0
\$\begingroup\$

I have used the Wikipedia FSPL formula for the farfield, and the one kindly provided by @mnsp for the near field:

To model the attenuation of the magnetic field across the spectrum. Low frequency magnets have a very quick attenuation by distance. For example a bar magnet easily loses it's influence after 3-4 cm. However as you increase the frequency, the attenuation is lowered, and it possibly reaches 0 right at the boundary of the near field.

In my thought experiment I have used a distance of 5 m to model this for a theoretical isotropic transmitter. The general formula is λ/(2*π) for the boundary.

So in our case we have pretty much close to 0 attenuation at 8.4 MHz, and it starts to go back up from there. Looks like this:

enter image description here

Here is the python3 code to remove any uncertainty how my calculations were done

import math

π  = math.pi    #pi
c0 = 299792458  #lightspeed
r = 5           #distance

#---------------------------CALCULATION---------------------------------------------------------

def FSPL(f):
 λ = (c0/f)
 k = (2 * π ) / λ

 PLANEWAVE= 20*math.log(f,10) + 20*math.log(r,10) + 20*math.log(4*π/c0,10)
 MAGNETIC  =abs(10*math.log(1/4  *  (  1/  (k*r)**2 + 1/  (k*r)**4     ) ,10))
 ELECTRIC  =abs(10*math.log(1/4  *  (  1/  (k*r)**2 - 1/  (k*r)**4 + 1/  (k*r)**6     ) ,10))

 if(r<(1/k)): #Fraunhofer Boundary
  return f,MAGNETIC,ELECTRIC  # Near Field
 else:
  return f,PLANEWAVE # Far Field

#------------------------------ITERATION--------------------------------------------------------

CYCLES=34
STEP=1
for i in range(0,CYCLES):
 print (FSPL(STEP))
 STEP=STEP*2

#------------------------------END--------------------------------------------------------------

Which begs the following questions:

  • What is exactly happening between the nearfield and farfield boundary?

  • Is the model theoretically accurate?

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6
  • \$\begingroup\$ Read about standing waves and properties of 1/4 wave vs 1/2 wave then you may understand H field at end points. (the 0 point near field) \$\endgroup\$ May 15 '17 at 16:59
  • \$\begingroup\$ You are not doing this question justice because you have given no clue as to which formula (or derivation of formula) in the linked article you are using. The bottom line is that this question is unanswerable without that detail. Show your formula and show a worked numerical example and don't expect miracles. \$\endgroup\$
    – Andy aka
    May 15 '17 at 17:56
  • \$\begingroup\$ @Andyaka obviously using the formula in the book for the near field. In this example I have used the H field formula added above. Where I assumed 1 for the gains. \$\endgroup\$
    – David K.
    May 16 '17 at 15:25
  • \$\begingroup\$ I see nothing in that formula relating to frequency. Listen, if you want this resolved you are going to have to come up with the exact way you got your numbers. As I said previously, a worked example would be good. \$\endgroup\$
    – Andy aka
    May 16 '17 at 18:12
  • 1
    \$\begingroup\$ @Andyaka you haven't read the doc, it say's right there that k = (2 * π ) / λ. But alright ,I have included the full python3 script that I used to calculate the values. I hope now all uncertainties about my question are nonexistent. \$\endgroup\$
    – David K.
    May 17 '17 at 16:10
0
\$\begingroup\$

That PDF provides various 1/distance^N contributors to field strength.

The Efield has [1/D^2 - 1/D^4 + 1/D^6]

The Hfield has [1/D^2 + 1/d^4]

enter image description here

I recall similar equations in Corson and Lorrain. I ain't a trained E&M jock, just trying to understand the mechanisms of shielding for reliable 16 and 24 bit systems.

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1
  • \$\begingroup\$ Well the formula indicates that the best reception of a signal of f frequency is right at λ/(2*π) distance from the transmitter, since there the attenuation is the smallest (ignoring of course the background noise) \$\endgroup\$
    – David K.
    May 17 '17 at 16:28
-1
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I believe I found the answer, I think I have used the formula incorrectly, since I used the absolute values of the decibels.

So in reality we have a path gain in the near field (+dB) and a path loss in the far field (-dB).

Of course if we add that up to a dB format power unit then it becomes an attenuation. So if we broadcast a 30 dBm signal, that will become: 30 dBm + - 65 db = - 35 dBm , for 8.6 GHz

So the correct outputs would look like this:

1       273    
2       261    
4       249    
8       237    
16      225    
32      213    
64      201    
128     189    
256     177    
512     165    
1024        153    
2048        141    
4096        129    
8192        117    
16384       105    
32768       93    
65536       81    
131072      68    
262144      56    
524288      44    
1048576     32    
2097152     21    
4194304     9    
8388608     -1    
16777216        -11    
33554432        -17    
67108864        -23    
134217728       -29    
268435456       -35    
536870912       -41    
1073741824      -47    
2147483648      -53    
4294967296      -59    
8589934592      -65    
\$\endgroup\$

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