M

Mr. Dubey • 52.30K Points
Coach

Q. For a Wilkinson power divider of insertion loss L and the coupler is matched to the external circuitry, and then the gain of the coupler in terms of insertion loss is:

(A) 2l
(B) 1/2l
(C) l
(D) 1/l
Correct : Option (B)

Explanation:
 to evaluate the noise figure of the coupler, third port is terminated with known impedance. then the coupler becomes a two port device. since the coupler is matched, Гs=0 and Гout=s22=0. so the available gain is │s21│2. this is equal to 1/2l from the available data.

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