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I have a problem on the understanding of a half-wavelength antenna. Namely, why is the half-wave antenna has a highest power emission. Considering Far Field factor F($\theta$), which is known from the following formula:

\begin{equation} |F(\theta)|^2= \frac{ (\cos{( \frac{kl} {2} \cos{\theta})} - \cos{ \frac{kl} {2} })^2 } {\sin^2{\theta}}, \end{equation} where $k$ is wavenumber, $l$ is a total dipole length (two equal length wires), $\theta$ is a an angle from the antenna axis.

That factor corresponds to the wave power. It seems like , that the highest Far field is for kl/2 = Pi. One can check that considering $\theta = 0$, and demanding the highest $|F(\theta)|^2$. By the way, the plot corresponding to different electrical length is shown on the following figure. One can see the highest $|F(\theta)|^2$ for $kl = 2\pi$.

Dipole antenna Far field factor

So could I ask you where is the problem in the latter? Should one condisder reflections of feed signal from the antenna or other things?

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The formula in the question is taken from page 44 of a book called The Theory of Electromagnetic Wave Propagation by CH Papas. The text says that the formula describes the radiation pattern for a center driven wire antenna. Radiation patterns normally show a graph of relative intensity which isn't the same thing as efficiency.

For transmitting, the efficiency of an antenna is described as the ratio between the power delivered to the antenna compared to the power radiated, and is determined mostly by the ratio of the radiation resistance to ohmic resistance, as per this formula :

$$ \eta = \frac{P_R}{P_{IN}} = \frac{R_R}{R_R+R_L} $$

The graph in the question it seems just shows the difference in the main lobes of the radiation patterns for antennas of different wavelengths. A full wave dipole does not have higher efficiency than a half wave dipole, and the length in wavelengths of an antenna doesn't determine its efficiency.

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  • $\begingroup$ Agree with that. This is only radiation pattern and do not say anything about the efficiency. The point is that one can take the pointing vector which is a factor multiplied by a radiation pattern. And in order to calculate directivity, one should divide the Pointing vector module on the $P_{iso}$ - the power from the equivalent isotropic radiator. The latter is calculated from $P_{iso} = \frac{P_{TRP}} {4\pi r^2}$ where $P_{TRP}$ is total radiated power and integrated pointing vector over a Sphere. $\endgroup$ Oct 13 at 7:32
  • $\begingroup$ That book is a classic, Andrew, what do you think? $\endgroup$ Oct 13 at 7:32
  • $\begingroup$ Ah yes, i can see from your formula that there is only energy integrated included. Thank you $\endgroup$ Oct 13 at 7:37
  • $\begingroup$ It seems that I just misunderstood the efficiency $\endgroup$ Oct 13 at 7:45
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I'm not sure if this is the correct answer or not: At varying electric lengths, at any moment in time, one part of the antenna is producing the wanted field(s) and another part might be producing a canceling field(s). With a half wavelength, all you get is the wanted field and no canceling field. The canceling field (I call it that because it works against the main field) is what creates lobes and direction. A full wavelength would be the worst because the top half of the antenna essentially cancels out the bottom half.

enter image description here

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    $\begingroup$ technical words: constructive and destructive interference $\endgroup$
    – user10489
    Oct 10 at 12:30
  • $\begingroup$ Thank you for the simple explanation. Should it be true for two wavelength dipole antenna? If antenna centrally feed, the current is basically asymmetric around the center. I want to say it is $I_0 sin(k(l/2-|z|))$. So, there is no annihilating current nodes for one wavelength antenna, while two wavelength one, there are. $\endgroup$ Oct 12 at 15:18
  • $\begingroup$ It probably doesn't matter if the antenna is fed from the center or from the end, you can probably think of it as a wire in space that is oscillating "magically" let's say, and then just focus on the voltage and current distribution. With the half-wave there is one place in the center with current, with a full-wave there are 2 places with current that oppose each other, and with 2 full-waves there are 4 places of current, 2 oppose the other 2. It's best to draw this, let me add a drawing to the answer regarding this comment. $\endgroup$
    – Jack0220
    Oct 12 at 15:48
  • $\begingroup$ Sorry I can't comment on the math you posted, I don't understand math :D $\endgroup$
    – Jack0220
    Oct 12 at 15:51
  • $\begingroup$ This answer has some interesting ideas but is technically incorrect, the applied sine wave source at the feed points results in a current distribution along the antenna elements which completely determines the radiation pattern, both positive and negative current flow contribute to antenna output, one does not cancel out the other, it just happens that a half wave length isn't long enough to allow for both positive and negative current. The terms constructive and destructive do not apply here, these terms are used to explain what happens when two wave forms add together. $\endgroup$
    – Andrew
    Oct 13 at 1:57
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It isn't.

  1. Gain in the broadside (θ=0) direction isn't "efficiency".
  2. A half-wave dipole doesn't maximize that parameter, even among dipoles; it increases until roughly 1.25 wavelengths, then decreases, then generally increases again, in an oscillating way.

There are nice things about half-wave dipoles (a convenient, easily-matched impedance and a simple pattern without lobes), but it doesn't maximize gain or directivity or anything like that.

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  • $\begingroup$ It seems like that a one wavelength dipole is more directive then a halfwave antenna. Also, I understand that the pattern of half wave dipole is for broad angles. $\endgroup$ Oct 12 at 15:10

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