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Chapter 4: Problem 45

A transmitter and receiver operate at \(10 \mathrm{GHz}\), are at the same level, and are \(4 \mathrm{~km}\) apart. The signal must diffract over a building that is half way between the antennas and is \(20 \mathrm{~m}\) higher than the line between the antennas. What is the attenuation (in \(\mathrm{dB}\) ) due to diffraction?

Short Answer

Expert verified
The attenuation due to diffraction is approximately 21.64 dB.

Step by step solution

01

- Find the wavelength

The frequency (u) is given as 10 GHz. The speed of light (c) is approximately \(c = 3 \times 10^8 \mathrm{m/s} \). Use the equation \(u = \frac{c}{\text{wavelength}}\) to find the wavelength (位).\[ 位 = \frac{c}{u} = \frac{3 \times 10^8}{10 \times 10^9} \mathrm{m} = 0.03 \mathrm{m}\]
02

- Calculate Fresnel's parameter (v)

Fresnel's parameter (v) is given by the formula \ v = \frac{h \sqrt{2}}{位 \sqrt{d_1 + d_2}} \, where h is the height difference and d_1, d_2 are the distances from the transmitter and receiver to the obstacle.\Given that the building is 20 m higher, h = 20 m, and both d_1 and d_2 are 2 km each: \[ v = \frac{20 \sqrt{2}}{0.03 \sqrt{2000 + 2000}} = \frac{20 \sqrt{2}}{0.03 \sqrt{4000}} = \frac{20 \sqrt{2}}{0.03 \times 63.245} 鈮� 14.98\]
03

- Determine diffraction loss

Diffraction attenuation (\text{L}) in dB for a single obstacle can be found using the approximate empirical formula: \[ L = 6.9 + 20 \log_{10}( \sqrt{(v - 0.1)^2 + 1} + v - 0.1 ) \] Using v 鈮� 14.98: \[ L = 6.9 + 20 \log_{10}(\sqrt{(14.98 - 0.1)^2 + 1} + 14.98 - 0.1) \] This equals: \[ 6.9 + 20 \log_{10}( \sqrt{(14.88)^2 + 1} + 14.88 ) 鈮� 6.9 + 20 \log_{10}( \sqrt{221.0544} + 14.88 ) 鈮� 6.9 + 20 \log_{10}( 14.87 + 14.88 ) 鈮� 6.9 + 20 \log_{10}( 29.75 ) \] Solving this gives: \[ \approx 6.9 + 14.74 \approx 21.64 \text{dB} \]

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Key Concepts

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

Fresnel's parameter
Fresnel's parameter is crucial in understanding how signals diffract around obstacles. It helps us quantify the impact of height and distance on signal diffraction. Fresnel's parameter, denoted by \( v \), is calculated using the formula: \( v = \frac{h \sqrt{2}}{\lambda \sqrt{d_1 + d_2}} \).
Here:
  • \( h \) is the height difference of the obstacle relative to the direct line of sight.
  • \( \lambda \) is the wavelength of the transmitted signal.
  • \( d_1 \) and \( d_2 \) are the distances from the transmitter and receiver to the obstacle, respectively.

In our example, \( h \) is 20 meters, \( d_1 \) and \( d_2 \) are 2 km each, and the wavelength \( \lambda \) is 0.03 meters. Substituting these values, we get:
\( v = \frac{20 \sqrt{2}}{0.03 \sqrt{4000}} \approx 14.98 \).
A higher Fresnel's parameter generally indicates greater diffraction loss.
Diffraction Loss
Diffraction loss is an important factor in radio frequency (RF) transmission, especially when signals need to navigate around obstacles. It quantifies the loss of signal strength due to diffraction. The diffraction attenuation (\( L \)) can be estimated using the empirical formula: \( L = 6.9 + 20 \log_{10}( \sqrt{(v - 0.1)^2 + 1} + v - 0.1 ) \).

For our scenario:
  • Using \( v \approx 14.98 \)
  • \[ L = 6.9 + 20 \log_{10}( \sqrt{(14.98 - 0.1)^2 + 1} + 14.98 - 0.1 ) \]
  • \( L \approx 6.9 + 20 \log_{10}( \sqrt{221.0544} + 14.88 ) \)
  • \( L \approx 6.9 + 20 \log_{10}( 29.75 ) \)
  • \( L \approx 21.64 \text{dB} \)

Understanding this loss helps in designing effective communication systems, ensuring minimal signal degradation.
Wavelength Calculation
The wavelength of a signal is fundamental in RF analysis, affecting parameters like diffraction loss. To find the wavelength \( \lambda \), we use the relationship between the speed of light (\( c \)) and frequency (\( u \)): \( \lambda = \frac{c}{u} \).

Steps:
  • The speed of light \( c \) is approximately \( 3 \times 10^8 \) m/s.
  • The frequency \( u \) is given as 10 GHz (\( 10 \times 10^9 \) Hz).
  • Substituting these into the formula, we get: \( \lambda = \frac{3 \times 10^8}{10 \times 10^9} = 0.03 \) meters.
A shorter wavelength will generally result in higher Fresnel's parameter, affecting the diffraction loss. Understanding wavelength is key to analyzing and predicting signal behavior.

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Most popular questions from this chapter

Describe the difference in multipath effects in a central city area compared to multipath effects in a desert. Your description should be approximately 4 lines long and not use a diagram.

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 \(1 \mathrm{~km}\)?

An antenna with an efficiency of \(50 \%\) has an antenna gain of \(12 \mathrm{dBi}\) and radiates \(100 \mathrm{~W}\). What is the EIRP in watts?

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?

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 \(10 \mathrm{~m} ?\)
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