1.5 The Short Dipole – FREE SAMPLE

The Hertzian Dipole as an Idealization

In deriving the field expressions for the Hertzian dipole, we assumed it to be an infinitesimally short filament of current. This implies that its length, $dl$, is vanishingly small compared to the wavelength, $\lambda$ (i.e., $dl \ll \lambda$). However, a critical assumption in this model is that the current, $I$, is uniform along its entire length.

This assumption introduces a physical contradiction. A fundamental boundary condition for any real conductor of finite length is that the current must be zero at its open ends. The Hertzian dipole model, which assumes a constant current along its length, violates this physical requirement. Consequently, the Hertzian dipole is an idealization that does not physically exist in open-ended wire antennas.

Its true value lies in its use as a mathematical building block. The Hertzian dipole is treated as a differential current element, $I dl’$, located at a source point $\vec{r}’$. This elemental source generates infinitesimal field contributions, $d\vec{E}$ and $d\vec{H}$, at an observation point $\vec{r}$. By integrating these contributions over the actual length and current distribution of a physical antenna, we can accurately determine its total radiated fields. This powerful technique forms the foundation for analyzing more realistic antenna models, including the short dipole discussed next.

The Transmission Line Analogy for Current Distribution

A real dipole of finite but electrically short length (i.e., length $L \ll \lambda$) must have a current distribution that decreases to zero at its ends. For a center-fed dipole, we can expect the current to be maximum at the central feed point. To deduce the shape of this current distribution, we can employ the transmission line analogy, an insightful model attributed to the work of Sergei Schelkunoff.

Consider a short, open-circuited, two-wire transmission line fed at one of its ends. When the line is much shorter than the wavelength, the standing wave of current represents only a small portion of a full sinusoid. In this case, the sinusoidal current distribution can be accurately approximated by a straight line. From the feed point to the open ends, where the current must be zero due to the open-circuit boundary condition, the current magnitude increases approximately linearly. Thus, the current on each wire rises from zero at the open end to a maximum at the feed point.

Now, imagine physically transforming this transmission line into a dipole antenna. As depicted in Fig. 1, we can flare out the parallel conductors by 180°, keeping the feed point fixed, until they form a single, linear element. The transmission line analogy posits that this physical change does not significantly alter the fundamental shape of the current distribution. It is therefore reasonable to model the current on a center-fed short dipole with a triangular distribution: maximum at the center and decreasing linearly to zero at the two ends.

A crucial distinction exists between the two structures. In the transmission line, the currents in the two conductors are anti-parallel (flowing in opposite directions), causing their radiated fields to largely cancel each other, which prevents significant radiation. In the dipole, the currents on the upper and lower arms are collinear and flow in the same direction. This alignment causes their radiated fields to add constructively, enabling the antenna to radiate power effectively.

Far-Field Radiation from a Short Dipole

To analyze the radiated fields of a short dipole, we begin by retaining only the $1/r$ terms from the Hertzian dipole expressions. These terms represent the radiating (far-field) components. By integrating them along the length of the dipole, we obtain the total radiated fields:

$\displaystyle E_\theta = \frac{j k}{4\pi} \, Z_{w} \, \int_{-L/2}^{L/2} I(z’) \, \frac{\exp{(-j k r)}}{r} \, \sin(\theta)\, dz’ \qquad (1)$

$\displaystyle H_\phi = \frac{j k}{4\pi} \, \int_{-L/2}^{L/2} I(z’) \, \frac{\exp{(-j k r)}}{r} \, \sin(\theta)\, dz’ \qquad (2)$

We assume the dipole of length $L$ is aligned along the z-axis and centered at the origin. Therefore, we can use the same coordinate system and expressions derived for the Hertzian dipole without any transformation. Since we are only considering the $1/r$ terms, the radial component of the electric field, $E_r$, vanishes. This is consistent with the behavior of radiated waves in the far-field region, where the wavefront is locally planar and the fields are transverse electromagnetic (TEM). In other words, no field component can point in the radial direction in the far field, this is a boundary condition at infinity.

By dividing equation (1) by equation (2), we obtain the wave impedance, $Z_w = E_\theta/H_\phi$, which is approximately equal to 377 Ω in free space.

In the far-field zone, the distance $r$ between the observation point and any point along the dipole is sufficiently large that we can treat $r$ as constant with respect to the integration variable $z′$, which runs along the length of the dipole. Since the dipole is electrically short compared to the wavelength, and $k = 2\pi/\lambda$, phase variations due to the exponential factor $\exp(-j k r)$ can also be neglected. Consequently, all multiplicative factors in equations (1) and (2) can be considered constant during the integration, and we are left with computing only the integral of the current distribution $I(z’)$ over the length of the dipole:

$\displaystyle E_\theta = \frac{j k}{4\pi} \, Z_{w} \, \frac{\exp{(-j k r)}}{r} \, \sin(\theta) \, \int_{-L/2}^{L/2} I(z’) \, dz’ \qquad (3)$

$\displaystyle H_\phi = \frac{j k}{4\pi} \, \frac{\exp{(-j k r)}}{r} \, \sin(\theta)\, \int_{-L/2}^{L/2} I(z’) \, dz’ \qquad (4)$

In the case illustrated in Fig. 1, there is no need to perform the integral explicitly. The current distribution along the short dipole is triangular, with a peak value $I_0$ at the feedpoint and tapering to zero at the ends. The integral of the current is thus equal to the area under the triangular distribution, which is straightforwardly given by:

$\displaystyle \int_{-L/2}^{L/2} I(z’) \, dz’ = I_0 \, \frac{L}{2}$

Substituting this into the field expressions yields the final far-field expressions for a short dipole:

$\displaystyle E_\theta = \frac{j k}{4\pi} \, Z_{w} \, \frac{I_0 L}{2} \, \frac{\exp{(-j k r)}}{r} \, \sin(\theta) \qquad (5)$

$\displaystyle H_\phi = \frac{j k}{4\pi} \, \frac{I_0 L}{2} \, \frac{\exp{(-j k r)}}{r} \, \sin(\theta) \qquad (6)$

Effective Length and Radiation Equivalence

By comparing equations (5) and (6) with the far-field expressions of a Hertzian dipole of the same physical length $L$, we observe that the fields in (5) and (6) are exactly half the values produced by a Hertzian dipole of length $L$, assuming the same feedpoint current $I_0$.

This leads to an important observation: a short dipole and a Hertzian dipole produce identical far fields if the Hertzian dipole has half the length of the short dipole, given the same current at the feedpoint. In other words, the far-field radiation from a short dipole can be modeled by a Hertzian dipole of a certain equivalent length, referred to as the effective length of the short dipole.

This effective length, denoted $L_e$, is defined as the integral of the current distribution $I(z)$ along the length of the dipole, normalized by the feedpoint current:

$\displaystyle L_e = \frac{1}{I_0} \, \int_{-L/2}^{L/2} I(z) \, dz \qquad (7)$

Thus, the effective length of a short dipole is half its physical length, $L_e = L/2$.

Equation (7) provides a general definition of the effective length and applies regardless of the specific current distribution shape, it does not need to be triangular. Thus, it serves as a fundamental concept in antenna theory, applicable to any short dipole with a known current distribution.

Once the short dipole is replaced conceptually by a Hertzian dipole of effective length $L_e$, subsequent calculations become more straightforward. We can reuse the expressions previously derived for the radiated power and radiation resistance of a Hertzian dipole, simply by substituting the infinitesimal length $dl$ with the effective length $L_e$.

However, it is important to note that this substitution does not hold when analyzing conductor losses, as we will discuss later.

Electric Field Intensity in the Far Field

As we have concluded earlier, the radiated fields of a short dipole can be modeled using the expressions derived for the Hertzian dipole. One key takeaway from this equivalence is that the short dipole exhibits the same donut-shaped $\sin^2(\theta)$ radiation pattern as the Hertzian dipole. This pattern is independent of the dipole’s length, as long as the length remains electrically small compared to the wavelength.

As a result, the directivity of the short dipole is the same as that of the Hertzian dipole:

• $D = 1.5$ (dimensionless), or
• $1.76 \, \text{dBi}$ (decibels relative to an isotropic radiator).

Likewise, the half-power beamwidth remains approximatelly 90 degrees. However, it is important to note that these are idealized values that serve as limiting cases. In practice, since a short dipole has a finite (though small) length, slight numerical deviations from these theoretical values may be observed in simulation or measurement. We will illustrate this with a simple simulation in the next section.

Another immediate result is that the RMS electric field intensity radiated by a short dipole is given by the same expression as that of a Hertzian dipole:

$\displaystyle E = \frac{\sqrt{30 \, P_{\text{in}} \, G}}{r} \, \sin(\theta) \qquad (8)$

Where:

• $E$ is the RMS electric field intensity (in V/m),
• $P_{\text{in}}$ is the input power (in Watts),
• $G$ is the antenna gain (dimensionless),
• $r$ is the distance from the antenna (in meters),
• $\theta$ is the zenith angle.

The calculator below can be used to compute $E$ by entering the required parameters:

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Electric Field Intensity (Short Dipole)

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It is essential to clarify that directivity and gain are not the same. Although the short and Hertzian dipoles have the same directivity, their gains may differ due to conductor losses. These losses depend on the square of the current distribution along the antenna. Therefore, even if both antennas are made of conductors with the same resistivity, their loss resistances will differ, leading to different radiation efficiencies, and hence, different gains.

Exercise:

Calculate the electric field intensity at a distance of 1 km from a short dipole radiating 1 kW of input power, in the H-plane ($\theta = 90^\circ$). Then, compare this value with the field radiated by an isotropic source under the same conditions.

If you express the result in dBμV/m, you will find a difference of 1.76 dB, which matches the directivity in dBi of the short dipole. This exercise provides a concrete demonstration of what is meant by directivity, the ability of an antenna to concentrate radiated energy in specific directions compared to an isotropic radiator.

Simulating the Short Dipole

As previously discussed, the radiation pattern of a short dipole approaches that of a Hertzian dipole as its length tends to zero. As the dipole becomes electrically longer, i.e. its physical length increases relative to the wavelength, deviations from the ideal pattern begin to emerge. One way to quantify this behavior is by observing changes in the directivity.

To explore this, let us simulate a short dipole in AN-SOF. A step-by-step guide for modeling and simulating a cylindrical antenna is provided in the article Modeling a Center-Fed Cylindrical Antenna with AN-SOF, which we recommend reviewing beforehand.

When working with antennas in terms of wavelength, it is convenient to choose a simulation frequency of 300 MHz, for which the wavelength is approximately 1 meter (more precisely, 0.999 m to three significant digits). This makes it easy to scale physical dimensions relative to the wavelength, for example, 1 meter corresponds to 1λ, and 0.01 meters corresponds to 0.01λ.

We begin by setting the frequency to 300 MHz, and then draw a dipole of length 0.01λ in free space. The conductor radius is set to 10⁻⁴λ (or 0.1 mm), making it very thin to minimize any thickness-related effects. Figure 2 summarizes the basic setup: drawing the dipole, dividing it into 3 segments, placing a source at the center, running the simulation, and viewing the results. As expected, the directivity shown in the Results tab is 1.76 dBi (to three significant figures), consistent with the theoretical value. Figure 3 shows the normalized radiation pattern, which exhibits the characteristic donut-shaped distribution of a Hertzian dipole.

To explore how the dipole behavior changes with electrical length, we can increase the frequency instead of the physical length. As frequency increases, the wavelength shortens, making the dipole electrically longer. For instance, performing a frequency sweep from 300 MHz to 3000 MHz results in a 10× frequency increase. At 3000 MHz, the same dipole becomes 0.1λ long, and the simulated directivity increases to 1.78 dBi.

AN-SOF also allows us to plot directivity as a function of frequency, as shown in Figure 4. The resulting upward trend illustrates the departure of the short dipole from the ideal Hertzian behavior. As the dipole becomes longer, the effect becomes more pronounced. However, we should avoid going much beyond 0.1λ in length, since at that point the dipole is no longer considered “short,” and the theoretical assumptions we used so far break down.

In particular, the current distribution begins to deviate from a triangular shape, and phase variations along the dipole length must be taken into account. A more accurate treatment of such cases will be presented in the chapter on linear antennas of arbitrary length.

Exercise:

We leave it as an exercise for the reader to increase the number of segments of the dipole and verify that the current distribution takes a triangular shape.