Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 1 down vote favorite

I found the transfer function for the spring mass damper system to be $$G(s)=\frac{1}{ms^{2}+bs+k},$$ and now I need to find the gain of this transfer function. I know that the gain is $G=|G(j\omega)|$, but I'm not really sure how to go about finding the gain of a transfer function with a quadratic term in the denominator.

share|improve this question
Get those $j$'s out of here! I won't abide it. Seriously, it took me several minutes to realize what was meant. – BebopButUnsteady Mar 27 '12 at 6:12

2 Answers

up vote 1 down vote accepted

$$G=\left|G(j\omega)\right|=\left|\frac{1}{m(j\omega)^{2}+bj\omega+k}\right|$$

$$=\left|\frac{1}{-m\omega^2+bj\omega+k}\right|$$

$$=\frac{1}{\left|bj\omega-m\omega^2+k\right|}$$

$$=\frac{1}{\sqrt{b^2\omega^2+(-m\omega^2+k)^2}}$$

Maybe?

share|improve this answer
That makes sense because I just read something where if the numerator is unity, like it is, then you just some the squares of the real and imaginary parts of the denominator and square root them. I think it would work – Greg Harrington Mar 27 '12 at 5:36
Is there a way to find the phase of this transfer function as well? – Greg Harrington Mar 27 '12 at 5:47
@GregHarrington This is a transfer function that transforms one function into another, so I take it you mean the change in phase. Wikipedia calls it "phase shift" and gives $\phi(\omega) = \arg(Y) - \arg(X) = \arg( G(j \omega))$, which isn't surprising. Since we speak of a single frequency input, the transfer function is reduced to a single complex number that transforms $X$ into $Y$ (multiply). This complex number would be simply $G(j\omega)$. – AlanSE Mar 27 '12 at 6:22
1  
The phase is the inverse tangent of the imaginary part divided by the real part, or atan2(imaginary part, real part). – nibot Mar 27 '12 at 11:50
up vote 0 down vote

By following the rules of arithmetic for complex numbers, you can evaluate the magnitude and phase of any rational transfer function. For a quick introduction to the arithmetic of complex numbers, the wikipedia article is decent:

Explicitly finding expressions for the magnitude and phase of a transfer function is something that's good to do once, but its not something we often do in practice.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.