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Chapter 4: Problem 17

A communication system operating at \(2.5 \mathrm{GHz}\) includes a transmit antenna with an antenna gain of \(12 \mathrm{dBi}\) and a receive antenna with an effective aperture area of \(20 \mathrm{~cm}^{2}\). The distance between the two antennas is \(100 \mathrm{~m}\). (a) What is the antenna gain of the receive antenna? (b) If the input to the transmit antenna is \(1 \mathrm{~W}\), what is the power density at the receive antenna if the power falls off as \(1 / d^{2},\) where \(d\) is the distance from the transmit antenna? (c) Thus what is the power delivered at the output of the receive antenna?

Short Answer

Expert verified
(a) The antenna gain is approximately 2.41 dBi. (b) The power density is approximately 1.26 x 10^{-4} W/m^2. (c) The power delivered is approximately 2.52 x 10^{-7} W.

Step by step solution

01

- Determine the wavelength

First, calculate the wavelength of the communication system using the speed of light equation. The frequency \( f \) is given as \(2.5 \text{ GHz} = 2.5 \times 10^9 \text{ Hz} \). The speed of light \( c \) is approximately \( 3 \times 10^8 \text{ m/s} \).\[ \lambda = \frac{c}{f} = \frac{3 \times 10^8 \text{ m/s}}{2.5 \times 10^9 \text{ Hz}} = 0.12 \text{ m} \]
02

- Calculate the antenna gain of the receive antenna

The antenna gain \( G_r \) can be determined using the effective aperture area \( A_e \) and wavelength \( \lambda \).\[ G_r = \frac{4\pi A_e}{\lambda^2} \]Where: \( A_e = 20 \text{ cm}^2 = 0.002 \text{ m}^2 \).\[ G_r = \frac{4\pi \times 0.002}{(0.12)^2} \approx 1.747 \text{ (unitless)}\text{ which converts to } 10 \log_{10}(1.747) \approx 2.41 \text{ dBi} \]
03

- Calculate the power density at the receive antenna

The power density \( S \) at a distance \( d \) from the transmit antenna can be calculated using the input power \(P_t\) and the distance \(d\).\[ S = \frac{P_t G_t}{4\pi d^2} \]Given: \( P_t = 1 \text{ W}, G_t = 12 \text{ dBi} = 10^{(12/10)} = 15.8489 \text{ (unitless)}, d = 100 \text{ m} \).\[ S = \frac{1 \times 15.8489}{4\pi \times (100)^2} \approx 1.26 \times 10^{-4} \text{ W/m}^2 \]
04

- Calculate the power delivered at the output of the receive antenna

The power delivered at the output of the receive antenna \( P_r \) can be calculated using the power density \( S \) and the effective aperture area \( A_e \).\[ P_r = S \times A_e \]\[ P_r = 1.26 \times 10^{-4} \times 0.002 = 2.52 \times 10^{-7} \text{ W} \]

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Key Concepts

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

Antenna Gain Calculation
In an RF communication system, antenna gain measures how effectively the antenna directs radio frequency energy compared to an isotropic antenna that radiates equally in all directions. The gain is often expressed in decibels isotropic (dBi). This means a higher gain indicates more focused energy. For calculation:
The formula used is: \[ G_r = \frac{4\pi A_e}{\lambda^2} \]
Where:
  • G_r is the antenna gain,
  • \( A_e \) is the effective aperture area,
  • \( \lambda \) is the wavelength.

In our exercise, after performing the calculations, the receive antenna gain is around 2.41 dBi. This helps understand how efficiently the received power is being captured by the antenna.
Effective Aperture Area
The effective aperture area represents an antenna's ability to intercept power from a passing radio wave. It's different from the physical size of the antenna and is defined as the area through which the received power density multiplied indicates the actual power collected.
The formula is:\( A_e = \frac{G_r \lambda^2}{4\pi} \)
  • \( A_e \) is the effective aperture area,
  • \( G_r \) is the antenna gain,
  • \( \lambda \) is the wavelength.

In the given problem, the effective aperture area given is 20 cm虏 or 0.002 m虏. This value is essential for understanding how much power the antenna can collect from the surrounding electromagnetic field.
Power Density
Power density describes how much power a radio wave carries per square meter. It's especially crucial when determining the amount of power that reaches the receiving antenna. The power density \( S \) can be derived from the transmit power and antenna gain as follows:
The formula is: \[ S = \frac{P_t G_t}{4\pi d^2} \]
Where:
  • \( P_t \) is the transmit power,
  • \( G_t \) is the gain of the transmit antenna,
  • \( d \) is the distance between transmit and receive antennas.

The result calculated from the exercise provides a power density of approximately \( 1.26 \times 10^{-4} \; W/m^2 \). This step is essential in determining how much power per unit area is available for the receiver.
Wavelength Calculation
Wavelength is a fundamental property of a radio wave, representing the distance between consecutive peaks of the wave. It can be calculated using the frequency of the system and the speed of light.
The formula used: \[ \lambda = \frac{c}{f} \]
Where:
  • \( \lambda \) is the wavelength,
  • \( c \) is the speed of light (\( 3 \times 10^8 \; m/s \)),
  • \( f \) is the frequency (in Hz).

In this specific exercise, the frequency is 2.5 GHz, leading to a calculated wavelength of 0.12 m. Understanding wavelength helps in visualizing how the antennas interact with radio waves.
Receiver Power Calculation
Receiver power calculation involves determining the actual power delivered to the output of the receive antenna. This is calculated by multiplying the power density at the receiver location by the effective aperture area.
The formula is: \[ P_r = S \times A_e \]
  • \( P_r \) is the received power,
  • \( S \) is the power density,
  • \( A_e \) is the effective aperture area.

From our given values, the resulting received power is found to be approximately \( 2.52 \times 10^{-7} \; W \). This final value helps in understanding the efficiency of the system and how much power the receiver is able to collect.

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Most popular questions from this chapter

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, \(45 \%\) 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? (b) Express the loss of the antenna in decibels. (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 milliwatts?

At \(60 \mathrm{GHz}\) the atmosphere strongly attenuates a signal. Discuss the origin of this and indicate an advantage and a disadvantage.

Stacked dipole antennas are often found at the top of cellphone masts, particularly for large cells and operating frequencies below \(1 \mathrm{GHz}\). These antennas have an efficiency that is close to \(90 \%\). Consider an antenna that has \(40 \mathrm{~W}\) of input power, an antenna gain of \(10 \mathrm{dBi},\) and transmits a signal at \(900 \mathrm{MHz}\) (a) What is the EIRP in watts? (b) If the power density drops as \(1 / d^{3},\) where \(d\) is the distance from the transmit tower, what is the power density at \(1 \mathrm{~km}\) if the power density is \(100 \mathrm{~mW} / \mathrm{m}^{2}\) at \(10 \mathrm{~m} ?\)

A communication system operating at \(10 \mathrm{GHz}\) uses a microstrip patch antenna as a transmit antenna and a dipole antenna as a receive antenna. The transmit antenna is connected to the transmitter by a \(20 \mathrm{~m}\) long cable with a loss of \(0.2 \mathrm{~dB} / \mathrm{m}\) and the output power of the transmitter is \(30 \mathrm{~W}\). The transmit antenna has an antenna gain of \(9 \mathrm{dBi}\) and an antenna efficiency of \(60 \% .\) The link between the transmit and receive antenna is sufficiently elevated that ground effects and multipath effects are insignificant. (a) What is the output power of the transmitter in \(\mathrm{dBm}\) ? (b) What is the cable loss between the transmitter and the antenna? (c) What is the total power radiated by the transmit antenna in \(\mathrm{dBm}\) ? (d) What is the power lost in the antenna as resistive losses and spurious radiation? Express your answer in \(\mathrm{dBm}\). (e) What is the EIRP of the transmitter in \(\mathrm{dBm}\) ? (f) The transmitted power will drop off as \(1 / d^{n}\) ( \(d\) is distance). What is \(n\) ? (g) What is the peak power density in \(\mu \mathrm{W} / \mathrm{m}^{2}\) at \(1 \mathrm{~km}\)?

On a resonant antenna a large current is established by creating a standing wave. The current peaking that thus results establishes a strong electric field (and hence magnetic field) that radiates away from the antenna. A typical dipole loses \(15 \%\) of the power input to it as resistive \(\left(I^{2} R\right)\) losses and has an antenna gain of \(10 \mathrm{dBi}\) measured at \(50 \mathrm{~m}\). Consider a base station dipole antenna that has \(100 \mathrm{~W}\) input to it. Also consider that the transmitted power density falls off with distance \(d\) as \(1 / d^{3}\). Hint, calculate the power density at \(50 \mathrm{~m}\). [Parallels Example 4.2\(]\) (a) What is the input power in \(\mathrm{dBm} ?\) (b) What is the power transmitted in \(\mathrm{dBm} ?\) (c) What is the power density at \(1 \mathrm{~km} ?\) Express your answer as \(\mathrm{W} / \mathrm{m}^{2}\). (d) What is the power captured by a receive antenna (at \(1 \mathrm{~km}\) ) that has an effective antenna aperture of \(6 \mathrm{~cm}^{2} ?\) Express your answer in first \(\mathrm{dBm}\) and then watts. (e) If the background noise level captured by the antenna is \(1 \mathrm{pW}\), what is the SNR in decibels? Ignore interference that comes from other transmitters.
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