Please help me solve the below Directional coupler problem
Three assumptions need to be made to solve this problem..

that the two coupled ports on the directional coupler are properly terminated.

the distance between the directional coupler and the far end of the transmission line where the short is at is electrically short, as in many orders of magnitude shorter than the wavelength being dealt with. Essentially the same as shorting the output port on the directional coupler directly.

That the directivity measured in dB is greater than the coupling also measured in dB. Keep in mind both these figures are the gain from coupled port to the transmission line. So a positive value implies the gain from transmission line to coupled port is the negative of this value. So by saying we assume the directivity is larger than the coupling this means we assume there is greater attenuation from reflected to forward coupled port then from forward to forward coupled port. This final assumption is pretty much always the case so its a safe one to make in any situation.


Taking these as assumptions then we can solve the problem. The input power is 0.01W, lets convert this to dBm to make it easier.
$$
  \begin{align}
  P_{dBm} &amp; = 10 \cdot log\left( 1000 \cdot \frac{P_W}{1W}\right) \tag{1} \\\\
  P &amp; = 10 \cdot log\left( 1000 \cdot \frac{0.01W}{1W}\right) \tag{2} \\\\
  P &amp; = 10\ dBm \tag{3}
  \end{align}
$$
Since we know the coupling is 20 dB then we know that the power seen at the forward coupled port would be \$-10\ dBm\$ if we had perfect directivity. Because we are working in dBm it is simple subtraction.
$$
  \begin{align}
  P_f &amp; = 10\ dBm - 20\ dB \tag{4} \\\\
  P_f &amp; = -10\ dBm \tag{5}
  \end{align}
$$
Now lets calculate the actual power see see at the forward port in dBm.
$$
  \begin{align}
  P_\hat{f} &amp; = 10 \cdot log\left( 1000 \cdot \frac{0.00000001W}{1W}\right) \tag{6} \\\\
  P_\hat{f} &amp; = -50\ dBm \tag{7}
  \end{align}
$$
this means the forward couple's port power without directivity effects in plain numbers is:
$$
  P_f = -10\ dBm = 0.0001\ W \tag{8}
$$
and for the power once directivity effects are applied as it leaks in from reflected we already know is.
$$
  P_\hat{f} = -50\ dBm = 0.00000001\ W \tag{9}
$$
Since we need to subtract and add and we cant do that via power linearly lets convert this into volts.
$$
  \begin{align}
  P &amp; = \frac{v^2}{R} \tag{10} \\\\
  V &amp; = \sqrt{P \cdot R} \tag{11}
  \end{align}
$$
Now we know the forward signal without directivity, minus the signal leaked in due to directivity, must equal the signal we actually see. Therefore.
$$
  \sqrt{P_f \cdot Z_0} - V_D = \sqrt{P_\hat{f} \cdot Z_0} \tag{12}
$$
Note the only reason we are subtracting here is because we know the reverse signal is 180 degrees out of phase with the forward, so it has a subtractive effect.
$$
  \begin{align}
  -V_D &amp; = \sqrt{P_\hat{f} \cdot Z_0} + -\sqrt{P_f \cdot Z_0} \tag{13} \\\\
  V_D &amp; = -\sqrt{P_\hat{f} \cdot Z_0} + \sqrt{P_f \cdot Z_0} \tag{14}
  \end{align}
$$
Then convert voltage back to power.
$$
  \begin{align}
  P_D &amp; = \frac{V_D^2}{Z_0} \tag{15} \\\\
  P_D &amp; = \frac{\left(-\sqrt{P_\hat{f} \cdot Z_0} + \sqrt{P_f \cdot Z_0}\right)^2}{Z_0} \tag{16}
  \end{align}
$$
Since we can assume all positive values here this simplifies to:
$$
  P_D = -2 \cdot \sqrt{P_f \cdot P_\hat{f}} + P_\hat{f} + P_f \tag{17}
$$
Plugin in our values and find the actual value:
$$
  P_D = 0.00009801\ W = -10.087296108\ dBm \tag{18}
$$
Now we know reflected wave in the transmission line is \$10\ dBm\$ and we know the actual power of reflected wave in the forward port due to directivity, so by simple subtraction we can arrive at the directivity in dB, remember dBm minus dBm equals dB.
$$
  \begin{align}
  10\ dBm - P_D &amp; = D \tag{19} \\\\
  D &amp; = 10\ dBm - (-10.087296108\ dBm) \tag{20} \\\\
  D &amp; = 20.087296108\ dB \tag{21}
  \end{align}
$$