I am using a 20m coaxial cable (5D-2V) and its far end is left open during the measurement.
The Smith Chart tells you everything, but it is easier to see its amplitude and phase separately.
First, the amplitude in dB.
% gnuplot gnuplot> f(x)=a*sqrt(x) gnuplot> fit f(x) "mydata.txt" using 1:5 via a iter chisq delta/lim lambda a 0 9.0593050625e+02 0.00e+00 8.02e-01 2.070472e-01 1 5.4539304323e+01 -1.56e+06 8.02e-02 3.445278e-01 2 5.4539209850e+01 -1.73e-01 8.02e-03 3.445736e-01 iter chisq delta/lim lambda a After 2 iterations the fit converged. final sum of squares of residuals : 54.5392 rel. change during last iteration : -1.7322e-06 degrees of freedom (FIT_NDF) : 3000 rms of residuals (FIT_STDFIT) = sqrt(WSSR/ndf) : 0.134832 variance of residuals (reduced chisquare) = WSSR/ndf : 0.0181797 Final set of parameters Asymptotic Standard Error ======================= ========================== a = 0.344574 +/- 0.0006355 (0.1844%) gnuplot> print a*sqrt(30) 1.88730727013936
The loss of 1.887[dB] at 30MHz agrees well with its nominal value of 0.44dB/10m, or 1.76dB/40m.
Next, the phase in radian.
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Let’s see at which frequencies the phase changes from -pi to +pi.
2.48, 1.3514, -0.299513, 1.3842, 0.469624, 36.9999, -3.1296, 0.947368 2.49, 1.39451, 0.083306, 1.397, 0.484627, 35.8549, 3.13826, 0.945733 7.53, 2.59869, -0.289392, 2.61476, 0.903662, 19.2411, -3.12999, 0.901191 7.54, 2.45728, 0.08333, 2.4587, 0.854434, 20.3477, 3.13825, 0.906313 12.6, 3.4673, -0.351551, 3.48508, 1.20654, 14.4212, -3.12746, 0.870308 12.61, 3.43064, 0.097069, 3.43201, 1.1938, 14.5746, 3.13769, 0.871586 ... 27.77, 5.43172, -0.091826, 5.43249, 1.89464, 9.20522, -3.13788, 0.804022 27.78, 5.31208, 0.100171, 5.31302, 1.85259, 9.41255, 3.13754, 0.807924
The period is (27.77MHz-2.48MHz)/5cyles, which is 5.06MHz/cycle.
The electrical length of the cable is thus (300m*MHz/5.06MHz)/2=29.64m.
Assuming the velocity factor of 66%, the physical length is known to be 19.56m.