The text below was sent to EDM for comment. In the meantime it might
be of interest here.
RF
+ + +
For your Engineers:
In your website FAQ I noted the text pasted following the line below. With
all due respect, you might want to re-visit the conclusions in your reply,
which may lead users of your equipment into an illegal situation.
In the US, Part 15 of of the FCC Rules does limit FM field strength to
250 uV/m at 3 meters in any direction from the transmit antenna. However the
power that this field strength will generate in a matched receiving dipole
is far, far less than the ~12.5 nW you state in your response below.
The reality is that the free space, peak field generated by a matched,
1/2-wave dipole TRANSMIT antenna will be ~250 uV/m at 3 meters, when radi-
ating ~ 12.5 nW.
A 30" whip used as a transmit antenna may not be as efficient a radiator as
a 1/2-wave dipole, but its gain is not sufficiently low enough to radiate a
legal field with anything remotely close to 10 mW of power applied to it --
as your FAQs imply.
Sincerely,
Richard Fry, Broadcast Engineer
Visit http://rfry.org for FM transmission system papers.
_____________
Q: I am confused. I read that my local authority allows only 10nW of RF
power, your weakest unit transmit 10mW of RF. In my book it is 1 Million
times over the limit!?
EDM Answer: Don't confuse RF power input to an antenna with actual
radiated power received or measured at a point in space. To try and
explain: Suppose you take a light bulb and suspend it in the middle of a
big sphere or ball. If that is say a 100W of light output, it is obvious
that if you can measure at a very small point on the inside surface the
light intensity, and convert it to equivalent power it will not be 100W any
more but, 100W divided by all the many millions of pin-points that makes up
the surface area of this sphere. The light source can be your antenna
radiating in all directions, and the point anywhere on the surface
representing the receiving antenna some distance from the transmit antenna.
Doing the math on a often enquired uV requirement, we calculated that:
250uV/m @ 3m ~ 12.5nW of received power intensity or in engineering
terms -59dBm using a 1m sampling antenna 3m distance away from the transmit
antenna. (This is only how WE see and interpret it as outsiders, you are
advised to get a 2nd opinion)
Also remember that apart from an antenna not be able to transmit or receive
without losses, there is free space attenuation that "eats" up some of the
signal as it travels from the transmit antenna to the receiver antenna.
<P ID="edit"><FONT class="small">Edited by rfry on 10/12/05 03:45 PM.</FONT></P>
be of interest here.
RF
+ + +
For your Engineers:
In your website FAQ I noted the text pasted following the line below. With
all due respect, you might want to re-visit the conclusions in your reply,
which may lead users of your equipment into an illegal situation.
In the US, Part 15 of of the FCC Rules does limit FM field strength to
250 uV/m at 3 meters in any direction from the transmit antenna. However the
power that this field strength will generate in a matched receiving dipole
is far, far less than the ~12.5 nW you state in your response below.
The reality is that the free space, peak field generated by a matched,
1/2-wave dipole TRANSMIT antenna will be ~250 uV/m at 3 meters, when radi-
ating ~ 12.5 nW.
A 30" whip used as a transmit antenna may not be as efficient a radiator as
a 1/2-wave dipole, but its gain is not sufficiently low enough to radiate a
legal field with anything remotely close to 10 mW of power applied to it --
as your FAQs imply.
Sincerely,
Richard Fry, Broadcast Engineer
Visit http://rfry.org for FM transmission system papers.
_____________
Q: I am confused. I read that my local authority allows only 10nW of RF
power, your weakest unit transmit 10mW of RF. In my book it is 1 Million
times over the limit!?
EDM Answer: Don't confuse RF power input to an antenna with actual
radiated power received or measured at a point in space. To try and
explain: Suppose you take a light bulb and suspend it in the middle of a
big sphere or ball. If that is say a 100W of light output, it is obvious
that if you can measure at a very small point on the inside surface the
light intensity, and convert it to equivalent power it will not be 100W any
more but, 100W divided by all the many millions of pin-points that makes up
the surface area of this sphere. The light source can be your antenna
radiating in all directions, and the point anywhere on the surface
representing the receiving antenna some distance from the transmit antenna.
Doing the math on a often enquired uV requirement, we calculated that:
250uV/m @ 3m ~ 12.5nW of received power intensity or in engineering
terms -59dBm using a 1m sampling antenna 3m distance away from the transmit
antenna. (This is only how WE see and interpret it as outsiders, you are
advised to get a 2nd opinion)
Also remember that apart from an antenna not be able to transmit or receive
without losses, there is free space attenuation that "eats" up some of the
signal as it travels from the transmit antenna to the receiver antenna.
<P ID="edit"><FONT class="small">Edited by rfry on 10/12/05 03:45 PM.</FONT></P>