Second Order Approximation to Transfer Function

In summary, the homework asks for the user to find a suitable compensator to control a system with no overshoot and stable under certain conditions. However, the given notes are insufficient and do not provide a solution.
  • #1
Illgresi
10
0
Hi all, hopefully this is in the correct section here. Any help is really gratefully received.

1. Homework Statement


I have a coursework, one question asks us to use a 2nd order approximation of the transfer function to..."estimate the settling time (5% of the settling value of output, peak time and rise time (10%-90% of the nalvalue of response) of the closed loop system with 25% of overshoot."

Unfortunately the notes given are completely insufficient and provide no examples.

Homework Equations



G(s) = (2360·K·s + 118000) / ((s + 160)·(s^2 - 1960))

The Attempt at a Solution



My first thought is to simply discard the (s + 160) term, however, this would leave only (s^2 - 1960) as the denominator, and without a middle term, the function has no damping coefficient. Without a damping coefficient the system is undamped, and therefore has no settling time!

Now, I realize I can calculate the damping ratio from the overshoot provided, however, this seems like a backward method.
 
Last edited:
Physics news on Phys.org
  • #2
Ok, I think I've figured it out. I think I was over complicating the problem in my head. Is this correct?

ξ = 0.404 (from %O = 125%)

ωn = √2360

→ Ts = 4 / (ξ·ωn) = 0.204 s

?
 
  • #3
If G(s) is a closed-loop transfer function then the response to a step input blows up (goes as ewt where w = 44.27 rad). So there can be no talk of overshoot etc.

If G(s) is an open-loop tansfer function, then the closed-loop transfer function is G/(1+G) which makes the math messy.

Sure you got the denominator right?
 
  • #4
Thanks for the reply!

The transfer function G(s) I described is the original system parameters. The coursework revolves around finding a suitable compensator to control the system. I must admit that question 7 is somewhat confusing, perhaps I could post the whole thing...

...
5. Suppose C(s) = K(s + 50). Find the range of K so that the closed loop system is stable using the Routh-Hurwitz Stability Criterion.
6. Verify your result in Step (5) using root locus.
7. By imposing the second order system approximation to the system, estimate the settling time (5% of the settling value of output, peak time and rise time (10%-90% of the final value of response) of the closed loop system with 25% of overshoot. [20%]
...

I have succesfully answered all questions bar question 7. The way in which it is worded, one could interpret that you must apply a second order approximation to the original system. However, as you rightly state, the original system is unstable and as such has no overshoot. Therefore the 2nd order approx must be applied to the PD system, i.e. C(s)*G(s). What I don't understand now is how to proceed when no value of K is provided?

Thanks!

Edit:

I should add, we've all been given an extension until the 21st because so many people struggled with Q8 which is to design a lead compensator for the system. I've done this with no problems personally, but question 7 still eludes me, as the notes we are given are really insufficient. As Q7 is probably only worth around 5% I would be happy to hand it in as is, but I'm craving to understand how the problem is solved.

Thanks again
 
Last edited:
  • #5
You should show all questions 1-8, I'm still confused as to what C(s) and G(s) are. A picture of the system would be good also.
 

Related to Second Order Approximation to Transfer Function

1. What is a second order approximation to transfer function?

A second order approximation to transfer function is a mathematical model used to approximate the behavior of a system or process. It is a simplified representation of a transfer function, which describes the relationship between the input and output of a system.

2. How is a second order approximation to transfer function calculated?

The second order approximation to transfer function is calculated by taking into account the system's natural frequency, damping ratio, and input signal. These parameters can be determined through experimentation or by analyzing the system's physical characteristics.

3. What are the advantages of using a second order approximation to transfer function?

One advantage of using a second order approximation to transfer function is its simplicity, which allows for easier analysis and understanding of a system's behavior. Additionally, it can provide a good approximation of a system's response, even for complex systems.

4. What are the limitations of a second order approximation to transfer function?

A second order approximation to transfer function is limited in its ability to accurately represent highly nonlinear systems. It also assumes that the system's input and output are linearly related, which may not always be the case. Additionally, the accuracy of the approximation may decrease as the frequency of the input signal increases.

5. How can a second order approximation to transfer function be applied in real-world situations?

A second order approximation to transfer function can be applied in various fields, such as engineering, physics, and economics, to model and predict the behavior of systems. It can be used to analyze the stability of control systems, design filters, and understand the dynamics of physical systems. It is also commonly used in signal processing and data analysis.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
12
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
15
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
931
  • Engineering and Comp Sci Homework Help
3
Replies
94
Views
11K
Replies
1
Views
3K
Back
Top