If that equation were just attenuation over distance, there would be no
frequency factor, or at least the attenuation would not be directly
proportional to frequency.  So the equation is somehow taking into account
antenna characteristics (i.e., it is not attenuation, but path loss, quite
different).  I don't know what type of antennas are assumed, but one thing is
for certain, AM XMIT/RCV antennas are very different than 2.4 GHz antennas. 
So you have to back out the 2.4 GHz antenna factors/gains/apertures and
substitute those for the AM link (high field intensity source, electrically
short whip or loop receive antenna).  The other thing to note is that I would
expect a 2.4 GHz link to operate over a range of one hundred feet or so, while
an AM broadcast link covers many miles.  The equation you cite cannot take
into account the various boundary conditions encountered over an AM link that
affect AM signal strength in a small local area (like passing under a bridge).



From: "George Stults" <[email protected]>
List-Post: [email protected]
Date: Thu, 2 Dec 2004 10:48:25 -0800
To: <[email protected]>
Subject: path attenuation in AM band




Hi Folks, 



I have an equation for the attenuation of a radiated signal as a function of
(distance, frequency, and environment.)  



ATTN = 37 + 20 Log (Freq)  + 20 Log (Dist),  Freq is in MHz, Distance in
Miles, Environment is free-space/line-of-sight (N=2). 



My information is that this equation works well for 2.4GHz frequency range.   



Could anyone tell me whether this equation would still make valid predictions
for attenuation in the AM band. (535 ­ 1650)Khz, and lower?  



Or could anyone suggest an equation that would predict path loss as a function
of distance, frequency, and environment for lower frequencies? 



Thanks in advance. 



George Stults 


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