# How can I calculate the voltage on the load with an UNUN (unbalanced-to-unbalanced) RF transformer?

I'm having hard time calculating how much RF voltage would be applied to the load in presence of an RF transformer (Unun with a specific ratio). Let me draw a simple circuit diagram. The AC power source is at the leftmost and its output impedance is (as normally) set to 50 Ohm. There is a load with an impedance $$\Z_L > 50\$$ Ohm at the other side. Therefore, the unbalanced-to-unbalanced (UNUN) RF transformer with a winding ratio $$\n\$$ is installed between them to match the impedance. The output impedance of the UNUN would be equal to $$\ 50 \times n^2 \$$ (for example, 450 Ohm when $$\ n = 3 \$$).

I was trying to approach this problem as following. Let $$\P_0\$$ the incident AC power generated from the source. If the UNUN output is not exactly the same with the load impedance $$\Z_L\$$, then some portion of the power should be reflected back. That is expressed by the reflection coefficient, $$\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0},$$ where $$\ Z_0 = 50 n^2\$$ is the output impedance of the UNUN transformer (without the transformer, it is just 50 Ohm). Thus the reflected and transmitted power are given by $$P_r = \Gamma^2 P_0, \qquad P_t = P_0 - P_r = (1-\Gamma^2)P_0.$$ At the same time, the relation between the transmitted power and the (rms) voltage on the load is $$P_t = \frac{V_{L, rms}^2}{Z_L}$$ Therefore, the voltage applied on the load should be $$V_{L, rms} = \sqrt{(1-\Gamma^2)P_0 Z_L}.$$

My question is following. Is the above calculation and reasoning are correct? If it's wrong, what is the correct way to evaluate the voltage applied on the load?

And my second question is, how should I evaluate the voltage applied to the loads $$\Z_L\$$ and $$\Z_L'\$$ in the following figure, where the power source output is splitted into two and each of them goes to a UNUN transformer respectively where the ratios $$\n, m\$$ are not necessarily the same.

In the second example, what happens if the three ports of the splitter are perfectly symmetric (so that no difference between inputs and outputs), so that a (reflected) current can flow bidirectionally the upper and lower parts back and forth?