I have a NanoVNA V2 PLUS4 v2.4.
The vna_qt app can export a csv file with content similar to the example below.
Can I make use of the four columns after frequency to calculate a complex impedance as a real and imaginary capacitance or inductance.
I’m looking for results in the format of 26.7 Ohms, 32.13 pF.
I feel like I should know this, but I have run out of time at the moment, so I’m hoping someone can make it easy for me! Lol!
Freq (MHz) re(S11) im(S11) re(Z) im(Z)
1 -0.35867 0.87787 1.924 33.549
1.05 -0.28729 0.90268 2.076 36.517
1.1 -0.21466 0.92313 2.186 39.661
1.15 -0.14315 0.93639 2.351 42.882
1.2 -0.06984 0.94436 2.537 46.375
1.25 0.00211 0.94771 2.689 50.039

Hi,

I am not sure if I understand well.

The resistance is equal to re(Z).

The im(Z) is a reactance.

If im(Z) > 0, im(Z) = L * omega = L * 2 * pi * f ==> L = im(Z) / (2 * pi *f)

If im(Z) < 0, im(Z) = 1 / (C * 2 *pi * f) ==> C = 1 / ( im(Z) * 2 * pi * f) )

For instance, for the first line, R = 1.924 ohm et L = 5.33957 µH

I hope this answers to your question.

> envoye : 9 fevrier 2025 a 06:17
> de : "Scot Mcclellan, AE0ZF via groups.io" <ism1=psu.edu@groups.io>
> a : NanoVNAV2@groups.io
> objet : [nanovnav2] Vna_qt app exports
>
>
> I have a NanoVNA V2 PLUS4 v2.4.
> The vna_qt app can export a csv file with content similar to the example
below.
> Can I make use of the four columns after frequency to calculate a complex
impedance as a real and imaginary capacitance or inductance.
> I'm looking for results in the format of 26.7 Ohms, 32.13 pF.
> I feel like I should know this, but I have run out of time at the moment, so
I'm hoping someone can make it easy for me! Lol!
> Freq (MHz) re(S11) im(S11) re(Z) im(Z)
> 1 -0.35867 0.87787 1.924 33.549
> 1.05 -0.28729 0.90268 2.076 36.517
> 1.1 -0.21466 0.92313 2.186 39.661
> 1.15 -0.14315 0.93639 2.351 42.882
> 1.2 -0.06984 0.94436 2.537 46.375
> 1.25 0.00211 0.94771 2.689 50.039
>
>
>

_._,_._,_

* * *

What you have is an impedance expressed as a resistance in series with a reactance, Z= R + jX. The re(Z) is the resistance and the im(Z) is the jX. For example, the value at 1 MHz is 1.924 +J33.549 ohms. This is a resistance of 1.924 ohms in series with an inductance of 33.549 ohms. If you want to know the value of the inductor, use the equation Xl= 2*3.14159*F*L. You know Xl=33.549, and at 1 MHz, 2 times 3.14159 (pi) time 1 MHz is 6.28318 million. Do the calculation and L= 5.34 micro-Henries.

If im(Z) is a negative number, it means it is a capacitor Z=R-jX. Remember Xc=1/(2*3.14159*F*C). So, to find the value of C, solve the equation. C=1/(2*3.14159*F*Xc).

I don't know if this is something you measured, but it looks like an inductor. The Q of the inductor is Xl/R. In this case the Q=17.44. Not a very high Q inductor.

You can convert a series resistance Rs and Xs to a parallel resistance and parallel Xp using the following equations.

Rp=Rs*(Qs*Qs+1) and Xp=Xs*(Qs*Qs+1)/(Qs*Qs) where Qs=Xs/Rs.

Hope this helps.

73s

Dick