What is the difference between a dipole (dipole antenna) and an antenna (whip antenna with wire counterweights)? Dipole antennas: Radial Dipole or dipole antenna.
To each wireless device need an antenna. This conductive mechanical device is a transducer that converts a transmitted radio frequency (RF) signal into the electric and magnetic fields that make up a radio wave. It also converts the received radio wave back into an electrical signal. An almost infinite number of configurations are possible for antennas. However, most of them are based on two main types: dipole and whip antennas.
The concept of "antenna"
A radio wave contains an electric field perpendicular to the magnetic field. Both are perpendicular to the direction of propagation (picture below). This electromagnetic field creates the antenna. The signal emitted by the device is generated in the transmitter and then sent to the antenna using a transmission line, usually a coaxial cable.
The lines are magnetic and electric lines of force that move together and support each other as they "move outward" from the antenna.
The voltage creates an electric field around the antenna elements. The current in the antenna creates a magnetic field. The electric and magnetic fields combine and regenerate each other according to Maxwell's famous equations, and the "combined" wave is sent out from the antenna into space. When a signal is received, an electromagnetic wave induces a voltage in the antenna, which converts the electromagnetic wave back into an electrical signal that can be further processed.
The primary consideration in the orientation of any antenna is polarization, which refers to the orientation of the electric field (E) with the ground. This is also the orientation of the transmitting elements relative to the ground. Vertical installed antenna, perpendicular to the ground, emits a vertically polarized wave. Thus, a horizontally located antenna emits a horizontally polarized wave.
Polarization can also be circular. Special configurations such as helical or helical antennas can emit a rotating wave, creating a rotating polarized wave. The antenna can create a rotation direction either to the right or to the left.
Ideally, the antennas on both the transmitting and receiving devices should have the same polarization. At frequencies below about 30 MHz, the wave is typically reflected, refracted, rotated, or otherwise modified by the atmosphere, ground, or other objects. Therefore, polarization matching on the two sides is not critical. At VHF, UHF and microwave frequencies, the polarization must be the same to ensure the highest quality signal transmission. And, note that the antennas exhibit reciprocity, that is, they work equally well for both transmission and reception.
Dipole or symmetrical dipole antenna
A dipole is a half-wave structure made of wire, tube, printed circuit board(PCB) or other conductive material. It is divided into two equal quarter wavelengths and fed by a transmission line.
The lines show the distribution of electric and magnetic fields. One wavelength (λ) is equal to:
half wave:
λ/2 = 492/f MHz
The actual length is usually reduced depending on the size of the antenna wires. Best approximation to electrical length:
λ/2 = 492 K/f MHz
where K is a coefficient relating the diameter of the conductor to its length. This is 0.95 for wire antennas with a frequency of 30 MHz or less. Or:
λ/2 = 468/f MHz
Length in inches:
λ/2 = 5904 K/f MHz
The K value is smaller for larger diameter elements. For a half-inch diameter tube, K is 0.945. The dipole channel for 165 MHz must have the length:
λ/2 = 5904(0.945)/165 = 33.81 inches
or two 16.9-inch segments.
The length is important because the antenna is a resonant device. For maximum radiation efficiency, it must be tuned to the operating frequency. However, the antenna works quite well over a narrow frequency range, like a resonant filter.
The bandwidth of a dipole is a function of its structure. It is generally defined as the range over which the antenna standing wave ratio (SWR) is less than 2:1. SWR is determined by the amount of signal reflected from the device back along the transmission line feeding it. It is a function of the antenna impedance in relation to the transmission line impedance.
The ideal transmission line is a balanced conducting pair with a resistance of 75 ohms. You can also use coaxial cable with a characteristic impedance of 75 Ohms (Zo). Coaxial cable with a characteristic impedance of 50 ohms can also be used, as it matches the antenna well as long as it is less than half the wavelength above ground.
Coaxial cable is an unbalanced line because RF current will flow outside the coaxial shield, creating some unwanted induced interference in nearby devices, although the antenna will work reasonably well. Best method supply - use a balun transformer at the supply point with a coaxial cable. A balun is a transformer device that converts balanced signals into unbalanced signals or vice versa.
The dipole can be mounted horizontally or vertically depending on the desired polarization. The supply line should ideally be perpendicular to the radiating elements to avoid distortion of the radiation, so the dipole is most often oriented horizontally.
The radiation pattern of an antenna signal depends on its structure and installation. Physical radiation is three-dimensional, but it is usually represented by both horizontal and vertical radiation patterns.
The horizontal radiation pattern of the dipole is a figure eight (Figure 3). The maximum signal appears at the antenna. Figure 4 shows the vertical radiation pattern. These are perfect specimens that are easily distorted by the ground and any nearby objects.
Antenna gain is related to directivity. Gain is usually expressed in decibels (dB) based on some "reference" such as an isotropic antenna, which is a point source of radio frequency energy that radiates the signal in all directions. Think of a point light source illuminating the inside of an expanding sphere. An isotropic antenna has a gain of 1 or 0 dB.
If a transmitter shapes or focuses the radiation pattern and makes it more directional, it has an isotropic antenna gain. The dipole has a gain of 2.16 dBi over an isotropic source. In some cases, the gain is expressed as a function of the dipole reference in dBd.
Vertical antenna with additional horizontal reflective elements
This device is essentially half a dipole mounted vertically. The term monopole is also used to describe this setup. The ground below the antenna, the conducting surface with the smallest λ/4 radius, or a pattern of λ/4 conductors called radials, make up the second half of the antenna (Fig. 5).
If the antenna is connected to a good ground, it is called a Marconi antenna. The main structure is the other λ/4 half of the transmitter. If the ground plane is of sufficient size and conductivity, then the grounding performance is equivalent to a vertically mounted dipole.
Quarter-wave vertical length:
λ/4 = 246 K/f MHz
The K-factor is less than 0.95 for verticals, which are typically manufactured with a wider tube.
The feedpoint impedance is half a dipole or approximately 36 ohms. The actual figure depends on the height above the ground. Like a dipole, the ground plane is resonant and typically has a reactive component to its fundamental impedance. The most common transmission line is 50-Ω coaxial cable, as it matches antenna impedance relatively well with an SWR below 2:1.
The vertical antenna with an additional reflective element is omnidirectional. The horizontal radiation pattern is a circle in which the device radiates the signal equally well in all directions. Figure 6 shows the vertical radiation pattern. Compared to the vertical radiation pattern of a dipole, the ground plane has a lower radiation angle, which has the advantage of wider propagation at frequencies below about 50 MHz.
conclusions
In addition, two or more vertical antennas can be configured with an additional reflective element to create a more directional and amplified signal. For example, a directional AM radio uses two or more towers to send a strong signal in one direction while canceling it in the other.
Standing wave ratio
Standing waves are patterns of voltage and current distribution along a transmission line. If the characteristic impedance (Zo) of a line matches the output impedance of the generator (transmitter) and the antenna load, the voltage and current along the line are constant. When the impedance is matched, maximum power transfer occurs.
If the antenna load does not match the linear impedance, not all of the transmitted power is absorbed by the load. Any power not absorbed by the antenna is reflected back down the line, interfering with the forward signal and creating variations in current and voltage along the line. These variations are standing waves.
A measure of this discrepancy is the standing wave ratio (SWR). SWR is usually expressed as the ratio of the maximum and minimum direct and reverse current or voltage values along the line:
SWR = I max /I min = V max /V min
Others more in a simple way Express SWR is the ratio of the characterizing impedance of the transmission line (Zo) to the impedance of the antenna (R):
SWR = Z o /R or R/Z o
depending on which impedance is greater.
The ideal SWR is 1:1. An SWR of 2 to 1 indicates a reflected power of 10%, meaning that 90% of the transmitted power arrives at the antenna. An SWR of 2:1 is generally considered the maximum allowable for the most efficient system operation.
Potential energy of a hard dipole
Let's consider the so-called rigid dipole - this is a dipole in which the distance between the charges does not change ($l=const$). Let us determine what the potential energy that a dipole has in an external electrostatic field. If the charge $q$, which is located at a field point with potential $\varphi $, has a potential energy equal to:
then the dipole energy is equal to:
where $(\varphi )_+;(\varphi )_-$ are potentials external field at the points where the charges $q$ and $-q$ are located. The electrostatic field potential decreases linearly if the field is uniform in the direction of the field strength vector. Let's direct the X axis along the field (Fig. 1). Then we get:
From Fig. 1 we see that the change in potential from $(\varphi )_+to\ (\varphi )_-$ occurs on the segment $\triangle x=lcos \vartheta$, therefore:
Electric dipole moment
Let's substitute (4) into (2), we get:
where $\overrightarrow(p)$=$q\overrightarrow(l)$ is the electric moment of the dipole. Equation (6) does not take into account the interaction energy of dipole charges. Formula (6) was obtained under the condition that the field is homogeneous; however, it is also valid for an inhomogeneous field.
Example 1
Assignment: Consider a dipole that is located in a non-uniform field that is symmetrical about the X axis. Explain how a dipole will behave in such a field from the point of view of the forces acting on it.
Let the center of the dipole lie on the X axis (Fig. 2). The angle between the dipole arm and the X axis is equal to $\vartheta \ne \frac(\pi )(2)$. In our case, the forces are $F_1\ne F_2$. A rotational moment will act on the dipole and
the force that tends to move the dipole along the X axis. To find the modulus of this force, we use the formulas:
In accordance with the equation for the potential energy of a dipole, we have:
we assume that $\vartheta=const$
For the X axis points we have:
\ \
At $\vartheta 0$, this means that the dipole is drawn into the region of a stronger field. For $\vartheta >\frac(\pi )(2)$ $F_x
Note that if $-\frac(\partial W)(\partial x)=F_x$, the derivative of the potential energy gives the projection of the force on the corresponding axis, then the derivative $-\frac(\partial W)(\partial \vartheta) =M_\vartheta$ gives the projection of the torque onto the $?$ axis:
\[-\frac(\partial W)(\partial \vartheta)=M_\vartheta=-pEsin \vartheta (1.4.)\]
In formula (1.4), the minus sign means that the moment tends to reduce the angle between the electric moment of the dipole and the field strength vector. A dipole in an electric field tends to rotate so that the electric moment of the dipole is parallel to the field ($\overrightarrow(p)\uparrow \uparrow \overrightarrow(E)$). At $\overrightarrow(p)\uparrow \downarrow \overrightarrow(E)$ the torque will also be zero, but such an equilibrium is not stable.
Example 2
Assignment: Two dipoles are located at a distance $r$ from each other. Their axes lie on the same straight line. The electric moments are equal, respectively: $p_1$ and $p_2$. Calculate the potential energy of any of the dipoles that will correspond to the stable equilibrium position.
The system will be in equilibrium when the dipoles are oriented as shown in Fig. 3, along the field, with charges opposite in sign to each other.
We will assume that the field creates a dipole with moment $p_1$; we will look for the potential energy of a dipole that has an electric moment $p_2$ at the field point (A) at a distance r from the first dipole. Let us assume that the arms of the dipole are small compared to the distance between the dipoles ($l\ll r$). Dipoles can be taken as point ones (so we assume that a dipole with moment $p_2\ is\ located\ at\ point\ A$). The field strength that creates the dipole on its axis at point A is equal in absolute value (at $\varepsilon =1$):
The potential energy of a dipole with moment $p_2$ at point A can be expressed by the formula:
where we took into account that the vectors of tension and electric moment of the dipole are co-directed in a state of stable equilibrium. In this case, the potential energy of the second dipole will be equal to:
Answer: The potential energies of the dipoles will be equal in value to $W=-p_2\frac(p_1)(2\pi (\varepsilon )_0r^3)$.
A. B. Rybakov,
, Military Space Cadet Corps, St. Petersburg
Dipole in field and dipole field
Basic questions of electrostatics: What field does this distribution of charges create and what force acts on these charges in an external field? Regarding a point charge, these questions are solved by well-known school formulas. The next important and simple object of electrostatics is, of course, a dipole. A dipole is two opposite, equal-sized point charges located at a fixed distance l from each other. A dipole is characterized by a dipole moment p = qL (1)Where l – a vector directed from a negative charge to a positive one.
Interest in the dipole is associated, in particular, with the fact that the molecules of many substances have a dipole moment, and in addition, the molecules of all substances acquire a dipole moment in an external electric field. And macroscopic bodies (both conducting and non-conducting current) in an external field are polarized, i.e. acquire a dipole moment. The most important applications of the results presented here are to fields in dielectrics.
Let's pose the most obvious questions in the stated topic and try to resolve them. We don’t need any special mathematics beyond the scope of the school course.
The derivative of the function Ф(х) will be denoted by dФ/dх. For the convenience of writing some results, we will use the scalar product of vectors.
Let us remind you that a b= a · b · cos α, where α is the angle between the vectors. We denote the dimensional constant in Coulomb's law
Dipole in the field (simple problems)
1 . What forces act on a dipole in a uniform electric field? 
Let the dipole p is in a field of tension E, let the dipole moment vector make an angle α with the field strength vector. It is easy to see that the dipole in this case is acted upon by a pair of forces with a moment
M = qElsin α = pEsin α, which tends to orient the dipole along the field lines. So if the dipole can rotate, then it will orient itself in the indicated way. Note that the dipole also has another equilibrium position when it is oriented in the opposite way, but this position is unstable.
2. What is the energy of a dipole in a uniform field?
As always, in problems where we are talking about potential energy, we must first agree on where we will measure this energy from. Let us count it from the equilibrium position indicated above. Then energy is the work that the field forces do when the dipole rotates around its center from the initial position, characterized by angle α (see figure to paragraph 1), to the equilibrium one. Recall that work is associated only with the movement of a charge along the direction E. The charges of the dipole with such rotation will shift along the field lines (in different directions) by l (1– cos α)/2. Therefore, the desired energy is W = qEl (1 – cos α) = pE (1 – cos α).
But more often in electricity textbooks they prefer to assume in this problem that W = 0 in the position of the dipole when the vector p perpendicular E. In this case
W = –qEl cos α = –pE.
The statement made at the end of paragraph 1 can now be formulated differently: the dipole now tends to occupy a position with minimal energy. Thus, the dipole molecules of a dielectric in an external field tend to all be oriented in the indicated manner (and thermal motion prevents them from doing this). 
3. Now let the dipole, oriented along the field lines, be in a non-uniform field. Then, as is easy to see, a force acts on it along the field lines, directed in the direction of increasing the field strength:
(the indices “+” and “–” mark the charge of the dipole to which the corresponding physical quantity belongs). It is this force that explains the simplest experiment in which a charged body (regardless of the sign of the charge) attracts small pieces of paper.
Dipole field 
4 . Before we start calculating the dipole field, let's look at some general points. Let, for example, we are interested in the gravitational field of some irregularly shaped asteroid. The field in the immediate vicinity of the asteroid can only be obtained by computer calculation. But the further we move away from the asteroid, the more accurately we can consider it as a material point (the field of which we know). In striving for greater mathematical rigor, it was necessary to say that we know the asymptotic behavior of the field at
We encounter a similar situation in an electrostatic field. The electrostatic field is very similar in its properties to the gravitational field (because the fundamental laws are similar: Coulomb’s law and the law of universal gravitation), but, so to speak, “richer” than it. After all, electric charges can be of two types, between them both attraction and repulsion are possible, and between “gravitational charges” (i.e. masses) only attraction is possible.
We will assume that positive and negative point charges q 1 , q 2 , … , q n are distributed in some limited area. Full system charge
(2)
We already understand that for Q ≠ 0 the field at large r transforms into the field of a point charge Q. But a very important question for us arises: what will the field be like at large distances if the total charge
Q = 0? The simplest distribution of point charges with Q = 0 is a dipole. That is why the study of the dipole field contains important fundamental points.
So, we will be mainly interested in such situations when all the characteristic dimensions r are very large compared to the distance l between the dipole charges. This situation can be described in two ways. Firstly, we can always keep in mind that the charges are located at a finite distance l from each other, and be interested in the behavior of the obtained solutions for But we can simply talk about a point dipole with a certain dipole moment p, then all our results are valid for any r > 0 (these two points of view are, of course, equivalent).
We will use formulas known to everyone for the fields of point charges and in the resulting expressions we will take into account that l is small. Therefore, let us recall the formulas for approximate calculations: if , then 
Everywhere in the calculations, the sign “≈” will indicate that we used these formulas in the case of a small parameter (the small parameter in the problems under consideration is l/r).
5 . A high-quality picture of the field lines of a dipole field is well known and is given in many textbooks, and we will not present it here. Although calculating the field at an arbitrary point is not difficult, we will still limit ourselves to calculating the potential and intensity along two selected directions. Let's align the origin of the coordinate system with the center of the dipole, and direct the x axis along the vector p
, and the Y axis is perpendicular (in this case, the dipole charges are separated from the origin by a distance ). We will assume that at an infinitely distant point
6. Calculate the dipole field strength on the Y axis.
According to the principle of superposition, E = E + + E –, Where E+ And E –– vectors of field strength of individual charges. From the similarity of triangles: 
which can be written as 
Now let's talk about the course of the potential along the Y axis. Since at any point on the Y axis the vector E
is perpendicular to the axis, then when some charge moves along this axis, the dipole field does not do any work, and therefore, at any point on this axis
7. Let's calculate the field potential j at an arbitrary point on the x axis. According to the principle of superposition, it is equal to the sum of the potentials created by positive and negative charges.
Let x > 0, then:
(3)
(expression for (x) for x< 0 будет c другим знаком). 
From the symmetry of the problem it is clear that on the x axis the field strength vector E has only the E x component. It can be calculated based on the well-known formula connecting field strength and potential: 
(4)
but in school courses formula (4) is usually ignored, so let’s calculate Ex directly: or 
So, when moving away from the dipole along the x-axis or along the y-axis, the field decreases as r –3. It can be proven that the field behaves in the same way in any direction.
We present the expression for the potential at an arbitrary point without derivation: 
(i.e. when deleting
In any direction except the Y axis, the potential decreases as r –2). Make sure that in special cases this formula leads to results already known to us.
8. Retreat. Let us recall that for an infinite uniformly charged plane the field strength does not depend on the distance from the plane (or, if you prefer, falls off as r 0). For a point charge, it decreases as r –2. For a dipole, as we found out, it decreases at infinity as r –3. Try to guess for which charge distribution the field strength decreases as r –1 ; r –4.
Interaction of a dipole with other charges
9. Now consider the interaction of a dipole and a point charge q′ (let q′ > 0). The figure largely repeats the figure in paragraph 5. There we calculated the dipole field strength and, therefore, already know what force acts on a point charge. Note that this interaction gives us the simplest example of non-central forces (remember where non-central forces between particles occur in the school course).
But there are still questions: what force acts on the dipole? where is it attached? You can answer these questions immediately, without hesitation. The required force F, according to Newton's third law, must be equal to – F ′ and must be applied on the same straight line with F ′. Perhaps someone will be surprised that the resultant of the two forces acting on the charges +q and –q of the dipole turned out to be applied somewhere away from the dipole. What does it mean? Does not mean anything. What does it mean that the resultant of the gravitational forces acting on the donut is applied at the center of the hole? The resultant of two forces does not have any special meaning; it simply replaces in all respects several (or even countless) forces in the fundamental equations of mechanics. (For the sake of objectivity, we note that there are very well-known authors for whom such a point of view is unacceptable. They prefer to say that the dipole from the side of a point charge is acted upon by a force applied to the dipole itself, and also by a moment of force).
10 . Find the force and energy of interaction of two dipoles whose vectors p 1 and p 2 lie on the same straight line. Distance between dipoles x.
Let's calculate the total energy of the charges of the second dipole in the field of the first (see paragraph 7): 
It is clear that dipoles facing each other with opposite poles (as in the figure) attract (this corresponds to the “–” sign in the expression for W); when one of the dipoles flips, the energy will change sign.
We will no longer reproduce rather monotonous calculations and immediately write out an expression for the magnitude of the interaction force of these dipoles (check it!): 
11. Find the energy of interaction of two dipoles for which p 1 lies on the straight line connecting the dipoles, and p 2 is perpendicular to it. Distance between dipoles x. (Check yourself - the answer is obvious.) 
12 . Find the interaction energy of two dipoles whose vectors p 1 and p 2 are parallel to each other and both are perpendicular to the x axis on which the dipoles are located. 
Additional Notes
13. So, a dipole gives us the simplest example of a system of charges with a total charge Q = 0. As we have seen, the field potential of a dipole at large distances from it decreases as r –2. Is it possible to generalize this result to a more general case?
The concept of dipole moment can be generalized so that it characterizes any charge distribution. In particular, for a system of n point charges, the dipole moment is determined as follows: 
.
(5)
It is easy to see that this quantity is additive. It can be proven that P at Q = 0 does not depend on the choice of origin. Make sure that in a particular case this formula turns into (1).
Calculate the dipole moment P of a number of simple charge distributions (in all cases the distance between nearest charges l).
One could also talk about continuous charge distributions, but then instead of the sums in (2) and (5), one would have to write integrals over the volume.
The above results tell us what the significance of the dipole moment is. And indeed, it is possible to general case prove that the further we move away from an arbitrary system of charges with a total charge Q = 0 and a dipole moment P ≠ 0, the closer its field will be to the field of an elementary dipole with a dipole moment P considered by us.
One could go further along this path and consider the field of a system of charges with Q = 0 and P = 0. One of the most simple examples such a system is shown in Fig. a is the so-called quadrupole. The quadrupole field potential decreases at infinity as r –3.
The series “point charge – dipole – quadrupole...” can be continued further. The general name for such objects is multipole. But we'll stop there. 
14. When an atom is placed in an electric field, the forces applied to the nucleus and to the electron shell are directed in different directions. Under the influence of these forces, the atom acquires a dipole moment R, coinciding in direction with the direction of the external field strength E
0 .
Of course, molecules also acquire a dipole moment in an external field (but for them, generally speaking, the previous statement about the direction of the vector R
).
But many molecules have dipole moments even in the absence of an external field. Moreover, these intrinsic dipole moments usually far exceed the induced moments (if we talk about ordinary fields achievable in the laboratory). For many processes in nature (in particular, for the existence of life), it is extremely important that a water molecule has a dipole moment.
“It is difficult to imagine what the world would be like if the atoms in the H 2 O molecule were arranged in a straight line, as in the CO 2 molecule; there would probably be no one to observe this” (E. Parcell. Electricity and Magnetism. - M., 1975).
Answers
To paragraph 8. A system of charges in which the field strength decreases at infinity as r –1 is an infinite uniformly charged thread.
To paragraph 11. When the first dipole moves along the x axis, its charges are acted upon by forces perpendicular to this axis from the second dipole, i.e. no work is done in this case, which means W = 0.
To paragraph 12. To simplify the calculation, we need to successfully choose a method for transferring one of the dipoles from infinity to the state of interest to us. It is convenient to first move it along the x-axis, orienting its dipole moment vector along the axis (in this case, the work of the dipole interaction forces is zero), and then rotate it by 90°. When rotating the second dipole, external forces must do work (see paragraph 2). 
This is the energy of interaction between dipoles.
To paragraph 13. The dipole moments are equal to: a) 0; b) 2qlj ;
c) 0; d) –3qli (here i and j are unit vectors in the directions of the X and Y axes, respectively).
Loop vibrators of the "D" series (the closest foreign analogue of ANT150D from Telewave) are made in disassembled form from three parts - the loop vibrator itself (1), the traverse (2) and the mounting unit (3) (see figure).
The loop vibrator is made of thick-walled aluminum pipe and has a length of about ?/2. The attachment point (4) to the traverse is welded using argon-arc welding, which guarantees reliable electrical contact at the current antinode. To match the 50-ohm cable, a 1/4-wave transformer is used; thanks to the power line laid inside the dipole, the antenna is balanced.
All contacts are soldered, and screw connections are painted over. The entire power supply unit is sealed: a PVC tube is used to impart rigidity, and a heat-shrinkable tube is used for sealing together with molecular adhesive-sealant (5). The entire antenna is protected from aggressive environments by a polymer coating. Antenna traverse - a pipe with a diameter of 35 mm is carefully adjusted to the dipole to facilitate installation of the antenna. The attachment point to the mast is cast silumin. Additional processing also ensures reliable docking with the traverse and easy attachment to a mast with a diameter of 38-65 mm at any angle. The antenna has a mark (6) for correct phasing, as well as a drainage hole (7) at the bottom of the vibrator.
The antenna uses domestic cable (8) RK 50-7-11 with low losses (0.09 dB/m at 150 MHz). The antennas are equipped with N-type connectors (9), which are carefully soldered and sealed.
Convenient cardboard packaging allows you to transport the antenna by any means of transport.
Loop dipoles of the "DP" series have some design differences from the dipoles of the "D" series.
Firstly, this antenna has a non-separable design - the dipole itself (10) is welded to a short crossarm (11). The dipole's power supply is asymmetrical, which, however, does not impair its characteristics at all. Due to the close location to the reflector mast, the band is somewhat narrower and amounts to 150-170 MHz, and the radiation level back is 10 dB lower. But in the main direction the gain is 3 dBd.
Secondly, the attachment to the mast is made with lightweight galvanized steel clamps (12) and allows you to attach the antenna to a mast (13) with a diameter of 25-60 mm. In all other respects, in terms of manufacturing technology, the “DP” series antennas do not differ from the “D” series dipoles.
Dipoles of the "DH" series are the cheapest antennas. They are a DIY kit where you can assemble a classic linear grounded vibrator with gamma matching within a few minutes using our instructions. The kit includes the emitter itself - a pin with a diameter of 12 mm (14), a traverse (15) with a hole for fastening and a welded bracket with a connector (16).
The gamma matcher parts allow you to adjust the dipole almost perfectly at any frequency you choose (using a conventional reflectometer).
Each dipole is equipped detailed instructions on settings and graphs of vibrator lengths.
In the hands of a master, this set will turn into a real connected, highly efficient antenna system!
Let us consider the field of the simplest system of point charges. The simplest system point charges is an electric dipole. An electric dipole is a collection of two point charges equal in magnitude but opposite in sign. –q And +q, shifted relative to each other by some distance. Let be the radius vector drawn from the negative charge to the positive one. Vector
is called the electric moment of the dipole or dipole moment, and the vector is called the dipole arm. If the length is negligible compared to the distance from the dipole to the observation point, then the dipole is called a point dipole.
Let's calculate the electric field of an electric point dipole. Since the dipole is a point one, it makes no difference, within the limits of calculation accuracy, from which point of the dipole the distance is measured r to the observation point. Let the observation point A lies on the continuation of the dipole axis (Fig. 1.13). In accordance with the principle of superposition for the intensity vector, the electric field strength at this point will be equal to
it was assumed that , .
In vector form
where and are the field strengths excited by point charges –q and + q. From Fig. 1.14 it is clear that the vector is antiparallel to the vector and its modulus for a point dipole is determined by the expression
It is taken into account here that under the assumptions made.
In vector form, the last expression will be rewritten as follows
Doesn't have to be perpendicular JSC passed through the center of a point dipole. In the accepted approximation, the resulting formula remains true even when beyond the point ABOUT any dipole point is accepted.
The general case is reduced to the analyzed special cases (Fig. 1.15). Let's lower it from the charge + q perpendicular CD to the surveillance line VA. Let's put it on point D two point charges + q And –q. This will not change the fields. But the resulting set of four charges can be considered as a set of two dipoles with dipole moments and . We can replace the dipole with the geometric sum of dipoles and . Now applying to dipoles the previously obtained formulas for the intensity on the extension of the dipole axis and on the perpendicular restored to the dipole axis, in accordance with the principle of superposition we obtain:
Considering that , we get:
here it is used that .
Thus, characteristic of the electric field of a dipole is that it decreases in all directions in proportion to , that is, faster than the field of a point charge.
Let us now consider the forces acting on a dipole in an electric field. In a uniform field charges + q And –q will be under the influence of forces equal in magnitude and opposite in direction (Fig. 1.16). The moment of this pair of forces will be:
The moment tends to rotate the dipole axis to the equilibrium position, that is, in the direction of the vector. There are two equilibrium states of a dipole: when the dipole is parallel to the electric field and when it is antiparallel to it. The first position will be stable, but the second will not, since in the first case, with a small deviation of the dipole from the equilibrium position, a moment of a pair of forces will arise, tending to return it to its original position; in the second case, the resulting moment takes the dipole even further from the equilibrium position.
Gauss's theorem
As mentioned above, it was agreed to draw the lines of force with such density that the number of lines piercing a unit of surface perpendicular to the lines of the site would be equal to the modulus of the vector. Then, from the pattern of tension lines, one can judge not only the direction, but also the magnitude of the vector at various points in space.
Let us consider the field lines of a stationary positive point charge. They are radial lines extending from the charge and ending at infinity. Let's carry out N such lines. Then at a distance r from the charge, the number of lines of force intersecting a unit surface of a sphere of radius r, will be equal. This value is proportional to the field strength of a point charge at a distance r. Number N you can always choose such that the equality holds
where . Since the lines of force are continuous, the same number of lines of force intersect a closed surface of any shape enclosing the charge q. Depending on the sign of the charge, the lines of force either enter this closed surface or go outside. If the number of outgoing lines is considered positive and the number of incoming lines negative, then we can omit the modulus sign and write:
| . | (1.4) |
Tension vector flow. Let us place an elementary pad with area . The area must be so small that the electric field strength at all its points can be considered the same. Let's draw a normal to the site (Fig. 1.17). The direction of this normal is chosen arbitrarily. The normal makes an angle with the vector. The flow of the electric field strength vector through a selected surface is the product of the surface area and the projection of the electric field strength vector onto the normal to the area:
where is the projection of the vector onto the normal to the site.
Since the number of field lines piercing a single area is equal to the modulus of the intensity vector in the vicinity of the selected area, the flow of the intensity vector through the surface is proportional to the number of field lines crossing this surface. Therefore, in the general case, the flow of the field strength vector through the area can be visually interpreted as a value equal to the number of field lines penetrating this area:
| . | (1.5) |
Note that the choice of the direction of the normal is conditional; it can be directed in the other direction. Consequently, the flow is an algebraic quantity: the sign of the flow depends not only on the configuration of the field, but also on the relative orientation of the normal vector and the intensity vector. If these two vectors form an acute angle, the flux is positive; if it is obtuse, the flux is negative. In the case of a closed surface, it is customary to take the normal outside the area covered by this surface, that is, to choose the outer normal.
If the field is inhomogeneous and the surface is arbitrary, then the flow is defined as follows. The entire surface must be divided into small elements with area , calculate the stress fluxes through each of these elements, and then sum the fluxes through all elements:
Thus, field strength characterizes the electric field at a point in space. The intensity flow does not depend on the value of the field strength at a given point, but on the distribution of the field over the surface of a particular area.
Electric field lines can only begin on positive charges and end on negative ones. They cannot begin or end in space. Therefore, if there is no electric charge inside a certain closed volume, then the total number of lines entering and exiting this volume must be zero. If more lines leave the volume than enter it, then there is a positive charge inside the volume; if there are more lines coming in than coming out, then there must be a negative charge inside. When the total charge inside the volume is equal to zero or when there is no electric charge in it, the field lines penetrate through it, and the total flux is zero.
These simple considerations do not depend on how electric charge distributed within the volume. It can be located in the center of the volume or near the surface that bounds the volume. A volume can contain several positive and negative charges distributed within the volume in any way. Only the total charge determines the total number of incoming or outgoing voltage lines.
As can be seen from (1.4) and (1.5), the flow of the electric field strength vector through an arbitrary closed surface enclosing the charge q, equal to . If inside the surface there is n charges, then, according to the principle of field superposition, the total flux will be the sum of the fluxes of field strengths of all charges and will be equal to , where in this case we mean the algebraic sum of all charges covered by the closed surface.
Gauss's theorem. Gauss was the first to discover the simple fact that the flow of the electric field strength vector through an arbitrary closed surface must be associated with the total charge located inside this volume.
