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I am trying to analyze data collected from an antenna experiment. I setup two cantennas next to each other and connected each to a Vector Network Analyzer. I performed a frequency sweep from 2.3 to 2.6 GHz, with step size of 150 KHz, and 2001 points. We did this for two different objects, one at 565mm away from the antennas and one from 895 mm away from the antennas. The received magnitude power (in log and linear form), and the phase data was collected at each frequency. I placed the magnitude and phase data into a complex signal and placed this signal through an IFFT. This converted into the time domain, where I expected to see a peak showing when the reflection off the object occurred. There theoretically should be a shift difference in the time domain between an object 565mm away and an object 895mm away. Unfortunately, when I place the signal through an IFFT, both of the objects have the peak at the same time and place. Can someone help me understand how exactly to find the shift between the two or what I am doing wrong in my code?

My MATLAB code: enter code here Data Files:
565mm: http://pastebin.com/Us972Rnn 895mm: http://pastebin.com/zUQfGppM

Cris Luengo
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cecchollett
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1 Answers1

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Your data does not show a frequency sweep from 2.3 GHz to 2.6 GHz. Instead it shows you are sweeping from 2.3 GHz to 2.319 GHz in 150 KHz steps and doing this 20 times for a total of 2001 total points.

I would start by making sure my setup is collecting what I expected.

Also you are trying to resolve a small range. A stepped frequency waveform down range resolution is directly related to c/2*BW (BW = full waveform bandwidth). Do some calculations to make sure you can resolve the differences you are trying to see with your setup. Even if your data swept .3 GHz you can only resolve .5 meter. So yes you will see both responses at the same range because they are inside the range resolution of you waveform.

Chandler G.
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