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!
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).
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.
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.
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.
![Beamforming directivity [Part 1, Fundamentals of mmWave communication]](http://i.ytimg.com/vi/WOE_nKBKUsE/mqdefault.jpg)
![Beamforming directivity [Part 1, Fundamentals of mmWave communication]](/img/tr.png)
Awesome video! Thanks for making it and making it available
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!
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.
Many thanks Prof. for sharing.
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).
As usual perfect explanation 🎓 Thank you for sharing knowledge!😎
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.
thank you! very informative video
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
🙏💐💐
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.
Are these slides available anywhere (pdf)?
Yes, I have added a link to it in the description to the video.
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.