 So, so, so this is we are going to begin by looking at I do not think what is this lectures number 35. So, basically it is a tutorial class, I am just putting down this so that I can know what to save. So, okay, so let us look at this guy here which is the problems on miscellaneous topics I will I think I should be able to cut and paste from here to Windows Journal also. So, I will just I will do that so that the problems are also recorded, okay. So, we will begin with convolutional codes because I think that is probably the area that is fresh in your mind, okay, okay, so let us just take this question, okay, so that is a question. So, now all of you can start at the question, start answering, selecting your, so what does this bracket F mean the question showed up in the final expression, so it must be an important question, right. So, it is easy enough, okay, so part A should be very, very trivial, right, how many D flip flops do you need three and then you connect it up, so for part B, you have to change it into systematic form, so what will you do? So, I will just do part B real quick, write down G of D S, okay, you can do it in many ways, so I think I like this better, okay, if you want you can write it in this way, right, then do an implementation, do, you know how to do this, the form in which I did that you put a plus before the first one and then you get W of D and then you do this, it is very easy to draw, so let me draw this, it is for fun, right, then you pull this guy out, this guy out, you get your V1B, okay, V0 is the same as usual, okay, so it is a simple, simple solution for that problem, let us go back and see if there is any other problem, okay, so you also have to draw one stage of the trellis for the systematic encoder, okay, so it is an H state trellis, right, so one can easily do that also, so the eighth question, right here, so ignore this part, I think, I do not know how to make you ignore this part, this part is not relevant, it is actually part of the previous question, okay, so the eighth question also has an F, which means if the game showed up in the fence, can you read it, it is okay, you want me to blow it up a little bit more, I don't know, it just refuses to select, I don't know, I think they have to undo the, undo the, undo the paste, so you guys have, problem is I did so many ink strokes after that, there it is, now I can do this, right, it is probably slightly better, so basically the picture is given to you and you have to do the reverse, it is not too difficult, I will do part B, I think once you do part B, you can also do one stage of the trellis, it is not too difficult, I will do part B because I can write down the answer very easily, what is G of D, same as the previous, 1 plus D part 3, 1 plus D plus D square plus D part 3, now how will you answer the part C, part C is an interesting question, encode the infinite, I am sorry, yeah, so U of D is basically 1 by 1 plus D, and you have to multiply U of D with G of D, so what will you get, so you should use some factoring knowledge you have, so U of D is 1 by 1 plus D, and then I am going to multiply G of D with U of D to get 1 plus D part 3 by 1 plus D, and then 1 plus D plus D square plus D part 3 by 1 plus D, okay, right, and the way I have chosen these polynomials, they will nicely divide, okay, what happens when you divide 1 plus D part 3 with 1 plus D, remember it is all modulo 2, right, so it is almost as if you are working with coefficients from G of 2, okay, so you know how to do this factoring, how will you divide, 1 plus D part 3 by 1 plus D, if all else fails, what will you do, long division, right, 1 plus D part 3, put it inside, put 1 plus D on this side and divide, you will get the answer, right, there is nothing else to worry about, but if you can see the factoring easily then you will cancel out very easily, but long division will work, okay, do not think long division is a bad idea, okay, it is a very good idea, it will work very nicely, and if you do that you will see you get 1 plus D plus D squared and 1 plus D squared, okay, so this is one of those catastrophic encodings, okay, you get an output of weight 5 and the input is of infinite weight, okay, that is that problem, okay, so the next question is also, but it has got only two states and then you have got trellis and then you have to decode, well that is just bitter bee, and there is one thing here which is decoding over the BSC, okay, so I did not talk about decoding convolutional codes over the binary symmetric channel, but it is easy enough, right, what will you do, how will you come, only thing that will change is branch metric, okay, so how will you compute branch metric now, in the BPSK, AWG in case your branch metric was the Euclidean distance between the received thing and the symbol vector, for binary symmetric channel what should be the branch metric, the hamming distance between the received bits corresponding to that stage and the output bits corresponding to each branch, okay, so you do that and then you do the exact same algorithm you will decode over the BSC, right, everything is exactly the same, okay, so nothing else will change except that the branch metric computation will change, branch metric will become hamming distance, maybe I will illustrate that for this example, okay, so this is problem 9, okay, so the trellis is very very easy, it is only a two state trellis, right, the output is itself an 1 plus D, okay, so G of D is what, 1 and 1 plus D, so if you have to do the trellis it is very easy, if you have input 0, output will be 0, 0, if you have input 1, output will be 1, 1, okay, and the next stage the trellis will become complete, so this is 0, this is state 0, state 1, okay, maybe you, maybe it is a good idea to write the state inside the, inside that box, then you can put the state metric below, okay, so 0, okay, so those are the states and if you are at state 1, then you put 0 as the input, you will get 0, 0, 0, 1, right, sorry, and then if you give 1 as an input, it will be 1, 0, okay, those are my transitions and I think there were 4 information bits, right, so I will have to repeat this twice more and then there is transition to 0, 0, 1, this is the trellis corresponding to 4 information bits and 1 extra bit for termination to the 0 state, okay, and I can repeat the same input output for each of those things, I do not have to do that once again, okay. So what are the outputs on a BSC, the outputs I have given are 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, I mean it is not really relevant but I can take any other output also, but I will take the same output just for fun, 1, 1, what does this say, 1, 1, 0, 1, 1, 0, then 0, 0, 1, 0, okay, so when you run the beta B the first thing you do is compute branch metrics for all the branches, in this case you can easily compute branch metrics for all the branches, you write down the branch metrics below 2, 0, 1, 1, 0, 2, and then you write 1, 1, 2, 0, 0, 2, 1, 1, this is 1, this is 2, okay, those are my branch metrics and then you keep doing the state metric from the left hand side, you start with 0 here, okay, right, you figured out how I computed the branch metrics, just hamming distance between those corresponding bits and things, okay, so here maybe I will use the red pen to show the survival paths, okay, if I do, if I look on the other side of stage 1, there is only one path entering each state, so there is no real choice to make, so I can take out the survival paths like this, okay, and my state metrics are 2 and 0, okay, so now here I have a choice, okay, 2 plus 1 is 3 for that path and 0 plus 0 is 0 for this path, so I will pick this path, okay, 0, here 0 plus 2 is 2, 2 plus 1 is 3, so I will pick this path, okay, 2, that is it, keep proceeding like this, you will quickly get to the other side, it is very, very easy, okay, 0 plus 1 is 1, 2 plus 2 is 4, so you pick 1, here you do 0 plus 1, that is 1 and 1 plus 0 is 1, 1 plus 1 is 2 and then you pick this guy, 1 plus 1, okay, so the survival path finally is this guy, okay, so now we can go back very easily and write down what u cap should be, what should u cap be, the path corresponding to this guy, right, so it should be 1, here you have 0 then 0, 0, 0, well the last 0 is irrelevant, so we will stop, okay, that is it, that was your u cap, very simple decision to make, okay, the same thing if you want to do for BPSKWGN, it will be a bit more complicated, all these numbers will be computed but one can do it, actually even for BPSKWGN, you do not have to do the Euclidean distance, you do not have to do R minus S square summation, you can do the dot product R dot S and then what will you do for the state metric, you will have to get the maximum metric, okay, so you have to maximize the correlation, not minimize, so you will pick the path with maximum metric always, if you do that with correlation it will be much faster, it will be exactly the same as what was there, okay, so that is one more idea you can use in an exam situation to speed up your computation, okay, so that is that, okay, so the question 10 is about some several observer, controller, canonical forms which we did not do, so question 10 is technically out of syllabus but let me point out one thing, the G of D you see on board for question 10 is what, it is a 2 by 3 thing, there is two input streams coming in and three output streams going out and one can easily do it, you do u1 of D, u2 of D, then all you have to do is v1 of D is summations coming out but you will see there are two different ways of thinking about drawing the implementation and you will see that those correspond to observer and controller, canonical forms and one of them will be better from a number of states point of view, okay, so that is the brief background to question 10 but I did not do that in this course at least, so you do not have to worry too much about it but that is how the, that is the general, more general example for a general convolutional code which is got more than one input stream coming in, okay, so that is how it works, okay, so those are all the questions I have for convolutional codes and I have had a tendency to either ask or not ask questions from convolutional codes, sometimes I ask, sometimes I do not ask, depends on how the other problems shape up, right, because convolutional codes on one side is quite trivial, nothing, from an exam point of view there is nothing non-trivial that I can ask, right, the only thing I can ask is give you a G of D, you do the real difficult parts in convolutional codes we did not see, we did not see in this class, so the only thing is drawing trellis and doing Viterbi decoding it is quite easy, okay, so there is nothing much to ask in that, any other question on convolutional codes, obviously I can assume safely that nobody has ever seen this set of questions before, right, so none of you are really familiar with these questions, so it is okay, by the way these questions are available, okay, and I have also added a few in the previous exams, I will send you the latest question, okay, so that is all about convolutional codes, let us go back to other fun stuff, okay, so you see the first question here asks you to do simulation of syndrome decoding, ML decoding and bitwise MAP decoding for the 7-4 Hamming code and the 3-1 repetition code and you have to find the coding gain in terms of EB over N0, you will have to do basically BER versus EB over N0 plot, okay, so even if you are, I think this is very useful, just doing this simulation will give you an idea but most of you may not do it, maybe you are doing other simulations but some simulation like this is very useful to do, okay, so question number two actually I did in class, okay, but I want to go through that very quickly, show that the soft ML and bitwise MAP decoders for the N1 repetition code are identical over an AWGN channel under BPSK modulation, okay, so I wanted to spend some time doing that, so basically set up the soft ML decoder for the N1 repetition code and the bitwise MAP decoder for the N1 repetition code and you will see both of them come up with the exact same criteria for deciding between 0 and 1, please do that, take some time and do that, BPSK AWGN and you can also find an exact expression for probability of bit and block error and you will see in terms of EB over N0 it will lie right on top of the uncoated curve, right, if you remember this, long time back we did this when I started off, showed you how N1 repetition code in EB over N0 does not give you any coding here, okay, so do that, okay, what are the code words? The code itself is all zeros and all ones, so if you do BPSK modulation the symbol vectors corresponding to all zeros, all plus one symbol vector corresponding to all ones is all minus one and suppose you receive a received vector R which is R1, R2, Rn, okay, if you do ML decoding, soft ML decoding, what will your C hat be? How will you decide C hat? It is argument of, yeah, so you can do it in the simplest way as dot product, right, I also showed you how minimum distance is the same as this for the argument of maximization over the symbol vectors R dot S, okay, right, so what condition do you get finally? Can you simplify this condition? So that is the condition, if you do this, you will see finally you will say C hat is all zeros if what? Summation of RI is greater than zero, okay, it is all ones else, okay, so the only thing you have to compute is summation RI, okay, if you want to decode maximum like, if you want to do maximum like decoding for the repetition code, you have to do summation RI, if it is greater than zero, then you conclude that what was transmitted was zero, otherwise you conclude what was transmitted was one, okay, what do you do for bitwise MAP? There is only one bit to decide, right, there is only one message bit, so we can take that as bit one, okay, so if you want you can take that as bit two also, all of them are same, they are supposed to be the same message bits, take just one bit, okay, and then you have to find what? Probability that the bit was zero given the entire received vector, how do you find the ratio of those two things and then there is a formula, right, I gave you a simple formula with capital L and small L which you can use, okay, so if you do that you will see capital L zero is the same as what? L zero, L one, L two, okay, or L one, L two, L three, I do not know, I think I have been numbering it as one, two, three, no, I do not want to put zero suddenly, do you get that, what is L one? L is e power two RI by sigma square, that is the likelihood ratio for each bit, final likelihood ratio though, a posteriori likelihood ratio will be the product of each of those things, okay, so in the numerator you will have all terms which have zero in the first bit, denominator you will have all code words which have one in the first bit and then you divide throughout by what? So you only consider zeros, right, if there is one it becomes, you replace that with one, why, why, because I will divide throughout by the product of those, the one probability, so all of those things will cancel to give me only likelihood ratios, so you can write the whole thing in terms of likelihood ratios, so only the zeros will matter, okay, so if you do that, if you remember that formula you will see capital L one becomes L one, L two, L three, and if you use that simplification you see this also becomes e power two by sigma square into summation RI, now what will your decision be? Here you have to say it is all zeros if L one is greater than one, okay, and if L one is less than one you say it is one, this is likelihood ratio, if you take log of that then you can say greater than zero, less than zero, and you see it matches exactly with the other condition because sigma square is positive, right, so everything depends on the same, okay, same as ML, okay, so both soft ML and bitwise MAP for the repetition code will do simply summation RI, if it is greater than zero you say it is all zeros, if it is less than zero you say it is all ones, okay, so to find probability of block error, I think I did that also, I will quickly tell you how to do that, you say find probability of error given that all zeros was transmitted, since the whole thing is symmetric you can say that will be the same as the overall probability of error, and given all zeros are transmitted each of the RAs become IID Gaussian with mean plus one and variance sigma square, summation RI becomes a Gaussian random variable with mean three and variance three sigma square, it is like people are very comfortable dealing with IID Gaussian random variables, okay, so once you come to three sigma square then probability that sigma RI is greater than zero is very easy, it will be Q function, Q of three by root three sigma, okay, so I will get Q of root three by sigma, but then if you adjust with EBO or N not, right sigma in terms of EBO or N not, you will simply get Q of root two EBO or N not, which is the same as the uncoded case, okay, so you will finally get probability of error to be Q of root two EBO or N not, which is the same as uncoded, so you can conclude repetition codes do not give you much coding gain, okay, so the next problem, yeah, so three and four are quite similar, as in you are given a code, then you have to come with soft ML and bitwise MAP decoders, so let me just do, let me do what should I do, let me do three, three looks like a little bit more, slightly simpler problem, okay, so you have to always start when you do ML, MAP and all that, what is the first step, what do you need to know for doing bitwise MAP, ML decoders, you have to know, you have to enumerate all the code words, okay, so if you do not enumerate all the code words, you cannot write down the expressions very easily, okay, so you enumerate all the code words, okay, go ahead, enumerate all the code words for this, so the way it is written down, it is systematic, where will you put the message bits and where will you put the parity bits, the way it is written down, first three are parity, the last two are message bits, right, so you put the message bits down and then you compute the code words, the first code word is the easiest code word to write down, okay, the other code words are slightly more difficult but in this case it is quite simple enough, okay, so one can write down very easily, okay, so check my calculations, make sure they are right, okay, so from here you can come convert to symbol vectors, okay, so if you convert to symbol vectors you will get all plus ones here, you will get minus one, minus one, plus one, plus one, minus one here, so on, okay, so those are the symbol vectors, I am going to do that in my head, I am not going to write down the symbol vector conversion, okay, suppose you are given a received vector R1, R2, R3, R4, R5, how will you implement a soft ML decoder, okay, the formula is the same, argument of the maximum of all the correlations, so you have to compute the four correlations, what are the four correlations you have to compute, okay, so the correlations you have to compute are, okay, so maybe I will call them A1, A2, A3, A4, A1 is what? First correlation is with the all zero code word, that will be simply R1, R2, R1 plus R2 plus R3 plus R4 plus R5, second correlation I have to compute is with this gate will be minus R1 minus R2 plus R3 plus R4 minus R5, the next correlation is with, okay, so it is quite simple, simple, right, okay, so one problem that I have asked before in one of the previous exams is how to, what is the minimum number of additions you have to do to compute these four correlations, okay, so not in an optimal way, but at least in a rough way, how many correlations, how many additions do you think you need to compute all the four correlations, one brute force way is what? 4 times 4 which is 16, if we do 16 additions, don't worry about the multiplication by minus 1, okay, so we will say that is trivial, we will assume that that is very trivial, doesn't cost anything, okay, additions cost a lot, okay, so I love to say 16 is one number which is brute force, can you reduce it in any way, what can you do? So just something, come on, can you see something, compute the full sum that is four additions and then you do what? Okay, I don't know, I will tell you what I observe which is very simple to do, okay, so you notice you can do R3 plus R4, that or minus of R3 plus R4 appears everywhere, another thing you will notice is R2 plus R5, that or minus of R2 plus R5 will appear everywhere, okay, so you can do simplifications like that, so it is enough if you add R3 to R4 once and R2 to R5 once, that is two additions, right, then after that you have how many additions per thing, two more additions after that, okay, so 2 times 4 is 8 plus 2, 10, I can do it in 10, okay, that is one way, I am not saying that is the simplest, maybe there are more smart ways of simplifying that further, okay, but definitely 16 is not the simplest way, there are simpler ways of implementing this, 10 additions one can imagine being able to do this, okay, is that clear, okay, so a question like this might be interesting to ask, okay, so how do you simplify the number of additions, so in fact I think you must have seen this question somewhere, the number of marks will depend on the number of additions you actually have, it will be something minus the number of additions, so you decrease the number of additions you will get more marks, okay, so that is the incentive for you, okay, alright, so that is the ML decoder, so the ML decoder is not complete, after this what do you have to do, you have to find the maximum, okay, that will involve three comparisons, can you find the maximum with three comparisons, yeah you can find, okay, so you do three comparisons, you find the maximum, once you find the maximum you can go back and output the most likely code, okay, so that is what you do for soft ML, okay, what do you do bit-wise MAP, can you write down the likelihood expressions, it is enough if you do likelihood expressions for what two bits, the two message bits which are the last two, okay, so you compute capital L4 and capital L5, go ahead and write down those expressions, okay, so let me copy the code word, codes, okay, so what is the expression for L4, okay, so the first term is the easiest term, right, in the numerator, L1, L2, L3, L4, L5 plus what is the other term, L3, okay, divided by the terms which have one there, L2, L5 plus L5, am I right, I am sorry, what did I do, second term is, L3, L4, alright, okay, I was mentally pulling the L4 out, okay, you are right, you are right, okay, so what is L5, capital L5, again the first term is very easy, L5 and then the next term would be L2, L5, the denominator you have, L3, L4 plus L5, am I right, am I right, okay, alright, so once again a question of how many multiplications you need and how many additions you need here, okay, and assuming addition, which is L1, oh the denominator it is L1, oh in both cases it is L1, yeah you are right, actually you are right, you are right, you are right, okay, so then the question is if you assume multiplication and addition are the same cost, which is better, whether MAP is better or or what, ML is better, so all those questions you can answer based on competition you can do that for these small codes, you cannot do it for large codes, for large codes they both are equally hopeless but for small codes maybe there is some gain that you will get from this, okay, there is also a question of doing the BSE thing, right, for BSE the syndrome decoder is optimal, you just do the syndrome, it is very easy to come up with the syndrome decoder, not going to do that, and then the question is can you come up with an example of a received word where the syndrome decoder, soft ML and bitwise MAP actually give you different outputs, okay, it is I think possible, but it is usually very difficult to come up with some examples, such examples, okay, at least soft ML and syndrome it is actually quite easy to come up with an example, for the bitwise MAP it is more of a pain, because the expressions do not simplify and you cannot keep evaluating e power and multiplying, but you can easily come up with an example where the soft ML and the syndrome decoder will differ, okay, so we can easily come up with an example, okay, so let us see where shall we go next, we have time here, okay, okay, so the next question, question number five that you see on the screen there is basically the, basically derivation of the LDPC update for the check notes, okay, so you remember if you look at the formula, it is very, very similar to the formula you have for the check note update in LDPC decoding and this is a derivation of that, this is a derivation in a different way, it shows you that this, the way I derived it, you remember the way I derived it was this tan H rule, right, showed you how the difference of the probability multiplies when you do these parities and then you get the tan H rule, okay, that seemed like little different, but you are calculating probabilities, which means you should be the same, it should be the same as doing some kind of bitwise MAP decoding for the even weight code, okay, it should be the same as that and that is what this derivation is all about, but it is not too critical, I am going to just leave it at that saying that this is similar to the check note update for LDPC codes, okay, so question number six is very similar to the question we saw just now and you can see the marks is 10 minus NA minus NC, okay, can you see that, so NA is the number of additions you did and NC is the number of comparisons you did, okay, so I think for this question you can get maximum of six marks, so the sixth question you can get six marks, I think you can do this whole decoding with two additions and two comparisons or some such thing, it is possible, it is a very simple code, okay, all right, so I think that is, okay, so we are ready to jump to problem 11, okay, look at that problem 11 and try to answer the question, maybe you can read it, you cannot read it, let me zoom a little bit, it is better, we will do one more zoom, okay, so this is a relatively simple thing to do, but I did not do it in class, so I am going to do that right now, so the question that is asked is you have to consider two codes, you consider a reach Solomon code, okay, N is 2 power M minus 1, T error correcting, so what should K be? 2 power M minus 2T and the minimum distance will be 2T plus 1, this is your NKM D, okay, so you can correct T symbol errors, okay, so you are using this over a binary symmetric channel with transition probability P, okay, so if you have a binary symmetric channel with transition probability P, what is the probability of a bit error, it is probability of a bit error, if I transmit a bit what is the probability it will be an error, P, it is true, what is the probability of a symbol error, if you have each symbol here has M bits, right, what is the probability that the symbol will be an error, 1 minus 1 minus P to the power M, so that is the first thing you compute, right, this is the probability that a symbol is an error, okay, so once you have this probability, say I am doing only part B, right, part A is coming up with parity check matrix, that is that random one structure, 1 alpha alpha square and then 1 alpha square alpha square, so on like that, so it is very easy to do, so you go for binary, the BCH code you do it slightly differently, but for reach Solomon code you go all the way from alpha to alpha power 2T, right, that is what you do, that is your construction and this is your probability of symbol error, then when will you make an error and bounded distance decoding, whenever there are more than, what is your symbol error correcting capability, T, okay, so if there are more than T symbols in error, then you will make an error with bounded distance decoding, okay, so what is the probability, see, what is, this is the probability that one symbol is an error, the errors in two symbols are completely independent, okay, so if you take n symbols together, the distribution for the number of errors will be what, number of symbol errors will be what type of distribution, you will get a binomial distribution with n trials and probability of error for each trial being PS, okay, so you can easily compute the probability that there will be more than T errors, okay, so that is the probability of block error, equals probability that greater than, strictly greater than T symbol errors in n symbols, okay, so this is a simple binomial expression, you sum from say j equals k plus 1 to, no, not k plus 1, sorry, T plus 1 to n, right, what do you have to do, n choose j PS per j 1 minus PS to the power n minus j, so that is a very simple expression that one can write down for probability of block error for the Reed-Solomon code, okay, is that clear, yes, it is a very simple expression, suppose I have had to worry about what would happen to this expression, for instance what happens to this expression as P tends to 0, okay, do not tell me it becomes 0, okay, yeah, of course it becomes 0 when P becomes 0, okay, what is the good approximation for very very low values of P, m times P, do you see that, see 1 plus x to the power something, a very good approximation for x being very very small is 1 plus mx, right, it will be 1 plus mx and then some terms involving x squared, x squared you can ignore compared to x, okay, so that is a good approximation, so here it becomes 1 minus mp, then when you do 1 minus that you will get PS becomes approximately m times P, okay, so that is a good approximation as P tends to 0, so now when P tends to 0 PS tends to m times P, further what can you say, can you approximate this any further, what will be the leading term, PS per j with, yeah PS power what, what is j, j varies, yeah you have to ignore compared to a particular power of PS you have to ignore all lower powers, you can do that when PS becomes very very small, so what is the dominating power PS, PS to the power t plus 1, so that term will dominate, okay, so every other term will actually go off to very very low value, one can one can imagine as you as it becomes lower and lower that is the only term that will dominate, okay, so of course you have some coefficients multiplying here, these n choose things also are a little bit dangerous, okay, so you have to be watchful of that but in general one can say PS power t plus 1 will kind of dominate this expression, so it is not a bad approximation, it is a reasonable approximation, okay, alright, so the other question is for the BCH code, so BCH code is different from the RS code mainly in two things, first is the number block length is in bits, secondly what is the issue, the dimension changes a little bit, right, what is the dimension, so the BCH code what is the dimension, n remains the same 2 power m minus 1, what is the dimension, roughly 2 power m minus 1 minus mt, okay, I know this is a rough thing because it is guaranteed to be greater than or equal to this, okay, but it is usually very equal to this and then here again minimum distance is greater than or equal to 2t plus 1, in most cases it will be equal to 2t plus 1, if you take higher rate examples, out of 2 power m minus 1 bits, okay which is n, how many bit errors can I correct, t bit errors I can correct, okay, so probability of block errors is actually relatively simple to write down here, it will be simply j equals t plus 1 to n, what, n choose j, p to the power j, 1 minus p to the power n minus j, okay, alright, so these two are different in one major way, okay, so the block length here is n symbols, okay, and symbols is how many bits, n times m bits, it is a lot of bits and the probability of error is actually controlled by the probability of symbol error as opposed to the probability of bit error and either case, okay, so that is how the things will differ, okay, so let us see if we can squeeze in one more problem, okay, so these problems are slightly more difficult, I want to take time over it, so let me not do that, okay, we will do problem 13, 13 is an interesting and simple enough problem, okay, so I have got a t error correcting rs code of length n over g of 2 power m, okay, I want you to determine the exact burst error correcting capability of c in bits, assuming you are doing bounded distance decoding, okay, what do we mean by burst error correcting capability, if there is a sequence of errors, consecutive errors, what length can I tolerate, okay, can you come up with the exact number, is it tm, am I right if I say tm, no, right, see it depends on where I start, no, I have to be very careful, okay, so what is the exact expression in terms of t and m, come up with the exact number for the maximum burst length that is correctable, what is the, yeah, just request some t minus, t minus 1 times m, is that good enough, see depends on where you place yourself, you can knock out t minus 2 consecutive symbols and then knock out 1 bit to the left and 1 bit to the right, that way you would have knocked out, well that is not right, so maybe t minus 1 into m, I do not know, think of it carefully, so remember what do I want, I want to come up with a number, right, what is burst error correcting capability, all bursts of length burst error correcting capability or lesser should be correctable, okay and there should be some burst of length burst error correcting capability plus 1 which is not correctable, both should be there, okay, so what is the minimum length where I can get a non-correctable thing, should be t minus 1 times m plus 2, okay, so I can always get t minus 1 times m plus 2 to be non-correctable and any length of t minus 1 times m plus 1 will be correctable, okay, is that right, think about it, okay, this is my answer, think about it for a while and let me know if you can find a burst which is not correctable, I am sorry, why do you need the plus 1, I am sorry, yes, why plus 1, that is the maximum number, you want to see the thing is if you have, each of these guys are symbols, maybe I am wrong, let me think about it carefully, right, each of these guys are symbols which is m bits long, right, m bits m bits, okay, right, I want to have a burst of errors affecting t plus 1 symbols, I want to affect t plus 1 symbols, what is the minimum burst length which will affect exactly t plus 1 symbols because if it affects t plus 1 symbols then I cannot correct it, okay, that will be t minus 1 times m plus 2, okay, you take t minus 1 m, suppose this is t minus 1 times m, okay and then you, then you, right, all these are in error, then you make an error here or make an error here which is a consecutive burst and you have affected t plus 1 symbols, right, so t minus 1 times m plus 2 is not the burst error correcting capability, t minus 1 times m plus 1 I can correct, any burst of length t minus 1 times m plus 1 I can correct or below that, okay, if it is plus 2 I cannot correct because I can come up with an example where this is hitting me, okay, so this whole burst, if it is only plus 1 nothing else will really work, you might say why did I exactly cover t minus 1 here, maybe that is shifted but if that is shifted then I lose only, right, I will only be able to do t, I cannot do more than that, this is the worst case possible, okay, part b is an interesting thing, you take m code words, capital M code words and do row column interleaving, so what is row column interleaving, you take code word 1, put it as first row, code word 2 put it as second row, likewise put m code words, row wise and then you shift out, column wise, okay, so if you take the first m symbols in your final output they will correspond to different code words, okay, so in that case what is the exact burst you are correcting capability, try to do a similar calculation, t, sorry, no capital M will play a role, I mean how can capital M not play a role, so why is capital M disappearing from your answer, you have to have capital M in your answer, right and you should have small m in your answer, how can you not have those things in your answer, you understand I am doing symbol level interleaving across different code words, right, across different code words, I am taking capital M code words, so whatever answer you have if you put m equals 1 you should get back this, okay, if you do not get m equals 1 then you are making some mistake, that is a sanity check, go back and check that, did you understand this interleaving I am talking about, okay, take a simple example, it is very easy to see, if you affect a burst of symbols, only say for instance if I do this interleaving, if m consecutive symbols are erased or are in error, okay, only one symbol per code word will be affected, but if I do not interleave, if m consecutive symbols are in error, then m symbols of a code word can be in error, okay, so you get a gain in burst error correcting capability if you interleave, okay, so what is that exact gain, just multiply this with m, are you sure, oh well it is not quite that simple, think in terms of the way I thought here, what is the worst case situation in which I will make t plus 1 error in at least one code word, okay, in any one code word I have to make t plus 1 errors, I do not have to make t plus 1 errors in all code words, any one code word if I make t plus 1 errors I am done, right, the whole decoding will fail, put down a worst case situation like this, okay, so the answer for some reason for me it seems like this should be the answer, okay, am I right, am I wrong, okay, so actually the exact answer is relevant in many applications, enough if you have the ballpark number, ballpark number is always mt, okay, and if you do m depth interleaving it is capital m into small m into t, that is the ballpark number, but it is good to know the exact expression in case you are doing the design, some cases you will have to do it properly, how many of you like this expression, right, does it sound right, okay, it sounds correct, right, so you take t minus 1 interleaved things, in each interleaving you have capital m symbols which are from different code words, capital m times m times t minus 1, you will hit t minus 1 in everything and then you can hit one guy on the right, one guy on the left, is that enough, that might affect different things, no, one guy on the right will be the first code word, one guy on the left will only be the last code word, no, so it is not quite right, there are more changes needed here, okay, see you are doing it in a sequence and then on the right hand side you will have a symbol corresponding to the first code word, on the left hand side you will have a symbol corresponding to the last code word, okay, so both of them will only get t plus m, right, so m looks like is the, right, m or m minus 1, it should be m, okay, it should be capital, it is m times, no, that is not no, no, I think it is more than m, I do not know, think about it, okay, think about what needs to be added, we are running out of time, so we will stop here, but think about what needs to be added to this, or maybe this is not the right way of doing it, capital m into m, why is one wrong, think about it, okay, let me stop here and we will, what do I stop the recording, oh, I have to do this one.