Question

Question:

CODE:

>> %% PID controller design

Kp = 65.2861;

Ki = 146.8418;

Kd = 4.0444;

Gc = pid(Kp,Ki,Kd);

% close-loop TF

T = feedback(G*Gc,1);

%% checking the design obejective

a_pid = stepinfo(T);

% Settling Time

tp_pid = a_pid.SettlingTime

% Overshhot

OS_pid = a_pid.Overshoot

%% steady-state error

[yout_pid,tout_pid] = lsim(T,stepInput,t);

% steady-state error

ess_pid = stepInput(end) - yout_pid(end);

>> %% Effect of P in G

Kp = 65.2861;

Ki = 0;

Kd = 0;

Gc = pid(Kp,Ki,Kd);

% close-loop TF

T_p = feedback(G*Gc,1);

[yout_p,tout_p] = lsim(T_p,stepInput,t);

%% Effect of I in G

Kp = 0;

Ki = 146.8418;

Kd = 0;

Gc = pid(Kp,Ki,Kd);

% close-loop TF

T_i = feedback(G*Gc,1);

[yout_i,tout_i] = lsim(T_i,stepInput,t);

%% Effect of D in G

Kp = 0;

Ki = 0;

Kd = 4.0444;

Gc = pid(Kp,Ki,Kd);

% close-loop TF

T_d = feedback(G*Gc,1);

[yout_d,tout_d] = lsim(T_d,stepInput,t);

%% plot the above response with PID control

figure(1)

plot(t,yout_p,'r')

hold on

plot(t,yout_i,'g')

hold on

plot(t,yout_d,'m')

hold on

plot(t,yout_pid,'b')

grid

legend('P','I','D','PID')

xlabel('Time(sec)')

ylabel('Amplitude')

title('Comparison study')

OUTPUT:

Figure 1 File Edit View Insert Tools Desktop Window Help I New Figure Comparison study 6 5 D PID 4 ليا 2. Amplitude -1 1 2 3 4 6 8 10 5 Time(sec)

PART A:

See the above matlab code and output explain your observation in a detailed in Theoratically basically its a design of PID controller

PART B :

Mathematically find the values of Kp ,Ki and Kd also find settling time , peak time ,rise time ,phase angle and overshoot your results must meet with the values given below in a picture

Motor transfer function will be:

s = tf('s');                                                                      

%% parameters

R = 1; % in ohm

L = 0.5; % in Henry

J = 0.01; % in kg.m^2

b = 0.1; % in Nms

K = 0.01; % in Nm/A

% compute transfer function

G = K/((J*s+b)*(L*s+R)+K^2)

G =

             0.01

----------------------------------------

0.005 s^2 + 0.06 s + 0.1001

your results will be like kp=65.2861, ki=146.8418 ,kd= 4.0444, rise time=0.174sec ,overshoot= 4.55% ,settling time=0.66 sec ,peak time =1.05

noted :Please solve mathematically in full steps that was easily understandable

Answer 1

0.01 Peak time time 0.66 su; 6.01 The TF of the System G(S) = -OT KP, kif Kd Such that rise times 00119 Soy yos: 4.554. f Bethling the we need to determine the values 1.05se; The Tf of the PID Controller is; Ge(s) = (kp + k + KBS); therefore, the Characteristics equation of the System would be; I+G(s) G((s) = 0; hence; 1+ 0.0055 + 0.065 +0.1001 -X (Kp + k + kds) -0. → 5(0.0055+ ·0055'+ 0.065 + 0.1001) + 0.01( kd kd5+ kpst ki) = 0. 70.005$+$(0.06+ 0.01 Kd) + (0.1001 + 0.01 Kb) + 0.016; = 0. = 5 + 5412+2 ka) + (2002+ 2kp) +26; =-0. - 0 Now, as per the given objectives, we determine the desired Poles by finding o Bogs the damping ratio (3) & natural Freq. of damping (Wh), then We can obtain the desired Characteristics equation. Now, as telp - 4.55%. as we unow, y up = e nafto-g2 x 100% ] = 88 Sto D4550 4:55, this, solving the above we get; 8 = 0.7 & the egne for Settling = 8.65 rad/ Su . time is ts3 on = 0.66 Solving we get; n = hence; the desired tocation Characteristics equa s² agonston 12.125+ 74-82 Comparing & & we get that all is and order System f® is and order System, hence, we need to add one more pole into & lauch that @ Sunday 26 3rd order: however; addition of pole does not change the becomes desire System dynamics Let us assume that the corresponding location of poles is at ug (greater than the dosred poles), hence, the characteristics equy would be; (Ste) (8'412.125+:14-83) - St 20.125 + 171-85+ 5986.→ 20.02 + 2 kp=1718 RK598-6 12+2 Ka - 20:12 Solving 0; get; Kd = 4:06. = 293.3. uation, ولا - up= 75.89 ki: 5986

The following matlab code can be used to validate the above answer

clear all
clc
s = tf('s');
%% parameters
R = 1; % in ohm
L = 0.5; % in Henry
J = 0.01; % in kg.m^2
b = 0.1; % in Nms
K = 0.01; % in Nm/A
% compute transfer function
G = K/((J*s+b)*(L*s+R)+K^2);
% PID controller
Kp = 75.89;
Ki = 299.3;
Kd = 4.06;
Gc = pid(Kp,Ki,Kd);

% close-loop TF
T = feedback(G*Gc,1);

%% checking the design obejective
a_pid = stepinfo(T);
% Settling Time
tp_pid = a_pid.SettlingTime
% Overshhot
OS_pid = a_pid.Overshoot

Computing the above code, we would get the folloiwng output

tp_pid =

    0.6165

OS_pid =

   15.1031