Hi!

I just obtained a nanoVNA and I am very happy about it.

Until very recently, I didn't know what a VNA was, and so the nano is a very good starting point to learn about this stuff.

I have tried some measurements (S11) and it's very accurate:

- I checked the return loss (called LogMag in the nano) with a Spectrum Analyzer with TG and a directional coupler and it's spot on.

- I checked the phase of the reflection with an oscilloscope and it's also very accurate.

Very good value for 50euros

I have some doubts, however, in the Smith chart trace of the nano.

For a concrete example, I have the following data point:

`Frequency: 7542500 Hz`

LogMag: -18.59 db

Phase: 143.61

Polar Diagram: -0.09 + 0.07j

Smith: 41.3 Ohm and 123nH

Starting from the top we have:

20*Log_10(|S11|) = -18.59

and this gives |S11| = 0.117625.

The second line gives us the phase of S11 (143.61deg or 2.50647rad).

So expressing S11 in cartesian form gives us

`-0.094688 + 0.0697841 i`

which is close to the reading (-0.09 + 0.07j). All good so far.

Now I can compute the input impedance Z of the DUT using the equation:

`S11 = (Z-50)/(Z=50)`

and we obtain

Z = 40.9805 + 5.79982 i

or, expressing it in Polar form,

|Z|= 41.3889 , with angle 0.140593 rad.

**Doubt:** this is where I am surprised. The Smith data reported by the nanoVNA is 41.3 Ohm and 123nH.

- 41.3 Ohm is the magnitude of Z (rather than its real part as I would expect).
- 123nH is the value L that satisfies: 5.79982 = 2*pi*L*f, with f = 7542500, so a value directly proportional to the imaginary parto of Z (at this specific f).

**Question:** is this way of presenting the Smith value correct/standard? I would expect to see the real+imaginary component of Z (modelling a pure resistor in series with an inductor) rather than what I actually get.

Can you help me understand what's going on?

Thanks!