I Get the time axis right in an inverse Fast Fourier Transform

AI Thread Summary
The discussion focuses on transforming S-parameter responses from a Vector Network Analyzer (VNA) into the time domain using the Inverse Fast Fourier Transform (IFFT) in MATLAB. The user, Thomas, is struggling with the correct time scaling for the transformed data, initially miscalculating the x-axis increment. The correct time step is identified as Δt = 1/(N Δf), with Δf determined to be 0.00125 GHz based on the provided data. Clarifications on the time scaling help Thomas understand the necessary adjustments for accurate representation. The conversation emphasizes the importance of precise frequency resolution in time-domain transformations.
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time (distance) axis of Inverse Fast Fourier Transform (IFFT) from Vector Network Analyzer (VNA) S-parameter data.
Hi
I would like to transform the S-parameter responce, collected from a Vector Network Analyzer (VNA), in time domain by using the Inverse Fast Fourier Transform (IFFT) . I use MATLAB IFFT function to do this and the response looks correct, the problem is that I do not manage to the time scaling correct. The VNA has a build in time domain toolbox so I know how it should look like, see the figure below.
Return-Loss, time domain, VNA.jpg

In the figure below I have plotted the MATLAB transformed data with the index on x-axis (also zoomed in between N=0 to 20, N_max is 1600)
Return-Loss, time domain, MATLAB.jpg

At first I assumed that the increment on x-axis would be 1/bw/N_max where bw is the band width (i.e. f_max-f_min) and N_max is the number of samples. This is obviously wrong with a factor of 1e-3.
I would much appreciate if someone would like to help to get this correct! I have attached the S-matrix data in the txt file. First column is the frequency in GHz, second is real part of the response and third is imaginary part of the response.
My thanks in advance
/Thomas
 

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  • Return-Loss, time domain, VNA.jpg
    Return-Loss, time domain, VNA.jpg
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The time step is given by
$$
\Delta t = \frac{1}{N \Delta f}
$$
From the data file, it appears that ##\Delta f = 0.00125\ \mathrm{GHz}## (the frequencies appear to be rounded values).
 
Hi
Thanks for the answer! Yes, this makes sense.
 
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