''Explain the advantage of using decibels rather than ratios.
In your discussion you should explain the term dBm.''
dBm is a power measurement, the reference level is 1mW.
So a transmitter with a power of 100mW is said to have an output of +20dBm.
Why use dBs rather than ratios?
Consider a point to point radio link, just making up a few numbers:
The transmitter has an output power of 10 watts, +40dBm.
The receiver needs a minimum signal of -125dBm (about 0.2µV into 50 ohms) for an acceptable signal.
At each end of the link the antenna has a gain of +12dB, and the feeder from the antenna to the Tx/Rx has a loss of -3dB.
The radio path loss is -160dB.
Will this link work?
Calculating:
At the transmitter end, the effective radiated power is
+40 -3 +12 = +49 dbm
The signal power at the receiver input is
+49 - 160 +12 -3 = -102dBm
The receiver needs at least -125dBm, so there is a (-102 - (-125)) = 23dB "Fade Margin".
The link will work very well.
Now you try the same calculation using ratios!
(Because I dont want to even try!).
If the teacher likes this answer, you own me a bacon sandwich!
JimB