P=V^2/R and if we normalise to a unit resistance (R=1) then the voltage needs to be set at a value that is the square root of the power we would like the signal to have.
Hello Professor Iain Collings, I discovered this site by accident and realized it is indeed a gold mine!! Thank you so much for the lucid descriptions, I am able to learn a lot and enjoy concurrently. I have been trying to grasp successive interference cancellation, and I am a bit confused. Assuming that you decode source 2 first, followed by source 1, the symbol x2 that is extracted first has an uncertainty determined by the summed complex gaussian (P1+N0). If you subtract that from y won't the remainder still retain that uncertainty? How do we ensure that x2 is exactly known at that point with certainty? Is this by passing the recovered x2 upstream past the channel decoder stage, then mapping the error-corrected bits back to the original complex symbol, multiplying that by the (presumably known) channel response of user 2 (g_2) to obtain the true contribution in the recovered baseband signal, and then subtracting from y? Am I on the right track, or going off on a tangent? Thank you for your advice. Also, when you decode the first symbol, how do you decide which UE it belongs to? Is that by correlating the received baseband waveform against the known channel response of the intended UE?
Yes, you're right. It all relies on decoding the data streams _exactly_ . If errors are made, then they will cause more errors to be made when decoding the subsequent data streams. In my video, I dealt with capacity. According to Shannon theory, as long as the data streams are sent at a rate that is lower than the capacity, then they can be decoded _exactly_ . And in terms of your question about working out which data stream is which, there is always an address header in each packet. I'm glad you like the channel. Let me know if there are other topics you'd like me to cover. For more details about "Capacity", check out: "What are Channel Capacity and Code Rate?" ua-cam.com/video/P0WY96WBUyA/v-deo.html
@@iain_explains Thank you so much for the confirmation (yes, I was also looking at it from the perspective of Shannon's capacity formula). I am sorry for I saw your reply much later, when I restarted my learning exercise following a hiatus. I am planning to go through all the lectures at your website in earnest, since the topics covered are all of keen interest for me. And I will be reaching out if/when I get stuck with doubts again🙂.
No, the time is just divided between the two users. There are no "time slots" when considering fundamental capacity. Time slots are an invention (or in other words, they are an imposed structure) that makes signalling easier to implement. They are not a fundamental resource. "Time" is the fundamental resource.
Sure, the ability to decode successfully depends on the relative powers, but it also depends on the rates. If you are using a sufficiently low rate, then it allows you to use a sufficiently powerful error correction code, to overcome whatever noise and interference there might be. That's what Shannon's formula tells us.
Yes, the SIR receiver can be implemented in the downlink too, but it's important to realise that the "interfering signal" is also coming through the same channel as the intended user's signal (since they are both coming from the base station), so there won't be a "strong" user and a "weak" user (of course there can still be a high-rate user and a low-rate user, depending on the data requirements of the users).
Sir, is treating interference as noise and SIC the same? Please tell me sir Sir, can you explain the rate region concept of two user interference channel by using the SIC scheme and treating interference as noise scheme and joint decoding scheme
It is always possible to treat interference as noise in any receiver (not just in SIC). If you don't know anything about the structure of the interference then you can't do anything about it anyway, no matter what receiver you are using. And thanks for the suggestion about the relationship between SIC and joint decoding - I've added it to my "to do" list.
What are benefits of cancellation of succcsive interference it would be reduced the energy or error occurred 🙏🙏 reply please bachelor degree holder request 🙏
Sir my research is based power domain noma. Some research articles say that power domain noma is dropped from 3gpp 17 release is true? Will it not be a part of 5g?
Simply decode the first data stream exactly as you would if the second data stream didn't exist (ie. treat the second data stream as being part of the noise), and then once the first data stream is decoded, subtract its effect out from the received signal, and go ahead and decode the second data stream.
I'm glad the video was helpful. I'm not exactly sure what you mean by your question. For a transmitter with constant power, when a signal only lasts half as long (eg. TDMA between two users), then it only has half the energy, since Energy = Power x time.
Sir why are we allocating square root of p(power) to user symbol in PD NOMA ?
Please answer this basic ..
P=V^2/R and if we normalise to a unit resistance (R=1) then the voltage needs to be set at a value that is the square root of the power we would like the signal to have.
Thank you sir
Hello Professor Iain Collings, I discovered this site by accident and realized it is indeed a gold mine!! Thank you so much for the lucid descriptions, I am able to learn a lot and enjoy concurrently.
I have been trying to grasp successive interference cancellation, and I am a bit confused. Assuming that you decode source 2 first, followed by source 1, the symbol x2 that is extracted first has an uncertainty determined by the summed complex gaussian (P1+N0). If you subtract that from y won't the remainder still retain that uncertainty? How do we ensure that x2 is exactly known at that point with certainty? Is this by passing the recovered x2 upstream past the channel decoder stage, then mapping the error-corrected bits back to the original complex symbol, multiplying that by the (presumably known) channel response of user 2 (g_2) to obtain the true contribution in the recovered baseband signal, and then subtracting from y? Am I on the right track, or going off on a tangent? Thank you for your advice.
Also, when you decode the first symbol, how do you decide which UE it belongs to? Is that by correlating the received baseband waveform against the known channel response of the intended UE?
Yes, you're right. It all relies on decoding the data streams _exactly_ . If errors are made, then they will cause more errors to be made when decoding the subsequent data streams. In my video, I dealt with capacity. According to Shannon theory, as long as the data streams are sent at a rate that is lower than the capacity, then they can be decoded _exactly_ . And in terms of your question about working out which data stream is which, there is always an address header in each packet. I'm glad you like the channel. Let me know if there are other topics you'd like me to cover. For more details about "Capacity", check out: "What are Channel Capacity and Code Rate?" ua-cam.com/video/P0WY96WBUyA/v-deo.html
@@iain_explains Thank you so much for the confirmation (yes, I was also looking at it from the perspective of Shannon's capacity formula). I am sorry for I saw your reply much later, when I restarted my learning exercise following a hiatus. I am planning to go through all the lectures at your website in earnest, since the topics covered are all of keen interest for me. And I will be reaching out if/when I get stuck with doubts again🙂.
Great tutorial. Helped a lot
Very lucid explanation.....thanks professor...
Glad you liked it
should there be 1/2 before the logarith term to characterize that the transmission is conducted over two timeslot in the case of tdma ?
No, the time is just divided between the two users. There are no "time slots" when considering fundamental capacity. Time slots are an invention (or in other words, they are an imposed structure) that makes signalling easier to implement. They are not a fundamental resource. "Time" is the fundamental resource.
Thank you, cleared it up for me. Explaining with paper is a good method, I feel.
Glad it helped!
Thanks for thé video. Is there any advantage of decoding the strong user first? I think doing such provides higher rate region, no?
Sure, the ability to decode successfully depends on the relative powers, but it also depends on the rates. If you are using a sufficiently low rate, then it allows you to use a sufficiently powerful error correction code, to overcome whatever noise and interference there might be. That's what Shannon's formula tells us.
@@iain_explains thanks
well explained video. Thank you
Glad it was helpful!
Dear prof, the SIC receiver, is implemented in uplink and dowlink?
Yes, the SIR receiver can be implemented in the downlink too, but it's important to realise that the "interfering signal" is also coming through the same channel as the intended user's signal (since they are both coming from the base station), so there won't be a "strong" user and a "weak" user (of course there can still be a high-rate user and a low-rate user, depending on the data requirements of the users).
Thank you very much sir for making this concept crystal clear!
I'm so glad you found it helpful.
Sir, is treating interference as noise and SIC the same? Please tell me sir
Sir, can you explain the rate region concept of two user interference channel by using the SIC scheme and treating interference as noise scheme and joint decoding scheme
It is always possible to treat interference as noise in any receiver (not just in SIC). If you don't know anything about the structure of the interference then you can't do anything about it anyway, no matter what receiver you are using. And thanks for the suggestion about the relationship between SIC and joint decoding - I've added it to my "to do" list.
@@iain_explains Thank you very much for clearing my doubt sir
Excellent Explanation in a smart way...Thanks.
Dear Iain, can you upload the video of MIMO-OFDM & V-BLAST. I shall very thankful to you.
Great suggestion, thanks, I've added it to my "to do" list.
What are benefits of cancellation of succcsive interference it would be reduced the energy or error occurred 🙏🙏 reply please bachelor degree holder request 🙏
Sir my research is based power domain noma. Some research articles say that power domain noma is dropped from 3gpp 17 release is true? Will it not be a part of 5g?
Sorry, I'm not sure. I haven't caught up with the latest 5G standards decisions.
This is the concept, how it is done??
Simply decode the first data stream exactly as you would if the second data stream didn't exist (ie. treat the second data stream as being part of the noise), and then once the first data stream is decoded, subtract its effect out from the received signal, and go ahead and decode the second data stream.
Great explanation. Thank you :)
why is it that when we are considering power to be half, energy reduces? (as explained in case of tdma)
I'm glad the video was helpful. I'm not exactly sure what you mean by your question. For a transmitter with constant power, when a signal only lasts half as long (eg. TDMA between two users), then it only has half the energy, since Energy = Power x time.
Thank you
thank you , great help
You're welcome!
Too good. Thanks.
Glad you liked it!
🔥🔥
Many thanks.
You're welcome.
Awesome