How to estimate the transfer function given only Bode plot data points?

Question

I'm trying to obtain transfer function given Bode plot data points. As an example I used the following RC filter and made AC sweep as follows:

I set the type of sweep to decade and for each decade the number of points as 100 points.

Then I exported the sweep to a txt file in Cartesian format as shown below:

Below is the first rows of Bode plot data saved to the text file from LTspice(first column is frequency the second column is the complex number):

1.00000000000000e-001   9.99999996052158e-001,-6.28318528237456e-005
1.02329299228075e-001   9.99999995866102e-001,-6.42953946745921e-005
1.04712854805090e-001   9.99999995671278e-001,-6.57930267936173e-005
1.07151930523761e-001   9.99999995467271e-001,-6.73255432451136e-005
1.09647819614319e-001   9.99999995253650e-001,-6.88937565895025e-005
1.12201845430196e-001   9.99999995029962e-001,-7.04984983141642e-005
1.14815362149688e-001   9.99999994795731e-001,-7.21406192743033e-005
1.17489755493953e-001   9.99999994550462e-001,-7.38209901440824e-005
1.20226443461741e-001   9.99999994293633e-001,-7.55405018782651e-005

Here is the Python code which reads and plots the exported LTspice Bode data.

import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

dataset=pd.read_csv("Draft224.txt",delimiter=r'[\t,]', header=None, engine='python')

f = dataset[0]
re = dataset[1]
ima = dataset[2]

mag = np.sqrt(np.power(re,2)+np.power(ima,2))

plt.subplot(2, 1, 1)
plt.semilogx(f,20*np.log10(mag),'.r')
plt.semilogx(f,20*np.log10(mag),'-b')
plt.xlabel('Frequency [Hz]')
plt.ylabel('Magnitude [dB]')
plt.show()
plt.grid()

plt.subplot(2, 1, 2)
deg = np.rad2deg(np.arctan2(ima,re))
plt.semilogx(f,deg,'-g')
plt.xlabel('Frequency [Hz]')
plt.ylabel('Angle [°]')
plt.show()
plt.grid()

So this is the data plotted on Python which is the same shown in LTspice plot:

My question is:

Above I used LTspice as an example to create such data points. But such Bode plot data could be from a network analyzer or a data-acquisition system.

Having only the Bode plot data points of a system/circuit but not the circuit diagram, how can we obtain the transfer function?

Answer 1

it is not that complicated, watch for the point the graph gets a slope downwards that is a pole. That is when the denominator of the transfer function is 0... later on that downward slope stops and becomes flat so there must be a 0 canceling out the pole earlier. This happens at 1kHz for the pole and at 1MHz for the 0.

You usually write bode functions in Laplace form so switch out the S for w*t where w is equal 2*pi*f and do not forget about the imaginary term.

t is time, pi is just pi, f is frequency you can solve for a given frequency.

now for the constant, your graph starts at 0 db and it is flat so it must be 1 normally you solve for how many Dbs it is at 0 hertz, in your case it starts at 0 so to summarize your bode plot looks like this

a constant A that is equal to the number of Decibels at 0 frequency since 20*log(A)=0 your frequency is 0 here so any S term is gone.

at 1Khz you have a pole so the denominator is 0 so s= 2*pi*f at 1kHz so s+2000*pi is your pole now your function looks like this the extra minus comes from the imaginary term. $$ F(s)=\frac{A}{S+2000\pi} $$ you can rewrite this as $$ \frac{A}{(\frac{S}{2000\pi})+1} $$ finally you have the remaining 0 at 1MHz so (S/1e6)+1

so the complete thing is F(s)= A times the pole times the zero

F(s)= A*((S/pi*1e6)+1 )/((S/2000*pi)+1) A is 1 so you can just ignore it in this case.

I did not do it perfectly but I think it explains the gist of it.