00:01
In this problem, we have to develop a logic circuit with four input variables, a, b, c, d that will produce a one output when exactly three input variables are one.
00:19
So the four inputs are a, b, c and d.
00:34
We draw the truth table b, c, d.
00:54
0000 the output is no output is only one when exactly 3 inputs variable 3 input variables are 1 in the second case also output will be 0 because 3 input variables are not 1 then we have 10 again 0 0 1 1 again 0 0 0 0 1 again 0 0 0 0 0 1 0 again 0 0 0 0 0 0 0 1 0 0 again 0 0 0 0 01 0 1 0 1 010 again 0 1 0 1 1 1 now 3 inputs are 1 so output will be 1 next one is 1 0 100 0 output is 0 1 0 1 0 1 0 0 0 00 0 0 out to 0 1 0 1 0 0 1 0 output is 0, 1, 1 again input is 1 because 3 inputs are 1.
02:32
Then we have 1, 1, 1 0, 1 1, 0 1, 1, 1, 1, 1, 1, again 3 inputs are 1, so it will be 1, again 1, 1 1 0, again 3 inputs are 1, and at last 4 inputs are 1, so output will be 0 because, we need exactly three inputs to be one for output to be one so we have one in three cases we make a carno map or k map corresponding to the two table so we have here ab cd ab a b zero zero one one after zero one one one c d c d zero one c d zero one so for abcd, 0 ,000, 0, it is 0.
04:59
Here it is 1, again 0, 0, 1...