A multi hole directional coupler is fed with single power of 2.8 mW at 10 GHz. The coupling factor is 3 dB and the directivity is better than 40 dB over X-band range. Find the distribution of power at all other ports.
The expression for coupling factor is
\(C=10\log{\left(\frac{P_1}{P_3}\right)}=3\)
\(\log{\frac{P_1}{P_3}}=0.3\)
\(\frac{P_1}{P_3}=2\)
\(P_3=\frac{2.8\times{10}^{-3}}{2}=1.4\ mW\)
The expression of directivity is
\(D=10\log{\left(\frac{P_3}{P_4}\right)}=40\)
\(\frac{P_3}{P_4}=10000\)
\(P_4=\frac{1.4\times{10}^{-3}}{10000}=0.14\ µW\)
Thus the power at port 2 is
P2=Input power-(Power in coupling port+Power in isolated port)
P2=2.8×10-3-(1.4×10-3 + 0.14×10-6)=1.4 mW
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