Why is there no attenuation in the magnetic field at the boundary of the near field?

Question

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

• http://grouper.ieee.org/groups/802/15/pub/04/15-04-0417-02-004a-near-field-channel-model.pdf

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:

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?

Answer 1

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]

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.

Answer 2

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