frequency and attenuation

Discussion in 'General Electronics Chat' started by muni, Jan 27, 2009.

  1. muni

    Thread Starter Active Member

    Jul 29, 2008
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    can any one explain what is relation between frequency and attenuation?
    is that relation has a linear behavior? or is a case of individual application
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    The output voltage of some circuits decreases as the input's signal frequency increases due to capacitances (parasitic or not) in the circuit. The attenuation of the output signal is not linear and depends upon the circuit.
     
  3. muni

    Thread Starter Active Member

    Jul 29, 2008
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    i'm sorry sir i think i've not quoted problem correctly
    actually in a discussion one of my friend argued that as the frequency increases the attenuation increases and that for lower frequencies the attenuation shall be very less. he quoted aan example as we are working on RADARs. as radar requires 2 KW avearage power to achieve a 200 NM range where as an FM/ short wave/ medium wave radio with little power it can communicate for 100s of km.
    in reply i gave the example of modulation in which low freq is super imposed on high carrier.
    that's why i would like to know is there any such relationship
     
  4. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    Is this in free space or in a transmission line, or what? It makes a big difference!

    Eric
     
  5. muni

    Thread Starter Active Member

    Jul 29, 2008
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    ok i would like to know in both cases. but presently please explain me in free space
     
  6. KL7AJ

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    Nov 4, 2008
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    Hi Muni:

    In free space (i.e. a perfect vacuum) the attenuation is independent of frequency....it follows the inverse square law (interestingly just like gravity!) In the real world however, losses due to absorption, dielectric heating, etc., generally increase with frequency. (You can have very severe attenuation near the molecular resonance frequency of water, on top of that, for example)

    In a transmission line, the loss is logarithmic....it is a given number of dB per distance of transmission line. This is also something that increases (sometimes drastically) with frequency, but it is not a simple relationship. However, loss in a transmission line ALWAYS goes up with frequency.


    Eric
     
  7. muni

    Thread Starter Active Member

    Jul 29, 2008
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    thank you sir,

    i've a doubt in vhf/uhf transmission it needs only 10 watts of power to cover a range of 200 km,but i think it needs more power incase of fm/sw/mw frequencies . why it is so?
     
  8. KL7AJ

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    Nov 4, 2008
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    Good question with a long answer!

    Most UHF/VHF communications are line-of-sight. For the most part, they experience near free-space attenuation. Thisi is why you can communicate with a satellite with just a few watts...the only additional attenuation is the short trip passing through the atmosphere...the rest is merely free space.

    At the other extreme, MW broadcasting, for example, the main path is GROUND WAVE. The attenuation of ground wave signals is ORDERS OF MAGNITUDE greater than free space attenuation. The immense power used by MW stations is to compensate for ground losses, not free-space attenuation. (Yes, A.M. broadcast stations burn up a lot of power heating earthworms!)

    In the middle is shortwave stations, where most of the path is through the ionosphere. There is a lot of loss here, as well....nowhere near as much as groundwave, such as i n MW broadcasting...but still significant.

    Finally, a lot of the power in lower frequencies is to overcome background noise...static and such, at the receiver. This can take a lot more power than would be necessary in a quiet receiving environment.

    Hope this helps!

    Eric
     
  9. Ron H

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    Apr 14, 2005
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    Eric, you know more about this than I do, but this document seems to have a different answer.
    http://didier.quartier-rural.org/implic/ran/sat_wifi/sigprop.pdf

    I don't know if it's correct or not, or perhaps I am interpreting it incorrectly.

    EDIT: Wikipedia confirms what you are saying, and which I remember from my long-lost education.
     
    Last edited: Jan 27, 2009
  10. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    Hi Ron:

    Gee...I wish you hadn't brought this up. :D. There has always been a bit of misunderstanding about "capture area", but hopefully I can clear this up.

    Indeed, the inverse square law applies to all electromagnetic radiation. Where the rub comes in is how you INTERCEPT that radiation...and this is where the "magic formula" comes in. See if you can follow this...it's a bit obscure until you get to the "aha" part. ^_^

    Let's look at a fixed amount of energy radiating from POINT SOURCE. As the signal spreads out, you have that fixed energy distributed across an ever-increasing "surface area". Now an antenna of a given length will intercept a smaller percentage of that surface area, the farther away from the source you get. Now, here's where it gets fun. A typical antenna increases in size in proportional to its wavelength....assuming a dipole, for example. Now an antenna that's "cut to length" is going to intercept a smaller percentage of that volume the shorter it is...in other words, the higher the frequency, the shorter the antenna. If we kept the RECEIVING ANTENNA SIZE in continuous proportion to the DISTANCE FROM THE POINT SOURCE, we would see the the signal intercepted by that antenna follow the inverse square law precisely. The fact of the matter is that we DON'T build antennas proportional to the distance...we build them proportional to the wavelength! (aha!)

    Hope this helps some.

    Eric
     
  11. muni

    Thread Starter Active Member

    Jul 29, 2008
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    Indeed, the inverse square law applies to all electromagnetic radiation. Where the rub comes in is how you INTERCEPT that radiation...and this is where the "magic formula" comes in. See if you can follow this...it's a bit obscure until you get to the "aha" part. ^_^

    Let's look at a fixed amount of energy radiating from POINT SOURCE. As the signal spreads out, you have that fixed energy distributed across an ever-increasing "surface area". Now an antenna of a given length will intercept a smaller percentage of that surface area, the farther away from the source you get. Now, here's where it gets fun. A typical antenna increases in size in proportional to its wavelength....assuming a dipole, for example. Now an antenna that's "cut to length" is going to intercept a smaller percentage of that volume the shorter it is...in other words, the higher the frequency, the shorter the antenna. If we kept the RECEIVING ANTENNA SIZE in continuous proportion to the DISTANCE FROM THE POINT SOURCE, we would see the the signal intercepted by that antenna follow the inverse square law precisely. The fact of the matter is that we DON'T build antennas proportional to the distance...we build them proportional to the wavelength! (aha!)

    Hope this helps some.

    Eric[/quote]

    THANK YOU SIR
    THIS POINT OF INVERSE SQUARE LAW IS VERY WELL UNDERSTOOD WITH YOUR EXAMPLES.
    BUT I'M STILL NOT CLEARED WITH THE FREQUENCY AND ATTENUATION
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    The other aspect deals with materials. We don't live in a vacuum. Just as RF doesn't travel in water where fish live, different frequencies react differently to our atmosphere. I suspect there is where part of the answer you are look for lies.
     
  13. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    THANK YOU SIR
    THIS POINT OF INVERSE SQUARE LAW IS VERY WELL UNDERSTOOD WITH YOUR EXAMPLES.
    BUT I'M STILL NOT CLEARED WITH THE FREQUENCY AND ATTENUATION[/quote]

    Hi Muni:

    What specific issue are you confused about?

    eric
     
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