Damping factor - match resistor value
I hope someone can help me with this


For a parallel R and C, quality factor is: -

$$Q = R\sqrt{\dfrac{C}{L}}$$

And damping ratio \$\zeta\$ is $$\dfrac{1}{2Q}$$ 

All formulas from this wiki page.

If you want an interactive tool that gives you the numbers try this. It tells me that to obtain a damping ratio of 0.9 requires a resistor of value 2.2 ohms: -



Solving the TF (assuming one resistor called R)

$$H(s) = \dfrac{R||X_C}{X_L + R||X_C} = \dfrac{\frac{R}{sC}}{sL(R + \frac{1}{sC})+ \frac{R}{sC}} = \dfrac{R}{s^2RLC +sL + R}$$

At this point you re-arrange to try and fit into the standard equation for a 2nd order low pass filter. Standard equation that I use is this: -

$$H(s) = \dfrac{\omega_n^2}{s^2 +s2\zeta\omega_n + \omega_n^2}$$

So, dividing the original equation's numerator and denominator by RLC gives this: -

$$H(s) = \dfrac{\frac{1}{LC}}{s^2 + \frac{s}{RC} +\frac{1}{LC}}$$

Hence \$\omega_n = \dfrac{1}{\sqrt{LC}}\$ and \$2\zeta\omega_n = \dfrac{1}{CR}\$.

If you rearrange things you get this: -

$$2\zeta = \dfrac{\sqrt{LC}}{RC} = \dfrac{1}{Q}$$

Hence: -

$$Q = R\sqrt{\frac{C}{L}}$$