Antenna system resonance and the nanoVNA contained the following:

## Relationship between angle of reflection coefficient and angle of impedance

It was stated above that the angle (or phase) of s11 or Γ is not the same as the angle (or phase) of Z.

Given Zo and Γ, we can find θ, the angle of Z.

\(

Z=Z_0\frac{1+\Gamma}{1-\Gamma}\)Zo and Γ are complex values, so we will separate them into the modulus and angle.

\(

\left | Z \right | \angle \theta =\left | Z_0 \right | \angle \psi \frac{1+\left| \Gamma \right | \angle \phi}{1-\left| \Gamma \right | \angle \phi} \\

\theta =arg \left ( \left | Z_0 \right | \angle \psi \frac{1+\left| \Gamma \right | \angle \phi}{1-\left| \Gamma \right | \angle \phi} \right )\)We can see that the θ, the angle of Z, is not simply equal to φ, the angle of Γ, but is a function of four variables: \(\left | Z_0 \right |, \psi , \left| \Gamma \right |, \& \: \phi\) .

It is true that if ψ=0 and φ=0 that θ=0, but that does not imply a wider simple equality. This particular combination is sometimes convenient, particularly when ψ=0 as if often the case with a VNA.

This article offers a simulation of a load similar to a 7MHz half wave dipole.

The load comprises L, L1, and C1 and the phase of s11 (or Γ) and phase of Z (seen at the source G) are plotted, along with VSWR.

Firstly, note that the two phase plots are very different, but in this case they cross over at phase=0 at 7.077MHz.

Secondly, note that even thought both phases are zero at 7.077MHz, the VSWR is 1.20. Neither phase demonstrates the best conditions for least feed line loss, minimum VSWR is slightly lower at 7.099MHz.

Maximum power in the load coincides with minimum VSWR at 7.099MHz.

Beware of claims that phase (of something) is the optimisation target, the author probably doesn’t really understand this stuff.