/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 37 At \(60 \mathrm{GHz}\) the atmos... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 37

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

Short Answer

Expert verified
At 60 GHz, oxygen absorption causes signal attenuation. An advantage is reduced interference; a disadvantage is limited range.

Step by step solution

01

Discuss the Origin of Attenuation

At 60 GHz, the primary cause of signal attenuation in the atmosphere is due to the absorption by oxygen molecules (O2). Oxygen exhibits a strong resonance at this frequency, which means that it absorbs the electromagnetic energy from the signal, converting it into heat and thus diminishing the signal strength as it travels through the atmosphere.
02

Indicate an Advantage

One advantage of this attenuation is that it provides natural isolation for communication channels. This means that communication links operating at 60 GHz are less likely to experience interference from other nearby systems operating on similar frequencies. This can be particularly beneficial in dense urban environments where many devices are operating simultaneously.
03

Indicate a Disadvantage

A disadvantage of this high level of attenuation is that it significantly limits the range of 60 GHz communication links. The signal can attenuate rapidly, which restricts the use of these frequencies to short-range applications. This can be a limitation for applications that require long-distance communication.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Oxygen Absorption
When discussing signal attenuation at 60 GHz, oxygen absorption is a key factor. At this frequency, oxygen molecules in the air absorb the energy from the electromagnetic waves. This process involves the oxygen (O2) molecules hitting a resonance point at 60 GHz. When this happens, the oxygen converts the energy from the signal into heat. As a result, the signal strength diminishes very quickly. This is why signals at this frequency do not travel far. This absorption plays a crucial role in communication systems, especially for those looking to use the 60 GHz band.
Frequency Isolation
One advantage of the strong attenuation at 60 GHz due to oxygen absorption is frequency isolation. This means that the signals are contained within a very limited range. In dense urban environments, where many devices are operating at similar frequencies, this can reduce interference.
  • It leads to clearer communication channels for devices using 60 GHz.
  • Reduces the chances of cross-talk between systems.
  • Allows more bandwidth availability since interference is minimized.
This is particularly beneficial for high-density networks and can improve overall communication reliability.
Communication Range Limitation
A significant disadvantage of high attenuation at 60 GHz is the communication range limitation. The rapid attenuation of the signal means that it is only effective for short-range communication.
  • This limits the use of 60 GHz frequencies to applications within a confined area.
  • Long-distance communication at this frequency would require many repeaters or amplifiers.
  • The technology is suitable for specific use cases like indoor wireless networks, point-to-point communication, or vehicle-to-vehicle communication.
This limitation means careful planning and deployment are needed to ensure functionality and coverage.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 900 MHz communication system uses a transmit antenna with an antenna gain \(G_{T}\) of \(3 \mathrm{~dB}\) and a receive antenna with an antenna gain \(G_{R}\) of \(0 \mathrm{~dB}\). If the distance between the antennas is \(200 \mathrm{~m},\) what is the link loss from the input to the transmit antenna and the output of the receive antenna if the power density reduces as \(1 / d^{2.5} ?\)

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 \(\mathrm{mW}\) ?

A communication system has a power density roll-off of \(1 / d^{2.5}\) between a transmit antenna and a mobile receive antenna which are separated by \(10 \mathrm{~km}\). At \(10 \mathrm{~m}\) from the transmit antenna, the power density is \(10 \mathrm{~W} / \mathrm{m}^{2}\). What is the power density at the receive antenna? [Parallels Example 4.5\(]\)

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?

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?
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Linear Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation

Nuclear Physics

Read Explanation

Thermodynamics

Read Explanation

Astrophysics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.