I am trying to get back into control systems and at the moment I am stuck on how to extract the damping ratio and gain from the Bode plot I generated.
Firstly the gain. As far as I can remember the gain is the value at f = 0, so 0 dB in my case. Is this correct?
The damping ratio can be calculated from the peak at the resonance frequency somehow, but I am unable to find a fitting formula here. So how can one find/calculate this value from the Bode plot?
A 2nd order low-pass peaking plot with magnitude information can be found on my website: -
If you need more details of how that formula is derived I can add that information. I've added that now - see derivation #1 lower down.
So, the trick here is reverse engineer the above formula: -
$$P = \dfrac{1}{2\zeta\sqrt{1 - \zeta^2}}$$
And, if you manipulate this into a quadratic solving for \$\zeta^2\$ you get: -
$$\zeta^2 = \dfrac{1\pm\sqrt{1-\frac{1}{P^2}}}{2}$$
So, if the peak is 20 times higher than the DC value (a peak gain of about 26 dB) you get this: -
$$\zeta^2 = \dfrac{1\pm\sqrt{0.9975}}{2}$$
Clearly the wrong answer is found when the numerator terms are added. If subtracted: -
$$\zeta^2 = 0.00062539111 $$
And \$\zeta =\$ 0.025.
Added derivation #1
Added web calculator
This is a calculator (also from my website) that shows the actual damping ratio (\$\zeta\$). It's a simple case of adding the two resistors in the original circuit into one lump value of 0.0025 Ω: -
You can find the calculator here.
Your circuit is too complicated to solve analytically. I provided some guidelines and analyzed your specific circuit at the bottom of my answer.
Well, let's analyze this circuit:
simulate this circuit – Schematic created using CircuitLab
The transfer function is given by:
$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\displaystyle\text{R}_2+\frac{1}{\displaystyle\text{sC}}}{\text{R}_1+\text{sL}+\text{R}_2+\frac{1}{\text{sC}}}=\frac{1+\text{CR}_2\text{s}}{\text{CL}\text{s}^2+\text{C}\left(\text{R}_1+\text{R}_2\right)\text{s}+1}\tag1$$
Now, we want to solve for the value that \$\displaystyle\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|\$ is at a maximum:
$$\frac{\partial\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|}{\partial\hat{\omega}}=0\space\Longleftrightarrow\space\hat{\omega}=\dots\tag2$$
Solving this gives:
$$\hat{\omega}=\frac{1}{\text{CR}_2}\cdot\sqrt{\frac{\displaystyle\sqrt{\left(\text{L}-\text{CR}_1\text{R}_2\right)\left(\text{L}+\text{CR}_2\left(\text{R}_1+2\text{R}_2\right)\right)}}{\text{L}}-1}\tag3$$
With the condition that \$\text{CR}_1\left(\text{R}_1+2\text{R}_2\right)<2\text{L}\$.
At the maximum we get:
$$\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|=\dots\tag4$$
So, the cut-off frequency can be solved:
$$\left|\space\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{1}{\sqrt{2}}\cdot\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|\space\Longleftrightarrow\space\omega=\dots\tag5$$
Which gives:
$$\omega_\pm=\dots\tag6$$
So, the bandwidth is given by:
$$\mathcal{B}:=\left|\omega_+-\omega_-\right|=\dots\tag7$$
And the quality factor is given by:
$$\mathcal{Q}:=\frac{\hat{\omega}}{\mathcal{B}}=\dots\tag8$$
For your specific circuit we find:
$$\hat{\omega}=1000000 \sqrt{\frac{\sqrt{1210366}}{11}-100}\approx122974.8895\space\text{rad/sec}$$ $$\left|\space\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|=\sqrt{\frac{3168 \sqrt{1210366}}{26375}+\frac{3485327}{26375}}\approx16.257$$ $$\omega_\pm=\frac{125000}{33} \sqrt{2 \left(6336 \sqrt{1210366}\pm\sqrt{48590017190881-44166063744 \sqrt{1210366}}-6970127\right)}$$ $$\omega_+\approx126707.9218\space\text{rad/sec}$$ $$\omega_-\approx119124.9685\space\text{rad/sec}$$ $$\mathcal{B}=\frac{250000}{33} \sqrt{6336 \sqrt{1210366}-528 \sqrt{174266351-158400 \sqrt{1210366}}-6970127}\approx7582.95\space\text{rad/sec}$$ $$\mathcal{Q}=12 \sqrt{\frac{11 \left(\sqrt{1210366}-1100\right)}{6336 \sqrt{1210366}-528 \sqrt{\frac{345895201}{158400 \sqrt{1210366}+174266351}}-6970127}}\approx16.2173$$
