Fundamentals of Wireless Channels
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- Опубліковано 27 сер 2024
- In this video, Professor Emil Björnson explains the basic principles of wireless communication channels, such as the impact of antennas and wave propagation. The free-space pathloss formula is derived and the impact of the antenna gains is clarified. Finally, the use of antenna arrays to create controllable antenna directivity is introduced.
The slides are available here: github.com/emi...
The video was produced for the course IK2560 Mobile Networks and Services at the KTH Royal Institute of Technology.
#mobilenetwork #cellular #digitalization #communication #mimo #mobileservices #wireless #5g #antennas #channels
This is the fundamental explanation of the wireless propagation channel we deserve. Succinct and interesting.
Thanks for this great video...
It answered many of my questions about radio access networks.
You explain so perfectly that it almost hides all the complexities behind these concepts!
thank you! very informative video
Many thanks Prof. for sharing.
As usual perfect explanation 🎓 Thank you for sharing knowledge!😎
Thanks, exciting video
This presentation suggests that channel quality is exclusively a function of signal strength. Isn't it actually a function of signal to noise ratio in the receiver. I think to be complete, this should be included.
This is a good point. The communication performance will be determined by the SNR. However, the noise power is determined by the receiver hardware design (it is the thermal noise limit * the noise figure), and is not affected by the communication channel. So for a particular receiver, the signal strength determines the SNR.
Thanks for creating great content.
I am curious about the coverage for 6G, and high frequencies. How is expected to have a "proper" coverage in 6G, since we will have very high bands? (As you know, in 5G with mmWave coverage is around 300m, for 6G with higher frequencies will be less).
On slide 5 we have A_iso=lambda^2/4*pi, how do we get this formular? It is not the area of a sphere (A=4*pi*r^2). Thank you very much.
There is a derivation of the area of an isotropic antenna here: en.m.wikipedia.org/wiki/Isotropic_radiator
Thanks, Ae is a theoretical quantity and it depends on the wavelength. Does it also depends on the kind of the antenna?
On the first glance it is intuitive that a dipol, monopol or patch have different eff. antenna areas.
The dependence on the wavelength is tightly connected to the antenna design. A dipole antennas has a length proportional to the wavelength, so you need to design it for your specific carrier frequency. It will then have the intended gain but the effective area shrinks as the wavelength becomes smaller. For patch or aperture antennas, you can design antennas for different frequencies that have the same physical area. This leads to more gain at higher frequencies, but also a more directional antenna so the effective area varies rapidly with the angle.
🙏💐💐
Are those 20,000 pieces spread evenly across the entire sphere? Across time? Which term gives 20,000?
42 dB ≈ 20000 (the equality holds exactly for 43 dB). It is a computation of the formula shown on the slide. Yes, we assume these pieces are evenly spread over the sphere. If you divide the surface area of the sphere with 1 m radio into that many pieces, each one has the size of the receiver and therefore only one of them can be captured by it.
Are these slides available anywhere (pdf)?
Yes, I have added a link to it in the description to the video.
Can anyone take this course? or should they be a student?
It is only available for students enrolled at our university. The three videos that are published so far are the only ones we intend to release for this year. Maybe more will be produced next year. It is meant as a complement to the normal lectures.