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An antenna operating at \(28 \mathrm{GHz}\) has an antenna gain of \(50 \mathrm{dBi}\). What is the effective aperture area of the antenna?

Short Answer

Expert verified
The effective aperture area of the antenna is approximately 0.91 square meters.

Step by step solution

01

Identify the given information

The frequency of the antenna is given as 28 GHz and the antenna gain is given as 50 dBi.
02

Convert gain from dBi to linear scale

The gain in linear scale can be calculated using the formula: \( G = 10^{\frac{G_{dBi}}{10}} \)Substitute 50 dBi into the formula: \( G = 10^{\frac{50}{10}} = 10^5 \)
03

Calculate the wavelength

The wavelength (λ) can be calculated using the speed of light (c) and the frequency (f): \( λ = \frac{c}{f} \)The speed of light, c, is approximately \(3 \times 10^8 \frac{m}{s}\). Substitute the frequency: \( λ = \frac{3 \times 10^8}{28 \times 10^9} = \frac{3 \times 10^8}{2.8 \times 10^{10}} = 1.07 \times 10^{-2} \text{ meters} \)
04

Use the formula for effective aperture area

The effective aperture area (A_e) of the antenna can be calculated using the formula: \( A_e = \frac{G \times λ^2}{4 \times \text{π}} \)Substitute the gain and wavelength into the formula: \( A_e = \frac{10^5 \times (1.07 \times 10^{-2})^2}{4 \times \text{π}} \)
05

Perform the final calculation

Continue with the calculation: \( (1.07 \times 10^{-2})^2 = 1.1449 \times 10^{-4} \)Then: \( A_e = \frac{10^5 \times 1.1449 \times 10^{-4}}{4 \times \text{Ï€}} = \frac{1.1449 \times 10^1}{4 \times \text{Ï€}} = \frac{11.449}{12.566} \text{ square meters} \approx 0.91 \text{ square meters} \)

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Key Concepts

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

Antenna Gain
Antenna gain is a crucial concept when discussing antenna performance. It measures the ability of an antenna to direct radio frequency energy in a particular direction compared to a theoretical isotropic antenna that radiates equally in all directions. Antenna gain is usually expressed in decibels relative to an isotropic radiator (dBi).

For example, in our exercise, the antenna has a gain of 50 dBi. To use this figure in calculations, it must be converted to a linear scale. This is done using the formula:

\[ G = 10^{\frac{G_{dBi}}{10}} \]

By substituting the given gain of 50 dBi into the formula, we get:

\[ G = 10^{\frac{50}{10}} = 10^5 \]

So, the antenna gain in a linear scale is 100,000. This value is vital for further calculations related to the antenna's effective aperture area.
Wavelength Calculation
The wavelength (\( \lambda \)) of a signal is another fundamental concept, especially in antenna theory. It’s important because the physical dimensions of the antenna are often related to the signal's wavelength. The wavelength can be calculated using the speed of light (\( \text{c} \)) and the frequency (\( \text{f} \)):

\[ \lambda = \frac{c}{f} \]

For our exercise, the antenna operates at 28 GHz. The speed of light is approximately \( 3 \times 10^8 \frac{m}{s} \). Substituting the frequency into the formula gives us:

\[ \lambda = \frac{3 \times 10^8}{28 \times 10^9} = \frac{3 \times 10^8}{2.8 \times 10^{10}} = 1.07 \times 10^{-2} \]

So, the wavelength is approximately 1.07 centimeters. This wavelength is used to determine the effective aperture area.
Effective Aperture Area
The effective aperture area (\( A_e \)) of an antenna represents the area through which the antenna captures electromagnetic energy. It is significant because it correlates with how well an antenna can receive signals. The formula to calculate the effective aperture area is:

\[ A_e = \frac{G \times \lambda^2}{4 \times \pi} \]

Using the values obtained in the previous steps, where the linear gain (\( G \)) is 100,000 and the wavelength (\( \lambda \)) is 1.07 centimeters (or 1.07 \times 10^{-2} meters):

\[ A_e = \frac{10^5 \times (1.07 \times 10^{-2})^2}{4 \times \pi} \]

Performing the calculations step-by-step:

\[ (1.07 \times 10^{-2})^2 = 1.1449 \times 10^{-4} \]

\[ A_e = \frac{10^5 \times 1.1449 \times 10^{-4}}{4 \times \pi} = \frac{1.1449 \times 10^1}{4 \times \pi} = \frac{11.449}{12.566} \text{ square meters} \]

Thus, the effective aperture area is approximately 0.91 square meters. A larger effective aperture area indicates better signal capturing ability, making this a fundamental metric in antenna design and analysis.

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Most popular questions from this chapter

An antenna with an antenna gain of \(8 \mathrm{dBi}\) radiates \(6.67 \mathrm{~W}\). What is the EIRP in watts? Assume that the antenna is \(100 \%\) efficient.

A microstrip patch antenna operating at \(2 \mathrm{GHz}\) has an efficiency of \(66 \%\) and an antenna gain of \(8 \mathrm{dBi}\). The power input to the antenna is \(10 \mathrm{~W}\). (a) What is the power, in \(\mathrm{dBm}\), radiated by the antenna? (b) What is the equivalent isotropic radiated power (EIRP) in watts? (c) What is the power density, in \(\mu \mathrm{W} / \mathrm{m}^{2},\) at \(1 \mathrm{~km}\) if ground effects are ignored? (d) Because of multipath effects, the power density drops off as \(1 / d^{4}\), where \(d\) is distance. What is the power density, in \(\mathrm{nW} / \mathrm{m}^{2},\) at \(1 \mathrm{~km}\) if the power density is \(100 \mathrm{~mW} / \mathrm{m}^{2}\) at \(10 \mathrm{~m}\) from the transmit antenna?

A hill is \(1 \mathrm{~km}\) from a transmit antenna and \(2 \mathrm{~km}\) from a receive antenna. The receive and transmit antennas are at the same height and the hill is \(20 \mathrm{~m}\) above the height of the antennas. What is the additional loss caused by diffraction over the top of the hill? Treat the hill as a knife-edge and the operating frequency is \(1 \mathrm{GHz}\).

A transmitter and receiver operate at \(100 \mathrm{MHz}\) are at the same level, and are separated by \(4 \mathrm{~km}\). The signal must diffract over a building half way between the antennas that is \(20 \mathrm{~m}\) higher than the direct path between the antennas. What is the attenuation (in decibels) due to diffraction?

The output stage of an RF front end consists of an amplifier followed by a filter and then an antenna. The amplifier has a gain of \(27 \mathrm{~dB}\), the filter has a loss of \(1.9 \mathrm{~dB},\) and of the power input to the antenna, \(35 \%\) is lost as heat due to resistive losses. If the power input to the amplifier is \(30 \mathrm{dBm},\) calculate the following: (a) What is the power input to the amplifier in watts? (b) Express the loss of the antenna in \(\mathrm{dB}\). (c) What is the total gain of the RF front end (amplifier + filter)? (d) What is the total power radiated by the antenna in \(\mathrm{dBm}\) ? (e) What is the total power radiated by the antenna in \(\mathrm{mW}\) ?

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