Lecture 16

[Music] okay welcome to the class what we are going to learn today is the receiver design okay for the transmit chain we have learned so what we have covered till now is basically the pdsh transmit chain and this becomes the pdsh receiver chain okay just for to complete the understanding okay where will pdss transmitter chain be implemented it will be implemented on the base station okay the down link transmitter chain who is the transmitter in the down link base station and who is the receiver in the down link UI okay so what we are learning here is basically the uee processing now so that was the base station processing whatever we have learned till now and we are going to learn the UEI processing okay there's only one part of the UEI processing it is much more complicated than not I'll show you okay because UI needs to do lot many more things before it can actually do this okay but hopefully when those modules have been implemented they have been executed okay so you have reached the stage where in like you have to process the data okay so we'll touch upon all that so this basically this receiver designed for the transmitter chain which is discussed that is pdsa chain okay and then we are going to see the PD receiver chain so similarly there is this base station transmitter if it is Uplink okay what happens that is called the P chain okay and physical Uplink shared Channel okay this part of the processing whatever we have learned is more or less similar in both These Chains okay right whether it's pdsa chain or P chain okay so this part whatever I showed you okay let's come here okay before we actually get into this pdsh receiver this part okay this is pdsh transmitter okay this becomes pdsh receiver okay so this basically is the split okay this is trans base station this is the UEI okay now if you flip it okay if you start talking about the Uplink okay then this part of the chain remains same there other modules which come which will be different okay so then if it is PS transmitter who will implement this UE will implement this and then this will be implemented by the base station okay right so please understand this now what is so where will the transmitter change begins from raw data it ends at the modulated symbols the pdsh receiver chain where will this begin begin from the Dem modation and then it has to give you this the raw data okay that's the process it's the reverse of what you have learned of the transmitter correct okay so we begin from here okay so we understand what kind of modulation schemes standard gives standard gives these modulation schemes starting from 4 to 256 quam okay now it has to be demodulated these s symbols need to be demodulated because that's what we are seeing Okay so and what is the job of this demodulation what should it do okay demodulation should basically look at these symbols because modat symbols okay we saw that okay if it is qpsk there will be four possible symbols if it is 16 quam there will be 16 possible symbols okay and now the question you could ask is okay if the mapping at the modulation is so clear what is the job of the demodulation okay right so that's we'll see that okay that so why demolation is complicated than modation okay because demolation has to decide okay and it has to decide based on what kind of received signal it is based on what kind of signal it is receiving okay so that's what the job deodor detects bits from these modulated symbols okay right and now you could ask me this okay that ldpc decoder okay works on the demodulated bits and not on the symbols we discuss this briefly in some form okay now you could actually have a ldpc decoder working on a symbols directly okay now why it is not done again because of the implementation constraint not practical to design decoder for different modulation schemes okay four different modulators you'll have four different demodulators four different decoders also okay so in practice people do that not in the standard but if you want to implement some other simpler communication systems okay you can take the directly the demodulate the symbols then you can feed them to the ldpc decoder okay it will work so but not practical to design so that's why basically this demodulation is there so that it gives you bits okay so now what is what is happening okay let's we'll take an example of Q pskd modulation these were the four constellation points you transmitted one of them if you're right okay if you look only at one symbol okay right let's say you Transit one of them what happens here you will not receive exactly one of them why because what is the reason okay how did your channel look like or how did your system look like okay right you transmitted if x is the modulated symbol what kind of transformation it under under goes it under goes it gets multiplied by H which is complex it gets multiply added by the noise gets added okay right so what happens let's say you transmitted one of these points you will okay now depending on what H is okay let's say h is you are in the basement of a building h becomes very poor H becomes zero it can okay right H becomes very what will happen you transited X it become zero right okay forget if H is also not there okay and then if if H is one just H is very strong okay if only noise is there okay right so you they're in a poor SNR regime what is SNR the signal power is much lower than ratio of signal power to the noise power noise power is very high when compared with signal power okay what will happen you transmitted this it will move here or it might move here also it might move here also fine that's how your Communication channel is okay now if this noise is not there you don't get your jobs as an engineer okay why because whatever you transmitting you're receiving without errors why will you need need ldpc encoders why will you need all these encoders or decoders sophisticated processing right so we have our jobs because noise is there and the channel is there okay right so now what is the job of this qpsk demodulator it looks at this received symbol okay this is what it has received okay now based on this it has to decide which of this point was transmitted correct okay if it can decide then then what will it do if it decide that this was the point which was transmitted then you know that this basically it is easy to demap this constellation point to these bits because that mapping you know right okay so what I'm saying here when I say that that mapping you know is basically this constellation is known at the receiver also okay it is not known what was transmitted okay correct so what I'm going to tell you a very simple rule okay this comes from in the communication Theory okay and what is this rule this is called in order to detect the bits from the symbols okay right what do you do you apply something all this nearest distance detection rule what is this okay in this in this case okay you will compute this distance from of this received point from all these transmit points okay right and whichever is point gives you the minimum distance you declare that as transmitter Point okay right and like one can show you that this is the best you can do okay right best in terms of reducing the error okay the best has to be defined appropriately okay so this decision rule basically gives you the minimizes the bit errors okay now once you have basically decided based on this minimum distance rule okay right that this was a point which was transmitted then you do the reverse mapping okay do we understand this okay first is basically apply the minimum distance Ru and then threshold so once you have decided this minimum distance you basically what will you say okay I'll you'll declare it you'll threshold it to that that this point was transmitted okay once this you have declared that this point is transmitted then it easy to do the reverse mapping now keep this in mind that okay this is a final Point d map the symbols into bits once this is decided okay now I'm taking a liberty and while telling you this okay this is something called the hard decoding okay where in what we are doing here is we receive this point and we are thresholding it to we declaring that this was the point which was transmitted okay or whether this was transmitted we are deciding okay at this stage itself okay this is called hard decision decoding if time permits I'll teach you about soft decision decoding also wherein you don't decide like this okay you calculate the probability of whether this symbol was transmitted or this symbol was transmitted or this symbol was transmitted or this symbol was transmitted okay right so in this class I'm not going to do that okay that like the soft decoding is better than this hard decoding it gives you performance gains okay we'll see that later okay now but for this class what are we doing what is demodulator doing okay it is taking in these symbols and then deciding which of this point was transmitted okay one of these points will be transmitted right okay correct and then based on that okay once this is decided let's say in this case because this is what will give you the nearest distance okay right and then you will do the demap the symbols into bits this mapping depping is known what is not known is what is deodor needs to decide is basically whether this point was transmitted or any of these four points were transmitted and there is a rule now I've given you the rule okay this is the nearest distance rule okay correct and this is called maximum likelihood decoding I'm not going to the details of this but that's what it is okay fine so that's how it is done that's the job of demodulator what has demodulator given you now bits correct okay and those bits you process and you do the reverse processing at the receiver okay what is the reverse processing you need to do the demolation is done fine okay then the instead of Scrambler you have a you have to descramble those bits okay how are you going to descramble those bits so what do you need to descramble one thing is that those these demodulated bits there's something else you need to descramble and what is that c pseudo random sequence okay so receiver has to generate that Pudo random sequence exactly the same pseudo random sequence which is generated by the transmitter and that information is known to the receiver also okay right otherwise it will not be see it will fail if receiver does not have that information it'll fail okay because if you use it a different pseudo random sequence it'll give you a completely different output okay so that information with like like which pseudo random sequence to use that is passed by the base station to the user okay through some other channel this called control Channel we'll study that also okay but for this processing we are what we are assuming here is that that sequence is known okay right so what is what is this there are scramble bits okay and scramble bit output was blda right this is what this demodulated output is going to give you it is not going to give you B it is going to give you B till what this is right but you need to get unscramble them and you need to get B from this how will you do that okay how will you do it you you scramble you have this blda right correct this is what we have this is what this Dem modated output will give you what will I do I'll again exort with the same sequence correct okay so what happens here so what is if you substitute we'll see how will you get B of because what is happening here okay B of I is B of I I'm writing it like this okay right we substitute it here I think if I have explained it in the next slide I'll not do it here okay I'll just write it here okay so what happens here if you substitute B over here okay why it how it how it becomes equal to B of I do we understand this okay I writing B of I is B of I exor with C of I okay now let's substitute what is B of I let's not assume that this is equal to B of I what is B of I this is B of I exor with C of I you are again exing it with C ofi correct okay now so you exing let's say so what is this I look at it like this C of I exal with C of I what will you get a bit stream of zeros because you exal the same sequence exal the sequence with itself you'll get a string of zeros correct okay so what is happening B of I exal with a string of zeros which is of the same length of B of I what happen B of I you exort anything with zero you get back the same sequence okay right so this is B of I that's how you get this B of not okay correct so this the idea okay you ex with the same sequence and that's why you need c of I to be known as the receiver fine okay so this is the example if you have understood that math this is very simple to understand okay so 8bit Cod is sequence B of I this is C okay so you get B TAA is this ex what is B TAA B of I exal with C ofi right okay so what you get here is 0 exal with 0 is 0 one exal with one is 0 0 exal with one is one there a scramble sequence how will you get Des scramble sequence you scramble it with again C of I and of course you'll get the get back B of I that's what I showed you in the last slide because C of I exard with C of I is zero and you'll get back the same sequence B of I simple whatever you did at the transmitter do the same thing at the receiver okay but then Scrambler has met his objective which is basically that you it maintains average energy constraint done the scrambling also done it is exactly the same as the transmitter okay right so what is the next module which follows after that so you look at this there's no notion of code blocks correct okay right but what follows after this is everything is basically happens in terms of code blocks there should be this code block segmentation module which is basically this right which should come here why should be there because what is segmentation doing it is deciding how many what is the size of each of the Cod block you need those numbers right okay you need those filler bits okay and the number all these numbers you should know okay so what did code block concatenation do okay rate match codes was it was they concatenated okay now you need to segment them okay and segmentation is basically you should know e right because what did you do you put C code blocks of size e together so you should know C you should know e you should know G who going to give you all these numbers segmentation right so receiver should exactly have the same control information okay right so with that hopefully that has been done because that's how it can segment okay so that information is needs to be passed on and that is basically all this information has to be there with the receiver you see how complicated it is okay receiver needs to know all that then only it can start this information this decoding that's what I said okay the receiver has done all this processing okay then it has this information then it needs to do this processing you see how complicated your mobile is okay right so now so with this basically you have these four rate matched code blocks correct okay and this rate is how much is the rate this rate is what was asked by the user correct okay fine so we have how many four such rate matched code blocks correct and then the this is basically 1 1 3 40 into 4 okay right so that is g and this is e once this e is there so what is happening we have to De inter leave okay once this code block has been segmented we have to De inter them okay so this is continues so basically what are we doing we have four rate match code block code block segmentation output is four code block of length 1 1 3 4 0 okay total output bit is 4 53 60 this is again the numbers remain same okay now this process happened you have how many code blocks four code blocks what do you need to do de intering right what is De interleaving you have to look at how inter leing is happening okay you to do the reverse processing of that how was this inter leing done it was intered using row column interl what did it do this correct so you need to know what modation scheme was used of course correct okay and then what will happen what are you going to do what will this Dever do okay it's going to just read them appropriately right you have to write four lines of code read them appropriately what are you going to do in this case you take every alternate bit put it like this every alternate bit put it like this then how you going to read it you're going to read them like this do the opposite of this what you did at the trans that's what is the intering right that's why it's not specified in the standard none of this is specified this is of course comes from communication Theory this is basic common sense right okay why will anybody specify that there's nothing to specify here okay right so this is basically you have done the de intering by reading them appropriately nothing else okay so de intering is done this is most interesting okay how is you have rate match how you do the rate recovery that requires some again Common Sense nothing more than that okay we'll see that okay so what are this Common Sense rate matching what did it do what did either punched the bits if the rate which was asked by the user is higher or repeated the bit if the rate which was asked by the user is lower than 1x3 that's what it did that's all right okay now let's say you look at this punching what are what is happening you're not transmitting certain number of bits right which were generated by the rate 1/3 ldpc encoder as I said why do you need this rate recovery because this ldpc decoder also works at rate 1x3 what does that mean it will Taken n Bits And transmit it'll give you K bits but you not transmitted n Bits correct how many bits you have transfered you have transfered e bits correct okay okay right so those e bits that information that rest of the bit information is not there at the receiver because you can not even transmit them correct what do that Common Sense tell you because you don't have that information about whether it was zero or one what kind of information you will give for n minus E bits some kind of neutral information right okay so I'll come back to it okay in a second what is this neutral information okay so ldpc decoder Works only for code rate of 1x3 right okay code rate should be reverted back to 1x3 because it works at 1x3 and when I say it reverted back to 1x3 it needs n Bits And so that it can give you K bits that's what it rate is K by n okay right how do you generate this rest of the N minus E bits do we understand this okay if it is redundancy version rv0 okay so you need to implement this circular buffer at the receiver also okay right so if it is redundancy version zero right okay what will happen here how many if it is let's I'll draw this okay how many bits you transmitted this is n right bottom is n this is K this is what you transmitted you do not transmit this okay because that's what was asked by the user okay now user will have to decode only using this information what will it do you don't have this information and that's what I introduce the notion of neutral bits okay right as a transmitter because what can you transmit as the transmitter either zero or one and what is anything in between if you say okay you fill in these bits will let's say zero what are you telling the receiver that zero was transmitted you're biasing the receiver in front of in favor of zero if you feed in one right if you write one okay what will happen here you're biasing the receiver in favor of one you're giving wrong information to the receiver receiver will understand that one was transmitted okay if you write zero over here receiver will understand that zero was transmitted but that's not the case what is the case then these bits were not transmitted okay right in order to not bias the receiver you give this neutral information what is this neutral information half in between right okay instead of feeding zero or one you write half correct okay ldpc decoder can work with non-binary information okay that's what you have to understand okay so ldpc decoder can work with non-binary information okay so instead of feeding zero and one you should feed half right and that's you then you are giving correct information to the ldpc decoder any other information is not correct because then you are biing the receiver in favor of that information okay so that's what I said okay so write neutral bits for bits puncher at the transmitter okay correct that is case number one what happens for case number two when you basically you repeated certain bits also right okay correct so what will happen here is I'll right what did you do when the rate was lower than what 1 by3 what did okay your buffer is only of this length correct okay this is basically what is that n right this is K okay but you're transmitting more bits than n that's what is rate lower than K by n right okay K by n is 1x3 but if you increase n right then the rate reduces that's what we have learned okay so but receiver can handle only n n Bits the ldpc decoder can handle only n Bits so what do you do you go back and add those bits to what you have already written there are two different extremes okay one is you did not transmit some information because the rate was higher other case was you transmitted extra parity bits okay which has reduced the rate but ldpc decoder cannot handle that reduced rate it can only work at 1x3 how will you revert it back that lower rate to 1x3 you just combine those bits again at the receiver right okay fine because user has asked for a lower rate if you don't do this combining then you're making your ldpc decoder work at rate 1x3 user has not asked for 1x3 it asked for a late lower than 1x3 you need to revert it back to that lower rate and how do you do that you simply add those bits that's the idea okay I hope you get it okay right that's so repeated bits are combined okay fine then what do you do you decode it because what is r done now okay let's go back okay so this feed this rate Mash bits fine to the ldbc decoder because it's again 1 by3 depending on what processing you did here okay so the feed these rate match code block to ldpc decoder fine it'll give you K bits okay validate CRC of each code block there a code block validation I've given you I've told you how to validate the CRC right CRC of each code block is validated then validate transport block CRC if transport block CRC gets validated f a job is done otherwise hybrid arq will be triggered right and then you'll go to the again processing that processing remains same okay signaling for hybrid arq is different okay in the sense that a knack a has to be sent Knack has to be sent that's a part of the control chain okay I'll teach you separately that okay but that's how the receiver processing will be done if everything goes fine okay goes to the Macer has got gotten a data you have gotten your video okay okay you gotten your voice correct okay if not if it is voice hybrid arq doesn't kick in right you'll ask yourself do the hybrid RQ manually okay you'll ask the transmitter to transmit it again right what did you speak okay otherwise for data I kicks in and then the same process happens okay so we'll stop here this is the receiver processing okay we'll continue in the next class thank you [Music] [Music] [Music]