00:01
All right, so here we will be designing the 3 -bit counter, right? so we are required to design a 3 -bit counter.
00:12
So we will use 3 -plops, right? and let the state diagram is let a, b, and c are state variables, right? so this will be 010 and 011 right and also 1 000 and 101 right and 110 right so this is the state variables right now there are so the counter counts up from the counter will counts up from 3 to 6 right so but as it is a 3 -bit counter so it can have 8 states so any state other than 2 other than 3 to 6 will be taken as don't care right so the state table will be let the state table will be looking like this so this is the present state right this is present state and this will be the next state and this is d sub a and d sub b and d sub c and this is the main term right now this will be this will be a b b and this is we know that this is zero zero zero zero one 0 – 010, right? and this is 0, 1 ,0, and this will be 011, and 0, 1, and 0, and this is 0, and this is 0, this is 0, this is 0 ,000, right? this will be this will be one zero and this will be a one zero one one zero right and one one one now the next state will be this will be x x x and x x x right and this will be the second will be the second will be the third will be 011 and 0 this is 1 0 right this is 1 0 and this is 1 0 and up to so on right now d sub a will be x x x and x and this term will be 0 now this is x x and x this term will be 1 this is 0 1 1 1 and this term will be 2 and up to so on right this will be up to seven and this will be up to x x x and x x right so the state table which is mentioned above this is so similarly d sub b and d sub c are obtained so then they are simplified using the knapp maps right so we will use the k maps right so here let let this is the k map k map knapp maps right right so let this is the knapp maps so this is x x and one zero and this is one one x x zero and this is one x zero and this is a and this is b c.
04:45
This is d sub a, right? d sub a.
04:49
So if we solve this, d sub a will be equal to b1 plus c, right? and let draw the k maps part d sub b.
05:05
So this will be part d sub b, right? so here we will take a and b.
05:15
Now this will be x x and 0 1 x 1 x 1 and 0 0 so d sub b will be equal to b c plus b c prime and d sub b will be equal to b plus c right now draw the carnap map's par d sub c d sub c right and this is a this is b c okay this is easy right these are just carnap maps these are easy so this is x x and 011 x0 1 x 01 and this is 0 0 1 1 1 1 1 1 0 and this is 0 0 0 0 and this will be this will be also 0011110 and 0 .0 .0 .01110...