I am trying to find RLC values from a Bode plot. I have a band-pass filter from a transfer function and have put arbitrary values for the RLC components in Matlab.
I am unsure how to mathematically prove the values of the components from using the Bode plot. I know I have to find the resonant frequency and cut-off values, but I am completely lost on how to extract data to input into a formula.
I'm also using an 800 Ω resistor as the voltage divider. Any help would be appreciated.
Your schematic is basically this:
simulate this circuit – Schematic created using CircuitLab
The analysis is pretty easy using SymPy:
var('R1 C1 L1 R Vin Vout s')
ZL1 = s*L1
ZC1 = 1/s/C1
KCL = Eq( Vout/R + Vout/(R1+ZL1+ZC1), Vin/(R1+ZL1+ZC1) )
tf2( simplify( solve( KCL, Vout )[0] / Vin ) )
{omega: 1/(sqrt(C1)*sqrt(L1)),
zeta: sqrt(C1)*(R/2 + R1/2)/sqrt(L1),
P: [{A: R/(R + R1), N: 1}]}
The N shown there confirms this is a bandpass. (If it were 0 then it would be a lowpass and if it were 2 then it would be a highpass.) The A is the gain.
Your circuit illustrates the following standard transfer function form for a 2nd order bandpass filter:
$$H_s = A\cdot\frac{2\zeta\left(\frac{s}{\omega_{_0}}\right)}{\left(\frac{s}{\omega_{_0}}\right)^2 + 2\zeta\left(\frac{s}{\omega_{_0}}\right) + 1}$$
And where \$\sigma=0\$, then:
$$H_{j\omega} = A\cdot\frac{j\,2\zeta\left(\frac{\omega}{\omega_{_0}}\right)}{1-\left(\frac{\omega}{\omega_{_0}}\right)^2 + j\,2\zeta\left(\frac{\omega}{\omega_{_0}}\right)}$$
In your case, the series RLC solution yielded (as already shown):
$$\begin{align*} \omega_{_0}&=\frac1{\sqrt{L_1\,\cdot\, C_1}} \\\\ \zeta&=\frac12\left(R+R_1\right)\sqrt{\frac{C_1}{L_1}} \\\\ A&=\frac1{1+\frac{R_1}{R}} \end{align*}$$
You already know that \$R=800\:\Omega\$ and your series RLC is specified using \$R_1\$, \$L_1\$, and \$C_1\$.
From your chart:
I can pick off two convenient symmetrical roll-off ends of the flat-top part as \$\omega_{_\text{L}}\approx 0.1\:\frac{\text{rad}}{\text{s}}\$ and \$\omega_{_\text{H}}\approx 10\:\text{k}\frac{\text{rad}}{\text{s}}\$.
The geometric center is \$\omega_{_0}=\sqrt{0.1\:\frac{\text{rad}}{\text{s}}\,\cdot\,10\:\text{k}\frac{\text{rad}}{\text{s}}}\approx 31.623\:\frac{\text{rad}}{\text{s}}\$.
I also see (approximate) that the top of the bandpass is at about \$-0.95\:\text{dB}=20\cdot\log_{10}\left(\frac1{1+\frac{R_1}{R}}\right)\$ and given that \$R=800\:\Omega\$ this means that \$R_1=R\cdot\left(10^{\frac{-0.95\:\text{dB}}{20}}-1\right)\approx 92.5\:\Omega\$.
(You may need to get a better bead on the top there. But that's my estimate from your graph.)
You can, if you want, read through some of this answer to find the following:
$$\zeta=\frac12\frac{\omega_{_\text{L}}+\omega_{_\text{H}}}{\sqrt{\omega_{_\text{L}}\cdot \omega_{_\text{H}}}}=\frac12\frac{\omega_{_\text{L}}+\omega_{_\text{H}}}{\omega_{_0}}=\frac12\frac{f_{_\text{L}}+f_{_\text{H}}}{f_{_0}}$$
In your case, then:
$$\begin{align*}\left(R+R_1\right)\sqrt{\frac{C_1}{L_1}}&=\frac{\omega_{_\text{L}}+\omega_{_\text{H}}}{\omega_{_0}} \\\\ &=\left(\omega_{_\text{L}}+\omega_{_\text{H}}\right)\sqrt{L_1\,\cdot\, C_1} \end{align*}$$
For \$\omega_{_\text{H}}\gg \omega_{_\text{L}}\$, which is definitely your case with more than 3 orders of magnitude separation, then:
$$\begin{align*}\omega_{_\text{H}}\sqrt{L_1\,\cdot\, C_1}&=\left(R+R_1\right)\sqrt{\frac{C_1}{L_1}}\end{align*}$$
But that reduces to:
$$\begin{align*}\omega_{_\text{H}}&\approx \frac{R+R_1}{L_1} \end{align*}$$
From the earlier estimate of \$\omega_{_\text{H}}=10\:\text{k}\frac{\text{rad}}{\text{s}}\$, then \$L_1\approx \frac{R+R_1}{\omega_{_\text{H}}}\approx \frac{800\:\Omega+92.5\:\Omega}{10\:\text{k}\frac{\text{rad}}{\text{s}}}\approx 89.25\:\text{mH}\$.
You should be easily able to use that to get \$C_1\$. Here, you know that \$\omega_{_0}^2=1000\:\frac{\text{rad}^2}{\text{s}^2}=\frac1{L_1 C_1}\$.
So \$C_1=\frac1{1000\:\frac{\text{rad}^2}{\text{s}^2}\,\cdot\,89.25\:\text{mH}}=11.2\:\text{mF}\$.
These appear to be creditably close to your comment that specifies \$L_1=100\:\text{mH}\$ and \$C_1=10\:\text{mF}\$. If you can pick off a better number for the gain than I did (\$-0.95\:\text{dB}\$) then that might change \$R_1\$ a little bit and thereby adjust the other values.
