EMC Question of the Week: August 19, 2024
A particular source delivers 2.0 mW of power to a 50-Ω receiver when they are directly connected. When they are connected through 20 cm of 50-Ω coaxial cable, the received power is only 1.8 mW due to cable loss. What would the received power be if the length of the connecting cable was 1 meter long?
- 1.0 mW
- 1.2 mW
- 1.4 mW
- 1.6 mW
Answer
The best answer is "b.” The power is reduced by 10% every 20 cm. That corresponds to a cable attenuation of 0.458 dB per 20 cm, or 2.29 dB/m. -2.29 dB is a factor of 0.59 in power, so 2 mW x 0.59 = 1.18 mW. To two significant figures, 1.2 mW is the best answer.
Or, viewing it another way, the source amplitude expressed in dBm is 3 dBm. The attenuation in the 1-meter cable is 2.29 dB. The power delivered to the load is therefore, 3 dB - 2.29 dB = 0.71 dBm, which is 1.18 mW.
Have a comment or question regarding this solution? We'd like to hear from you. Email us at