Farthest Away from COL

[1L6E6VHF]

How many mV/m is 60 dBu?

One. (1mV/m = 60dBu)

Generally mV/m is used for AM and dBu for FM/TV.

W9WI is absolutely correct, but I'm compelled to add for the benefit of newer readers that since the power ratio of one signal over another is the square of the voltage ratio, that a tenfold increase in voltage equals a hundredfold increase in the power, thus:

1μV/m - 0dBμ
3.16~uV/m = 10dBμ
10μV/m = 20dBμ
100μV/m = 40dBμ (aka 0.1 mv/m - protected day GW contour of class A AMs)
.5mV/m = 54dBμ (original protected day contour of regionals - often waivered)
5mV/m = 74dBμ (AM city-grade contour)

 

[NHRadio]

W9WI is absolutely correct, but I'm compelled to add for the benefit of newer readers that since the power ratio of one signal over another is the square of the voltage ratio, that a tenfold increase in voltage equals a hundredfold increase in the power, thus:

Can you restate that in a language approaching English for us non-engineers? I have only enough engineering knowledge to be dangerous.

 

[Hotseat]

WSTH 106.1 licensed to Alexander City, AL. Tower is in Opelika 42 miles away. (Barely) serving Columbus, GA.

 

[w9wi]

W9WI is absolutely correct, but I'm compelled to add for the benefit of newer readers that since the power ratio of one signal over another is the square of the voltage ratio, that a tenfold increase in voltage equals a hundredfold increase in the power, thus:

Can you restate that in a language approaching English for us non-engineers? I have only enough engineering knowledge to be dangerous.

I'll try

One mV/m means one millivolt per meter. (to oversimplify a bit, if you have 1 mV/m of signal, there's enough floating around in the air that if you had an antenna one meter -- about 3 feet -- long, there'd be one millivolt of signal on that antenna.)

One uV/m means one microvolt per meter. That's 1/1000 of one millivolt. The "u" in dBu stands for uV.

Or, to look at the other side, one mV/m is 1,000 uV/m.


dB, or "decibels", are a logarithmic way of expressing ratios. If you're talking about voltage, the formula is:

dB = 20 * log(voltage ratio)

The ratio between one mV/m and one uV/m is 1,000. The log of 1,000 is 3. (try it, with Windows Calculator in "Scientific" mode) 20 times 3 is of course 60 -- so one mV/m is 60 decibels stronger than one uV/m.

Again, decibels are a *ratio* -- a way of comparing two quantities. It would be meaningless to say "I have a 6dB battery"; a "6dB battery" has twice the voltage of... what? You have to define what you're comparing to. I could certainly say "I have a battery that has 6dB more voltage than a D cell". A D cell has 1.5 volts, so 6dB more would be twice that, or 3 volts.

When you're talking about radio signals, the reference for comparison is one uV/m. That's where the "u" in "dBu" comes from. If we say 100.3 The Beaver delivers 60dBu of signal to Ashland City, we mean it delivers 60 decibels more than one uV/m -- that's 1,000 times.


In practical terms...

0.1mV/m = 40dBu = how much signal a clear-channel AM station must deliver to a location during the day if it's to be protected from interference.
Also, how much signal an FM station can deliver 10% of the time to a given location without interfering with another station that already serves that location.

1mV/m = 60dBu = how much signal an FM station must deliver to a location to be considered to "serve" that location.

5mV/m = 74dBu = how much signal an AM station is required to provide across its city-of-license during the day.

Does that help?

 

[NHRadio]

It helps. Thanks. The decibel thing isn't sinking in but I understand the rest.
So, if Class A AMs are protected out to 40dbu, why? Few if any radios are good enough anymore to be able to get a listenable signal over all the man made hash...or am I making a math error?

 

[1L6E6VHF]

It helps. Thanks. The decibel thing isn't sinking in but I understand the rest.
So, if Class A AMs are protected out to 40dbu, why? Few if any radios are good enough anymore to be able to get a listenable signal over all the man made hash...or am I making a math error?

No math error. The rules were written several decades ago, and reflected the Radio Commission's desire to assure that as many people in rural areas as possible would not have to wait until sunset to hear at least one radio station.

In the early 1930's, nobody had a plasma TV, a personal computer, a compact florescent light bulb, or a cell phone charger with a switching power supply. In fact, a significant part of rural America did not have electricity (though the REA was changing that fast). Farm households could buy farm radios that used 32-volt batteries. A farm did not have an HOA telling them they could not throw a 50' wire into a tree. Connect that wire to the farm radio, and that .1 mV/m signal could actually sound pretty good, provided there were no thunderstorms about.