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Dumb antenna question

  1. Apr 24, 2008 #1
    In making antennas, the higher the frequency the smaller the length of the antenna. The length of the antenna is somewhat equal to the wavelength.
    But why? I mean, can't we just grab any piece of metal and put in AC current and there would be EM wave correspond to the AC signal whatever the frequency is 1Khz or 100Ghz? What makes the frequency depend on length?
  2. jcsd
  3. Apr 24, 2008 #2


    Staff: Mentor

    It is all about efficiency. Yes, you can put any frequency signal into any piece of metal, but the energy can go into heating the metal or other losses besides the radiation that you actually want. Having the antenna proportional to the wavelength gives the best efficiency with most of the energy radiating away and very little being lost to other undesired effects.
  4. Apr 24, 2008 #3
    Thanks for the reply. So it's about efficiency. But what determines the efficiency at different frequency?

    Specifically at a certain frequency what determines how much input energy got coupled into EM wave and how much got "wasted"? Why is it antenna length dependent?

    If this is too much math to cover here can anyone point me to a source that explains it? I really need to understand this. Thanks.
  5. Apr 24, 2008 #4


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    a simple horizontal dipole antenna is a teeny bit less than 1/2 wavelength. and a simple vertical antenna is half that length (1/4 wavelength) where the ground plane is used as the missing half.

    what is happening in a radiating element of a transmitting antenna is that charge is being forced to slosh back and forth. from the POV of the receiving antenna, when the charge of the transmitting antenna moves, that causes the free charges in the conducting element of the receiving antenna to likewize move. the forces acting on those charges will be proportional to amount of charge moving in the transmitting antenna and inversely proportional to the square of the distance between the antenna. force acting on charges is "electromotive force" (e.m.f.) and can be measured as a changing voltages (which is what the first stage of the receiver does).

    so transmitter electronics forces charge to move and slosh back and forth in the transmitting antenna. because like charges repel and unlike charges attract, this sloshing charge acts on charges in the receiving antenna, those moving charges are a current that is amplified in the receiver.

    now imagine a sinusoidal current applied to the middle of a simple dipole antenna (a straight conductor). integrate that sinusoidal current and you get a sinusoidal function for the net charge (as a function of time) that crosses that middle point in the conductor. and when the current (defined in some direction) is negative, that means charge is flowing in the opposite direction as when it is positive. that mean charge on one side of the conductor is decreasing while on the other side it is increasing. then the current changes sign (becomes positive in the original direction) and charge starts sloshing back to where it was before.

    now, ask yourself, if the charge moves at some speed that is a little slower than the speed of light, how far will it go before the sinusoidal current changes polarity and that charge is drawn back to go to the other side?
  6. Apr 25, 2008 #5
    Now I see, thanks for the explanation! Charge carriers move a lot slower than c, but E field does, now it makes a lot of sense.
  7. Apr 25, 2008 #6


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    yeah, that's something i keep forgetting. i know that the wave velocity in a transmission line is somewhere around c/2, but it's not charge carriers.

    BTW, you can have shorter antenna than 1/2 wavelength, but both the effective "aperture" of the antenna is smaller (reducing the effectiveness of radiation) and the the driving impedance changes and usually needs to be compensated with an adjustable loading capacitor and inductor. when i was a ham radio operator (about 4 decades ago), i had a simple "long wire" antenna (about 40 meters long) with a loading circuit that i could use to "tune" the antenna to nearly any frequency i was using. it worked pretty good.
  8. Apr 25, 2008 #7
    You need two capacitors to control the impedance (the ompedance is a complex number and you want to make the real part equal to the impadence to be the complex conjugate of the inpedance of the radio).

    I have a long wire antenna for my shortwave radio. The antenna tuner I use consists of a coil and two capacitors. The capacitors are in series. The coil is parallel to the two capacitors. The antenna is connected to one of the two points where the capacitor is connected to the coil, the connection to the radio is the other such point and the earth is connected to the point where the two capcitors are conneccted.

    To tune the antenna, you select a coil, then you turn one capacitor so that you get maximal signal strength. Then you turn the other capacitor to maximize further. Then you go back to the first capcitor and repeat the process until you cannot make improvements to the signal.

    It's a bit cumbersome but there is no other way since we are optimizing in two variables. :smile:
  9. Apr 25, 2008 #8


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    sounds like three variables: two continuous and one discrete.

    what i had when i was a ham radio person long ago was a 52 [itex]\Omega[/itex] coax cable coming out of the transceiver going into an SWR meter, another 52 [itex]\Omega[/itex] coax cable coming out of the SWR meter going into the antenna load box which had inside a coil in series followed by a capacitor shunted to ground. from the node where the coil and cap are connected, out went the antenna wire and the box chassis was connected directly to earth ground. the coil was literally a rolling thing that was continuously adjustable (and the capacitor was variable also).

    i could disconnect it from the box and connect to a 50 [itex]\Omega[/itex] dummy load i got from Heathkit. i could kick the transmitter into "tune" and the output loading of the transceiver were adjusted to maximize the output to the dummy load. the SWR was precisely 1:1 (despite the 52 to 50 ohm mismatch) and there was no reflected power. then i could reconnect to the antenna box and i had to adjust both the coil and the cap to minimize SWR and could usually get it to the same 1:1 for any frequency i was operating at. (two degrees of freedom is necessary because impedance is a complex quantity and i needed to match both real and imaginary components to my 52 ohm driver.) eventually, i learned to get along without the dummy load.

    the SWR meter told me there was no reflected power, it was all going out, and if there was neglegible resistance, then, i think that conservation of energy means that all of the energy was radiated. the antenna was the correct length for the 75 meter band (around 4 MHz, the "voice end" of the 80 meter band) and was then close to an integer number of 1/2 wavelengths for any of the ham radio frequencies that the transceiver was good for.
  10. Apr 25, 2008 #9
    The capacitors are there to soak in the current so as to allow shorter antenna right? Can the antenna be viewed as a capacitor element in the circuit or antenna has an inductor element too? So by having an external LC circuit you can tune the whole thing to minimize impedance.
  11. Apr 26, 2008 #10
    Solving the differential equation might help maybe?
  12. Apr 26, 2008 #11


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    the reason the inductive and capacitive loading elements exist is to match the impedance driving the antenna to the characteristic impedance of the transmission line. and the loading capacitor at the output of the transmitter is for matching the output impedance of the transmitter to the characteristic impedance of the transmission line. usually the characteristic impedance of a transmission line is considered to be virtually real (imaginary part is considered to be zero)

    an adjustable capacitor shunted across the antenna terminals changes both the real and imaginary parts of the impedance going into the antenna. you adjust that until the real part of the antenna impedance matches the real part of the characteristic impedance of the line. there will be some leftover imaginary part which is capacitive and the adjustable inductive element in series is there to cancel that part.

    if [itex]\omega_0[/itex] is your operating frequency, then

    [tex] Z_0 = \frac{1}{\frac{1}{Z_A} + j \omega_0 C} + j \omega_0 L [/tex]

    you have two constraints: match the real and imaginary part of the transmission line characteristic impedance, [itex]Z_0[/itex] to the real and imaginary parts of the right-hand side of the equation. you have two degrees of freedom, C and L that you can twist all sorts of ways to match it.
  13. Apr 27, 2008 #12
    Thanks for helping me out guys! I'm in good shape to pick up a book now :)
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