# Second order transfer function

Discussion in 'Homework Help' started by TeODP, Apr 18, 2015.

1. ### TeODP Thread Starter New Member

Apr 18, 2015
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0
Hello guys.So i have this transfer function G(s)= 7.8 / ( 14.99*(s^2) + 13.33*s+8.8 ) and i need to find the gain K and the damping ratio ζ .What should i do? When i had an one order tranfer function i compared it with the the classic type k/(τs+1) . Should i do the same now ? I mean , i have to compare it with κ*(ω^2) / (s^2) + 2*ζ*ω*s+(ω^2) ???

2. ### Papabravo Expert

Feb 24, 2006
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Is your G(s) the open loop transfer function, or the closed loop transfer function?

3. ### WBahn Moderator

Mar 31, 2012
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Assuming that this is the total transfer function (i.e., closed loop, if that's relevant), then why wouldn't you put it into standard form so that you can pull the various pieces out?

4. ### TeODP Thread Starter New Member

Apr 18, 2015
11
0
its a closed loop tranfer function. thats what i have.
Gm(s) = 10/ 14.99*(s^2)+13.33*s+1 , Gs(s) = 1 , Gc(s) = 0.78

5. ### TeODP Thread Starter New Member

Apr 18, 2015
11
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what do u mean by standard form ?

6. ### WBahn Moderator

Mar 31, 2012
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You called it "classic form".

7. ### TeODP Thread Starter New Member

Apr 18, 2015
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oooh , so i have to use the type (b^2-4*a*c) (i dont know how its called in english) on the denominator in order to separate the fraction and after that i can use the standard form ?

8. ### WBahn Moderator

Mar 31, 2012
19,125
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Why do you need to do that (that is called the 'discriminant', by the way)?

Just manipulate the coefficients so that they match the standard form. The most important being that the s^2 term in the denominator has a coefficient of 1.

Before you start, you should look at what the value of K is going to HAVE to be by inspection. Hint, it's a little less than 1.

9. ### TeODP Thread Starter New Member

Apr 18, 2015
11
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When i compared the G(s) with the standard form i found that K is 0.8 . Well, im not sure but is the denominator going to be like that s*(14.99*s+13.33)+8.8 ?

10. ### WBahn Moderator

Mar 31, 2012
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How did you get 0.8?

Does the standard form look like s(a·s+b) + c?

11. ### TeODP Thread Starter New Member

Apr 18, 2015
11
0
no it doesnt...
im kinda confused

12. ### WBahn Moderator

Mar 31, 2012
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Well, how did you do it for a first order circuit?

$
H(s) = \frac{A}{Bs+C} = \frac{K}{\tau s + 1}
$

You see that you need the constant term to be 1, so you divide top and bottom by C

$
H(s) = \frac{$$\frac{A}{C}$$}{$$\frac{B}{C}$$s+1} = \frac{K}{\tau s + 1}
$

And now you can match things up and see that

$
K = \frac{A}{C}
\tau = \frac{B}{C}
$

Just do the same thing.

13. ### TeODP Thread Starter New Member

Apr 18, 2015
11
0
sorry but i didnt know that i always must have the +1 ,i thought that it was just a number ...
thank u so much about that
i need to ask something more , how can i find the damping ratio ζ?

14. ### Papabravo Expert

Feb 24, 2006
10,688
1,983
When put into the proper form the parameter K appears in the denominator which is what makes the root locus interesting. That is why I asked my original question. The form you used in your original post looked like the the open loop forward transfer function. When placed in a unity feedback configuration, without a transfer function in the feedback loop, the form of the closed loop function becomes

G(s) / (1 + G(s)) or, if K is shown explicitly K*G(s) / (1 + K*G(s))

and now the gain parameter occurs in both the numerator and the denominator.

You also have to make the term for s^2 equal to 1 as well for a second order system.

Check out the following:
http://www.egr.msu.edu/classes/me451/zhug/2009F/support_docs/RL_rules.pdf
Check the block diagram on page 2 and tell me if that is the picture you have of asystem with an adjustable gain K, with H(s) = 1, or H(s) = something else.

For K=0 the open loop function is 0 and the closed loop function is 0/1 = 0
For K=∞, the open loop function → ∞, but the closed loop function approaches 1
The locus of the roots as a function of K begins at the open loop poles, (for K=0) and ends at the open loop zeros, ( as K → ∞), including any zeros at infinity.

In the more complicated case of a transfer function in the feedback loop, call it H(s), you have

K*G(s)*H(s) / (1 + K*G(s)*H(s)) for the closed loop function

You are still interested in the open loop poles and the open loop zeros. Once you have the root locus you can find the intersection of the root locus with other points of interest like a particular damping factor or settling time or percent overshoot.

Last edited: Apr 19, 2015
15. ### WBahn Moderator

Mar 31, 2012
19,125
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If you get it into standard form you know what the natural frequency is, right? You know what the product of the damping factor and the natural frequency is, right? Well....

16. ### TeODP Thread Starter New Member

Apr 18, 2015
11
0
thanks a lot for the help and sorry that i bothered u with my dumb questions.
btw k=0.88? ;p

17. ### TeODP Thread Starter New Member

Apr 18, 2015
11
0
i still cant figure out how ill create the standard form through the second order transfer function. =/
mby im just sleepy...

18. ### Papabravo Expert

Feb 24, 2006
10,688
1,983
This was the transfer function from your first post

G(s)= 7.8 / ( 14.99*(s^2) + 13.33*s+8.8 )
I asked you if it was open loop or closed loop and you answered that it was closed loop, with the following components.

Gm(s) = 10/ 14.99*(s^2)+13.33*s+1
Gs(s) = 1
Gc(s) = 0.78​

The Gs(s) = 1 is the transfer function in the feedback path, so this is a unity feedback system.
The Gc(s) = 0.78 looks like a fixed gain block, as opposed to a parameterized gain block which I have been calling K and which is generally used to describe root locus plots.
The Gm(s) = ... is the transfer function of the "plant".

Nowhere in what you have written down is a parameter K shown explicitly. I have not done the algebra myself, but I'm sure you can demonstrate the correspondence between the closed loop function G(s) and the three components of G(s), namely Gm(s), Gs(s), and Gc(s). If the block labeled Gc(s) is to be replaced by the parameter K, then that is one kind of problem. If on the other hand the Gc(s) = 0.78 is fixed then, that is another kind of problem. I think your case is the latter case of Gc(s) = 0.78 being fixed and you have to transform the closed loop function from your original post into a standard form to extract the information you seek. I apologize for introducing confusing information if that is the case.

19. ### WBahn Moderator

Mar 31, 2012
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You didn't bother anyone with dumb questions. If we just gave you the answer, or showed you how to do it, then you probably wouldn't learn as much. You have to fight with this stuff.

Yes, K=0.88 (for the transfer function in your original post). You know that K is the DC gain and so you set s=0 and see what survives. Then you perform your manipulations on it and if K doesn't end up being what you know it HAS to be, then you know you have messed up somewhere (which we all do on an all-too-frequent basis).

20. ### WBahn Moderator

Mar 31, 2012
19,125
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I missed the post where he gave that information entirely.

Yes, those components yield the closed loop transfer function given in the original post.