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I am very confused how ham operators deal with all this, because most of the information in internet is controversial.

I have coaxial cable with loss of 34dB at 100 meters at 2400mhz. People say that if you send 1 mW of power (0dBm) trough this cable @ 2400mhz at the other end of it you will get -34dBm. Is this true ? (because of 0dBm - 34dB = -34dBm)

If this is true, then why people say that a source signal of 20dBm will be attenuated with the same calculation ( 20dBm - 34dB = -14dBm) when 20dBm is actually 100 times more watts of power than 0dBm ? My logic is that if you have 100 times more transmitted power you have 100 times the distance in order to get the same -34dB... 100 times the distance is 100meters * 100 = 10km !!? Obviously that's too much distance to believe myself but that's more logical than what many other say that -14dBm will attenuate to -34dBm after couple more meters of additional cable length (because of 34dB * 1.59 = 54dB. 1.59 is 1.59 times the distance of the cable length, which is 159 meters. So 20dBm - 54dB = -34dBm again at 159 meters cable length, which is obviously not 10 km!). So to recap - 0dBm goes 100 meters and 20dBm goes 159 meters. Makes no sense...

Similar, (but not identical as this scenario is wireless) I also see some guys say that the difference between 31dBm and 20dBm is only one brick wall, because brick wall attenuates 11dB for example. Again the difference between 31dBm and 20dBm is 12 times the power. If 20dBm can penetrate 1 brick wall and have 8dBm remainder, why wouldn't 31dBm penetrate 12 times thicker brick wall and have the same remainder of 8dBm ?

Please tell what is right and what is wrong in these statements ?

P.S. Let's don't talk about that WiFi is bidirectional, we're talking about TX only / signal delivery only scenario and how it propagates. I really hope for simple answers than complicated math/physics calculations

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    $\begingroup$ We don't do calls here. Asking and answering stays in written, so I'm removing that request from your question $\endgroup$
    – Marcus Müller
    Commented Jun 16, 2023 at 22:08

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I have coaxial cable with loss of 34dB at 100 meters at 2400mhz. People say that if you send 1 mW of power (0dBm) trough this cable @ 2400mhz at the other end of it you will get -34dBm. Is this true ? (because of 0dBm - 34dB = -34dBm)

If your cable is 100 m long, yes.

If this is true, then why people say that a source signal of 20dBm will be attenuated with the same calculation ( 20dBm - 34dB = -14dBm) when 20dBm is actually 100 times more watts of power than 0dBm ?

Because that's how it works. No matter how much you put in, you get 34 dB less out - in other words, 1/2500 of what you put in.

My logic is that if you have 100 times more transmitted power you have 100 times the distance in order to get the same -34dB... 100 times the distance is 100meters * 100 = 10km !!?

Your logic is wrong, sorry. The following things you write about distance really make no sense to me, as hard as I tried (and I'm a communications engineering electrical engineer). Your logic has no good relationship with physics, it seems!

some guys say that the difference between 31dBm and 20dBm is only one brick wall

That isn't correct without context. The difference between 31 dBm and 20 dBm is an attenuation by a factor of 11 dB, roughly a factor of 12.6 in linear terms (in power).

How much a physical object attenuates some transmission depends on the size, thickness and material of that object, and the frequency of the transmission. For 3 MHz, these 11 dB, for a dry hollow ceramic brick wall, seem too pessimistic, for 300 MHz maybe realistic, for 2.4 GHz very optimistic and for 300 GHz totally wrong.

why wouldn't 31dBm penetrate 12 times thicker brick wall and have the same remainder of 8dBm ?

Because for every unit of distance you get attenuation, so for two units of the distance, you get the attenuation squared, not doubled.

Because dB are a logarithmic unit, the attenuation in dB grows proportionally to distance, as the attenuation in linear terms grows exponentially.

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  • $\begingroup$ About the physics - I think it is quite right actually. If you have a 20mm hose attached to fountain with flow rate 20 liters/minute how far would the water go in the air ? Now imagine the same 20mm hose with flow rate of 100 liters/minute ? I bet it would go about 4-5 times more further in the air (although we have gravity considerations in here). I see this as similar example. There's no reason 100mw of power to reach 159 meters and 1mw 100 meters on that coax cable. $\endgroup$
    – Svetoslav
    Commented Jun 16, 2023 at 22:54
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    $\begingroup$ A coax cable is not a fountain. Your physics is wrong. I really can't help you with that - it's something you will have to accept. $\endgroup$
    – Marcus Müller
    Commented Jun 16, 2023 at 23:05
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    $\begingroup$ Maybe this helps: take a piece of cable that for your frequency of interest has 3dB of attenuation. So, half the power you put in on one end comes out at the other. That's easy to picture, right? Then, take a second, identical cable and attach it to the output of the first. Now, half of what the first puts out comes out of the second. Because what the first one puts out is already only half of the original input, in total the output is one quarter of the original input. Now attach another identical cable: you get one half of one quarter, that is, one eighth of the input for three times length. $\endgroup$
    – Marcus Müller
    Commented Jun 16, 2023 at 23:12
  • $\begingroup$ I can't haha. Ok, do you know if you have antenna at the very end of this coax, what will be the radiated power in comparison between source of signal 20dbm and 0dbm at 100 meters? 100 times more is that correct ? If correct how come the dB is now not strong enough to be transmitted much better than 0dBm source signal ? $\endgroup$
    – Svetoslav
    Commented Jun 16, 2023 at 23:12
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    $\begingroup$ @Svetoslav do you understand what I wrote in my long comment above? Your assumption that attenuation must be proportional to length is simply not founded in reality: every new piece of cable of the same length attenuates by the same factor. So, the factors "stack" multiplicatively, not "additively". $\endgroup$
    – Marcus Müller
    Commented Jun 16, 2023 at 23:16

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