1 00:00:14,160 --> 00:00:19,950 In the previous lecture you were looking at the crossbar and how the cross points are 2 00:00:19,950 --> 00:00:28,300 operated. So if you remember then I had actually drawn a state diagram there by which we can 3 00:00:28,300 --> 00:00:35,280 build up a control circuitry for a cross point. So let us look at that and see how we can 4 00:00:35,280 --> 00:00:39,620 design a cross point circuitry afford is a very trivial exercise, but we will still do 5 00:00:39,620 --> 00:00:40,620 it. 6 00:00:40,620 --> 00:00:41,620 Unable to capture the image because lecturer writes and erases the content immediately. 7 00:00:41,620 --> 00:00:49,071 So the state diagram was something like this, so there was an OFF state when the cross point 8 00:00:49,071 --> 00:00:54,920 should not be operated or in fact it will always remain in cross point is OFF and ON 9 00:00:54,920 --> 00:01:01,430 state I am using another state called intermediate state only to ensure that R goes first and 10 00:01:01,430 --> 00:01:07,840 then the row actually goes first and then the column is activated. So that it goes into 11 00:01:07,840 --> 00:01:14,100 ON state if the column goes first and row goes second, then it will not go into ON state. 12 00:01:14,100 --> 00:01:19,270 So that is the reason why we actually have also an intermediate state and we also have 13 00:01:19,270 --> 00:01:25,750 an ON state and you can actually map or compare this figure what I had drawn in the earlier 14 00:01:25,750 --> 00:01:36,020 one. So in fact this is RC the row and column control which is will lead to this state is 15 00:01:36,020 --> 00:01:48,330 00, 01 and 11 it will go into the intermediate state if the row is activated column is not 16 00:01:48,330 --> 00:01:53,070 then will it can go into intermediate state and it can only go to ON state only by intermediate 17 00:01:53,070 --> 00:02:02,350 one okay. And of course if when it is in the intermediate 18 00:02:02,350 --> 00:02:12,500 state if R goes back to 0 so you have 00 or 10 directly moves to 01 in that case it will 19 00:02:12,500 --> 00:02:21,720 come back to the OFF state. It can only go to ON state when after this the column also 20 00:02:21,720 --> 00:02:31,000 goes up. And of course, after this even if row actually remains becomes 0 it will remain 21 00:02:31,000 --> 00:02:42,500 in the same state the column has to be maintained at one. And of course, if you have 10 basically 22 00:02:42,500 --> 00:02:47,540 the row is activated you will come back to again the intermediate one and for all other 23 00:02:47,540 --> 00:02:58,950 combinations. So 01 and 11 you will retain here and if you 24 00:02:58,950 --> 00:03:06,070 have a 00 you will come back to this particular state okay. So with this actually now we can 25 00:03:06,070 --> 00:03:11,690 build up a logic circuit I am going to do it with a toggle flip-flop you can actually 26 00:03:11,690 --> 00:03:18,650 use anything else. So since there are three states I require at least two flip-flops so 27 00:03:18,650 --> 00:03:24,620 that I can each flip-flop flip can possibly represent two states, so I need to have for 28 00:03:24,620 --> 00:03:30,560 at least two represent these three ones the fourth state I will not be using okay. 29 00:03:30,560 --> 00:03:36,740 So I will be using two flip-flops I will draw the diagram picture here itself, I call it 30 00:03:36,740 --> 00:03:47,470 T1, I call it T0, Q0 and Q1 these are the outputs in fact there will be inverted output 31 00:03:47,470 --> 00:03:53,330 also which will be coming this can be used by us. And of course, I will take these state 32 00:03:53,330 --> 00:03:58,590 outputs and based on that I will decide whether I will be, I am going to actually activate 33 00:03:58,590 --> 00:04:07,959 the cross point or not. So cross point is structure something like this I had drawn 34 00:04:07,959 --> 00:04:21,690 it earlier also. This is one, so I will, I am actually going 35 00:04:21,690 --> 00:04:26,961 to take and short these two and this is my activation I call it signal p, whenever P 36 00:04:26,961 --> 00:04:32,380 is high this cross point is active, when P is low it is not going to be active. So I 37 00:04:32,380 --> 00:04:39,070 am going to generate a p from here and the way I can represent it is that this intermediate 38 00:04:39,070 --> 00:04:47,940 state is represented as 10 ON is 01 you can do it other way around then the whole computation 39 00:04:47,940 --> 00:04:53,030 actually will change. And I can put it as 00 which is the OFF state 40 00:04:53,030 --> 00:04:58,130 there is no 11 state of course in this case so that will never be happening in this system 41 00:04:58,130 --> 00:05:08,440 okay. So I have to generate this p, so p should be only activated when you are in ON state 42 00:05:08,440 --> 00:05:36,160 which actually means your Q1 has to be 0 and Q0 has to be 1. So only in that case 43 00:05:36,160 --> 00:05:41,480 this is what is going to be the controlling the P so this will only be one when Q1 is 44 00:05:41,480 --> 00:05:45,610 0 and Q0 is 1 rest all the time it will be 0. 45 00:05:45,610 --> 00:05:50,770 And that is what I want here, so it is if you are in ON state this will remain ON otherwise 46 00:05:50,770 --> 00:05:55,400 this cross point will not be ON if you are not in the ON state. So all other states are 47 00:05:55,400 --> 00:06:02,151 immaterial, so where 001011 does not matter of course 11 will never be happening. And 48 00:06:02,151 --> 00:06:09,270 of course, now I can actually build up I need to ask just estimate what will be toggle values 49 00:06:09,270 --> 00:06:14,570 depending on what is the current state of these flip-flops and what is my input. 50 00:06:14,570 --> 00:06:15,740 Unable to capture the image because lecturer writes and erases the content immediately. 51 00:06:15,740 --> 00:06:25,590 So I can actually draw a karnaugh map for this it is actually elementary, but let us 52 00:06:25,590 --> 00:06:34,940 do it. So I am assuming that you know that how the karnaugh map operates. 53 00:06:34,940 --> 00:06:38,900 Unable to capture the image because lecturer writes and erases the content immediately. 54 00:06:38,900 --> 00:06:51,320 So let us see if it is you are in 00 a state so Q1 and Q0 both are 00 that 0 state means 55 00:06:51,320 --> 00:07:00,210 you are in OFF. So if your RC is 00 you come back to OFF state okay, from OFF you come 56 00:07:00,210 --> 00:07:09,120 back to the OFF state, so what does it mean, that from OFF you are coming back to OFF state 57 00:07:09,120 --> 00:07:14,550 so you are in 00 state you are in 00 state so there should not be any toggling. So all 58 00:07:14,550 --> 00:07:21,050 toggle T1 and T0 both have to be 00. So I require two of them, so one is for P1 59 00:07:21,050 --> 00:07:46,130 other one is P0, so I have to put a 0 here a 0 because there is no toggling required 60 00:07:46,130 --> 00:07:52,350 okay, when it is 01 then also no toggling is required. So here I have to put a 0, when 61 00:07:52,350 --> 00:08:03,080 it is 11 no toggling is required. So I have to put a 0 here okay, and when it is 10 then 62 00:08:03,080 --> 00:08:11,180 it should go from OFF to intermediate state okay, so Q1 has to be toggled. So it is 00 63 00:08:11,180 --> 00:08:17,570 RC is 10. So it has to go from 00 to 10 state which 64 00:08:17,570 --> 00:08:27,770 means this has to be toggled and this should not be toggled okay. So once you are this 65 00:08:27,770 --> 00:08:32,820 basically means I have taken care of when you were in 00 state, so what will be the 66 00:08:32,820 --> 00:08:41,190 next state that I have decided. There is no 11 state so that becomes a do not care condition. 67 00:08:41,190 --> 00:08:50,089 So I can use it for optimizing by basically logic circuit here, the community or logic 68 00:08:50,089 --> 00:08:56,300 which I will be using here as a input to generate inputs to these flip-flops. 69 00:08:56,300 --> 00:09:07,170 So if you are in 01 state in that case sorry, you are in 10 state which is intermediate 70 00:09:07,170 --> 00:09:20,769 one. So if you are in 10 state you get a 00 then what is going to happen. So I am talking 71 00:09:20,769 --> 00:09:28,089 about this particular row, so 10 with a 00 you have to go to OFF state which actually 72 00:09:28,089 --> 00:09:34,390 means only the Q1 has to be toggled. So this will be toggled and this will not be toggled 73 00:09:34,390 --> 00:09:44,620 okay. And if you are going to have a 01 as a input in 10 state then also the same result 74 00:09:44,620 --> 00:09:51,230 so I am going to put a 1 here and a 0 here okay. 75 00:09:51,230 --> 00:10:00,249 Now if you put a 10 is being inputted then what is going to happen, so if you are in 76 00:10:00,249 --> 00:10:05,790 the same state put 10 you go back to the same one, so no toggling will be required so it 77 00:10:05,790 --> 00:10:21,050 has to be 00. And if it is 11 you go from 10 to 01 so both of them have to be toggled. 78 00:10:21,050 --> 00:10:29,269 So I will put a 1 here and a 1 here. Now if you are in ON state what is going to happen, 79 00:10:29,269 --> 00:10:40,709 you are in state 01 if your RC is going to be 00 okay, basically these so it goes from 80 00:10:40,709 --> 00:10:46,389 01 state to 00. So Q1 is not toggled, but Q0 is toggled, so 81 00:10:46,389 --> 00:10:54,929 Q0 is toggled Q1 is not toggled. Then if you are in get a 01 you remain in same state so 82 00:10:54,929 --> 00:11:06,470 no toggling, so both of them will remain 00 when you get a 10 then you go to 10 state, 83 00:11:06,470 --> 00:11:14,889 so both of them have to be toggled. So this is what is going to happen. And when you have 84 00:11:14,889 --> 00:11:22,249 11 you remain in the same state so no toggling. So this is what will be your karnaugh map. 85 00:11:22,249 --> 00:11:32,449 And of course, now you can do the optimization. So in this case you will have this will do 86 00:11:32,449 --> 00:11:49,820 three terms and of course, if you solve for this you should get P1 = RCQ bar plus Q1R 87 00:11:49,820 --> 00:12:03,319 bar this actually term corresponds to this value. And then you will have Q1C, so there 88 00:12:03,319 --> 00:12:09,860 are one two and three terms so T1 will be given by this and similarly, from here I can 89 00:12:09,860 --> 00:12:18,549 find out what will be T0. So this requires only two terms the one is combination of this 90 00:12:18,549 --> 00:12:23,230 another one is this combination. So this should be Q0C bar plus Q1RC. 91 00:12:23,230 --> 00:12:29,309 Unable to capture the image because lecturer writes and erases the content immediately. 92 00:12:29,309 --> 00:12:36,779 So essentially you will take Q1 Q0 you will take R and C row and column controls, you 93 00:12:36,779 --> 00:12:45,200 take these as the inputs and you will take this as a input and you will implement the 94 00:12:45,200 --> 00:12:52,309 logics which are here and will generate T1 and you are also generate T0. So these the 95 00:12:52,309 --> 00:13:00,230 two logics will be implemented these ones will be implemented in this box. And you get 96 00:13:00,230 --> 00:13:41,799 your controller for a cross point okay, and that is how your crossbar is going to be implemented.