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Gain difference between half-wave and infinitesimal dipole?
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In discussion around another Q&A here, Glenn W9IQ commented:
The directive gain difference between a half wavelength dipole and an infinitesimal dipole is only 0.39 dB
I'm curious where this number comes from. I recall from e.g. the FCC amateur licensing tests the relationship between dBi and dBd; to summarize that Wikipedia link:
- dB(isotropic) – the forward gain of an antenna compared with the hypothetical isotropic antenna, which uniformly distributes energy in all directions.
- dB(dipole) – the forward gain of an antenna compared with a half-wave dipole antenna. 0 dBd = 2.15 dBi [emphasis mine]
Based on a naïve understanding of "half wavelength dipole" and "infinitesimal dipole" I would expect the the gain difference to be 2.15 dB, not 0.39 dB. What am I missing?
antenna-theory
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natevw - AF7TB
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In discussion around another Q&A here, Glenn W9IQ commented:
The directive gain difference between a half wavelength dipole and an infinitesimal dipole is only 0.39 dB
I'm curious where this number comes from. I recall from e.g. the FCC amateur licensing tests the relationship between dBi and dBd; to summarize that Wikipedia link:
- dB(isotropic) – the forward gain of an antenna compared with the hypothetical isotropic antenna, which uniformly distributes energy in all directions.
- dB(dipole) – the forward gain of an antenna compared with a half-wave dipole antenna. 0 dBd = 2.15 dBi [emphasis mine]
Based on a naïve understanding of "half wavelength dipole" and "infinitesimal dipole" I would expect the the gain difference to be 2.15 dB, not 0.39 dB. What am I missing?
antenna-theory
asked 7 hours ago
natevw - AF7TB
2,2211731
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up vote
2
down vote
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up vote
2
down vote
favorite
In discussion around another Q&A here, Glenn W9IQ commented:
The directive gain difference between a half wavelength dipole and an infinitesimal dipole is only 0.39 dB
I'm curious where this number comes from. I recall from e.g. the FCC amateur licensing tests the relationship between dBi and dBd; to summarize that Wikipedia link:
- dB(isotropic) – the forward gain of an antenna compared with the hypothetical isotropic antenna, which uniformly distributes energy in all directions.
- dB(dipole) – the forward gain of an antenna compared with a half-wave dipole antenna. 0 dBd = 2.15 dBi [emphasis mine]
Based on a naïve understanding of "half wavelength dipole" and "infinitesimal dipole" I would expect the the gain difference to be 2.15 dB, not 0.39 dB. What am I missing?
antenna-theory
asked 7 hours ago
natevw - AF7TB
2,2211731
In discussion around another Q&A here, Glenn W9IQ commented:
The directive gain difference between a half wavelength dipole and an infinitesimal dipole is only 0.39 dB
I'm curious where this number comes from. I recall from e.g. the FCC amateur licensing tests the relationship between dBi and dBd; to summarize that Wikipedia link:
- dB(isotropic) – the forward gain of an antenna compared with the hypothetical isotropic antenna, which uniformly distributes energy in all directions.
- dB(dipole) – the forward gain of an antenna compared with a half-wave dipole antenna. 0 dBd = 2.15 dBi [emphasis mine]
Based on a naïve understanding of "half wavelength dipole" and "infinitesimal dipole" I would expect the the gain difference to be 2.15 dB, not 0.39 dB. What am I missing?
antenna-theory
antenna-theory
asked 7 hours ago
natevw - AF7TB
2,2211731
asked 7 hours ago
natevw - AF7TB
2,2211731
asked 7 hours ago
natevw - AF7TB
2,2211731
asked 7 hours ago
natevw - AF7TB
2,2211731
asked 7 hours ago
natevw - AF7TB
2,2211731
2,2211731
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5 Answers
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An infinitesimal dipole is not the same thing as an isotropic antenna.
An isotropic antenna is a mathematical notion for comparing antenna patterns. It cannot exist — not just as a matter of physical construction, but also of arrangement of electromagnetic fields — because there is no possible arrangement of polarization of the wave which covers the sphere of all directions away from the antenna (mathematically, this is the hairy ball theorem).
An infinitesimal dipole (also known as a Hertzian dipole) has a dipole radiation pattern — for a dipole of any length whatsoever, the radiated field goes to zero along the axis of the dipole. (This pair of zero points means that a consistent polarization pattern can exist.)
Therefore, more of the radiation must go out in other directions, compared to an isotropic antenna, and so the design must have gain over an isotropic antenna.
I don't know the theory to explain where the figure of 2.15 dBi comes from, but I hope this explains why there is a difference at all.
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
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This is well explained on the Wikipedia entry for dipole antenna.
The "infinitesimal" dipole to which you refer is also called a Hertzian dipole. As shown in the Wikipedia entry, this theoretical construct has a gain of 1.5. The gain of the half-wave dipole is 1.64. The dB ratio of (1.64/1.5) is 0.39dB.
answered 6 hours ago
Brian K1LI
4338
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I sense that the question may be along the line of "If a dipole becomes infinitely small, isn't it essentially an isotropic antenna?" since this would be the only way an antenna could have 0 dBi gain at 100% efficiency. So I will construct this answer around a discussion of an isotropic antenna and the notion of antenna gain.
An isotropic antenna is a theoretical construct for an antenna that radiates equally in all directions. The isotropic antenna is a point source and is always considered to be 100% efficient (all power applied is radiated as electromagnetic radiation). It is helpful to picture this antenna at the center of a large sphere and the radiant flux (radiant power) from the antenna illuminating the sphere. The isotropic antenna would therefore uniformly illuminate the sphere - there would be no brighter or darker spots anywhere on the sphere.
Antenna gain can be generally defined as the peak radiant flux divided by the average radiant flux of the antenna. If we go back to the illuminated sphere concept, we would find the brightest spot on the sphere, measure the radiant flux at this spot, and divide this by the average radiant flux around the entire sphere.
So if we consider that the isotropic antenna uniformly illuminates the sphere, the peak radiant flux is the same as the average radiant flux. The linear gain of the isotropic antenna is therefore 1. Recalling that dB is a comparison of powers and that radiant flux is a power term, we can convert this linear gain of 1 compared to an isotropic antenna (itself), and the result is the familiar 0 dBi term.
Now consider that any other antenna construct is not capable of illuminating the sphere equally. That is to say that there will always be one or more spots on the sphere that are brighter (a higher radiant flux) than other spots on the sphere. Thus if we compute the maximum radiant flux divided by the average radiant flux of such an antenna, the linear gain of this 100% efficient antenna will always be greater than 1. If the linear gain is greater than 1, then it follows that the dBi gain must be greater than 0. Thus any antenna, other than an isotropic antenna, has a gain of > 0 dBi.
Up to this point, I have asserted that the referenced gains are for 100% efficient antennas. Antenna gain without consideration of efficiency (i.e. independent of efficiency) should more properly be called directivity. So in the above discussion, directivity may rightfully be substituted for gain when considering all antennas to be 100% efficient.
We could now dive into equations to compute the specific gain for various dipole constructions but my guess is this is not at the heart of the question. If the OP would like to explore specific formulae, please comment and I will enhance my answer accordingly.
answered 4 hours ago
Glenn W9IQ
12.3k1738
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
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Infinitesimal and isotropic are not the same thing.
Infinitesimal: an indefinitely small quantity; a value approaching zero. In the case of an infinitesimal dipole, we're talking about the length of the dipole. Kinda. More on that below.
Isotropic: having a physical property that has the same value when measured in different directions. In the case of antennas, we're talking about radiant intensity.
An isotropic antenna, that is, one which radiates with equal intensity in any direction, is not realizable. The mathematical proof comes from the hairy ball theorem.
Someone with a radio background will hear "dipole" and think "two wires". But someone with a physics background thinks a different thing. An electric dipole is two equal but opposite electric point charges separated by some distance. An infinitesimal dipole is the limiting case when the distance between these charges approaches, but does not reach zero.
The electric field looks like this:
By Geek3 CC BY-SA 3.0, from Wikimedia Commons
In practice, these separated charges come from the charge accumulation at the ends of the antenna as current takes charge away from one end while charge is being deposited in the other end. Of course in a real antenna the charge is distributed throughout the antenna: the infinitesimal dipole is a simplification with just two points, defined only by a direction and a moment.
If an infinitesimal dipole oscillates as an antenna does, then it can be shown with a lot of math derived from Maxwell's equations that the electric far-field is proportional to:
$$ E(theta, phi) = 1 over r sin(theta) $$
where the antenna is along the z axis, and:
$theta$ (theta) is the angle away from the z axis,
$phi$ is the angle away from the x axis (irrelevant here, since the dipole is omnidirectional),
$r$ is the distance away from the antenna
So the field is strongest when $theta = pi/2$, corresponding to the horizon, if the antenna is vertical. The electric field decreases with $1/r$, and since power is proportional to the square of voltage this means power would decrease with $1/r^2$, the inverse square law.
The corresponding formula for a half-wave dipole is:
$$ E(theta, phi) propto
cosleft( pi cos theta over 2 right)
over
r sintheta
$$
As you can see graphically, the half-wave dipole is just a little more "pointed":
Thus, if the two antennas have equal efficiency, the half-wave dipole will have just a little more gain.
Why so? No one actually builds infinitesimal dipoles: theoretically their gain is about the same but in practice their efficiency is extremely low. However, they are mathematically simple, and a very good approximation of real antennas can be made by modeling them as a collection of infinitesimal dipoles. The caveat is the infinitesimal dipole model assumes uniform current throughout, so we must divide the antenna into pieces small enough that each individual piece has mostly uniform current.
Thinking of a half-wave dipole in this way, you can now consider it sort of a colinear array of infinitesimal dipoles. It isn't physically large enough to yield a lot of gain, but it does yield just a little: 0.39 dB. We see that in the graph above.
answered 2 hours ago
Phil Frost - W8II
25.7k142113
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The gain difference is because the short dipole has a triangular current distribution, while the half wave dipole is sinusoidal.
(the triangle is actually the end bits of a sinusoidal shape, but if its very short, that's a triangle).
For the same feed current, the sinusoidal distribution has slightly more moment - area under the current graph if you like - so slightly higher gain.
The Wikipedia article on dipoles is really good too, just skim over the equations like I do and read the conclusion.
answered 6 hours ago
tomnexus
2,032413
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
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5 Answers
5
active
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
An infinitesimal dipole is not the same thing as an isotropic antenna.
An isotropic antenna is a mathematical notion for comparing antenna patterns. It cannot exist — not just as a matter of physical construction, but also of arrangement of electromagnetic fields — because there is no possible arrangement of polarization of the wave which covers the sphere of all directions away from the antenna (mathematically, this is the hairy ball theorem).
An infinitesimal dipole (also known as a Hertzian dipole) has a dipole radiation pattern — for a dipole of any length whatsoever, the radiated field goes to zero along the axis of the dipole. (This pair of zero points means that a consistent polarization pattern can exist.)
Therefore, more of the radiation must go out in other directions, compared to an isotropic antenna, and so the design must have gain over an isotropic antenna.
I don't know the theory to explain where the figure of 2.15 dBi comes from, but I hope this explains why there is a difference at all.
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
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up vote
1
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An infinitesimal dipole is not the same thing as an isotropic antenna.
An isotropic antenna is a mathematical notion for comparing antenna patterns. It cannot exist — not just as a matter of physical construction, but also of arrangement of electromagnetic fields — because there is no possible arrangement of polarization of the wave which covers the sphere of all directions away from the antenna (mathematically, this is the hairy ball theorem).
An infinitesimal dipole (also known as a Hertzian dipole) has a dipole radiation pattern — for a dipole of any length whatsoever, the radiated field goes to zero along the axis of the dipole. (This pair of zero points means that a consistent polarization pattern can exist.)
Therefore, more of the radiation must go out in other directions, compared to an isotropic antenna, and so the design must have gain over an isotropic antenna.
I don't know the theory to explain where the figure of 2.15 dBi comes from, but I hope this explains why there is a difference at all.
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
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up vote
1
down vote
up vote
1
down vote
An infinitesimal dipole is not the same thing as an isotropic antenna.
An isotropic antenna is a mathematical notion for comparing antenna patterns. It cannot exist — not just as a matter of physical construction, but also of arrangement of electromagnetic fields — because there is no possible arrangement of polarization of the wave which covers the sphere of all directions away from the antenna (mathematically, this is the hairy ball theorem).
An infinitesimal dipole (also known as a Hertzian dipole) has a dipole radiation pattern — for a dipole of any length whatsoever, the radiated field goes to zero along the axis of the dipole. (This pair of zero points means that a consistent polarization pattern can exist.)
Therefore, more of the radiation must go out in other directions, compared to an isotropic antenna, and so the design must have gain over an isotropic antenna.
I don't know the theory to explain where the figure of 2.15 dBi comes from, but I hope this explains why there is a difference at all.
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
An infinitesimal dipole is not the same thing as an isotropic antenna.
An isotropic antenna is a mathematical notion for comparing antenna patterns. It cannot exist — not just as a matter of physical construction, but also of arrangement of electromagnetic fields — because there is no possible arrangement of polarization of the wave which covers the sphere of all directions away from the antenna (mathematically, this is the hairy ball theorem).
An infinitesimal dipole (also known as a Hertzian dipole) has a dipole radiation pattern — for a dipole of any length whatsoever, the radiated field goes to zero along the axis of the dipole. (This pair of zero points means that a consistent polarization pattern can exist.)
Therefore, more of the radiation must go out in other directions, compared to an isotropic antenna, and so the design must have gain over an isotropic antenna.
I don't know the theory to explain where the figure of 2.15 dBi comes from, but I hope this explains why there is a difference at all.
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
edited 7 hours ago
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
answered 7 hours ago
Kevin Reid AG6YO♦
14.7k22665
14.7k22665
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This is well explained on the Wikipedia entry for dipole antenna.
The "infinitesimal" dipole to which you refer is also called a Hertzian dipole. As shown in the Wikipedia entry, this theoretical construct has a gain of 1.5. The gain of the half-wave dipole is 1.64. The dB ratio of (1.64/1.5) is 0.39dB.
answered 6 hours ago
Brian K1LI
4338
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up vote
1
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This is well explained on the Wikipedia entry for dipole antenna.
The "infinitesimal" dipole to which you refer is also called a Hertzian dipole. As shown in the Wikipedia entry, this theoretical construct has a gain of 1.5. The gain of the half-wave dipole is 1.64. The dB ratio of (1.64/1.5) is 0.39dB.
answered 6 hours ago
Brian K1LI
4338
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is well explained on the Wikipedia entry for dipole antenna.
The "infinitesimal" dipole to which you refer is also called a Hertzian dipole. As shown in the Wikipedia entry, this theoretical construct has a gain of 1.5. The gain of the half-wave dipole is 1.64. The dB ratio of (1.64/1.5) is 0.39dB.
answered 6 hours ago
Brian K1LI
4338
This is well explained on the Wikipedia entry for dipole antenna.
The "infinitesimal" dipole to which you refer is also called a Hertzian dipole. As shown in the Wikipedia entry, this theoretical construct has a gain of 1.5. The gain of the half-wave dipole is 1.64. The dB ratio of (1.64/1.5) is 0.39dB.
answered 6 hours ago
Brian K1LI
4338
answered 6 hours ago
Brian K1LI
4338
answered 6 hours ago
Brian K1LI
4338
answered 6 hours ago
Brian K1LI
4338
4338
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1
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I sense that the question may be along the line of "If a dipole becomes infinitely small, isn't it essentially an isotropic antenna?" since this would be the only way an antenna could have 0 dBi gain at 100% efficiency. So I will construct this answer around a discussion of an isotropic antenna and the notion of antenna gain.
An isotropic antenna is a theoretical construct for an antenna that radiates equally in all directions. The isotropic antenna is a point source and is always considered to be 100% efficient (all power applied is radiated as electromagnetic radiation). It is helpful to picture this antenna at the center of a large sphere and the radiant flux (radiant power) from the antenna illuminating the sphere. The isotropic antenna would therefore uniformly illuminate the sphere - there would be no brighter or darker spots anywhere on the sphere.
Antenna gain can be generally defined as the peak radiant flux divided by the average radiant flux of the antenna. If we go back to the illuminated sphere concept, we would find the brightest spot on the sphere, measure the radiant flux at this spot, and divide this by the average radiant flux around the entire sphere.
So if we consider that the isotropic antenna uniformly illuminates the sphere, the peak radiant flux is the same as the average radiant flux. The linear gain of the isotropic antenna is therefore 1. Recalling that dB is a comparison of powers and that radiant flux is a power term, we can convert this linear gain of 1 compared to an isotropic antenna (itself), and the result is the familiar 0 dBi term.
Now consider that any other antenna construct is not capable of illuminating the sphere equally. That is to say that there will always be one or more spots on the sphere that are brighter (a higher radiant flux) than other spots on the sphere. Thus if we compute the maximum radiant flux divided by the average radiant flux of such an antenna, the linear gain of this 100% efficient antenna will always be greater than 1. If the linear gain is greater than 1, then it follows that the dBi gain must be greater than 0. Thus any antenna, other than an isotropic antenna, has a gain of > 0 dBi.
Up to this point, I have asserted that the referenced gains are for 100% efficient antennas. Antenna gain without consideration of efficiency (i.e. independent of efficiency) should more properly be called directivity. So in the above discussion, directivity may rightfully be substituted for gain when considering all antennas to be 100% efficient.
We could now dive into equations to compute the specific gain for various dipole constructions but my guess is this is not at the heart of the question. If the OP would like to explore specific formulae, please comment and I will enhance my answer accordingly.
answered 4 hours ago
Glenn W9IQ
12.3k1738
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
add a comment |Â
up vote
1
down vote
I sense that the question may be along the line of "If a dipole becomes infinitely small, isn't it essentially an isotropic antenna?" since this would be the only way an antenna could have 0 dBi gain at 100% efficiency. So I will construct this answer around a discussion of an isotropic antenna and the notion of antenna gain.
An isotropic antenna is a theoretical construct for an antenna that radiates equally in all directions. The isotropic antenna is a point source and is always considered to be 100% efficient (all power applied is radiated as electromagnetic radiation). It is helpful to picture this antenna at the center of a large sphere and the radiant flux (radiant power) from the antenna illuminating the sphere. The isotropic antenna would therefore uniformly illuminate the sphere - there would be no brighter or darker spots anywhere on the sphere.
Antenna gain can be generally defined as the peak radiant flux divided by the average radiant flux of the antenna. If we go back to the illuminated sphere concept, we would find the brightest spot on the sphere, measure the radiant flux at this spot, and divide this by the average radiant flux around the entire sphere.
So if we consider that the isotropic antenna uniformly illuminates the sphere, the peak radiant flux is the same as the average radiant flux. The linear gain of the isotropic antenna is therefore 1. Recalling that dB is a comparison of powers and that radiant flux is a power term, we can convert this linear gain of 1 compared to an isotropic antenna (itself), and the result is the familiar 0 dBi term.
Now consider that any other antenna construct is not capable of illuminating the sphere equally. That is to say that there will always be one or more spots on the sphere that are brighter (a higher radiant flux) than other spots on the sphere. Thus if we compute the maximum radiant flux divided by the average radiant flux of such an antenna, the linear gain of this 100% efficient antenna will always be greater than 1. If the linear gain is greater than 1, then it follows that the dBi gain must be greater than 0. Thus any antenna, other than an isotropic antenna, has a gain of > 0 dBi.
Up to this point, I have asserted that the referenced gains are for 100% efficient antennas. Antenna gain without consideration of efficiency (i.e. independent of efficiency) should more properly be called directivity. So in the above discussion, directivity may rightfully be substituted for gain when considering all antennas to be 100% efficient.
We could now dive into equations to compute the specific gain for various dipole constructions but my guess is this is not at the heart of the question. If the OP would like to explore specific formulae, please comment and I will enhance my answer accordingly.
answered 4 hours ago
Glenn W9IQ
12.3k1738
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
add a comment |Â
up vote
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up vote
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I sense that the question may be along the line of "If a dipole becomes infinitely small, isn't it essentially an isotropic antenna?" since this would be the only way an antenna could have 0 dBi gain at 100% efficiency. So I will construct this answer around a discussion of an isotropic antenna and the notion of antenna gain.
An isotropic antenna is a theoretical construct for an antenna that radiates equally in all directions. The isotropic antenna is a point source and is always considered to be 100% efficient (all power applied is radiated as electromagnetic radiation). It is helpful to picture this antenna at the center of a large sphere and the radiant flux (radiant power) from the antenna illuminating the sphere. The isotropic antenna would therefore uniformly illuminate the sphere - there would be no brighter or darker spots anywhere on the sphere.
Antenna gain can be generally defined as the peak radiant flux divided by the average radiant flux of the antenna. If we go back to the illuminated sphere concept, we would find the brightest spot on the sphere, measure the radiant flux at this spot, and divide this by the average radiant flux around the entire sphere.
So if we consider that the isotropic antenna uniformly illuminates the sphere, the peak radiant flux is the same as the average radiant flux. The linear gain of the isotropic antenna is therefore 1. Recalling that dB is a comparison of powers and that radiant flux is a power term, we can convert this linear gain of 1 compared to an isotropic antenna (itself), and the result is the familiar 0 dBi term.
Now consider that any other antenna construct is not capable of illuminating the sphere equally. That is to say that there will always be one or more spots on the sphere that are brighter (a higher radiant flux) than other spots on the sphere. Thus if we compute the maximum radiant flux divided by the average radiant flux of such an antenna, the linear gain of this 100% efficient antenna will always be greater than 1. If the linear gain is greater than 1, then it follows that the dBi gain must be greater than 0. Thus any antenna, other than an isotropic antenna, has a gain of > 0 dBi.
Up to this point, I have asserted that the referenced gains are for 100% efficient antennas. Antenna gain without consideration of efficiency (i.e. independent of efficiency) should more properly be called directivity. So in the above discussion, directivity may rightfully be substituted for gain when considering all antennas to be 100% efficient.
We could now dive into equations to compute the specific gain for various dipole constructions but my guess is this is not at the heart of the question. If the OP would like to explore specific formulae, please comment and I will enhance my answer accordingly.
answered 4 hours ago
Glenn W9IQ
12.3k1738
I sense that the question may be along the line of "If a dipole becomes infinitely small, isn't it essentially an isotropic antenna?" since this would be the only way an antenna could have 0 dBi gain at 100% efficiency. So I will construct this answer around a discussion of an isotropic antenna and the notion of antenna gain.
An isotropic antenna is a theoretical construct for an antenna that radiates equally in all directions. The isotropic antenna is a point source and is always considered to be 100% efficient (all power applied is radiated as electromagnetic radiation). It is helpful to picture this antenna at the center of a large sphere and the radiant flux (radiant power) from the antenna illuminating the sphere. The isotropic antenna would therefore uniformly illuminate the sphere - there would be no brighter or darker spots anywhere on the sphere.
Antenna gain can be generally defined as the peak radiant flux divided by the average radiant flux of the antenna. If we go back to the illuminated sphere concept, we would find the brightest spot on the sphere, measure the radiant flux at this spot, and divide this by the average radiant flux around the entire sphere.
So if we consider that the isotropic antenna uniformly illuminates the sphere, the peak radiant flux is the same as the average radiant flux. The linear gain of the isotropic antenna is therefore 1. Recalling that dB is a comparison of powers and that radiant flux is a power term, we can convert this linear gain of 1 compared to an isotropic antenna (itself), and the result is the familiar 0 dBi term.
Now consider that any other antenna construct is not capable of illuminating the sphere equally. That is to say that there will always be one or more spots on the sphere that are brighter (a higher radiant flux) than other spots on the sphere. Thus if we compute the maximum radiant flux divided by the average radiant flux of such an antenna, the linear gain of this 100% efficient antenna will always be greater than 1. If the linear gain is greater than 1, then it follows that the dBi gain must be greater than 0. Thus any antenna, other than an isotropic antenna, has a gain of > 0 dBi.
Up to this point, I have asserted that the referenced gains are for 100% efficient antennas. Antenna gain without consideration of efficiency (i.e. independent of efficiency) should more properly be called directivity. So in the above discussion, directivity may rightfully be substituted for gain when considering all antennas to be 100% efficient.
We could now dive into equations to compute the specific gain for various dipole constructions but my guess is this is not at the heart of the question. If the OP would like to explore specific formulae, please comment and I will enhance my answer accordingly.
answered 4 hours ago
Glenn W9IQ
12.3k1738
answered 4 hours ago
Glenn W9IQ
12.3k1738
answered 4 hours ago
Glenn W9IQ
12.3k1738
answered 4 hours ago
Glenn W9IQ
12.3k1738
12.3k1738
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
add a comment |Â
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
Given the mention of an illuminated sphere, which after all also involves electromagnetic radiation, it might be worth mentioning that the difference between a visible light source and a radio antenna that allows the light source to be isotropic is that the light is incoherent (arbitrary phase and polarization) and the radio signal is coherent.
– Kevin Reid AG6YO♦
4 hours ago
add a comment |Â
up vote
1
down vote
Infinitesimal and isotropic are not the same thing.
Infinitesimal: an indefinitely small quantity; a value approaching zero. In the case of an infinitesimal dipole, we're talking about the length of the dipole. Kinda. More on that below.
Isotropic: having a physical property that has the same value when measured in different directions. In the case of antennas, we're talking about radiant intensity.
An isotropic antenna, that is, one which radiates with equal intensity in any direction, is not realizable. The mathematical proof comes from the hairy ball theorem.
Someone with a radio background will hear "dipole" and think "two wires". But someone with a physics background thinks a different thing. An electric dipole is two equal but opposite electric point charges separated by some distance. An infinitesimal dipole is the limiting case when the distance between these charges approaches, but does not reach zero.
The electric field looks like this:
By Geek3 CC BY-SA 3.0, from Wikimedia Commons
In practice, these separated charges come from the charge accumulation at the ends of the antenna as current takes charge away from one end while charge is being deposited in the other end. Of course in a real antenna the charge is distributed throughout the antenna: the infinitesimal dipole is a simplification with just two points, defined only by a direction and a moment.
If an infinitesimal dipole oscillates as an antenna does, then it can be shown with a lot of math derived from Maxwell's equations that the electric far-field is proportional to:
$$ E(theta, phi) = 1 over r sin(theta) $$
where the antenna is along the z axis, and:
$theta$ (theta) is the angle away from the z axis,
$phi$ is the angle away from the x axis (irrelevant here, since the dipole is omnidirectional),
$r$ is the distance away from the antenna
So the field is strongest when $theta = pi/2$, corresponding to the horizon, if the antenna is vertical. The electric field decreases with $1/r$, and since power is proportional to the square of voltage this means power would decrease with $1/r^2$, the inverse square law.
The corresponding formula for a half-wave dipole is:
$$ E(theta, phi) propto
cosleft( pi cos theta over 2 right)
over
r sintheta
$$
As you can see graphically, the half-wave dipole is just a little more "pointed":
Thus, if the two antennas have equal efficiency, the half-wave dipole will have just a little more gain.
Why so? No one actually builds infinitesimal dipoles: theoretically their gain is about the same but in practice their efficiency is extremely low. However, they are mathematically simple, and a very good approximation of real antennas can be made by modeling them as a collection of infinitesimal dipoles. The caveat is the infinitesimal dipole model assumes uniform current throughout, so we must divide the antenna into pieces small enough that each individual piece has mostly uniform current.
Thinking of a half-wave dipole in this way, you can now consider it sort of a colinear array of infinitesimal dipoles. It isn't physically large enough to yield a lot of gain, but it does yield just a little: 0.39 dB. We see that in the graph above.
answered 2 hours ago
Phil Frost - W8II
25.7k142113
add a comment |Â
up vote
1
down vote
Infinitesimal and isotropic are not the same thing.
Infinitesimal: an indefinitely small quantity; a value approaching zero. In the case of an infinitesimal dipole, we're talking about the length of the dipole. Kinda. More on that below.
Isotropic: having a physical property that has the same value when measured in different directions. In the case of antennas, we're talking about radiant intensity.
An isotropic antenna, that is, one which radiates with equal intensity in any direction, is not realizable. The mathematical proof comes from the hairy ball theorem.
Someone with a radio background will hear "dipole" and think "two wires". But someone with a physics background thinks a different thing. An electric dipole is two equal but opposite electric point charges separated by some distance. An infinitesimal dipole is the limiting case when the distance between these charges approaches, but does not reach zero.
The electric field looks like this:
By Geek3 CC BY-SA 3.0, from Wikimedia Commons
In practice, these separated charges come from the charge accumulation at the ends of the antenna as current takes charge away from one end while charge is being deposited in the other end. Of course in a real antenna the charge is distributed throughout the antenna: the infinitesimal dipole is a simplification with just two points, defined only by a direction and a moment.
If an infinitesimal dipole oscillates as an antenna does, then it can be shown with a lot of math derived from Maxwell's equations that the electric far-field is proportional to:
$$ E(theta, phi) = 1 over r sin(theta) $$
where the antenna is along the z axis, and:
$theta$ (theta) is the angle away from the z axis,
$phi$ is the angle away from the x axis (irrelevant here, since the dipole is omnidirectional),
$r$ is the distance away from the antenna
So the field is strongest when $theta = pi/2$, corresponding to the horizon, if the antenna is vertical. The electric field decreases with $1/r$, and since power is proportional to the square of voltage this means power would decrease with $1/r^2$, the inverse square law.
The corresponding formula for a half-wave dipole is:
$$ E(theta, phi) propto
cosleft( pi cos theta over 2 right)
over
r sintheta
$$
As you can see graphically, the half-wave dipole is just a little more "pointed":
Thus, if the two antennas have equal efficiency, the half-wave dipole will have just a little more gain.
Why so? No one actually builds infinitesimal dipoles: theoretically their gain is about the same but in practice their efficiency is extremely low. However, they are mathematically simple, and a very good approximation of real antennas can be made by modeling them as a collection of infinitesimal dipoles. The caveat is the infinitesimal dipole model assumes uniform current throughout, so we must divide the antenna into pieces small enough that each individual piece has mostly uniform current.
Thinking of a half-wave dipole in this way, you can now consider it sort of a colinear array of infinitesimal dipoles. It isn't physically large enough to yield a lot of gain, but it does yield just a little: 0.39 dB. We see that in the graph above.
answered 2 hours ago
Phil Frost - W8II
25.7k142113
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Infinitesimal and isotropic are not the same thing.
Infinitesimal: an indefinitely small quantity; a value approaching zero. In the case of an infinitesimal dipole, we're talking about the length of the dipole. Kinda. More on that below.
Isotropic: having a physical property that has the same value when measured in different directions. In the case of antennas, we're talking about radiant intensity.
An isotropic antenna, that is, one which radiates with equal intensity in any direction, is not realizable. The mathematical proof comes from the hairy ball theorem.
Someone with a radio background will hear "dipole" and think "two wires". But someone with a physics background thinks a different thing. An electric dipole is two equal but opposite electric point charges separated by some distance. An infinitesimal dipole is the limiting case when the distance between these charges approaches, but does not reach zero.
The electric field looks like this:
By Geek3 CC BY-SA 3.0, from Wikimedia Commons
In practice, these separated charges come from the charge accumulation at the ends of the antenna as current takes charge away from one end while charge is being deposited in the other end. Of course in a real antenna the charge is distributed throughout the antenna: the infinitesimal dipole is a simplification with just two points, defined only by a direction and a moment.
If an infinitesimal dipole oscillates as an antenna does, then it can be shown with a lot of math derived from Maxwell's equations that the electric far-field is proportional to:
$$ E(theta, phi) = 1 over r sin(theta) $$
where the antenna is along the z axis, and:
$theta$ (theta) is the angle away from the z axis,
$phi$ is the angle away from the x axis (irrelevant here, since the dipole is omnidirectional),
$r$ is the distance away from the antenna
So the field is strongest when $theta = pi/2$, corresponding to the horizon, if the antenna is vertical. The electric field decreases with $1/r$, and since power is proportional to the square of voltage this means power would decrease with $1/r^2$, the inverse square law.
The corresponding formula for a half-wave dipole is:
$$ E(theta, phi) propto
cosleft( pi cos theta over 2 right)
over
r sintheta
$$
As you can see graphically, the half-wave dipole is just a little more "pointed":
Thus, if the two antennas have equal efficiency, the half-wave dipole will have just a little more gain.
Why so? No one actually builds infinitesimal dipoles: theoretically their gain is about the same but in practice their efficiency is extremely low. However, they are mathematically simple, and a very good approximation of real antennas can be made by modeling them as a collection of infinitesimal dipoles. The caveat is the infinitesimal dipole model assumes uniform current throughout, so we must divide the antenna into pieces small enough that each individual piece has mostly uniform current.
Thinking of a half-wave dipole in this way, you can now consider it sort of a colinear array of infinitesimal dipoles. It isn't physically large enough to yield a lot of gain, but it does yield just a little: 0.39 dB. We see that in the graph above.
answered 2 hours ago
Phil Frost - W8II
25.7k142113
Infinitesimal and isotropic are not the same thing.
Infinitesimal: an indefinitely small quantity; a value approaching zero. In the case of an infinitesimal dipole, we're talking about the length of the dipole. Kinda. More on that below.
Isotropic: having a physical property that has the same value when measured in different directions. In the case of antennas, we're talking about radiant intensity.
An isotropic antenna, that is, one which radiates with equal intensity in any direction, is not realizable. The mathematical proof comes from the hairy ball theorem.
Someone with a radio background will hear "dipole" and think "two wires". But someone with a physics background thinks a different thing. An electric dipole is two equal but opposite electric point charges separated by some distance. An infinitesimal dipole is the limiting case when the distance between these charges approaches, but does not reach zero.
The electric field looks like this:
By Geek3 CC BY-SA 3.0, from Wikimedia Commons
In practice, these separated charges come from the charge accumulation at the ends of the antenna as current takes charge away from one end while charge is being deposited in the other end. Of course in a real antenna the charge is distributed throughout the antenna: the infinitesimal dipole is a simplification with just two points, defined only by a direction and a moment.
If an infinitesimal dipole oscillates as an antenna does, then it can be shown with a lot of math derived from Maxwell's equations that the electric far-field is proportional to:
$$ E(theta, phi) = 1 over r sin(theta) $$
where the antenna is along the z axis, and:
$theta$ (theta) is the angle away from the z axis,
$phi$ is the angle away from the x axis (irrelevant here, since the dipole is omnidirectional),
$r$ is the distance away from the antenna
So the field is strongest when $theta = pi/2$, corresponding to the horizon, if the antenna is vertical. The electric field decreases with $1/r$, and since power is proportional to the square of voltage this means power would decrease with $1/r^2$, the inverse square law.
The corresponding formula for a half-wave dipole is:
$$ E(theta, phi) propto
cosleft( pi cos theta over 2 right)
over
r sintheta
$$
As you can see graphically, the half-wave dipole is just a little more "pointed":
Thus, if the two antennas have equal efficiency, the half-wave dipole will have just a little more gain.
Why so? No one actually builds infinitesimal dipoles: theoretically their gain is about the same but in practice their efficiency is extremely low. However, they are mathematically simple, and a very good approximation of real antennas can be made by modeling them as a collection of infinitesimal dipoles. The caveat is the infinitesimal dipole model assumes uniform current throughout, so we must divide the antenna into pieces small enough that each individual piece has mostly uniform current.
Thinking of a half-wave dipole in this way, you can now consider it sort of a colinear array of infinitesimal dipoles. It isn't physically large enough to yield a lot of gain, but it does yield just a little: 0.39 dB. We see that in the graph above.
answered 2 hours ago
Phil Frost - W8II
25.7k142113
answered 2 hours ago
Phil Frost - W8II
25.7k142113
answered 2 hours ago
Phil Frost - W8II
25.7k142113
answered 2 hours ago
Phil Frost - W8II
25.7k142113
25.7k142113
add a comment |Â
add a comment |Â
up vote
0
down vote
The gain difference is because the short dipole has a triangular current distribution, while the half wave dipole is sinusoidal.
(the triangle is actually the end bits of a sinusoidal shape, but if its very short, that's a triangle).
For the same feed current, the sinusoidal distribution has slightly more moment - area under the current graph if you like - so slightly higher gain.
The Wikipedia article on dipoles is really good too, just skim over the equations like I do and read the conclusion.
answered 6 hours ago
tomnexus
2,032413
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
add a comment |Â
up vote
0
down vote
The gain difference is because the short dipole has a triangular current distribution, while the half wave dipole is sinusoidal.
(the triangle is actually the end bits of a sinusoidal shape, but if its very short, that's a triangle).
For the same feed current, the sinusoidal distribution has slightly more moment - area under the current graph if you like - so slightly higher gain.
The Wikipedia article on dipoles is really good too, just skim over the equations like I do and read the conclusion.
answered 6 hours ago
tomnexus
2,032413
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The gain difference is because the short dipole has a triangular current distribution, while the half wave dipole is sinusoidal.
(the triangle is actually the end bits of a sinusoidal shape, but if its very short, that's a triangle).
For the same feed current, the sinusoidal distribution has slightly more moment - area under the current graph if you like - so slightly higher gain.
The Wikipedia article on dipoles is really good too, just skim over the equations like I do and read the conclusion.
answered 6 hours ago
tomnexus
2,032413
The gain difference is because the short dipole has a triangular current distribution, while the half wave dipole is sinusoidal.
(the triangle is actually the end bits of a sinusoidal shape, but if its very short, that's a triangle).
For the same feed current, the sinusoidal distribution has slightly more moment - area under the current graph if you like - so slightly higher gain.
The Wikipedia article on dipoles is really good too, just skim over the equations like I do and read the conclusion.
answered 6 hours ago
tomnexus
2,032413
edited 1 hour ago
answered 6 hours ago
tomnexus
2,032413
answered 6 hours ago
tomnexus
2,032413
answered 6 hours ago
tomnexus
2,032413
2,032413
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
add a comment |Â
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
The "short" dipole - short implying a length less than 1/10 wave - is assumed to have a triangular current distribution. The "Hertzian" or "infinitesimal" dipole is assumed to have constant current over its length.
– Brian K1LI
4 hours ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
Yes, that's right... correcting now
– tomnexus
1 hour ago
add a comment |Â
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