A signal travels from point A to point B. At point A, the signal power is 100 W, and at point B, it is 50 W. What is the attenuation in dB?
Q1. A signal travels from point A to point B. At point A, the signal power is 100 W, and at point B, it is 50 W. What is the attenuation in dB?
Answer: -3 dB
Explanation: Attenuation (dB) = 10 log10(P2/P1) = 10 log10(50/100) = 10 log10(0.5) = -3 dB.