Second Order Approximation to Transfer Function

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The discussion revolves around a coursework question requiring the use of a second-order approximation of a transfer function to estimate system parameters like settling time, peak time, and rise time with a specified overshoot. The original transfer function is deemed unstable, complicating the application of the second-order approximation. The participant realizes that the approximation must be applied to a compensator-controlled system rather than the original unstable system. There is confusion regarding the lack of a provided gain value (K) for calculations. Despite the challenges, the participant expresses a desire to understand the problem fully, indicating a willingness to submit incomplete work if necessary.
Illgresi
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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.
 
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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

?
 
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?
 
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
 
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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.
 

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