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Is the following operation valid or invalid? The question is to calculate receiver gain and this is one of the steps I got confused.

The receiver sensitivity is given -45dBm and transmitted power is 70dBm.

70dBm - (-45dBm) = 115dB. So, 70dBm + 45dBm = 115dB?

But isn't dBm + dBm operation invalid? Can someone please explain if I'm missing something?

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But isn't dBm + dBm operation invalid?

x dBm is a power. It's perfectly reasonable to add together two powers to get another power, powers add after all. However, if you took a signal of -45dBm and added it to a signal of +70dBm, then you would get a final power indistinguishable from +70dBm, as the tiny power of -45dBm would barely add anything to it. But that's not what you're trying to do.

If instead you want to find the gain or attenuation to get between two power levels, for instance between -45dBm and +70dBm, you are not really doing an addition or subtraction, but taking a ratio. You can easily express that ratio as dB. The ratio between -45dBm and +70dBm is 115dB.

You would express the path loss of a +70dBm transmitted signal as +70dBm - 115dB = -45dBm.

As the powers are already expressed in log form as dBm, you can find the dB ratio by simple subtraction of the dBm numbers.

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  • \$\begingroup\$ 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 \$\endgroup\$
    – eldenlord9394
    Commented Oct 24, 2022 at 9:49
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    \$\begingroup\$ Given that your expression for Pr uses a Pt of about +40, PL of about -40, and the G's are negligible, I would expect the answer to be about zero (this is the 'back of the envelope, sanity check, sum'). When I do it without those gross approximations, I get +2.25 dBm. How is the given answer about -30? How is your answer about +30? \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 10:10
  • \$\begingroup\$ Actually, I converted dB to dBm using this relation: dBm = dB + 30. That is how I got +30 value. I think the professor made a mistake and I should check with him for a correction. \$\endgroup\$
    – eldenlord9394
    Commented Oct 24, 2022 at 11:11
  • \$\begingroup\$ @eldenlord9394 No, dBW (dB watts) and dBm (short for dBmW) (dB milliwatts) are both absolute powers, and are in a ratio of 1000 in power = 30dB different. dBm = dB + 30 is nonsense, it's an ill-formed expression, the left hand side is an absolute power, and the right hand side is a ratio. In general, dBx means dB with respect to x. It's an absolute power with respect to the given x. If it's just dB, it's a ratio, or an error if it's meant to represent an absolute level (the plane made 125dB on takeoff - ugh!) \$\endgroup\$
    – Neil_UK
    Commented Oct 24, 2022 at 11:17
  • \$\begingroup\$ I see. Thanks for the detailed explanation. \$\endgroup\$
    – eldenlord9394
    Commented Oct 24, 2022 at 13:19
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You're missing the point of using decibel: multiplication in linear terms becomes addition in logarithmic terms.

So, since the question was "what is the ratio of transmitted power to sensitivity" (and not: what is the difference), subtracting the two decibel values was logical.

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You need to be aware of the context of the addition:

dBm+dBm=dBm is illegal. Here you are actually multiplying powers. The result is power squared, not power so the operation does not make sense.

But if you are finding the ratio of powers to obtain a gain and one power is less than 1mW, then dBm-(-dBm)=dB. Notice that the units cancel in the ratio so the equation is valid.

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  • \$\begingroup\$ 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. \$\endgroup\$
    – carloc
    Commented Oct 23, 2022 at 17:35
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    \$\begingroup\$ @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. \$\endgroup\$
    – user319836
    Commented Oct 23, 2022 at 18:08
  • \$\begingroup\$ @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 \$\endgroup\$
    – eldenlord9394
    Commented Oct 24, 2022 at 9:49

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