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What is the distance at which an antenna can receive signal?

  1. Dec 3, 2014 #1
    1. The problem statement, all variables and given/known data
    Assume that the power radiated by the television transmitter uniformly fills the upper hemisphere. A UHF television with a single-turn circular loop antenna of radius 8 cm requires a maximum induced voltage above 24 mV for operation.

    The speed of light is 2.99792 × 108 m/s.

    Find the distance d at which reception is lost from a 569 kW transmitter operating at 0.16 GHz.

    Answer in units of km

    2. Relevant equations
    dE/dx = -dB/dt
    E_induced = -N(dMFlux/dt)
    E = E_0 cos(k(x - vt)) (where v = c)
    k = 2Pi / lambda
    S = P_tran / (2Pir^2) (not 4Pir^2 because its a hemisphere, so only half)
    S = E_0^2 /(mu_naught * c * 2)
    lambda = c/f

    3. The attempt at a solution
    I found -dB/dt with the wave function, took the derivative of E = E_0 cos(k(x - vt)), got dE/dx to be -.5*k*E_0.
    Set them equal, solved for E_0, and got (-dB/dt * lambda) / Pi.
    I then set the poynting vectors equal so P_tran / (2Pid^2) = E_0^2 /(mu_naught * c * 2), which when solved for r is

    d = ((P_tran * mu_naught * c)/(Pi * E_0^2))^(1/2)

    I ended up with a value or arbitrary units, no where near what it should be.
    What am i doing wrong?
     
    Last edited: Dec 3, 2014
  2. jcsd
  3. Dec 3, 2014 #2

    mfb

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    Where does the .5 come from?

    What did you get for E0?

    Where did you use the size of the antenna?

    Where did the units start to get inconsistent? It would help if you write down the actual equations instead of describing them.
     
  4. Dec 3, 2014 #3
    .5 gives the average of the sinusoidal function.

    E_0 = (V*lambda)/(Pi^2*r_ant^2) (according the my calculation of using the wave equation to derivate E_0).

    The size of the antenna is used in E_0, because you're calculating the induced voltage through the loop.

    The units are consistent until the very end when i set there Poynting values equal to each other and solve for d.
     
  5. Dec 3, 2014 #4

    mfb

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    It does not. The average first derivative of a sinus function is zero. And why do you want to take the average?

    Sure, but that is not clear from the first post.

    How do you get inconsistent units there? Looks like power/area in both formulas.
     
  6. Dec 3, 2014 #5
    Sorry about it not being clear.
    But I'm obviously lost then!
    That's what I mean, I don't know WHY the units are wrong, I just know that when I plugged them in to Wolfram Alpha to double check, they were very wrong!

    Where would you recommend I start? Did I start on the right path and veer off? If so, where?
     
  7. Dec 3, 2014 #6

    mfb

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    Start with showing your actual work here (every step), otherwise it is impossible to tell what went wrong.
    Why don't you copy the WolframAlpha query to this thread?
     
  8. Dec 3, 2014 #7
    ImageUploadedByPhysics Forums1417645397.080268.jpg

    Is my 4th step correct, before I continue? Hopefully everything is clearer now though!
     
  9. Dec 3, 2014 #8
    ImageUploadedByPhysics Forums1417645579.653822.jpg ImageUploadedByPhysics Forums1417645597.754145.jpg

    Sorry, that came out a lot less clear than I wanted.
    If you sill can't read it I'll upload it to imgur or drop box or something.
     
  10. Dec 3, 2014 #9
    Here is the wolfram alpha interpretation,
     
    Last edited: Dec 3, 2014
  11. Dec 3, 2014 #10

    mfb

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  12. Dec 3, 2014 #11
    Technically mu_0 * c gives you 377 Ohms, i thought wolfram alpha would deal with it, but it did not.
    I see what i did, I didn't square V/m. Ok, back to business then, is my 4th step correct, or no?
     
    Last edited: Dec 3, 2014
  13. Dec 3, 2014 #12

    mfb

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    Hmm, nice relation with ##\mu_0 c##.

    Where did you square E_0?
     
  14. Dec 3, 2014 #13
    The units were stupid on my part! The problem is still knowing what E_0 is though.
     
  15. Dec 3, 2014 #14

    mfb

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    dE/dx is right, the relationship to E_0 is the last missing step. And it is just a constant factor you have to fix.
     
  16. Dec 3, 2014 #15
    So, are you saying dE/dx = -κ* Eo is incorrect?
     
    Last edited: Dec 3, 2014
  17. Dec 3, 2014 #16
    For all of those in the future who come seeking a solution, this is it.
    keep going from where i left off, plug and chug by setting energy flux of the transmitter (P_tran / 2 Pi d^2 ) equal to (E_0^2 / 2 mu_0 c).
    Then solve for d!
     
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