sir, i have a doubt if the channel is low pass in nature than how it allows passing the high-frequency carrier signal and how it allows passing of the whole passband as it comprises of very high frequencies speaking of the 4G communication
I took a course in advanced electronic communications at FAU of Boca Raton FL in mid 2006, with Dr. Alo, and I just came here for review. Although I got A in this course it is always good to review this fabulous material. Thank you Iain for your excellent explanation of Rayleigh Fading.
I'm glad to hear that. It has always been my intention for people to be able to watch my videos multiple times, and hopefully pick up extra things each time.
@@Sahil_Antil It means this video lecture reached, one of the biggest populated countries. So keep doing like this, fans from India will appreciate for your effort (i.e) irrespective of subscriber count
thanq professor, it was soo clear and understandable, you are unique of explaining things in the mode of "why" how" " what" . can we have some videos on cooperative networking
I'm glad you like the explanations and style of my videos. I'll give some thought to "cooperative networking", but it is a big topic and not well defined in what it exactly covers. Do you have specific aspects you'd suggest to cover?
Thanks for the suggestion. Great timing. I've got a video coming out later today that explains why an equaliser is needed. I'm planning the next one to be on equalisation. And I'll add the rake receiver to my "to do" list.
Excellent explanation of how the received vectors from multiple paths sum to the Rayleigh distribution. Would be good to tie this into fading process experienced by receiver when in motion.
Thanks for the suggestion. I'm not sure if you've seen them, but I have these two videos on the channel related to your suggestion: "What are Doppler Shift, Doppler Spread, and Doppler Spectrum?" ruclips.net/video/LLr3-kotbz4/видео.html and "What are Fast Fading and Slow Fading?" ruclips.net/video/Tm-Uyajcuqs/видео.html
Great explanation! About the confusion point which you mentioned students sometimes have about the dip in the centre of the rayleigh curve and its overall shape: its easy to interpret that by just remembering that it is the radial Jacobian of the 2D gaussian. So at the centre the 2D gaussian is a maximum, so it's derivative (by extension Jacobian) would be zero. That rises quickly to the max as you go further away from the origin in the gaussian (until I believe you reach the gaussian's inflection point, at which point it would BE the max), and then it falls off towards zero as you reach the edges of the gaussian. What I'm still contemplating though is why the rayleigh distribution is the jacobian of the 2D gaussian 😀
Yes, you're right, that's another way to think about the hole in the middle. In terms of your question at the end of your comment: that relationship is just a direct mathematical result of the change of variables from cartesian to polar. Perhaps the question you were intending to pose was: Why are the quadrature components of the noise independent? Now that's a good question. In reality, since we can't measure and track individual electrons, we need to resort to models of overall group behaviours, and this model just seems to work well.
We don't tend to describe them as "fading losses". I guess on a dB (ie. log) scale they appear as losses (if you compare to some reference power level, perhaps an imaginary line-of-sight path), but we tend to describe them more in terms of the "fading gain" or "fading path gain". And I wouldn't say there are any "typical" figures. It all depends on the local scattering environment.
Thank you for your amazing explanation. I have a question. At the beginning of the video, you explained sinusoid wave with a delay can be seen as phase rotated point in a constellation plot. I don't clearly understand how the delayed signal can be like that. could you explain about that more detail please?
@Lain: question: the sinusoid I believe is simplified version of transmitted signal for explanation but for higher modulated signals like 64/256 QAM, the signal has lot of randomness and PAR , and it almost appears like a noise on signal , how does phase delayed signal add up constructively or destructively since all the signals have high amplitude randomness..?
The issue is not the order of the QAM modulation, it is the bandwidth of the signal/channel. The single sinusoid discussion relates to a narrow band situation (low symbol rate). When a signal occupies a wider bandwidth (higher symbol rate) then the picture is more complicated. However, the same principles apply to all the frequencies within the bandwidth of the signal.
Hello Iain, Your videos are great and impart a fundamental understanding of communication systems. I have a basic question. Why are the channel and signals multiplied at the receiver? I am trying to understand intuition, especially from the perspective of convolution.
Great question. Perhaps the relation to convolution becomes clear if I tell you that the impulse response of the channel (in these non-ISI cases) is a delta function, with a value (ie. "height") given by the gain of the channel. The convolution of the input signal with the "delta function channel" equals the signal multiplied by the "value" of the delta function. This video explains that point: "Convolution with Delta Function" ruclips.net/video/TIcfY19dk0c/видео.html and for a video explaining what "ISI" is, check out this one: "What is Intersymbol Interference ISI?" ruclips.net/video/I087FUvW2ys/видео.html
Yes, as long as there are enough "multipaths". In other words it can't just be three or four reflected paths. It needs to be a "rich scattering environment".
I don't know if this video's comment section would be appropriate for this question, but given the symbols being received at a delayed rate, how the receiver differentiates the symbol received if it's a "current" one (i.e. non delayed) or a delayed one in order to properly reconstruct this given symbol?
Good question. Actually this is something that needs to be solved for all digital communication systems, whether they are fading or not. There is always a delay between the transmitter and the receiver. And the clocks at the transmitter and receiver are always seperate and independent (by definition, since they are at different locations), so they always need to be "locked". This is called timing recovery.
hello lain @ 0:33 the received waveform will also suffer a delay as it will not receive instantly but rather with delay and thus its phase will be also rotated, so why have you considered it without delay signal and why in AWGN channel in contrast to rayleigh fading channel there is no phase shift rather it should have one due to the time delay between Tx and Rx and hence resulting in to rotated constillations
The oscillator in the receiver doesn't know the phase of the oscillator in the transmitter, so there will always be uncertainty about the "absolute phase". So the receiver oscillator locks onto an estimate of the phase. When we draw these baseband pictures, we are assuming that the "overall phase rotation due to the end-to-end propagation delay of the earliest arriving path" is already compensated for.
Have you tried typing "Rayleigh Fading" into Wikipedia? If you do, you'll find the following equation: {\displaystyle p_{R}(r)={\frac {2r}{\Omega }}e^{-r^{2}/\Omega },\ r\geq 0}
sir, i have a doubt if the channel is low pass in nature than how it allows passing the high-frequency carrier signal and how it allows passing of the whole passband as it comprises of very high frequencies speaking of the 4G communication
Sorry, I'm not sure what you mean when you say that the channel is "low pass in nature". Do I say that somewhere in the video? Perhaps I meant to say "narrow band".
R is the random variable for the range (ie. the distance from the origin). This video might help to understand the notation: "What is a Probability Density Function (pdf)?" ruclips.net/video/jUFbY5u-DMs/видео.html
There is nothing fundamental (ie. from the laws of physics) about the Rayleigh p.d.f.'s relation to fading. It is just that in many situations the Rayleigh p.d.f. has been found to be a good model for real-world channel measurements.
![Rayleigh Fading Channels [Video 8]](/img/1.gif)
The best Rayleigh Fading explanation I have ever seen! Thank you!
Thanks. I'm glad it was helpful.
You are right!
These are pitched at just the right level. I am finding all of the videos to be both helpful and interesting. Thanks!
Glad you like them!
I have reviewed other similar lectures on RUclips and Iain is definitely one of the best.
Finally a bright and clear explanation! Too bad it wasn't on RUclips during my final project back in 2014
Glad you liked it.
Outstanding explanation. Thank you so much, Professor!!!
Glad you liked it!
sir, i have a doubt if the channel is low pass in nature than how it allows passing the high-frequency carrier signal and how it allows passing of the whole passband as it comprises of very high frequencies speaking of the 4G communication
I took a course in advanced electronic communications at FAU of Boca Raton FL in mid 2006, with Dr. Alo, and I just came here for review. Although I got A in this course it is always good to review this fabulous material. Thank you Iain for your excellent explanation of Rayleigh Fading.
I'm glad you've found the videos useful. It's always great to hear when people who've left university are still keen and interested.
Thanks a lot for a very conceptual lecture, watched it two times to understand deeply. Once again infinite thanks..
I'm glad to hear that. It has always been my intention for people to be able to watch my videos multiple times, and hopefully pick up extra things each time.
Hai, I'm from India ...... you're doing good..... keep continue
what India has to do with the lecture?
@@Sahil_Antil It means this video lecture reached, one of the biggest populated countries. So keep doing like this, fans from India will appreciate for your effort (i.e) irrespective of subscriber count
that's such an amazing work, thanks prof!
ô anh học gì ở đây thế? chỉ cho em tý kiến thức được không
Amazing explanations! Thank you so much for your contributions. From Vietnam :)
Glad it was helpful!
thanq professor, it was soo clear and understandable, you are unique of explaining things in the mode of "why" how" " what" . can we have some videos on cooperative networking
I'm glad you like the explanations and style of my videos. I'll give some thought to "cooperative networking", but it is a big topic and not well defined in what it exactly covers. Do you have specific aspects you'd suggest to cover?
@@iain_explains AF and DF relaying, with swipt, ofdm, noma,
I love people that made me understand, thanks a lot !
Glad I could help!
Great! Thank You.
Sir it would be very helpful if you made videos on the equalizer and rake receiver.
Thanks for the suggestion. Great timing. I've got a video coming out later today that explains why an equaliser is needed. I'm planning the next one to be on equalisation. And I'll add the rake receiver to my "to do" list.
@@iain_explains Thank You Sir..:)
Excellent explanation of how the received vectors from multiple paths sum to the Rayleigh distribution. Would be good to tie this into fading process experienced by receiver when in motion.
Thanks for the suggestion. I'm not sure if you've seen them, but I have these two videos on the channel related to your suggestion: "What are Doppler Shift, Doppler Spread, and Doppler Spectrum?" ruclips.net/video/LLr3-kotbz4/видео.html and "What are Fast Fading and Slow Fading?" ruclips.net/video/Tm-Uyajcuqs/видео.html
Great explanation!
About the confusion point which you mentioned students sometimes have about the dip in the centre of the rayleigh curve and its overall shape: its easy to interpret that by just remembering that it is the radial Jacobian of the 2D gaussian. So at the centre the 2D gaussian is a maximum, so it's derivative (by extension Jacobian) would be zero. That rises quickly to the max as you go further away from the origin in the gaussian (until I believe you reach the gaussian's inflection point, at which point it would BE the max), and then it falls off towards zero as you reach the edges of the gaussian.
What I'm still contemplating though is why the rayleigh distribution is the jacobian of the 2D gaussian 😀
Yes, you're right, that's another way to think about the hole in the middle. In terms of your question at the end of your comment: that relationship is just a direct mathematical result of the change of variables from cartesian to polar. Perhaps the question you were intending to pose was: Why are the quadrature components of the noise independent? Now that's a good question. In reality, since we can't measure and track individual electrons, we need to resort to models of overall group behaviours, and this model just seems to work well.
Amazing Explanation
Wowww such a good explanation of the process, you made it easy for us ! Thankss
Glad it was helpful!
Thank you for such a great explaination.
Glad it was helpful!
Thanks for your precious effort, all of your videos are very useful and helpful.
Thanks for your comment. It's my pleasure.
Superb explanation!!! Thank you for sharing!!!
Glad you liked it
Thank you so much.. great explanation!
Glad it was helpful!
Very simple and very helpful.
Glad it was helpful!
Very very nice explanation
I'm glad you liked it.
My head is still dizzy. I'll work on breaking this apart. Lol. Thank you for a good start.
Glad you found it useful.
helpful and useful, thank you so much professor!
wow this is so nice, could you do Nakagami fading also ?
Thanks for your comment and the suggestion. I'll add it to my to-do list.
Here's the link to the Nakagami video: ruclips.net/video/ztpNbE-Vpaw/видео.html
Sir, Can you explain what is the inverse Rayleigh distribution? Thank you
Thanks for the suggestion. However I'm not familiar with the inverse Rayleigh distribution. I'll have to look into it.
nice explanation, thank you
Glad it was helpful!
Thanks so much for the insightful explanation. But may ask, what are the typical values of Rayleigh fading losses?
We don't tend to describe them as "fading losses". I guess on a dB (ie. log) scale they appear as losses (if you compare to some reference power level, perhaps an imaginary line-of-sight path), but we tend to describe them more in terms of the "fading gain" or "fading path gain". And I wouldn't say there are any "typical" figures. It all depends on the local scattering environment.
thanks a lot for the video ; it was very helpful especially the last part explanation of the video
Glad it was helpful!
Thank you for your amazing explanation.
I have a question. At the beginning of the video, you explained sinusoid wave with a delay can be seen as phase rotated point in a constellation plot.
I don't clearly understand how the delayed signal can be like that.
could you explain about that more detail please?
It might help to watch this video that I posted recently: "What is a Constellation Diagram?" ruclips.net/video/kfJeL4LQ43s/видео.html
Thank you Professor! Very useful!
Glad it was helpful!
thanks for the clear explanation
Glad it was helpful!
@Lain: question: the sinusoid I believe is simplified version of transmitted signal for explanation but for higher modulated signals like 64/256 QAM, the signal has lot of randomness and PAR , and it almost appears like a noise on signal , how does phase delayed signal add up constructively or destructively since all the signals have high amplitude randomness..?
The issue is not the order of the QAM modulation, it is the bandwidth of the signal/channel. The single sinusoid discussion relates to a narrow band situation (low symbol rate). When a signal occupies a wider bandwidth (higher symbol rate) then the picture is more complicated. However, the same principles apply to all the frequencies within the bandwidth of the signal.
you're the man. tnx
You're welcome!
Hello Iain, Your videos are great and impart a fundamental understanding of communication systems. I have a basic question. Why are the channel and signals multiplied at the receiver? I am trying to understand intuition, especially from the perspective of convolution.
Great question. Perhaps the relation to convolution becomes clear if I tell you that the impulse response of the channel (in these non-ISI cases) is a delta function, with a value (ie. "height") given by the gain of the channel. The convolution of the input signal with the "delta function channel" equals the signal multiplied by the "value" of the delta function. This video explains that point: "Convolution with Delta Function" ruclips.net/video/TIcfY19dk0c/видео.html and for a video explaining what "ISI" is, check out this one: "What is Intersymbol Interference ISI?" ruclips.net/video/I087FUvW2ys/видео.html
@@iain_explains Thank you. It's clear now. ☺
Thanks a lot..very clear explanation..so, for each multipath environment without direct path we can say that the channel has a rayleigh dstribution?
Yes, as long as there are enough "multipaths". In other words it can't just be three or four reflected paths. It needs to be a "rich scattering environment".
@@iain_explains Thanks a lot sir
I don't know if this video's comment section would be appropriate for this question, but given the symbols being received at a delayed rate, how the receiver differentiates the symbol received if it's a "current" one (i.e. non delayed) or a delayed one in order to properly reconstruct this given symbol?
Good question. Actually this is something that needs to be solved for all digital communication systems, whether they are fading or not. There is always a delay between the transmitter and the receiver. And the clocks at the transmitter and receiver are always seperate and independent (by definition, since they are at different locations), so they always need to be "locked". This is called timing recovery.
hello lain @ 0:33 the received waveform will also suffer a delay as it will not receive instantly but rather with delay and thus its phase will be also rotated, so why have you considered it without delay signal and why in AWGN channel in contrast to rayleigh fading channel there is no phase shift rather it should have one due to the time delay between Tx and Rx and hence resulting in to rotated constillations
The oscillator in the receiver doesn't know the phase of the oscillator in the transmitter, so there will always be uncertainty about the "absolute phase". So the receiver oscillator locks onto an estimate of the phase. When we draw these baseband pictures, we are assuming that the "overall phase rotation due to the end-to-end propagation delay of the earliest arriving path" is already compensated for.
Great explaination! Thank you!
Glad it was helpful!
thank you so much for your great explanation
Glad it was helpful!
you're my hero
I'm glad you like the videos.
What is the mathematical expression for this fading model please
Have you tried typing "Rayleigh Fading" into Wikipedia? If you do, you'll find the following equation: {\displaystyle p_{R}(r)={\frac {2r}{\Omega }}e^{-r^{2}/\Omega },\ r\geq 0}
Clear explanation. Thank$!
Glad it was helpful!
Thanks professor
You are welcome
sir, i have a doubt if the channel is low pass in nature than how it allows passing the high-frequency carrier signal and how it allows passing of the whole passband as it comprises of very high frequencies speaking of the 4G communication
Sorry, I'm not sure what you mean when you say that the channel is "low pass in nature". Do I say that somewhere in the video? Perhaps I meant to say "narrow band".
Thank you, sir
Most welcome
Very helpful, thanks!
Glad it was helpful!
You save my meeting ...
Glad I could help.
thank you sir
You're welcome. Glad you like the video.
Thank you
You're welcome
Hi mr
I need answer for my question which is what does rayleigh flat fading means?
You might like to watch my video: "What are Flat Fading and Frequency Selective Fading?" ruclips.net/video/KiKPFT4rtHg/видео.html
Wow!
Thanx alot
Most welcome
nice
THis is good, thanks.
Whats R in the equation
R is the random variable for the range (ie. the distance from the origin). This video might help to understand the notation: "What is a Probability Density Function (pdf)?" ruclips.net/video/jUFbY5u-DMs/видео.html
How can we protect ourselves from emf
Also here what how is fading involved here and what is special about Rayleigh fadit?
There is nothing fundamental (ie. from the laws of physics) about the Rayleigh p.d.f.'s relation to fading. It is just that in many situations the Rayleigh p.d.f. has been found to be a good model for real-world channel measurements.
여기가 통신 맛집이네
Great concept - I like it! (assuming Google's translation is accurate: "This is a communication restaurant").