Wave propagation 1

[WhoJuntao]

Hi,

I read somewhere that a given radio signal has a wavelength of 1Km and since the wall in front of it is only 1m thick, the wave has no problem penetrating it. It is only when the wavelength starts to compare with the ditance between the iron ribs in the concrete wall that reflection problems start to arise.

I don't know much about this so if anyone can explain in two lines i'd be grateful.

Thanks.

 

[blahdroneblah]

Hi WhoJuntao: you've got two main effects here:
1/ if the obstacle is << wavelength then the wave is only slightly perturbed by its presence and tends to wrap around it.
2/ If the obstacle is a dielectric with low loss and with embedded mesh or grid of conductors then when the wave length _inside the obstacle_ is << distance between conductors it tends to penetrate through the gaps , otherwise the mesh looks like a continuous surface. Skin depth can play a role in how well the mesh shields low frequencies .

 

[WhoJuntao]

Hi & 10x

it's about one antenna in GF and its coverage in the upper floor.

Obstacle is the ceiling: 450mm thick.
Iron rods spacing inside the ceiling concrete slab: 150mm
Ceiling is infinitely long (1Km (airport concourse))
TX/RX frequencies: 383/420Mhz (Terrestrial Trunked Radio (TETRA)) => wavelength 750mm

from what you say, 750>150mm => the ceiling is like:
1. a 100% mirror ? or
2. an x% refractor? so that x% of the power passes and (1-x) is reflected back into the same space?

if 2.) is correct, the antenna above should have different channel allocation, right? or better have the same chaneels so that someone above is caught For Sure by This antenna ...?

10x

 

[johnwiss]

Think of it this way: Low frequency = Long wavelength and waves only interact with objects that are in size proportional to the wavelength (actually a fraction of the wavelength). The wave diffracts around small objects and only start to reflect when the object impinged on by the wave can &quot;pin&quot; the wave by allowing a reflection that interferes with the wave's propagation and as such must be big enough to &quot;influence&quot; it. Note that you can see inside a microwave although there is enough EM radiation to cook your head--the small conduction slots relative to the cm waves allow the door to look opaque to the microwaves--thank goodness!!

 

[WhoJuntao]

johnwiss - i get you 80%

that microwave trick seems interesting, can you please elaborate?

 

[BOMBOVA]

Xf is frequency of 1 km/1 wavelength
Zf is frequency of mesh that is equal to the square ribbing
in the wall, it is one wavelength of the square grid.
It is apature coupled and passes through the grid.
The grid metal as mesh is most effective in shielding, attenuating the rf signal at approximately 100 times the Zf
, frequency, and probably not more than 20db of attenuation
at best. So the waves pass right through. Now if you put up a metal wall, the attenuation at vhf and uhf and celullar
fx goes up to 80 to 120 db ie. Copper Wall.
RTG. Burnaby