Searching \ for '[EE]: Calculating RF power at a distance' in subject line. ()
Help us get a faster server
FAQ page: www.piclist.com/techref/io/serials.htm?key=rf
Search entire site for: 'Calculating RF power at a distance'.

Exact match. Not showing close matches.
'[EE]: Calculating RF power at a distance'
2002\10\07@195529 by

I'm trying to set up a network of PICs that chatter to each other
via RF. Everything is ticketyboo so far, but I'd like to know how
to calculate the transmitter power to meet the requirements of
between PICs is 100m - what minimum wattage would the
transmitter be ? I'm guessing the inverse square law is the guide,
and V/m too. I've looked up RF calcs with no practical success,
so if anyone can offer real-world help that would be appreciated

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listservmitvma.mit.edu with SET PICList DIGEST in the body

Pathloss(in dB) = 36.6 + 20*Log(F) + 20*log(D)

where D is in statute miles and F is in MHz.

(I can spit up something suitable for meters given a
minute or two!)

The loss figure is for propagation ostensibly in free
space, but works for most line-of-sight paths in the
VHF range (and up) too.

Gain for each antenna on each end are then *added* to
the loss figure to arrive at the total loss between
a transmitter RF output port and a receiver input
port.

This formula has served me well every time I have used
it, most recently for a 60 MHz wireless link.

The only thing it *doesn't* take into account is loss
due to obstacles, trees ...

RF Jim

{Original Message removed}
PathLoss(in dB) = 92.45 + 20*log(d) + 20*log(F)

d in kilometers and F in MHz.

Just don't use this formula where significant ground
wave exits  like in the AM broadcast band over open
sea (salt) water!

Note that these seeemingly simplistic formulas take into
account 1) inverse square law and 2) 'antenna aperature'/capture
area.

Transmitter power-> Xmit
Ant gain -> Path loss -> Receive
Ant gain -> power at Receiver

RF Jim

{Original Message removed}
What power is required is determined by what antenna(s) are used....

*
|  __O    Thomas C. Sefranek   tcscmcorp.com
(*)/ (*)  Bicycle mobile on 145.41, 448.625 MHz

http://www.harvardrepeater.org

{Original Message removed}
2 uV is approx. -107 dBm
100 meter path loss at 400 MHz: 64.5 dB

Assuming 10 mW (.01 W) transmit power and using
dipoles (approx 2 dB gain each end):

Power input to matched receiver:  8.95  nano-watts  = -50.5 dBm
Matched receiver (50 Ohm) input volts:   668.86  micro-volts

Looks liie you would be in the green ...

Above resutls derived from program: LINOSITE

90 GHz. Aperture, beam widths, etc."

http://www.webbworks.com/crstrode/calcs/rfcalcs.htm

RF Jim

{Original Message removed}
> What power is required is determined by what antenna(s) are used....

6" or shorter stubby much preferred. Looking through a couple of
catalogues some are quoted as having gains of 3dB or 4.5dB. Now
I'll have to pick a frequency to suit

Thanks to RF Jim I've got a practical resource to hopefully fill in
a few blanks

http://www.webbworks.com/crstrode/calcs/rfcalcs.htm

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listservmitvma.mit.edu with SET PICList DIGEST in the body

> Pathloss(in dB) = 36.6 + 20*Log(F) + 20*log(D)
>
> where D is in statute miles and F is in MHz.

OK, Jim, I'm not an RF expert like you, but this just seems contrary to the
laws of physics to me.

The 36.6 term is a scale factor to account for the specific units chosen.
I'll trust you figured this out correctly.  The 20*Log(D) term says the
signal is attenuated a fixed multiple for each unit distance.  I was
expecting the inverse square law here.  The 20*Log(F) term says that there
is an additional fixed attenuation multiple for each unit increase in the
frequency.  That makes no sense at all.

Did you perhaps copy the equation for something like signal loss in a
RF communication which I took to mean propagation between antennas.  I was
expecting equations for the propagation of electromagnetic waves in free
space.  What am I missing here?

*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: To leave the PICList
piclist-unsubscribe-requestmitvma.mit.edu

> OK, Jim, I'm not an RF expert like you, but this just seems contrary to
the
> laws of physics to me.
>
> The 36.6 term is a scale factor to account for the specific units chosen.
> I'll trust you figured this out correctly.  The 20*Log(D) term says the
> signal is attenuated a fixed multiple for each unit distance.  I was
> expecting the inverse square law here.

Recall that 10 log x^2 = 20 log x
The inverse term is effectively linearised by the log.

>The 20*Log(F) term says that there
> is an additional fixed attenuation multiple for each unit increase in the
> frequency.  That makes no sense at all.

This is to do with effective aperture size of the antenna relative to the
wavelength..
Think of an isotropic radiator painting a sphere with a signal radiating in
all directions. As you concentrate that signal onto a segment of the sphere
you get "gain". The physical size of the antenna plays a part in how large
(or small) a part of the sphere the aerial directs the signal onto. For a
given aerial size (eg a 1 metre dish) the gain is related to how many
wavelengths the dish is across. If you double the frequency and halve the
wavelength you get 4 times the area and the area illuminated by the dish
reduces accordingly. The area effect again gives you a 20 log term rather
than 10 log which you would otherwise expect.

> Did you perhaps copy the equation for something like signal loss in a
> transmission line or resistive media?  The original poster was asking
> RF communication which I took to mean propagation between antennas.  I was
> expecting equations for the propagation of electromagnetic waves in free
> space.  What am I missing here?

Free space is effectively a resistive media for these purposes once you are
beyond the near field (ie once the E & M fields have "settled down" to a
constamnt relationship. Interestingly (and irrelevantly to this particlular
problem) for the large satellite and deep space dishes, the near field
exytends to beyond the earth's atmosphere, so far field measurements are
"difficult".

Hmm - 2am - bedtime - wonder if brain is still working - AVR circular buffer
code seems to run so maybe it is - if any of the above is gobbldey gook
please take 2 asprin and call me in the morning. G'night.

Russell McMahon

--
http://www.piclist.com hint: To leave the PICList
piclist-unsubscribe-requestmitvma.mit.edu

> Recall that 10 log x^2 = 20 log x
> The inverse term is effectively linearised by the log.

Yeah, you're right.  I'm not sure what I was thinking.

> This is to do with effective aperture size of the antenna relative to the
> wavelength..

But I thought this was loss due to propagation in free space, and that the
antenna gain was added in separate from this equation.

> Think of an isotropic radiator painting a sphere with a signal radiating
in
> all directions. As you concentrate that signal onto a segment of the
sphere
> you get "gain". The physical size of the antenna plays a part in how large
> (or small) a part of the sphere the aerial directs the signal onto. For a
> given aerial size (eg a 1 metre dish) the gain is related to how many
> wavelengths the dish is across. If you double the frequency and halve the
> wavelength you get 4 times the area and the area illuminated by the dish
> reduces accordingly. The area effect again gives you a 20 log term rather
> than 10 log which you would otherwise expect.

But this says gain should *increase* with frequency.  The original equation

> Free space is effectively a resistive media for these purposes ...

Huh!?  Free space does not dissipate power when radiation passes thru it.
Therefore the only attenuation of that radiation as it travels thru free
space is because it spreads out radially from the transmitter.  This is
where the inverse square of distance effect comes from.

*****************************************************************
Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

--
http://www.piclist.com hint: To leave the PICList
piclist-unsubscribe-requestmitvma.mit.edu

Another analogy of gain is to take your maglite or similar focusable
flashlight and aim it at a wall in fairly bright conditions.  When you
first turn it on, and the pattern is wide on the wall, you won't be able
to see the reflected light... or very little of it.

Focus it down to a dot, and you can then see it, even with other light
(noise) around it and mixed into it, but only at that one location the
flashlight is aimed at.

There's the same amount of light leaving the flashlight, but now it's
focused on a point and you have a "usable" signal that a "light

Same thing happens with RF, and antenna "gain" is actually antenna
"focusing"... once you get over that hump mentally (many people think
that when calculating Effective Radiated Power that there really will be
that many watts of RF at the antenna... nope...) then it starts to make
more sense.

So as someone put it very succinctly, the question, as posed, can't
truly be answered, because you design RF systems as a whole -- INCLUDING
the antenna information and expected site noise floor, etc.  Wherever
that "site" may be, a small box with a PIC running (making RF noise) in
it, space, an outdoor antenna, an antenna on the outside of that little
PIC box, you get the idea.... (GRIN).

Nate

On Tue, 2002-10-08 at 08:10, apptech wrote:
{Quote hidden}

--
http://www.piclist.com hint: To leave the PICList
piclist-unsubscribe-requestmitvma.mit.edu

Response within text below.

----- Original Message -----
From: "Olin Lathrop" <olin_piclistEMBEDINC.COM>
To: <PICLISTMITVMA.MIT.EDU>
Sent: Tuesday, October 08, 2002 7:49 AM
Subject: Re: [EE]: Calculating RF power at a distance

> > Pathloss(in dB) = 36.6 + 20*Log(F) + 20*log(D)
> >
> > where D is in statute miles and F is in MHz.
>
> OK, Jim, I'm not an RF expert like you, but this just seems contrary to
the
> laws of physics to me.
>
> The 36.6 term is a scale factor to account for the specific units chosen.

Correct.

> I'll trust you figured this out correctly.  The 20*Log(D) term says the
> signal is attenuated a fixed multiple for each unit distance.  I was
> expecting the inverse square law here.  The 20*Log(F) term says that there

And it is inverse square law:

20*Log(F) => 10*Log(F^2)

(following from the old rules of:
1) add logs to multiply numbers and
2) multiply a log to raise a number to a power)

(10 deci-Bels = 1 Bel) this further reduces to:

2*Log(F) => Log(F^2)

> expecting the inverse square law here.  The 20*Log(F) term says that there
> is an additional fixed attenuation multiple for each unit increase in the

Correct

> frequency.  That makes no sense at all.

Sure it does.

This formula assumes an antenna's 'size' or capture area is
linked to the frequency of interest and it makes perfect
sense - as one progrsses *up* in frequency one progresses
*down* in wavelengh (as they are, of course, inversely
related) and this translates into, for a *fixed* distance,
a reduction in 'captured' (incident energy) seen at the

This portion of the equation adjusts the 'path loss' for
this effect that is dependent on frequency.

>
> Did you perhaps copy the equation for something like signal loss in a
> transmission line or resistive media?  The original poster was asking

Negative. Straight out of the "Reference Data For Radio Engineers Handbook".

> RF communication which I took to mean propagation between antennas.  I was
> expecting equations for the propagation of electromagnetic waves in free

You would prefer to take it back to Maxwell's equations and three-space
geometry?

> space.  What am I missing here?
>

Dunno ... application of the basic rules of math maybe?

RF Jim

>
> *****************************************************************
> Embed Inc, embedded system specialists in Littleton Massachusetts
> (978) 742-9014, http://www.embedinc.com
>

--

Response within text below.

----- Original Message -----
To: <PICLISTMITVMA.MIT.EDU>
Sent: Tuesday, October 08, 2002 9:10 AM
Subject: Re: [EE]: Calculating RF power at a distance

> > OK, Jim, I'm not an RF expert like you, but this just seems contrary to
> the
> > laws of physics to me.
> >
> > The 36.6 term is a scale factor to account for the specific units
chosen.
> > I'll trust you figured this out correctly.  The 20*Log(D) term says the
> > signal is attenuated a fixed multiple for each unit distance.  I was
> > expecting the inverse square law here.
>
> Recall that 10 log x^2 = 20 log x
> The inverse term is effectively linearised by the log.
>
> >The 20*Log(F) term says that there
> > is an additional fixed attenuation multiple for each unit increase in
the
> > frequency.  That makes no sense at all.
>
> This is to do with effective aperture size of the antenna relative to the
> wavelength..
> Think of an isotropic radiator painting a sphere with a signal radiating
in
> all directions. As you concentrate that signal onto a segment of the
sphere

Better to picture, for practical purposes, a set of 'crossed
dipoles' that radiate a nearly "isotropic pattern" ... especially
when considering the calculations involving a typical 'path loss'
equation. This way the mind may visualize the same antenna for
'receive' and 'transmit' and the physical size (physical dimensions)
of the dipoles can be seen to change as the frequency of operation
is changed (as will be explained in more detail below) ...

(So-called theoretical, infinetely-small "isotropic point sources"
are so hard to visualize and harder yet to explain their operation

> you get "gain". The physical size of the antenna plays a part in how large

We term it 'gain' - this seems to confuse a lot of folks as
the layman and newcomer to electronics only expect *active*
devices to have 'gain'.

> (or small) a part of the sphere the aerial directs the signal onto. For a
> given aerial size (eg a 1 metre dish) the gain is related to how many
> wavelengths the dish is across. If you double the frequency and halve the
> wavelength you get 4 times the area and the area illuminated by the dish
> reduces accordingly. The area effect again gives you a 20 log term rather
> than 10 log which you would otherwise expect.

Restated in part, for clarity, to:

** If you double the frequency you halve the wavelength
^^^ ----- --- ----------

>
> > Did you perhaps copy the equation for something like signal loss in a
> > transmission line or resistive media?  The original poster was asking
> > RF communication which I took to mean propagation between antennas.  I
was
> > expecting equations for the propagation of electromagnetic waves in free
> > space.  What am I missing here?
>
> Free space is effectively a resistive media for these purposes once you
are

Uhhh - free space, to date, appears to be *lossless*.

Otherwise - you have to explain into *what* form the RF (E and
H fields) energy converts into (normally, any losses seen
in any *lossy* medium are converted into a thermal form known
as 'heat' energy).

> beyond the near field (ie once the E & M fields have "settled down" to a
> constamnt relationship. Interestingly (and irrelevantly to this
particlular
> problem) for the large satellite and deep space dishes, the near field
> exytends to beyond the earth's atmosphere, so far field measurements are
> "difficult".

You must mean the large, widely spaced arrays arrays - as opposed to
just a single dish 10, 20 or 50 meter dish ...

>
> Hmm - 2am - bedtime - wonder if brain is still working - AVR circular
buffer
> code seems to run so maybe it is - if any of the above is gobbldey gook
> please take 2 asprin and call me in the morning. G'night.
>

Not a bad essay for 2 AM!

RF Jim

>
>             Russell McMahon
>

--