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The optimum length for a dipole antenna is a multiple of half the wavelength that it is designed to receive or emit. Why is this? If an electromagnetic wave has E in the x-axis, B in the y-axis, and propagates along the z-axis, it will be generate the optimal current in a dipole antenna positioned along the x-axis.

However, the wavelength of the EM wave is measured in the z-axis. Of course, the fluctuations of the E field occur along the x-axis, which is why the antenna is positioned this way, but then why does the length of the antenna in the x-axis matter?

Qmechanic
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7 Answers7

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Electromagnetic waves are transverse, so the electric field in the wave is parallel to the antenna. It excites electrons into longitudinal motion in the dipole antenna rather like air in an organ pipe. The electrons cannnot get beyond the end of the antenna so the condition for resonance is the same as sound waves in a closed-end tube. The lowest frequency is therefore when the length of the tube is half a wavelength. At resonance you get a big efffect in the antenna even when the wave amplitude is small.

Note that the speed of the longitudinal "sound-like" waves is still that of light, and the frequency is unchanged, so the wavelength of the longitudinal waves in the wire of the antenna is the same as the wavelength of the electromagnetic wave, even though the propagation directions of the waves are perpendicular.

mike stone
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Here is an animation showing how a receiving antenna works. Wikipedia - dipole antenna
The Wikipedia page also contains substantial related explanatory material.

enter image description here

The "3D" animation illustrates the statement that you have made in your post, If an electromagnetic wave has E in the x-axis [North-West to South-East], B in the y-axis [North to South], and propagates along the z-axis [South-West to North-East], it will be generate the optimal current in a dipole antenna [two conductors] positioned along the x-axis.

Charges are moved in the conductor, ie a current flows, and this sets up a potential difference across the resistor which could be the input stage of a radio receiver.

A transmitting antenna works the same way except that an alternating voltage is applied where the resistor is and the alternating voltage causes the charges in the metal conductor to accelerate.
Accelerating charges emit electromagnet radiation which in this case will be of the same frequency as that of the alternating voltage.

A key feature is the wavelength of the incoming electromagnetic wave, $\lambda$, and its relationship to the length of the horizontal conductors, $\approx \frac \lambda 2$, which then means that a standing wave is set up or put another way, resonance occurs, which means that the response of the system, the dipole antenna, to the incoming electromagnetic wave, the driver, is maximised.

Further resonances are possible for lengths of the dipole being approximately $\frac{3\lambda}{2},\,\frac{5\lambda}{2}$, etc. but the system is not as efficient/responsive as for a half-wave dipole.

Farcher
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The length of a dipole antenna is chosen so that the antenna is resonant at the frequency to be emitted. Since the wave speed in a good conductor is close to the speed of light in vacuum, the length should be close to the half the wavelength of the radiowaves in vacuum. See https://en.wikipedia.org/wiki/Dipole_antenna#Dipole_antennas_of_various_lengths

A different length can also be used if the resonance frequency is adjusted by use of a matching network.

my2cts
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In addition to the usual explanations in terms of resonance or impedance matching, here is a more wordy/graphic physical explanation for why certain lengths are better for both receiving and emitting dipole antenna, without formulae.

For a receiving dipole, we want to achieve as high voltage between the center feeding lines as possible. This voltage isn't directly proportional to external electric field, but, instead, to conservative component of net electric field, which is proportional to degree of separation of charges on the dipole line. Charge density will be oscillating between positive and negative value, and thus we want the highest amplitude of this possible. The higher the current, the higher the charge density amplitude.

For an emitting dipole, to get lots of emitted power, we need lots of charges accelerating in the same direction, in sync. In dipole antenna, this is accomplished by an oscillating electric current which makes charges there to accelerate back and forth (on the diagrams below, left and right). The higher the current, the higher the acceleration.

Thus in both cases, reception and emission, we want to maximize the current amplitudes in the dipole.

Current amplitude can't be the same everywhere in the dipole; it can be large at the center (or in between the center and one end), but it has to be zero at the free ends (because charges usually don't jump out of the conductor end, so current stops there). This means that unless the current is zero everywhere, the current magnitude has to vary along the dipole and decrease to zero near the ends.

If dipole length is close to an integer multiple of $\lambda/2$, the current will oscillate in a standing wave pattern on the dipole, where the free ends are two end nodes of the standing wave.

If dipole length is in between two integer multiples of $\lambda/2$, this prevents such a standing wave, so current makes a more complicated oscillation, possibly with waves running in both directions.

When the antenna is much shorter than $\lambda/2$, it behaves similarly to a capacitor with small capacitance, with current charging and discharging the ends, but being present (and thus charges accelerating) only in a very short piece of conductor. Then, like from a capacitor with oscillating voltage, emission of radiation is very weak.

Increasing the dipole length makes more of the conductor carry the oscillating current, and thus more accelerating charges are present and radiating; this increases emission (assuming current does not change appreciably - but in fact, the current should increase, so emission should increase even more).

When further increasing the length, eventually a special value is reached, close to $\lambda/2$, where the current oscillation achieves a standing wave pattern on the dipole. The current pattern looks like this(the horizontal lines are arms of the dipole, arrows show direction of current, and density of arrows shows amplitude of intensity of current - more arrows, greater amplitude):

 ->    ->->->->    ->  
__________  __________ 
          ||

The emission is strongest where the current amplitude is the highest, at the center of the dipole; but all dipole parts, including those near the ends, contribute. In this shortest standing wave, all parts of the dipole still emit in phase, and thus add up constructively.

If we increase the dipole length still more, it is again not possible for the current to oscillate in phase in all parts of the dipole. The dipole is too long - the current wave has half wavelength close to $\lambda/2$ (because it travels in conductor, not in vacuum), so when there is more length available on the dipole arms, somewhere on them, the current has to become zero (zero current) and thus change sign. One part of the dipole then produces radiation electric field with opposite phase (and thus, direction) than the other. These fields, far from the dipole, add up destructively, and net radiation intensity far from the dipole is decreased.

For example, in dipole of length $\lambda$, in one arm of the dipole, current moves towards the center, and in the other half, current moves also towards the center. It looks like this:

 ->    ->->->->    ->    <-    <-<-<-<-    <-
______________________  ______________________ 
                      ||

Notice that for every charge accelerating to the right, we have a mirror charge on the other side, accelerating to the left. Their radiation electric fields point in opposite directions, and cancel most of each other, so we get a very decreased emission.

By further increasing the dipole length, the cancellation ceases to be so strong, and emission increases. When we get to $\frac{3}{2}\lambda$, we get yet again a standing wave of current, with two thirds of the dipole having charges accelerating in one direction, and the remaining third having charges accelerating in the opposite direction.

 ->    ->->->->    ->  <-    <-<-<-<-    <-  ->    ->->->->    ->
________________________________  ________________________________ 
                                ||    

So emission of the two thirds on the ends will add up constructively, and will be partially cancelled by the center third, but not completely, because a section of a dipole can only cancel radiation of another section of same emission intensity, that is, one third can cancel only one another third. So very approximately, we have emission worth of one dipole third remaining. We could continue increasing the length, and by the same principle, we can expect strong emission for all lengths that are integer multiples of $\lambda/2$, and decreased emission for any lengths in between.

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Consider a practical dipole antenna as a series of infinitesimal dipoles laid end-to-end. Each one of them radiates with the pattern of an infinitesimal dipole — that is, they radiate most strongly in the plane perpendicular to the axis of the antenna, but they radiate a nonzero amount of energy in every direction except end-on.

The radiation received from the antenna at any given point will be the vector sum of the radiation received from all of those infinitesimal dipoles (that is, you have to consider the phase as well as amplitude differences). The contributions tend to interfere constructively in the perpendicular directions, while the phase shifts cause destructive interference further off-axis, which is why "long" dipoles have greater gain than the infinitesimal dipole.

Plus of course there are the dramatic effects of length on input impedance, caused by the increased radiation resistance of a longer antenna ("more infinitesimal dipoles"), the self-inductance of the antenna, and the reflective boundary condition at the end of the wire. At a certain theoretical level these effects aren't too important, but they make a great difference to the efficiency and practicality of antennas made out of real-world materials.

hobbs
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I will consider an antenna receiving signal from a propagating electromagnetic field, though the reverse, an antenna emitting a propagating electromagnetic field will just be the reverse.

When a propagating EM field impinges upon an antenna the electric field (which is perpendicular to the propagation direction) puts forces on the electrons in the antenna and drives them at the same frequency as the frequency of the incident EM field. This means that the electrons oscillate at the same frequency as the incident field.

Now, the antenna, plus the cable plugged into the antenna, act as a waveguide for a new guided wave. This waveguide has a certain frequency/wavelength dependent impedance. It turns out, that for some standard antennas, the impedance exhibits resonances (local minima) at wavelengths corresponding to approximately $$ \lambda_{\text{guided}} = 2L/n $$ where $\lambda_{\text{guided}}$ is the wavelength of the guided wave, $L$ is the length of the antenna, and $n$ is a positive integer.* Given other types of resonant cavities we're familiar with in physics, it is not so surprising that there is a simple relationship between the length of the antenna and the resonant wavelengths for the guided wave. This relationship is explained well in the other answers.

However, the above (and many of the other answers) doesn't address the heart the OPs question which is "why does the wavelength of the propagating wave in $z$ related to the wavelength of the captured wave in $x$?" The answer is that these two wavelengths are related by the corresponding frequencies. We have

\begin{align} f_{\text{propagating}} =& c_{\text{air}}/\lambda_{\text{propagating}}\\ f_{\text{guided}} =& c_{\text{waveguide}}/\lambda_{\text{guided}}.\\ \end{align} But, recall above, that I explained that, because the electrons are directly driven by the propagating wave, that $$ f_{\text{guided}} = f_{\text{propagating}}, $$ so from this we can quickly conclude that \begin{align} \lambda_{\text{guided}} = \frac{c_{\text{waveguide}}}{c_{\text{air}}} \lambda_{\text{propagating}} = r \lambda_{\text{propagating}}. \end{align} And we have resonances when $$ \lambda_{\text{propagating}} = \frac{1}{r} \frac{2L}{n}. $$ Based on properties of the antenna/receiver we probably have $r\approx 0.6$, so a factor of order unity.

So we can see that

  • the resonant wavelengths of the guided wave are related to the length of the antenna because the antenna is like a cavity for the guided wave
  • the frequencies of the guided and propagating waves are equal because the guided wave is directly driven by the propagating wave
  • the wavelength of the propagating/guided wave is directly related to the frequency of the propagating/guided wave by the corresponding speed of light
  • Therefore the resonant wavelengths of the propagating wave are simply related to the lengths of the antenna.

I think it is an interesting observation the OP makes about the direction of energy propagation being possibly redirected when the propagating wave is captured by the antenna. But, of course, this is exactly what a waveguide does. More generally, when converting a propagating wave to a guided wave we need to satisfy certain phase/momentum matching conditions. In the case of e.g. a dipole antenna this is best satisfied when the antenna is parallel with the electric field.

*Very possibly I'm off by a factor of two or something here or have $n$ in the wrong spot in the equation. If so let me know in the comments and I'll correct it.

Jagerber48
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Here is how I think of this. Your mileage may vary, and will be lower in California:

A generic antenna is a device intended to match the impedance of a transmission line carrying the RF signal to the impedance of free space. The matched-impedance case furnishes the maximum possible useful energy transfer from the line to free space, in which there are no reflections of incident power at the connection between the feed line and the antenna. All the incident electrical power in the line is transformed into electromagnetic waves departing into free space.

The voltage-to-current characteristic (i.e., the impedance) of a real-life antenna depends on its physical dimensions relative to the wavelength of the driving signal. This means that the matched-impedance case dictates the length of the antenna. The prototypical real-life antenna is the center-fed dipole and the matched-impedance condition yields the length of the dipole to be a half-wavelength.

niels nielsen
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