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Design a two-element dipole array that will radiate equal intensities in the \(\phi=0, \pi / 2, \pi\), and \(3 \pi / 2\) directions in the \(H\) plane. Specify the smallest relative current phasing, \(\xi\), and the smallest element spacing, \(d\).

Short Answer

Expert verified
Short Answer: To design a two-element dipole array that radiates equal intensities at given directions, the smallest relative current phasing (ξ) is 0, and the smallest element spacing (d) is \(\lambda / 2\).

Step by step solution

01

Array Factor

A two-element dipole array's array factor in the H-plane is given by the following equation: Array Factor (AF) = \(1 + \cos(\beta d \sin \phi + \xi)\) Where \(\beta = 2\pi / \lambda\) is the phase constant, \(d\) is the element spacing, and \(\xi\) is the phase difference between the currents of the two elements.
02

Setting Array Factor values for given directions

We need to find the values of \(d\) and \(\xi\) for which the array factor would have equal intensities in the directions \(\phi = 0, \pi / 2, \pi\), and \(3 \pi / 2\). For \(\phi = 0\): AF(\(0\)) = \(1 + \cos(\beta d \sin 0 + \xi)\) For \(\phi = \pi / 2\): AF(\(\pi / 2\)) = \(1 + \cos(\beta d \sin(\pi / 2) + \xi)\) For \(\phi = \pi\): AF(\(\pi\)) = \(1 + \cos(\beta d \sin \pi + \xi)\) For \(\phi = 3\pi / 2\): AF(\(3 \pi / 2\)) = \(1 + \cos(\beta d \sin(3 \pi / 2) + \xi)\)
03

Solving for ξ and d

To find ξ and d that satisfy the above conditions, we will equate the array factors and then solve the equations. Equating array factors for \(\phi = 0\) and \(\phi = \pi\): \(1 + \cos(\beta d \sin 0 + \xi) = 1 + \cos(\beta d \sin \pi + \xi)\) \(\cos(\xi) = \cos(\beta d - \xi)\) This gives us the first equation: \(\xi = \beta d - \xi\) Now, equating array factors for \(\phi = \pi / 2\) and \(\phi = 3\pi / 2\): \(1 + \cos(\beta d \sin(\pi / 2) + \xi) = 1 + \cos(\beta d \sin(3 \pi / 2) + \xi)\) \(\cos(\beta d + \xi) = \cos(-\beta d + \xi)\) This gives us the second equation: \(\beta d + \xi = -\beta d + \xi\) Let's solve the two equations: 1. \(\xi = \beta d - \xi\) 2. \(\beta d + \xi = -\beta d + \xi\) Adding equations 1 and 2: \(2\xi = 0\) Solving for ξ, we obtain \(\xi = 0\). Substituting ξ back into either Equation 1 or 2, we get: \(\beta d = 0\) As the phase constant \(\beta\) is non-zero, it means that \(d = 0\). However, in a two-element dipole array, the element spacing (d) cannot be zero. Therefore, we should look for the "smallest non-zero" value of d. This occurs when the phase difference between the adjacent elements is an integral multiple of \(2\pi\). Hence, \(\beta d = n (2\pi)\), where \(n\) is an integer and \(n \neq 0\). For the "smallest" value of \(d\), let's take \(n = 1\). Therefore, \(d = \frac{2\pi}{\beta} = \frac{\lambda}{2}\) So, the smallest relative current phasing, ξ, is 0, and the smallest element spacing, d, is \(\lambda / 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Array Factor
The array factor is a crucial concept when designing antenna arrays, particularly for achieving specific radiation patterns. In a two-element dipole array, the array factor describes how the combination of multiple individual antennas, or elements, alter the overall radiation pattern produced by the array.

Mathematically, the array factor for a two-element dipole array in the H-plane can be expressed as \[AF = 1 + \cos(\beta d \sin \phi + \xi)\] where
  • \(\beta\) is the phase constant,
  • \(d\) is the spacing between elements, and
  • \(\xi\) is the phase difference between the two currents driving the antennas.
This factor helps to determine how the elements' constructive and destructive interference at different angles achieves desired radiation intensities.

By tweaking these parameters \(\beta\), \(d\), and \(\xi\), engineers can direct antenna beams towards or away from specific directions.
Phase Constant
The phase constant is an essential parameter in understanding wave propagation and antenna behavior. Denoted by \(\beta\), it defines the phase shift per unit length in the medium the wave is traveling through.

For free space, the phase constant is determined by the equation \[\beta = \frac{2\pi}{\lambda}\]where
  • \(\lambda\) is the wavelength of the signal being transmitted.
The phase constant is integral in the formula for the array factor and influences how light or radio waves propagate from each antenna element.

Understanding \(\beta\) is crucial because it directly affects how antenna array elements are spaced and phased to produce the desired pattern of radiation.
Element Spacing
Element spacing \(d\) refers to the distance between the centers of the two dipole elements. This distance plays a significant role in shaping the radiation pattern of the antenna array.

In the context of our exercise, it is important to find the smallest element spacing \(d\) that would ensure equal intensities in certain desired directions, specifically \(\phi = 0, \pi / 2, \pi\), and \(3 \pi / 2\).

Through calculations, the smallest non-zero value for \(d\) is determined when it equals half of the wavelength, or \[d = \frac{\lambda}{2}\]This spacing allows the array to maintain equal intensity in all specified directions.

Spacing of \(\lambda/2\) is a standard value used in many antenna designs since it offers good performance while avoiding grating lobes, which are unwanted, repeated patterns in the radiation.
Current Phasing
Current phasing \(\xi\) is a critical factor in controlling the directivity of the dipole array's radiation pattern. It refers to the relative difference in phase between the currents driving two elements in an antenna array.

To achieve specific radiation patterns where equal intensity is required in particular directions, properly setting the current phasing is crucial.

In this exercise, the smallest relative current phasing necessary to achieve the desired pattern is found to be \[\xi = 0\]This means that the currents feeding both dipole elements are in phase with each other.

Selecting the correct \(\xi\) while considering other array parameters like phase constant \(\beta\) and element spacing \(d\) ensures the tailored design of the radiated pattern.

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Most popular questions from this chapter

A dipole antenna in free space has a linear current distribution with zero current at each end, and with peak current \(I_{0}\) at the enter. If the length \(d\) is \(0.02 \lambda\), what value of \(I_{0}\) is required to \((a)\) provide a radiation-field amplitude of \(100 \mathrm{mV} / \mathrm{m}\) at a distance of \(1 \mathrm{mi}\), at \(\theta=90^{\circ} ;(b)\) radiate a total power of \(1 \mathrm{~W} ?\)

Write the Hertzian dipole electric field whose components are given in Eqs. (15) and (16) in the near zone in free space where \(k r<<1\). In this case, only a single term in each of the two equations survives, and the phases, \(\delta\) and \(\delta_{\theta}\), simplify to a single value. Construct the resulting electric field vector and compare your result to the static dipole result (Eq. (36) in Chapter 4). What relation must exist between the static dipole charge, \(Q\), and the current amplitude, \(I_{0}\), so that the two results are identical?

A short current element has \(d=0.03 \lambda\). Calculate the radiation resistance that is obtained for each of the following current distributions: \((a)\) uniform, \(I_{0} ;(b)\) linear, \(I(z)=I_{0}(0.5 d-|z|) / 0.5 d ;(c)\) step, \(I_{0}\) for \(0<|z|<0.25 d\) and \(0.5 I_{0}\) for \(0.25 d<|z|<0.5 d\).

Consider a lossless half-wave dipole in free space, with radiation resistance, \(R_{\mathrm{rad}}=73\) ohms, and maximum directivity \(D_{\max }=1.64\). If the antenna carries a 1-A current amplitude, \((a)\) how much total power (in watts) is radiated? \((b)\) How much power is intercepted by a \(1-\mathrm{m}^{2}\) aperture situated at distance \(r=1 \mathrm{~km}\) away? The aperture is on the equatorial plane and squarely faces the antenna. Assume uniform power density over the aperture.

In a linear endfire array of \(n\) elements, a choice of current phasing that improves the directivity is given by the Hansen-Woodyard condition: $$ \xi=\pm\left(\frac{2 \pi d}{\lambda}+\frac{\pi}{n}\right) $$ where the plus or minus sign choices give maximum radiation along \(\phi=180^{\circ}\) and \(0^{\circ}\), respectively. Applying this phasing may not necessarily lead to unidirectional endfire operation (zero backward radiation), but it will do so with the proper choice of element spacing, \(d\). (a) Determine this required spacing as a function of \(n\) and \(\lambda .(b)\) Show that the spacing as found in part \((a)\) approaches \(\lambda / 4\) for a large number of elements. \((c)\) Show that an even number of elements is required.

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