Timeline for Valid or invalid logarithm operation?
Current License: CC BY-SA 4.0
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| Oct 24, 2022 at 9:49 | comment | added | eldenlord9394 | @RussellH Thank you for reply. Can you help me to find Pr(dBm) using this expression? Pr(dBm) = Pt(dBm) + Gt(dBi) + Gr(dBi) + PL(dB). Given: Pt = 37dBm, Gt = 1.8dBi, Gr = 1.85dBi and PL = -38.4dB. The answer given is Pr = -29.75dBm but I'm getting +32.25dBm. Thanks | |
| Oct 23, 2022 at 18:08 | comment | added | user319836 | @carloc: I haven't said any thing different. dBm is always used to represent a power level, relative to 1mW sure, but a power level none the less. dB without a suffix is always used to represent a gain or loss. \$P1dBM+P2dBm=10log(P1P2/1\mu W^{2})\$ is meaningless. The suffix is appended to give the decibel a physical attribute, to associate is with a physical quantity and assign a zero reference. dBm, dBW, dBmV, dBV, dBi, dBmA, dBrnC, shall I go on. Sure all of these are pure unitless numbers but have real physical identifiable meaning, with units and a zero reference. | |
| Oct 23, 2022 at 17:35 | comment | added | carloc | I'm afraid dBm is not a power at all, it's a unit less quantity, a kind of power ratio if you like. It's defined as dBm=10Log(P/1mW) and logarithms, and other transcendental functions, only "work" on pure numbers. That's why they can be summed and manipulated as we usually do. | |
| Oct 23, 2022 at 16:56 | history | answered | user319836 | CC BY-SA 4.0 |