Final answer:
The actual signal power in dBm, when the background noise is -100 dBm and the total power is -97 dBm, is approximately -100.4 dBm. This calculation requires converting dBm to milliwatts, subtracting noise power from the total power, and converting back to dBm.
Step-by-step explanation:
The student has asked about the actual signal power in dBm when the background noise level is -100 dBm and the total power is at -97 dBm. The total power is the sum of the signal power and the noise power; since these are given in dBm, which is a logarithmic scale, we cannot simply add or subtract these values directly.
To find the signal power, we use the following formula for adding decibels:
Ptotal = 10 log10(Psignal + Pnoise)
First, we convert the noise and total power from dBm to milliwatts:
- Convert the -100 dBm noise level to milliwatts: Pnoise = 10(-100/10) mW
- Convert the -97 dBm total level to milliwatts: Ptotal = 10(-97/10) mW
The signal power in milliwatts (Psignal) is then the difference between the total power and the noise power. We then convert this value back to dBm using Psignal(dBm) = 10 log10(Psignal).
Using this procedure, we find that the signal power is approximately -100.4 dBm.