Damping factor - match resistor value

Question

I have the following problem.

Consider the circuit below, where the bottom of the arrow is the +-pole of the voltage drop:

What resistor value for \$R_2\$ gives a damping factor of \$d=0.9\$ for the filter transfer function \$\frac{V_{out}}{V_{in}}\$.

Okay so first thought is to find the transfer function. We can write a node equation for \$V_{out}\$.

\$ \frac{V_{out}}{R_2}+\frac{V_{out}}{600 \Omega}+\frac{V_{out}}{330 \text{nF}}+\frac{V_{out}-V_{in} }{5.1 \text{uH}} =0\$

Rearranging this equation with Maple gives me following transfer function.

\$ \frac{V_{out}}{V_{in}}=\frac{196080 \cdot R_2}{1+3.22638\cdot 10^6 \cdot R_2}\$

But from here I'm kind of stuck, since I don't know where to go. However, I have these equations from my notes that might help me.

\$ d=\frac{1}{Q_{factor}} \$

\$ B_{bandwidth}=\frac{f_c}{Q_{factor}} \$

Where \$ f_c\$ is the resonance frequency.

I hope someone can help me with this.

Answer 1

The transfer function of this circuit can be found in two simple ways: an impedance divider where you write that \$\frac{V_{out}}{V_{in}}=\frac{{\frac{1}{sC_2}||R_{eq}}}{\frac{1}{sC_2}||R_{eq}+sL_1}\$ and you develop this expression. Or you can use the fast analytical techniques or FACTs described in the book I wrote and obtain the transfer function in a few steps, without writing a line of algebra. 5 small sketches is all you need in which you determine the resistance \$R\$ driving the capacitor or the inductor when the source is zeroed (replaced by a short circuit). \$R_{eq}\$ is the equivalent resistance made of the existing 600-\$\Omega\$ already in place and the one you will parallel with it:

Assemble the time constants as in the below sheet and you have the transfer function:

Now, how do obtain the resistance you want? The thing is to re-write the transfer function in the normalized second-order form which is: \$H(s)=\frac{1}{1+\frac{s}{\omega_0Q}+(\frac{s}{\omega_0})^2}\$. Then, you identify the variables between this expression and the one we have determined to link the equivalent resistance and the quality factor \$Q\$ you want (or damping ratio \$\zeta\$ since the two are linked):

And if you do the maths, you find a resistance of 2.192 \$\Omega\$ needs to be paralleled with the 600-\$\Omega\$ one to obtain the wanted damping ratio.

Answer 2

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: -

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