Talk:DBm

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This article contains material from the Federal Standard 1037C (in support of MIL-STD-188), which, as a work of the United States Government, is in the public domain.
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The dBm is still a ratio between two quantities, as are all decibel measurements, this on relative to 1 mW, so shouldn't it also have no units, simular to Sound intensity level? --LeakeyJee 14:57, 31 May 2006 (UTC)

what do you mean by "it also have no units"? dBm is a shorthand for the power in Decibel relative to 1mW. it *does* have units in the name. that's what the "m" is for in dBm.

[edit] Impedance dependent?

The first paragraph states that dBV is impedance dependent:

dBm (or dBmW) and dBW are independent of impedance (as opposed to dBV which is dependent, for example).

I disagree. As it says on the Decibel page:

dBV - voltage relative to 1 volt, regardless of impedance.

Surely the impedance is only important when converting between dBm and dBu/dBV? Pediacycle 04:48, 14 August 2007 (UTC)

A dBm is referenced to 600 ohms and is a power measurement. 0 dBm = 1mW across 600 ohms. (as defined by the I.E.E.E.) Combining Kirchoff's voltage law (V=IR)and the power formula (P=IV) you get: V = sqrt(PR); for a 600 ohm circuit, V = sqrt (0.001 * 600) = 0.7746 volts For a 50 ohm circuit, V = sqrt (0.001 * 50) = 0.2236 volts A dBu is not reference to impedance and is a voltage measurement. 0 dBu = 0.7746 volts. dBu's came later in electronics so that is why you see the same voltage as 0 dBm, but it's only the same at 600 ohms!

70.70.3.5 03:09, 24 August 2007 (UTC) John Fulton - sorry don't have any web links the two formulas are laws and the rest is math.

The 600Ω sounds like it is in the context of telephone lines. However dBm is used for RF, inlcuding power levels on transmission lines and also in free space. Here the dBm is a power referenced to milliwatt and not to any particular impedance. When it comes to a radio wave the concept fo a volt is not relevant, you would have to measure volts per meter or watts per square meter, and even this is a different concept. Graeme Bartlett 04:20, 24 August 2007 (UTC)

[edit] "Typical power"

The typical power of an FM broadcast transmitter is given in the article as 100kW. While this is true for a transmitter with a desired range of 30-40 miles, many local radio stations have an Effective radiated power of 1kW or less, since this is enough to cover an area a few miles across. Also, FM transmitters with mixed polarization may have an ERP higher than this. The wording has been altered to reflect this. See also [1]. --♦IanMacM♦ (talk to me) 09:30, 11 November 2007 (UTC)

[edit] Signal strength in dBm?

I understand the concept of transmitted power, and I can understand power at the receiver end too, once it has been converted to electricity. But what is the "power" of a radio signal? Take for example the description (from the table of typical values) "Typical maximum signal strength (−10 to −30 dBm) of wireless network". I am similarly puzzled by "Typical RF power inside a microwave oven". Can someone explain this to me? Thunderbird2 12:25, 11 November 2007 (UTC)

Any statement involving the use of the word "typical" is open to debate. Radio transmitters, microwave ovens etc have variable power and they are subject to the inverse square law governing the strength of electromagnetic radiation at a distance. Broadcast radio transmitters have a wide range of power outputs, and microwave ovens can range from 500 to 2000 Watts. The figure that I gave for the typical maximum received strength of a wireless network is based partly on research from PassMark Wireless Monitor [2] and also from my own tests of a wireless router with the receiver next to the transmitter (maximum received signal around -25dBm for a 100mW transmission.) PassMark gives the figure of maximum strength for a wireless network transmission as ranging from -10dBm to -45dBm, and this may vary as the maximum strength for a wireless network is dependent on a range of factors. Any further comments here would be welcome. --♦IanMacM♦ (talk to me) 18:38, 11 November 2007 (UTC)

I wasn't questioning the values. Rather, my question was a conceptual one. I will try to phrase it differently. Imagine a radio transmitter of power W in free space. At a distance R, the intensity of an expanding spherical wave, neglecting absorption, would be W/(4 pi R^2). It then seems natural (to me) to measure signal strength at that position in watts per square metre rather than watts. Or for the microwave oven, I would think more in terms of energy density (joules per cubic metre). So, another way of asking the same question is: How do I convert the intensity for the radio transmitter (or energy density for the microwave) into power? Thunderbird2 19:02, 11 November 2007 (UTC)

I don't think it's an indication of the field strength at any given location inside the oven. It's simply the total output power of the microwave elements. Oli Filth(talk) 19:08, 11 November 2007 (UTC)
I see. That clears up the question about the oven. For the wireless network is it also the transmitted power? Thunderbird2 19:15, 11 November 2007 (UTC)
The transmission power of a wireless router for home and small office use is usually around 100mW (20dBm). There are several ways of measuring the received signal strength of a wireless network, and there is a good explanation at [3] (PDF format). --♦IanMacM♦ (talk to me) 09:01, 12 November 2007 (UTC)
I still don't understand. The pdf tells me how to convert power (in watts) to power level (in dBm). But how do I get to power (from intensity) in the first place? Thunderbird2 09:31, 12 November 2007 (UTC)
In general, it's not a simple relationship. It's a function of power density (power per square metre), antenna aperture, effective receiver resistance, etc. Luckily, this article doesn't mention "intensity". Oli Filth(talk) 11:01, 12 November 2007 (UTC)
OK, but we're getting closer because power per square metre is the same thing as intensity. You say the signal strength (in dBm) is a function of receiver resistance. Does that mean it's the power in the electrical circuit after reception? That would make more sense to me. Thunderbird2 11:32, 12 November 2007 (UTC)
With certain assumptions, the received power (i.e. the signal strength in the receiver) is given by the free-space path loss equation. Obviously, this won't be the same as the signal strength at the transmitter, which is what the numbers in this article are geared towards. Oli Filth(talk) 11:38, 12 November 2007 (UTC)