Understanding the Resonance of Dipol Antennas

[Anton Alice]

Alright.

I have also a few questions concerning dipol-antennas:

first:

It has often been said, that the dipol antenna, in order to receive a certain frequency, needs to have a length of λ/2.
I don't see a clear reason for that. If one would not know that, one had to calculate the Capacity and Inductance of that piece of antenna (which certainly depend on the thickness of the antenna, too)
And each of these quantities can only be approximated. So how does it come, that we can hit the resonance by just letting the length be λ/2?
Is there a simple explanation?

 

[nsaspook]

Alright.
I have also a few questions concerning dipol-antennas:
Is there a simple explanation?

You don't need to be at exactly λ/2 to radiate but that's one point where we have a standing wave on the antenna so the fields reinforce each other to maximize EM radiation in some direction.

 

[Anton Alice]

First:
Referring to the video:
Generally, the antenna is a resonant circuit. So it should be single-mode system, and that single mode (ground state) is then oscillating at resonance frequency.
Is that correct? The lecturer then continues with antennas, with length larger than λ/2. Such antennas can then be exited by the same frequency with which the λ/2-antenna was exited, but the standing waves look different (they have more node-points). Actually those longer antennas can be thought of as being a join of many λ/2-antennas. Am am correct so far?

You don't need to be at exactly λ/2 to radiate but that's one point where we have a standing wave on the antenna so the fields reinforce each other to maximize EM radiation in some direction.

Ok, but why is it at that point? Or to put it differently:

Why is it not possible to create the same current pattern of a λ/2-antenna inside a λ/3-antenna, for example?

As I said earlier,my approach was: The resonance frequency of a piece of wire with length λ/2 is f=c/λ. And the resonance frequency should also be calculable by f = 1/ (2Pi Sqrt(LC)) . And one should then find, that λ = 2Pi/c * Sqrt(LC). Problem here is, that L and C are difficult to find. Therefore I was either searching for a way to find L and C, or for a more instructive explanation.

EDIT: How do you quote, so that the name appears?

 

[nsaspook]

We use resonant (where the elements have electrical min/max nodes at the ends) antennas because it usually makes the matching of the antenna to the transmission line and RF source much easier as a mainly resistive impedance for max power transfer. Non-resonant antenna elements when combined with the reactive tuning elements needed to null antenna reactance and match the resistive source impedance can work just as well to radiate and receive.

It's hard to explain in words the physical relationship of how the fields move from the transmission line to the ends on the elements so I'll use another video first to see if you can see the connection. Electrical signals do travel down wires (as fields) at nearly the speed of light but the average electron velocity is very small so a rf frequency voltage is generated between the feed point and the end of the antenna from current flow.

 

[Anton Alice]

that video does not explain why an emitting antenna with length l radiates EM-waves with wavelength λ = 2l.

EDIT:

What happens, if I have a sending antenna in free space, and a receiving antenna somewhere in a substance with ε>1? Can I still receive with the same performance as if the receiver were in free space?
Or do I have to adjust the length of the antenna? The wavelength of the EM-wave in the substance would decrease. But the frequency would remain the same. From my point of view, the antenna would receive with same performance, because only the frequency is relevant.

 

[nsaspook]

that video does not explain why an emitting antenna with length l radiates EM-waves with wavelength λ = 2l.

The emitting antenna radiates EM-waves with a wavelength equal to the free-space wavelength of the rf source signal. The length and matching components of the antenna governs the coupling of the EM energy surrounding the antenna conductors to the free space impedance of about 300 ohms as waves.

 

[Anton Alice]

Therefore the receiving antenna in the substance would need to be adjusted, because the coupling would be different, because of the different impedance of the surrounding substance? And this is why the length of the antenna need to be adjusted to the new wavelength inside the medium?

Ok, but does that answer why an emitting antenna with length l radiates EM-waves with wavelength λ = 2l? You have been saying, that the antenna is a kind of impedance-transformation. Well, if the antenna is alone, with no other components attached to it, then the impedance of the "non existent" components would be zero.
But it is impossible to match the 300 ohms to the zero ohm. I am really confused...

 

[davenn]

Why is it not possible to create the same current pattern of a λ/2-antenna inside a λ/3-antenna, for example?

because as explained in the video or here on Wiki ...
https://en.wikipedia.org/wiki/Antenna_(radio)

Current and voltage distribution[edit]
The antenna conductors have the lowest feed-point impedance at the resonant frequency where they are just under 1/4 wavelength long; two such conductors in line fed differentially thus realizes the familiar "half-wave dipole". When fed with an RF current at the resonant frequency, the quarter wave element contains a standing wave with the voltage and current largely (but not exactly) in phase quadrature, as would be obtained using a quarter wave stub of transmission line. The current reaches a minimum at the end of the element (where it has nowhere to go!) and is maximum at the feed-point. The voltage, on the other hand, is the greatest at the end of the conductor and reaches a minimum (but not zero) at the feedpoint. Making the conductor shorter or longer than 1/4 wavelength means that the voltage pattern reaches its minimum somewhere beyond the feed-point, so that the feed-point has a higher voltage and thus sees a higher impedance, as we have noted. Since that voltage pattern is almost in phase quadrature with the current, the impedance seen at the feed-point is not only much higher but mainly reactive.

It can be seen that if such an element is resonant at f0 to produce such a standing wave pattern, then feeding that element with 3f0 (whose wavelength is 1/3 that of f0) will lead to a standing wave pattern in which the voltage is likewise a minimum at the feed-point (and the current at a maximum there). Thus, an antenna element is also resonant when its length is 3/4 of a wavelength (3/2 wavelength for a complete dipole). This is true for all odd multiples of 1/4 wavelength, where the feed-point impedance is purely resistive, though larger than the resistive impedance of the 1/4 wave element. Although such an antenna is resonant and works perfectly well at the higher frequency, the antenna radiation pattern is also altered compared to the half-wave dipole.

The use of a monopole or dipole at odd multiples of the fundamental resonant frequency, however, does not extend to even multiples (thus a 1/2 wavelength monopole or 1 wavelength dipole). Now the voltage standing wave is at its peak at the feed-point, while that of the current (which must be zero at the end of the conductor) is at a minimum (but not exactly zero). The antenna is anti-resonant at this frequency. Although the reactance at the feedpoint can be canceled using such an element length, the feed-point impedance is very high, and is highly dependent on the diameter of the conductor (which makes only a small difference at the actual resonant frequency). Such an antenna does not match the much lower characteristic impedance of available transmission lines, and is generally not used. However some equipment where transmission lines are not involved which desire a high driving point impedance may take advantage of this anti-resonance.

its isn't the way the current and voltage distribution is when at resonance. See the animation in the section above the bit I quoted
It primarily has to do with the voltage distribution on the dipole. For a low feedpoint impedance at a given frequency the voltage at the
feedpoint needs to be at a minimum. As the voltage at the feedpoint increases, so does the impedance. This results in an impedance mismatch between antenna and transmitter/transmission line resulting in significant losses because reflected power of the mismatched impedance at the feedpoint.Dave

 

[davenn]

EDIT: How do you quote, so that the name appears?

hilite the bit of text you want to quote and a little blue tab appears with +quote/reply, click on reply

Therefore the receiving antenna in the substance would need to be adjusted, because the coupling would be different, because of the different impedance of the surrounding substance? And this is why the length of the antenna need to be adjusted to the new wavelength inside the medium?

I can't make sense of that ??

Ok, but does that answer why an emitting antenna with length l radiates EM-waves with wavelength λ = 2l?

The antenna emits EM AT the wavelength of the freq being generated by the transmitter

 

[Anton Alice]

Making the conductor shorter or longer than 1/4 wavelength means that the voltage pattern reaches its minimum somewhere beyond the feed-point,

why? Why can't the minimum stay at the feedpoint? The voltage on the antenna is a standing wave, so it is the superposition of two waves traveling in opposite direction.
These voltages waves, how fast do they propagate in the antenna? lightspeed?EDIT:

and another question is: If we use plane waves, where E and B are in phase, how does the antenna deal with that? Because for a resonantly driven antenna, E and B are 90° shifted? Is this a problem?

and what about the antenna, which is inside a substance? Do I have to adjust the length of the antenna to the new lambda inside the substance? Sorry if I have overlooked the answer

 

[davenn]

why? Why can't the minimum stay at the feedpoint?

I answered that in my post above post #8

These voltages waves, how fast do they propagate in the antenna? lightspeed?

slightly less if it is bare wire or other conductor ( eg. aluminium tube) around 95 - 98% of c

and another question is: If we use plane waves, where E and B

just a small correction it is the magnetic H field we work with not the B field

and what about the antenna, which is inside a substance? Do I have to adjust the length of the antenna to the new lambda inside the substance?

yes we do. putting the antenna inside, say, a fibreglass radome, increases the antenna to air capacitance, increasing the capacitance
decreases the resonant freq of the antenna. Therefore we have to cut the antenna slightly shorter to bring its resonant freq back up to
where we want it to beDave

 

[Anton Alice]

Ah ok, thank you.

why? Why can't the minimum stay at the feedpoint?

Maybe that question should be put differently: Why is it for the voltage wave not possible to keep his minimum at the feedpoint, while the phase velocity, i.e. driving frequency is being changed.

I answered that in my post above post #8

slightly less if it is bare wire or other conductor ( eg. aluminium tube) around 95 - 98% of c

Sorry that I don't see the answer in #8. But the fact, that the propagation speed of voltage/current wave is nearly c, isn't this the answer?

 

[tech99]

Alright.

I have also a few questions concerning dipol-antennas:

first:

It has often been said, that the dipol antenna, in order to receive a certain frequency, needs to have a length of λ/2.
I don't see a clear reason for that. If one would not know that, one had to calculate the Capacity and Inductance of that piece of antenna (which certainly depend on the thickness of the antenna, too)
And each of these quantities can only be approximated. So how does it come, that we can hit the resonance by just letting the length be λ/2?
Is there a simple explanation?

Any conductor will respond to a passing EM wave, it does not depend on length. However, a half wave dipole is one of those antennas which has a resistive source impedance, arising from its resonance, which is convenient and efficient for the design of the feeder and receiver.
An EM wave travels along a conductor at nearly the speed of light because of the electrical properties of the conductor in vacuo. LIke a tuning fork supporting a sound wave, where inertia and springiness determine the wave velocity and hence the wave length, the permeability and permittivity of the vacuum determine the speed of the wave on a conductor, and it works out to be the speed of light. LIke a string which is half a wavelength long and supporting a standing sound wave, a conductor will do the same with an EM wave.