we know on paper it takes 11nW to the terminals of a 1/2 wave dipole to produce 250uV / m @ 3m, but how much power does it take to produce canada's standard of 100uV / m @ 30m?
have not (or missed it) seen a topic covering this for canada's license exempt standard.
44 nw
Answer = 182.9 nW for free space conditions, however environmental reflections of the radiation of an antenna installed near the earth may produce a net field higher or lower than 100 µV/m at 30 meters.
The question has been answered but for those interested the calculation goes as follows:
Neglecting near field effects the field strength varies with distance according to FS = 1/D. To find the field strength at 3 meters knowing the FS at 30 meters, FS(3) = (30/3)x FS(30). In this case FS(3) = (30/3)*100 = 1000uV/m at three meters.
The field strength is proportional to the square root of the power so to get the power ratio we square the field strength ratio and the power is P = (FS1/FS2)^2 times the power for FS2. If 11.43 nW produces 250 uV/m (FS2 here) at 3 m then using the numbers above P = (1000/250)^2 X 11.43 = 16 x 11.43 = 182.9 nW.
Another interesting case is if we are given the power P then the field strength at 3 meters is FS = SQRT(P/11.43 nW) X 250 uV/m. So for 1 mW, as an example, the field strength will be SQRT(1E-3W/11.43E-9W) x 250uV/m = 3381 uV/m at three meters.
Old professors never stop trying to explain things.
Neil
This is really good for guys like
me, who didn't go to school for
this stuff. I admire you all greatly.
Bruce, DOGRADIO
Neil wrote: "Old professors never stop trying to explain things."
While I never have been a professor, I am old, have taught quite a few courses to broadcast engineers before I retired, and also have a need to explain things 🙂
Neil's development leads to the correct answer (of course).
Another, maybe quicker way of producing this answer is to use this approach.
For the power needed to be radiated by a center-fed, 1/2-wave dipole in the direction of maximum radiation to produce a given field intensity at a given distance, over a free space path:
P = (E * D)^2 / 49.2
where
P = Applied power in watts
E = Desired field intensity in V/m
D = Distance to the above E field in meters
So, if we want know what power it takes to produce a free-space field of 100 µV/m at a distance of 30 meters, we have...
P = (0.0001 * 30)^2 / 49.2
= 0.003^2 / 49.2
= 0.000009 / 49.2
= 0.0000001829 watts (approx) ANSWER
That power expressed in units of nanowatts is about 182.9.
Of course the answer could be calculated in one step using the first equation above, but I wanted to show the intermediate results leading to that answer.
I'm old.
In the days of the old Ramsey FM-10,
it was said that, if the antenna was
extended to the correct length (which
the operator had to figure out) the
transmitter would comply with Part 15.239.
Or... maybe it was never really said... Maybe
it was "suggested" in a very remote way.
In those days, there were dicussion groups
based around the FM-10. There were a lot
of funny modifications. I think one guy built
a little heater in the oscillator - to keep it
on channel. Sort of like a crystal oven. Anyway, there
was some kind of feedback loop that kept the "VFO" where it was supposed to be. I think a
lot of people had fun trying to get the FM-10
to work right. I put a cooling fan on mine!
HAHAHAHAHA.
So somebody said that the Ramsey FM-10 put
out 5 mW. I had mine in an upstairs bedroom
where I tried to comply with the regs. With
the antenna partly collapsed, the FM-10 didn't
go very far. It seemed to be in the ballpark
of 15.239.
Then again - that room was a junk room. It
was full of all kinds of stuff that must have
been attenuating the signal.
Sometimes I really miss the old days
of not knowing. But then learning what is
really correct years later - has a greater value.
That's why this board is such a good place to
learn about electronics.
Bruce, DOGRADIO
Side by side we have knowledge from men who know and profess, with those among us who wonder and confess.
Part 15 is the junction of science and philosophy.
It initially takes scientists to figure out all of
what is needed to make all of this wonderful
technology.
Then we apply it to the rules as best as we can.
"To get a NOUO or to not get a NOUO!" That
is the question! Do I understand the reality of
RF even if I can't see it? If an antenna tower
falls down in the middle of the woods, does the
receiver S meter go down even if you are not there??
Is the receiver really there? And WHY WHY WHY
doesn't the BFO work? And if the receiver is a Heathkit
HR-10B, that question isn't even necessary in the first place!!
And I'm at work!! Are my dogs still at home?? "HOOVER!!"
Bruce, DOGRADIO
Robert's question is a very good one.
Especially when somebody (me) is just understanding
what is going on.
Bruce, DOGRADIO