,1,1)

plot(t,y1,'.'), grid on

title('Uncompensated system ramp response')

subplot(2,1,2)

plot(t,y2,'-'), grid on

title('Compensated system ramp response')

3 运行结果

4 参考文献

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