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:

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: Correct Answer 1/2L

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.

Related Questions

The Wilkinson power divider is a:
Statement (I): To use a sensor, we generally need to add signal conditioning circuitry, such as circuits which amplify and convert from analog to digital, to get the sensor signal in the right form, take account of any non-linearities, and calibrate it. Statement (II): A smart sensor is integrated with the required buffering and conditioning circuitry in a single element and provides functions beyond that of just a sensor.
The analysis of Wilkinson coupler is done using: