00:01
Hi, in this question we have a combinational logic circuit with 2 bit input and 5 bit output.
00:08
So here we have so input is ab and output is s4 s3 s2 s1 s0.
00:16
So here 2 bit is 00 so s4 is 00000.
00:23
Then we have 01 so this will be 00001.
00:27
Then we have 10 this will be 01000 and we have 11 so this will be 11011.
00:36
So this is nothing but 3 cube is 27 base 10 this is 2 cube is 8 base 10 and this is 1 cube and this is 0.
00:44
So now we have the k map so k map for so now we have to draw the k map so first case is for s0.
00:53
So s0 we have this is the k map so here it's 01 and 10 so 01 this is a and this is b this will be 00 so s0 will be equal to b.
01:10
Then for s1 we have so this will be the k map this is 01 and this is 01 so this is 000 so on plotting this type we have s1 is a .b.
01:24
Then we have third for s2 it is so this is 000 this is 01 this is 01 a and b so this is a 4 group so s2 is 0.
01:38
Then we have for s3 so again taking the k map so here 01 01 this is a b so this is 00 so s3 is a.
01:50
Then for fifth part we have so again taking the k map 01 01 so this will be 000 so again taking for 3 we have s4 is equal to a b.
02:04
So these are the outputs...