Answer:
[tex]P_O = 0.989 watt = 19.9 dBm[/tex]
Explanation:
Given data:
P_1 power = 20 dBm = 0.1 watt
coupling factor is 20dB
Directivity = 35 dB
We know that
coupling factor [tex] = 10 log \frac{P_1}{P_f}[/tex]
solving for final power
[tex]20 = 10 log\frac{P_1}{P_f}[/tex]
[tex]2 = log \frac{P_1}{P_f}[/tex]
[tex]100 = \frac{0.1}{P_f}[/tex]
[tex]P_f = 0.001 watt = 0 dBm[/tex]
Directivity [tex]D = 10 \frac{P_f}{P_b}[/tex]
[tex]35 = 10 \frac{0.001}{P_b}[/tex]
[tex]P_b = 3.162 \times 10^{-7} wattt[/tex]
output Power [tex]= P_1 -P_f - P_b[/tex]
[tex] = 0.1 - 0.001 - 3.162 \times 10^{-7}[/tex]
[tex]P_O = 0.989 watt = 19.9 dBm[/tex]