# How to find the value of the parameter $a$ in this transfer function?

I am given a transfer function of a second-order system as: $$G(s)=\frac{a}{s^{2}+4s+a}$$

and I need to find the value of the parameter $a$ that will make the damping coefficient $\zeta=.7$. I am not sure how to do this but I might have found something that might have helped so I am going to take a stab at it. I found a transfer function in the book of a second order spring-mass-damper system with an external applied force in the book as: $$G(s)=\frac{a}{m\omega_{n}^{2}}(\frac{\omega_{n}^{2}}{s^{2}+2\zeta \omega_{n}+\omega_{n}^{2}})$$

I was thinking that I could just write $\omega_{n}^{2}=a$ and $2\omega_{n}\zeta=4$. Then I could just solve for $a$. Is this possible?

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Is this really a homework question? It doesn't read like one at all. – David Zaslavsky♦ Mar 1 '12 at 1:35
Yes it's a homework question. I am given a transfer function and then asked to find the value of $a$ that would make $\zeta=.7$ – Greg Harrington Mar 1 '12 at 2:05
I guess you're just really good at writing questions then ;-) – David Zaslavsky♦ Mar 1 '12 at 2:19
Well i gave the question and then gave my solution and asked if it was right. Maybe it didn't look like a question to the common reader. Oops. Do you have any knowledge about this question though? – Greg Harrington Mar 1 '12 at 2:28
No "oops" needed, you certainly haven't done anything wrong here. I guess it's just different enough from the usual "how do I find velocity" type stuff that I wouldn't peg it as a homework question if you hadn't included the tag. I don't have any particular knowledge about this stuff, though. – David Zaslavsky♦ Mar 1 '12 at 2:49

## 1 Answer

You have the answer.

Consider $2*\zeta*\omega_n = 4$. $\zeta = 0.7$. $\omega_n^2 = a$ What value of $\omega_n$ (or $a$ in your case) satisfies this?

It wouldn't be 2.86 because $\omega_{n}=\sqrt{a}$. I got $a=8.163$. But thank you for letting me know that I did it right. I didnt think that I could equate it like that – Greg Harrington Mar 1 '12 at 2:05
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