A recent post online provides an interesting demonstration of the value of the Smith chart in analysing a measurement problem.
I have 5.175m of “JSC 1320 300 Ohm Ladder Line 300 Ohm 20 AWG / 7 Strands Bare Copper”. … The first step is to sweep it to determine the velocity factor. Yet, when I sweep from 12-17MHz, I get the Smith chart attached. There’s no point when the impedance is close to zero.
It helps to understand the nature of what one is measuring, indeed the expected outcome if possible.
Let’s model the scenario in Simsmith.
Ok, it doesn’t look like a short circuit as the OP observed. At VF=0.8 and 12MHz a quarter wavelength is just under 5m, and at 17MHz, three quarters wavelength is over 10m, so the line section is not long enough to observe input impedance near zero within the stated sweep range. The line loss used is and estimate of this type of line.
But wait, the curve is not really like the one quoted… lets try some model calibration.
Above, the model line loss adjusted up to get a similar trace.
The line loss is higher than I would have expected. A common cause is an inadequate test fixture that encourages common mode current giving rise to radiation, and more loss. Kinks or bumps in the Smith chart spirals are a strong hint of fixture / common mode issues.
BTW, if the Smith chart was plotted normalised to the line Zo, 300Ω, we get a section of a classic spiral centred on the chart centre.
The Smith chart is so informative!