Converting Bridge Voltages to Load Impedance: Method 2

There is an alternative way of doing Step 2 as described in Converting Bridge Voltages to Load Impedance: Method 1

The RHS of Eq4 with $Z_l = R+jX$ is

$$\vert \frac{R-50+jX}{R+50+jX}\vert $$

Squaring the modulus $\vert a+jb\vert $ gives $a^2+b^2$, so squaring top and bottom of the equation above gives

Eq6	$$\frac{(R-50)^2+X^2}{(R+50)^2+X^2}=\frac{R^2-100R+2500+X^2}{R^2+100R+2500+X^2}$$

But $R^2+X^2 = \vert Z_l\vert ^2$ so

Eq7	$$\frac{(R-50)^2+X^2}{(R+50)^2+X^2}=\frac{\vert Z_l\vert ^2-100R+2500}{\vert Z_l\vert ^2+100R+2500}$$

Which can be solved to give

$$R = \frac{\vert Z_l\vert ^2+50^2}{100}\frac{1-(\frac{V_r}{V_f})^2}{1+(\frac{V_r}{V_f})^2}$$

Now that we know $R$ as well as $\vert Z_l\vert $, we can use $ X=\sqrt{\vert Z_l\vert ^2-R^2} $ to get the complex impedance and VSWR.

We can rewrite these equations to use the bridge voltages directly:

$$R = \frac{(50\frac{V_z}{V_a})^2+50^2}{100} \frac{1-(\frac{V_r}{V_f})^2}{1+(\frac{V_r}{V_f})^2}$$

$$X = \pm\sqrt{(50\frac{V_z}{V_a})^2 - R^2}$$