What is the minimum frequency that can be transmitted by a PCB?
Of course this question is size and shape dependent. So I read somewhere that the antenna lenght must be about λ/4 to transmit those frequencies.
So I was not sure if you take the longest or the shortest side, I believe you take the longest one. So in the case of a rectangular PCB, say 10cm x 5 cm, I took 10 cm for example.
Now we need to define the near field of the antenna, so one precise way, described by ednmag.com, is:
r = (2* l^2 ) / λ , where l is the longest size of the antenna
another general definition is (by Ott & White):
r = λ / (2 * π )
In fact the Ott & White definition, can actually be applied to PCB's too.
So going with the first formula, in case of our example (10x5 cm PCB),
At 500 meters, using the first formula for the fraunhofer boundary definition, using the path loss equations (Friis , Schantz) :
The first formula tells us that the minimum frequency that can be transmitted by this PCB is roughly
47713 Hz, since the path loss is positive before this, so this is our minimum reference distance I believe.
Also using the general formula
λ / (2 * π ), the min frequency is
72273 Hz Electric and
76303 Hz Magnetic field, roughly.
So there is some discrepancy between the two formulas, but it's roughly in the same range.
However the first formula already puts the limit in the far field where we have planewaves with 377 Ω impedance, while the second one is still in the range where the magnetic and electric field has discrepancies between their impedance.
So I assume the first formula is more accurate in our case since I'd expect the impedance to be 377Ω.
The question is as, stated in the beginning, whether my thoughtprocess is correct, and given that the first formula is more accurate, is the minimum frequency in this case roughly 47713 Hz?
I have read that you need a λ/4 length antenna to transmit those waves. So in our case that would be 12km long antenna, which is obviously larger than 10cm.