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Antenna: reception range

  1. Feb 22, 2010 #1
    A very basic question: the signal reception requires that the receiver be at the communication region of the sender? Or what is important is the intersection of the comm. range of both sender and receiver? A practical example: Assume that both sender and receiver have the same hardware, antenna, polarization, etc. At a distance X from the sender, a spectrum analyzer detects a -100dBm signal. Suppose that the receiver sensibility of the receiver is -50dBm. Can the receiver decode this signal at the position X? If the answer of my first question is yes, then this answer is no. However, if the primary answer is no, then the answer for this question is yes. Please, let me clarify this point.
     
  2. jcsd
  3. Feb 22, 2010 #2
    If the receiver input noise (do you mean "sensitivity"?) is ~-50 dBm (is all kTB noise?), then it will not be able to receive a ~-100 dBm signal at its antenna output. A better directional antenna, narrower bandwidth, or a very cold GaAS to reduce noise figure in the first stage may help.

    You should be able to achieve ~-110 dBm per MHz input noise at receiver. Where does the -50 dBm come from?

    Bob S
     
  4. Feb 22, 2010 #3
    Assume NO noise floor to consider in this problem. Note that, in this example, the RX limit of the receiver is -50dBm and it is exactly the double of -100dBm and we can quickly conclude that such reception will not occur at the point X. However, if what matters is the intersection of the radiation patterns, instead of having the receiver AT the communication range of the sender, the answer can be different. I am asking this because, based on a experience with a directional antenna, I noticed that I can receive a signal with a RSS -80dBm from a distant sender. With the spectrum analyzer, the signal is very small and can't be detected properly (-98dBm). If the additional of the directional antenna is "improving" the signal strength, it seems to me that what is working is the intersection of the radiation patterns of sender and receiver. Am I wrong? If yes, why a directional antenna works for a receiver? I can clearly understand the concept of "gain" caused by the directional antenna from the perspective of the sender, but what about the perspective of the receiver?
     
  5. Feb 22, 2010 #4

    sophiecentaur

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    Are we assuming a certain operating frequency here?
    Is this line of site? Is it subject to rain attenuation? What are the noise statistics and the signalling system to be used?
    What the OP is asking is for someone to work out a power budget for an unspecified link working under unspecified conditions. It can't be done.
    You have to specify modulation and coding, transmitter power, Tx antenna gain, transmission distance, interference sources and path attenuation, receiving antenna gain and sidelobe pattern, Rx Noise figure. You put these, and probably something I have forgotten and you get a performance figure (oh, yes there's bandwidth as well).
    Whole books have been written on this subject and it's a bit hopeful to expect the answer to emerge 'just like that'.
    Links don't tend to be 'reversible' either.
     
  6. Feb 22, 2010 #5
    Please, let me rewrite my doubt. Actually, the question is not about path loss attenuation or any other kind of attenuation. It is only a conceptual question.

    I will make a drawing. Assume a perfect scenario: no loss, except the air path loss, same hardware for sender and receiver, symmetric link, etc. Also assume perfect isotropic antennas. Forget the "bad/real" details.

    Picture 1:

    *********************** .........****************************
    Sender................................* ......*.........................Receiver..............
    .....S*..................................*...*...............................*R...................
    .........................................*.......*...................................................
    ***********************..........****************************
    ......A.....................................B...................................C..................

    The distance AB=BC. Using a spectrum analyzer, it is measured at point B a signal strength of -50dBm. At point C, -100dBm.
    Suppose that the signal strength of -50dBm is enough for the receiver, but not -100dBm.
    Does the receiver "get" the signal? YES/NO? Observe that the point B is the intersection point of both communication ranges of the devices. That is, if the receiver is now transmitting, at the point B we will also detect -50dBm with a SA.

    If the answer is still NO, so the receiver would only get the signal at least if it is in position B, such as in Picture 2?

    Picture 2:
    ***********************
    Sender................................*...Receiver
    ...S*.....................................* R
    .........................................*
    ***********************
    ....A........................................B

    Thanks in advance.
     
    Last edited: Feb 22, 2010
  7. Feb 22, 2010 #6
    The difference between -50 dBm and -100 dBm is 50 dB, or a power ratio of 100,000 to 1. Use

    SNRdB = 10 Log10[P2/P1]

    See http://en.wikipedia.org/wiki/Signal-to-noise_ratio

    So where is all the extra signal going? How do you compare RSS and the input noise level of your receiver? What receiver bandwidth are you talking about?

    Bob S
     
  8. Feb 22, 2010 #7
    If the receiver needs at least -50dBm and it's only getting -100dBm then it won't receive the signal. The answer is so simple that I wonder if I understood your question correctly.
     
  9. Feb 22, 2010 #8
    The answer is the following:
    kTB = -115 dBm
    k= 1.38 x 10-23 Joules/ deg K
    T= 273 kelvin
    B= 1 MHz
    What is your bandwidth?
    Add 3 dB for amplifier noise figure
    add 20 dB for 20 db SNR
    total = -92 dBm
    What is your antenna gain?

    Why is your receiver RSS -50 dBm?

    Bob S
     
  10. Feb 22, 2010 #9
    Maybe this will help

    Suppose a signal from a transmitter is -100 dBm at distance x

    At distance 2x, excluding ground and atmospheric effects, the signal is 4 times lower

    dB ratio is 10 Log10[0.25] = -6 dB

    So signal strength at distance 2x is -106 dB

    Bob S
     
  11. Feb 23, 2010 #10

    sophiecentaur

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    Is this question just about the inverse square law?
    The power will spread out to cover the face of a (notional) sphere.
    Wherever the transmitter is, that's the centre of the sphere and the received power flux will depend on the surface area of the sphere where the receiver is.
    That original diagram and question seem to imply things about the situation which may not apply.
    There are two extremes of model for a comms link:
    For cable transmission, the loss is calculated in dB per km so, half way along the cable, the transmission loss IN dB is halved. 100dB loss becomes 50dB loss.
    For free space transmission, the loss at the half way point is only 6dB different. 100dB loss becomes 94dB loss.
    Does this resolve your problem?
    This shows that, for really long distance comms, radio beats cable but, for short distances, the cable wins.

    Apologies if anyone thinks I have just restated what they have already said.
     
  12. Feb 23, 2010 #11
    If the signal power is -100 dBm (10-13 watts), and it halved ( 5 x 10-14 watts), the signal power is -103 dBm.

    Look at the attenuation per 100 feet (in the rightmost column) for a 10 MHz signal in various types of RG-8 cable in

    http://www.generalcable.com/NR/rdonlyres/AA7490F5-67CA-4456-A0EB-3847276B65D8/0/Pg094_095_RG8U.pdf

    In 1 Km, about 3280 feet, the typical attenuation for a 10-MHz signal is ~32.8 x 0.57 dB = 18.7 dB.

    Bob S
     
  13. Feb 23, 2010 #12

    sophiecentaur

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    Bob S
    I'm not sure what our point is. Are you agreeing or disagreeing with my last post?
    1. Double the distance of a free space transmission link and you get 1/4 power flux density. That's 6dB of anyone's money.
    2. If you double the length of the cable you quote, you will get twice your 18.7dB of loss (37dB). If you double the new length, you will get 74dB loss - but doubling the free space distance just gives you an extra 6dB of loss each time. However low loss your cable happens to be, there will always be a link distance where the losses are equal and, beyond that, the free space link wins. For short distances, the cable is the better bet - because, with even the narrowest of beams, it soon is too far away for a receive antenna to intercept all the power and the inverse square law kicks in to give a very rapid initial rate of attenuation compared with the few tens of dB of a good cable.
    Plotting a graph of the two received signal powers is very instructive.
     
  14. Feb 23, 2010 #13
    I apologize. You are correct. I misread your post. Bob S.
     
  15. Feb 24, 2010 #14

    sophiecentaur

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    I think it's all down to an initial, possible, misconception in the OP, which equates (by implication) all forms of transmission path.
    Deep space probes just LOVE the inverse square law.
     
  16. Feb 24, 2010 #15
    Also, because the receiver input noise power is kTB (plus a small noise figure), low temperatures and low bandwidth reduce the needed signal threshold. Pluto's mean surface temperature is ~ 40 kelvin.

    Bob S
     
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