Second order transfer function

Thread Starter

TeODP

Joined Apr 18, 2015
11
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) ???
 

WBahn

Joined Mar 31, 2012
30,075
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) ???
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?
 

Thread Starter

TeODP

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

WBahn

Joined Mar 31, 2012
30,075
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.
 

Thread Starter

TeODP

Joined Apr 18, 2015
11
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 ?
 

WBahn

Joined Mar 31, 2012
30,075
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 ?
How did you get 0.8?

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

WBahn

Joined Mar 31, 2012
30,075
no it doesnt...
im kinda confused
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.
 

Thread Starter

TeODP

Joined Apr 18, 2015
11
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.
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?
 

Papabravo

Joined Feb 24, 2006
21,227
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:

WBahn

Joined Mar 31, 2012
30,075
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....
 

Thread Starter

TeODP

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

Papabravo

Joined Feb 24, 2006
21,227
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.
 

WBahn

Joined Mar 31, 2012
30,075
thanks a lot for the help and sorry that i bothered u with my dumb questions.
btw k=0.88? ;p
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).
 

WBahn

Joined Mar 31, 2012
30,075
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​
I missed the post where he gave that information entirely.

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