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For a dipole antenna of overall length \(2 \ell=\lambda\), evaluate the maximum directivity in decibels, and the half-power beamwidth.

Short Answer

Expert verified
The maximum directivity of the dipole antenna is approximately 2.15 dB, and the half-power beamwidth is approximately 120 degrees.

Step by step solution

01

Calculate the Maximum Directivity

The maximum directivity of a dipole antenna can be calculated using the following formula: \(D_{max} = \frac{4 \pi A_{eff}}{\lambda^2}\), where \(A_{eff}\) is the effective aperture of the antenna. For a dipole antenna, the effective aperture for maximum directivity is given by: \(A_{eff} = \frac{\lambda^2}{4\pi}\) The directivity can be expressed in decibels as: \(D_{dB} = 10\log_{10}(D_{max})\) Now let's plug in the values into the formula for directivity: 1. \(A_{eff} = \frac{\lambda^2}{4\pi}\) 2. \(D_{max} = \frac{4 \pi A_{eff}}{\lambda^2} = \frac{4 \pi \cdot \frac{\lambda^2}{4\pi}}{\lambda^2} = 1.64\) 3. \(D_{dB} = 10\log_{10}(1.64) \approx 2.15\,\text{dB}\)
02

Calculate the Half-power Beamwidth

The half-power beamwidth (HPBW) of a dipole antenna can be determined by using the trigonometric relationship between the beamwidth and the overall length of the antenna, which can be given by: \(\sin(\theta_{HPBW}/2) = \frac{\lambda}{2 \ell}\) Let's compute the HPBW: 1. \(\sin(\theta_{HPBW}/2) = \frac{\lambda}{2 \ell} = \frac{\lambda}{2\lambda} = 0.5\) (since \(2\ell = \lambda\)) 2. \(\theta_{HPBW} = 2 \arcsin(0.5) \approx 120^\circ\) The maximum directivity of the dipole antenna is approximately \(2.15\,\text{dB}\), and the half-power beamwidth is approximately \(120^\circ\).

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Key Concepts

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

Maximum Directivity Calculation
Directivity is a crucial concept in antenna theory, relating to how focused the energy radiation pattern is. The maximum directivity of an antenna, expressed as \(D_{max}\), describes the peak value of directivity. It is the ratio of the intensity in a given direction to the intensity averaged over all directions.

For a dipole antenna, maximum directivity is a function of the effective aperture \(A_{eff}\) and wavelength \(\lambda\). Specifically, \(D_{max} = \frac{4 \pi A_{eff}}{\lambda^2}\). When dealing with a dipole antenna with an overall length equal to the wavelength (\(2 \ell = \lambda\)), the \(A_{eff}\) can be simplified to \(\frac{\lambda^2}{4\pi}\), leading to a \(D_{max}\) value of 1.64. This figure represents how much more effectively the antenna radiates power in its favored direction versus an isotropic radiator.
Effective Aperture
The effective aperture, \(A_{eff}\), of an antenna is a measure of its effectiveness in receiving power from a plane wave propagating through space. Essentially, this value tells us the size of an area that an antenna can 'collect' energy from. For a dipole antenna, this is theoretically calculated using the formula \(A_{eff} = \frac{\lambda^2}{4\pi}\).

An increased \(A_{eff}\) means a better ability to receive or transmit energy, which directly impacts the directivity. This relationship is why calculating effective aperture is essential for understanding an antenna's directivity performance. It's worth noting that the actual received power is also influenced by other factors like antenna efficiency and impedance matching.
Half-Power Beamwidth
In antenna terminology, the half-power beamwidth (HPBW) is an angle depicting the width of the main lobe of the radiation pattern, within which the radiated power is at least half the maximum value. For a dipole antenna, this is often the range over which the antenna emits most effectively.

The exercise provided uses a relation involving trigonometry to calculate HPBW: \(\sin(\theta_{HPBW}/2) = \frac{\lambda}{2 \ell}\). Intuitively, HPBW gives a practical insight into the antenna's coverage area. For example, a narrow beamwidth result in concentrated energy in a particular direction, which enhances the directivity but reduces the coverage area. Knowing the HPBW is essential for alignment and performance optimization of the antenna system.
Directivity in Decibels
Directivity can also be expressed in a logarithmic scale as decibels (dB), allowing for easier comparison between different antenna performances on a convenient scale that compresses the range of values. The formula \(D_{dB} = 10\log_{10}(D_{max})\) converts the dimensionless unit of directivity into decibels.

Decibels provide an intuitive way to express power ratios as addition and subtraction rather than multiplication and division. This conversion to a dB scale is particularly useful when dealing with quantities that span several orders of magnitude, which is often the case in radio frequency applications. In the given exercise, a \(D_{max}\) of 1.64 translates into approximately \(2.15\,\text{dB}\), providing a convenient way to report and understand the antenna’s performance.

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Most popular questions from this chapter

A large ground-based transmitter radiates \(10 \mathrm{~kW}\) and communicates with a mobile receiving station that dissipates \(1 \mathrm{~mW}\) on the matched load of its antenna. The receiver (not having moved) now transmits back to the ground station. If the mobile unit radiates \(100 \mathrm{~W}\), what power is received (at a matched load) by the ground station?

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.

Two short antennas at the origin in free space carry identical currents of \(5 \cos \omega t \mathrm{~A}\), one in the \(\mathbf{a}_{z}\) direction, and one in the \(\mathbf{a}_{y}\) direction. Let \(\lambda=2 \pi\) \(\mathrm{m}\) and \(d=0.1 \mathrm{~m}\). Find \(\mathbf{E}_{s}\) at the distant point where \((a)(x=0, y=1000\), \(z=0) ;(b)(0,0,1000) ;(c)(1000,0,0) .(d)\) Find \(\mathbf{E}\) at \((1000,0,0)\) at \(t=0 .\) (e) Find \(|\mathbf{E}|\) at \((1000,0,0)\) at \(t=0\).

Consider an \(n\) -element broadside linear array. Increasing the number of elements has the effect of narrowing the main beam. Demonstrate this by evaluating the separation in \(\phi\) between the zeros on either side of the principal maximum at \(\phi=90^{\circ} .\) Show that for large \(n\) this separation is approximated by \(\Delta \phi \doteq 2 \lambda / L\), where \(L \doteq n d\) is the overall length of the array.

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?

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