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| 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 |
| Answer» B. 1/2l | |
| 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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