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.

enter image description here

enter image description here

  • 1
    \$\begingroup\$ Do you know how to develop the standard form of the bandpass transfer function? If so, you should have the DC gain from it and this directly relates to the attenuation of the center of your bandpass region. So you can compute the series resistance directly from that. Do you understand this much? \$\endgroup\$
    – jonk
    Nov 3, 2022 at 5:56
  • \$\begingroup\$ Thanks, that helps to clarify Bode plot data, I'm aware how to find the bandwidth, quality factor and w1 and w2 for a band pass, yet those values are dependant on the L and C values which are still lost on me, i'm sure i'm missing something simple... \$\endgroup\$
    – meak21
    Nov 4, 2022 at 6:48
  • \$\begingroup\$ Also the set values in Matlab are 0.1 Henry and 0.01 Farads, and the values you've hinted at are 0.016 Henry and 0.0015 Farad, is this value discrepancy due to frequency/impedance of the circuit behaviour? \$\endgroup\$
    – meak21
    Nov 4, 2022 at 7:58
  • \$\begingroup\$ No. It's because I didn't read the units on your graph, imagining it as Hz when it was rad/s. So I need to make some adjustments to my answer. Sorry about that. \$\endgroup\$
    – jonk
    Nov 4, 2022 at 8:02
  • \$\begingroup\$ If you adjust my values for the inductor and capacitor by multiplying it by \$2\pi\$ then we are in the right place. Does that make sense? \$\endgroup\$
    – jonk
    Nov 4, 2022 at 8:09

1 Answer 1



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.

Standard Form of 2nd Order Bandpass

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\$.

Extracting Info from the Chart

From your chart:

enter image description here

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.


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