img
Question:
Published on: 10 August, 2022

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.

Answer:

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

Random questions