mirror of
https://github.com/gabrielkheisa/control-system.git
synced 2024-11-23 20:03:22 +07:00
Add assignment five
This commit is contained in:
parent
5f1fa203da
commit
fc83b04f43
93
Assignment 5 - PD/README.md
Normal file
93
Assignment 5 - PD/README.md
Normal file
@ -0,0 +1,93 @@
|
|||||||
|
# Derivative Effect on Control System
|
||||||
|
This dir is belong to Control System class contains with Derivative Effect on Control System. This code 100% original made by my hand :), please leave some notes if you're going to use it. Thanks!
|
||||||
|
|
||||||
|
## Software
|
||||||
|
This program ran in Matlab
|
||||||
|
|
||||||
|
## Variables
|
||||||
|
`s = tf('s');` defines `s` as 'frequency domain' for transfer function and will be used further.
|
||||||
|
```
|
||||||
|
J = 0.01;
|
||||||
|
b = 0.1;
|
||||||
|
K = 0.01;
|
||||||
|
R = 1;
|
||||||
|
L = 0.5;
|
||||||
|
```
|
||||||
|
Those variable comes from BLDC control system.
|
||||||
|
```
|
||||||
|
Kp = 1;
|
||||||
|
Kd = 1;
|
||||||
|
% Kd = 3;
|
||||||
|
% Kd = 5;
|
||||||
|
% Kd = 7;
|
||||||
|
% Kd = 9;
|
||||||
|
```
|
||||||
|
Variable above is the constant from PD control, we're trying to varies the constant to analyze derivative effect on control system
|
||||||
|
|
||||||
|
## Process
|
||||||
|
The BLDC motor control system should be defined as transfer function by initialize its numerator-denumerator and *tf()* function.
|
||||||
|
```
|
||||||
|
num_motor = [K];
|
||||||
|
den_motor = [J*L J*R+b*L R*b+K*K];
|
||||||
|
|
||||||
|
motor = tf(num_motor,den_motor)
|
||||||
|
```
|
||||||
|
Besides the plant function, the PD-control system defined by `C = tf([Kd Kp 0],[0 1 0])`. The vector is set according to PD formula which `PD = Kp + Kd * s`. After that, both of system are multiplied each others without feedback by `complete = feedback(motor*C,1);`
|
||||||
|
|
||||||
|
That system will be test with step, ramp, and impulse input by call below lines
|
||||||
|
```
|
||||||
|
subplot(311), impulse(complete); % Impulse reponse
|
||||||
|
subplot(312), step(complete); % Step Response
|
||||||
|
subplot(313), step(complete / s); % Ramp response
|
||||||
|
stepinfo(complete)
|
||||||
|
```
|
||||||
|
|
||||||
|
Since Matlab doesn't provide any steady-state error calculation, we process it by call below lines
|
||||||
|
```
|
||||||
|
[y,t] = step(complete); % Calculate Steady-State error
|
||||||
|
sse = abs(1 - y(end))
|
||||||
|
```
|
||||||
|
Last line works to limit the graph
|
||||||
|
```
|
||||||
|
xlim([0 50])
|
||||||
|
ylim([0 3])
|
||||||
|
```
|
||||||
|
|
||||||
|
|
||||||
|
## Testing
|
||||||
|
For Kp = 1
|
||||||
|
| Param | Kd = 1 | Kd = 3 | Kd = 5 | Kd = 7 | Kd = 9 |
|
||||||
|
|--- |--- |--- |--- |--- |--- |
|
||||||
|
| Rise Time | 0.0540 | 0.0140 | 0.0081 | 0.0057 | 0.0044 |
|
||||||
|
| Settling Time | 2.1356 | 3.2085 | 3.9313 | 4.6494 | 5.3646 |
|
||||||
|
| Overshoot | 50.9930 | 232.5791 | 359.4791 | 452.0385 | 522.2002 |
|
||||||
|
| SSE | 0.9088 | 0.9077 | 0.9075 | 0.9075 | 0.9069 |
|
||||||
|
|
||||||
|
### Kp = 1, Kd = 1
|
||||||
|
|
||||||
|
![Kp = 1, Kd = 1](https://user-images.githubusercontent.com/77116615/189711219-a3d00c73-0902-4a3f-899c-aca785cf0e6c.png)
|
||||||
|
### Kp = 1, Kd = 3
|
||||||
|
|
||||||
|
![Kp = 1, Kd = 3](https://user-images.githubusercontent.com/77116615/189711238-016657a1-8b3d-4209-8033-192464c00bed.png)
|
||||||
|
### Kp = 1, Kd = 5
|
||||||
|
|
||||||
|
![Kp = 1, Kd = 5](https://user-images.githubusercontent.com/77116615/189711244-acaa9905-21dc-4e80-915e-c628775a6665.png)
|
||||||
|
### Kp = 1, Kd = 7
|
||||||
|
|
||||||
|
![Kp = 1, Kd = 7](https://user-images.githubusercontent.com/77116615/189711248-cdac4205-6511-460d-8eb8-96378564774e.png)
|
||||||
|
### Kp = 1, Kd = 9
|
||||||
|
|
||||||
|
![Kp = 1, Kd = 9](https://user-images.githubusercontent.com/77116615/189711253-c2a2208c-fdb8-4640-b419-b2909250ed0e.png)
|
||||||
|
|
||||||
|
## Conclusion
|
||||||
|
Based on previous tests, we conclude that by adding Integral constant :
|
||||||
|
* Risie time is **increased**
|
||||||
|
* Settling time is **increased**
|
||||||
|
* Overshoot is **decreased**
|
||||||
|
* SSE is **decreased**
|
||||||
|
|
||||||
|
### Notes
|
||||||
|
Contact nanda.r.d@mail.ugm.ac.id for more information
|
||||||
|
### Links
|
||||||
|
You can access the source code here
|
||||||
|
[github.com/nandard/control-system.git](https://github.com/nandard/control-system.git)
|
33
Assignment 5 - PD/week_5.m
Normal file
33
Assignment 5 - PD/week_5.m
Normal file
@ -0,0 +1,33 @@
|
|||||||
|
s = tf('s');
|
||||||
|
J = 0.01;
|
||||||
|
b = 0.1;
|
||||||
|
K = 0.01;
|
||||||
|
R = 1;
|
||||||
|
L = 0.5;
|
||||||
|
|
||||||
|
Kp = 1;
|
||||||
|
%Kd = 1;
|
||||||
|
%Kd = 3;
|
||||||
|
%Kd = 5;
|
||||||
|
%Kd = 7;
|
||||||
|
Kd = 9;
|
||||||
|
|
||||||
|
num_motor = [K];
|
||||||
|
den_motor = [J*L J*R+b*L R*b+K*K];
|
||||||
|
motor = tf(num_motor,den_motor)
|
||||||
|
|
||||||
|
C = tf([Kd Kp 0],[0 1 0])
|
||||||
|
|
||||||
|
complete = feedback(motor*C,1);
|
||||||
|
|
||||||
|
subplot(311), impulse(complete); % Impulse reponse
|
||||||
|
subplot(312), step(complete) % Step Response
|
||||||
|
subplot(313), step(complete / s); % Ramp response
|
||||||
|
title("Ramp Response");
|
||||||
|
stepinfo(complete)
|
||||||
|
|
||||||
|
[y,t] = step(complete); % Calculate Steady-State error
|
||||||
|
sse = abs(1 - y(end))
|
||||||
|
|
||||||
|
xlim([0 3])
|
||||||
|
ylim([0 3])
|
Loading…
Reference in New Issue
Block a user