2nd Order Control system PD controller

In summary: You can try plotting your solution and see if it meets the design criteria. Alternatively, you can run a simulation to see if your KP and KI values meet the criteria for rise time and settling time.
  • #1
DC2R
2
0
Hi,

This question on PD control is from a practice quiz.

1. Homework Statement
upload_2016-4-28_10-3-23.jpeg

If you can't see it- the question asks to find values for Kp and Kd such that the system achieves 5% OS and has a settling time Ts of 3s.
Cs = 3
Cd = 2
m = 5

Homework Equations


ω_n^2/(s^2 + 2ζw_n + ω_n^2) - 2nd order form

Overall Transfer Function I found:
Y/R = CG/1+CG = [ (Kp + Kds)/m ] / [ s^2 + ((Cd + Kd)/m) + ((Cs + Kp)/m)]

Ts = 4/σ (2% of final value); where σ = ζω_n

The Attempt at a Solution


From the transfer function I found above i noticed that it is not exactly in the 2nd order form. However I was told that we still set

ω_n = SQRT[ ((Cs + Kp)/m) ] {1}
because when we add a derivative controller it only adds a zero which effects the numerator, not the poles on the denominator.

firstly we know for 5%OS → ζ≈0.7
from Ts = 3 = 4/ζω_n
→ ω_n = 4/3ζ = 40/21

from eqn {1} - solving for Kp
Kp = (ω_n^2)m - Cs = (5)(40/21)^2 - 3 = 15.14

At this point I stopped because their answers were Kp = 17.93 and Kd = 12.12.
Could you explain if the process i used is incorrect because I can't understand what I am doing wrong.

Thank you
(btw I just joined Physics Forums like an hour ago)
 

Attachments

  • problem.JPG
    problem.JPG
    32.2 KB · Views: 1,156
Physics news on Phys.org
  • #2
There are often many ways to solve control problems. Plot your solution. Does it meet the design criteria?
 
  • #3
hi, can someone please have a look at my working to see if I am on the right track, I spent many hours on it and I am not even sure if this is the correct way.

donpacino, I am a beginner in Matlab, I am not sure how a plot would verify my results.
 
  • #4
DC2R said:
hi, can someone please have a look at my working to see if I am on the right track, I spent many hours on it and I am not even sure if this is the correct way.

donpacino, I am a beginner in Matlab, I am not sure how a plot would verify my results.
run a simulation to see if your KP and KI values meet the criteria for rise time and settling time.

There are often multiple solutions that will work for problems like this
 

1. What is a 2nd Order Control system PD controller?

A 2nd Order Control system PD controller is a type of control system used in engineering and automation that uses a Proportional-Derivative feedback mechanism to control a system's output. It is based on a second-order differential equation and is commonly used in systems where high precision and fast response times are required.

2. How does a 2nd Order Control system PD controller work?

A 2nd Order Control system PD controller works by continuously monitoring the system's output and calculating the error between the desired output and the actual output. It then uses this error to adjust the system's control input using both proportional and derivative control actions, which help to minimize the error and stabilize the system's output.

3. What are the advantages of using a 2nd Order Control system PD controller?

There are several advantages to using a 2nd Order Control system PD controller, including its ability to provide fast response times and accurate control in complex systems. It also has the ability to eliminate steady-state errors and is robust to disturbances, making it a popular choice for many control applications.

4. Are there any limitations to using a 2nd Order Control system PD controller?

While a 2nd Order Control system PD controller has many advantages, it also has some limitations. One limitation is that it is sensitive to changes in system parameters, which can affect its performance. It also does not take into account any future changes in the system, making it less effective in dynamic systems.

5. How is a 2nd Order Control system PD controller different from other types of controllers?

A 2nd Order Control system PD controller is different from other types of controllers because it uses a combination of proportional and derivative control actions, whereas other controllers may only use one or the other. It also takes into account the rate of change of the error, allowing for faster and more precise control of the system's output.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
707
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
765
  • Electrical Engineering
Replies
2
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
3
Replies
94
Views
10K
Back
Top