Questions: Reg EM wave transmission

[sarveshkumarv]

Why microwave frequencies are more affected due to rain? How does water actually affect microwaves?And lower frequencies have more diffraction.Why is it so?

 

[Claude Bile]

And lower frequencies have more diffraction.Why is it so?

One can analyse single-slit diffraction in terms of the Heisenberg Uncertainty principle. A photon passing through a slit has a positional uncertainty, dx equal to the width of the slit, d. The uncertainty in momentum component across the width of the slit, dp_x must obey the HUP, i.e.;

dx.dp_x = h (Taking the best case scenario)

Now dp_x can be related back to the total momentum, p and thus the wavelength. Simple trig gives dp_x = 2.p.sin(theta), where theta is the angular range of the photons emerging from the slit. Combining everything gives;

d.2.p.sin(theta) = h

Given that p = h/L - where L is the wavelength.

d.2.h.sin(theta)/L = h
d.2.sin(theta)/L = 1
2.sin(theta) = L/d

Hence theta is dependent on the ratio of the wavelength and slit width. Longer wavelengths will therefore diffract into a larger range of angles.

You can get the same result rigorously from Maxwell's equations.

Claude.

 

[G01]

Microwave frequencies are affected by rain since water absorbs microwaves really well. The rain absorbs the microwaves and makes the signal weaker in the process. This is actually what makes a microwave oven work. The microwaves in the oven are absorbed by the water in the food and heat it up in the process.

Lower frequencies have more diffraction because they're wavelengths are bigger. Since the wavelengths are larger, the wave is disrupted more when it passes through an opening than a higher frequency, shorter wavelength wave. When an opening disrupts a wave, it acts like a point source, spreading the waves spherically outward from it. High frequency waves are less disturbed by this opening, and pass through with less of that point source effect.

EDIT: Posted same time as Claude

 

[sarveshkumarv]

Can a narrow beam termed as a beam with small wavelength?

 

[mgb_phys]

(CROSS POST)
is it correct to term a narrow beam as a beam with small wavelength?

Also what is the mathematical relationship between
1.)attenuation and frequency,
2.)power and frequency( power transmitted and power consumed) and
3.) transmitted power and distance

No a narrow beam is one that doesn't spread out as much. Think of a narrow beam of light from a flashlight,
although it is possible to make a narrower beam from a small wavelength than a a larger as described above.

Attenuation and frequency will depend on the material you are transmiting in. Water in the air strongly absorbs some frequencies for instance.

Power and frequency aren't linked, you can transmit any power level at any frequency ( depending on your equipement). My cordless keyboard uses a very high frequency (2.4Ghz) but very low power so the signal does not travel very far, only across my desk. The BBC world service uses a very low frequency but very high power so the signal travels a long way around the world.

Ignoring any complicated effects and the Earth's curvature, for a signal transmitted in all directions distance is approximately the square root of power. So to go twice as far you need 4x as much power. If you think of the signal spreading out like the surface of a sphere you can see that the area goes up as r^2.

 

[sarveshkumarv]

Questions:

If you think of the signal spreading out like the surface of a sphere you can see that the area goes up as r^2?

I can't understand that part.Please explain.

Also if i take into account the atmospheric effects then does that mean the relation between power and distance change?If so how does that take place?

So based on the previous post if i want to tx something my choice of freq will depend on my medium and attenuation that my medium offers to that freq and the transmission power will depend on the distance to which i got to transmit.Is that right?

Also i learned microwaves cannot penetrate buildings.So a line of sight communication is needed.Why is it so?Can you explain in physical terms why microwaves cannot penetrate?

 

[mgb_phys]

Ok - we have to guess about the the posters level of education, especially if they don't speak English. It's hard to know how complicated an answer to write. So don't be offended if you know all this!

A simple transmitter wil send a signal out in all directions. So imagine the pwoer spreading out from the centre of a sphere. The receiver collects the power over a small area on the surface.
As the sphere gets bigger the area of the surface gets bigger so the signal is more spread out and the amount that goes into the small area of the receiver is less.
But if you make a sphere twice as big the surface area is 4x as big because area = pi r^2.
Atmospheric effects usually make the range less because they absorb some power - just like you can't see as far on a foggy day because the fog absorbs light.

But there are tricks for doing very long distance radio signals where the signal bounces of layers high in the atmosphere which means you can send a signal to somewhere over the horizon - even on the other side of the world. This is more complicated and you don't have to worry about it.

Yes, to send a signal a given distance is a complicated mixture of the frequency, the weather, the transmitter and receiver design.

Generally smaller wavelelngths don't penetrate as far. A simple picture is to image a long wavelength radio wave 10m long. Now a wall of a building is very small 0.1m compared to this wave so it is very easy for the wave to pass straight through it, but microwaves are only 0.1m long so the wall looks much thicker to them.
There are also complications depending on what the wall is made of, so microwaves will go through wood but not through metal, they will go further through a brick wall than a concrete wall with metal rods.

 

[sarveshkumarv]

What is the difference between attenuation and fading?Do they mean the same?

 

[mgb_phys]

What is the difference between attenuation and fading?Do they mean the same?

In general yes but I suspect fading might be used for when a signal changes in strength due to changing conditions and attenuation is used for a fixed decrese in signal strength.

 

[sarveshkumarv]

but attenuation need not mean fixed decrease?

 

[Loren Booda]

Do microwaves primarily interact with the nuclei, hydrogen bond or other valences of H₂O?

 

[mdelisio]

Do microwaves primarily interact with the nuclei, hydrogen bond or other valences of H₂O?

I think at these frequencies it is rotation of the entire water molecule.

 

[mgb_phys]

Do microwaves primarily interact with the nuclei, hydrogen bond or other valences of H₂O?

As a rough guide:
X-ray = interact with nuclei
UV-visible = electron levels
IR = molecular bonds
MicroWave-Radio = rotation of molecule

 

[Loren Booda]

mgb_phys

As a rough guide:
X-ray = interact with nuclei
UV-visible = electron levels
IR = molecular bonds
MicroWave-Radio = rotation of molecule

A handy rule of thumb!