📘 ❞ Dc Motor M-File ❝ كتاب

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█ _ 0 حصريا كتاب Dc Motor M File 2024 File: تصميمDc باستخدام الماتلاب     مخبر التصميم بإستخدام الحاسب  تأليف: أيهم الصالح بناء النموذج الرقمي لمحرك التيار المستمر طريقة أويلر العكسية Key MATLAB commands used in this tutorial are: tf , step feedback Contents Proportional control PID control Tuning the gains From main problem, dynamic equations Laplace domain and open loop transfer function of DC are following (1)$$ s(Js + b)Theta(s) = KI(s) $$ (2)$$ (Ls R)I(s) V(s) KsTheta(s) $$ (3)$$ P(s) frac{dot{Theta}(s)}{V(s)} frac{K}{(Js b)(Ls R) K^2} qquad [frac{rad sec}{V}] $$ The structure control system has form shown figure below For original problem setup derivation above equations, please refer to Speed: System Modeling page For a 1 rad sec reference, design criteria Settling time less than 2 seconds Overshoot 5% Steady state error 1% Now let's controller using methods introduced Introduction: PID Controller Design Create new m file type J 01; b 1; K 01; R 1; L 5; s tf('s'); P_motor K ((J*s+b)*(L*s+R)+K^2); Recall that for is: (4)$$ C(s) K_{p} frac {K_{i}} {s} K_{d}s frac{K_{d}s^2 K_{p}s K_{i}} $$ Proportional control Let's first try employing proportional with gain 100, is, 100 To determine closed function, we use feedback command Add code end your Kp 100; C pid(Kp); sys_cl feedback(C*P_motor,1); Now examine response run it window You should generate plot can view some system's characteristics by right clicking on choosing Characteristics from resulting menu In below, annotations have specifically been added Settling Time, Peak Response, Steady State t 0:0 01:5; step(sys_cl,t) grid title('Step Response Proportional Control') From see both steady overshoot too large Recall increasing Kp will reduce However, also recall often results increased overshoot, therefore, appears not all requirements be met simple This fact verified experimenting different values Specifically, you employ SISO Tool entering sisotool(P_motor) then opening Analysis Plots tab Control Estimation Tools Manager With Real Time Update box checked, vary Compensator Editor effect A little experimentation verifies what anticipated, is insufficient meeting given requirements; derivative or integral terms must كتب الإلكترونيات والطاقة مجاناً PDF اونلاين قسم يختص بالكتب المتعلقة بالالكترونيات والاجهزة يحتوي القسم مجموعة من اساسيات الالكترونيات ومبادئها وعناصرها للمبتدئين والهواه هندسة تمكنك إكتساب المهارات اللازمة لهذا التخصص هذا الجزء تجد تدور حول أساسيات إلكترونيات القوى أو القدرة بالتفصيل والصور اَلْإِلِكْترُونِيَّات (بالإنجليزية: Electronics) هو مجال بدراسة الشحنات الكهربية (الإلكترونات المتحركة) الموصلات اللافلزية (غالبًا ما يُطلق عليها أشباه موصلات) حين يشير مصطلح الكهرباء إلى تدفق خلال موصلات فلزية

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Dc Motor M-File
كتاب

Dc Motor M-File

Dc Motor M-File
كتاب

Dc Motor M-File

عن كتاب Dc Motor M-File:
تصميمDc Motor M-File باستخدام الماتلاب


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تأليف:

أيهم الصالح


بناء النموذج الرقمي لمحرك التيار المستمر بإستخدام طريقة أويلر العكسية



Key MATLAB commands used in this tutorial are: tf , step , feedback

Contents

Proportional control
PID control
Tuning the gains
From the main problem, the dynamic equations in the Laplace domain and the open-loop transfer function of the DC Motor are the following.

(1)$$ s(Js + b)Theta(s) = KI(s) $$

(2)$$ (Ls + R)I(s) = V(s) - KsTheta(s) $$

(3)$$ P(s) = frac{dot{Theta}(s)}{V(s)} = frac{K}{(Js + b)(Ls + R) + K^2} qquad [frac{rad/sec}{V}] $$

The structure of the control system has the form shown in the figure below.


For the original problem setup and the derivation of the above equations, please refer to the DC Motor Speed: System Modeling page.

For a 1-rad/sec step reference, the design criteria are the following.

Settling time less than 2 seconds
Overshoot less than 5%
Steady-state error less than 1%
Now let's design a controller using the methods introduced in the Introduction: PID Controller Design page. Create a new m-file and type in the following commands.

J = 0.01;
b = 0.1;
K = 0.01;
R = 1;
L = 0.5;
s = tf('s');
P_motor = K/((J*s+b)*(L*s+R)+K^2);
Recall that the transfer function for a PID controller is:

(4)$$ C(s) = K_{p} + frac {K_{i}} {s} + K_{d}s = frac{K_{d}s^2 + K_{p}s + K_{i}} {s} $$

Proportional control

Let's first try employing a proportional controller with a gain of 100, that is, C(s) = 100. To determine the closed-loop transfer function, we use the feedback command. Add the following code to the end of your m-file.

Kp = 100;
C = pid(Kp);
sys_cl = feedback(C*P_motor,1);
Now let's examine the closed-loop step response. Add the following commands to the end of your m-file and run it in the command window. You should generate the plot shown below. You can view some of the system's characteristics by right-clicking on the figure and choosing Characteristics from the resulting menu. In the figure below, annotations have specifically been added for Settling Time, Peak Response, and Steady State.

t = 0:0.01:5;
step(sys_cl,t)
grid
title('Step Response with Proportional Control')


From the plot above we see that both the steady-state error and the overshoot are too large. Recall from the Introduction: PID Controller Design page that increasing the proportional gain Kp will reduce the steady-state error. However, also recall that increasing Kp often results in increased overshoot, therefore, it appears that not all of the design requirements can be met with a simple proportional controller.

This fact can be verified by experimenting with different values of Kp. Specifically, you can employ the SISO Design Tool by entering the command sisotool(P_motor) then opening a closed-loop step response plot from the Analysis Plots tab of the Control and Estimation Tools Manager window. With the Real-Time Update box checked, you can then vary the control gain in the Compensator Editor tab and see the resulting effect on the closed-loop step response. A little experimentation verifies what we anticipated, a proportional controller is insufficient for meeting the given design requirements; derivative and/or integral terms must be added to the controller.









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