# Second Order Approximation to Transfer Function

1. Dec 13, 2015

### Illgresi

Hi all, hopefully this is in the correct section here. Any help is really gratefully received.

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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 realise I can calculate the damping ratio from the overshoot provided, however, this seems like a backward method.

Last edited: Dec 13, 2015
2. Dec 13, 2015

### Illgresi

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. Dec 15, 2015

### rude man

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. Dec 15, 2015

### Illgresi

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...

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: Dec 15, 2015
5. Dec 16, 2015

### rude man

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.