00:01
Hello students, in this question we need to solve gates by using boolean algebra.
00:11
In the first gate we can use first and and gate in which inputs are b, c, d.
00:18
The output will be multiplication of this which will give b, c, d.
00:23
Then again this is going to input as a inverting to the second gate along with that two inputs are given that is a and e.
00:38
Therefore, the result of this input will be multiplication of these inputs.
00:47
Again it is inverted then bar whole bar gate cancel it gives like this.
00:59
The final expression y will be b, c, d bar a, e.
01:06
So, we use and gate in the first nand gate the product is b, c, d whole bar and again a and e is given to the second nand gate.
01:18
Nand gate will invert the product then the product will become like this.
01:23
Then the final product is again inverted that means whole bar is applied bar and bar gate cancel it will give a, e.
01:29
B, c, d bar will be there.
01:32
Therefore, y will be equal to this much in the first gate.
01:36
Then in the second gate y1 input is given to xor gate.
01:42
Y1 is expressed with boolean algebraic expression a b bar into a bar b.
01:52
Here a is a and another one is or gate that is c plus d plus e is given.
02:01
Substitute this a and b in the above expression and solving we get y1 will be equal to a dot c bar plus a dot d bar plus a dot e bar plus a bar dot c plus then a bar dot d plus a bar dot e.
02:32
This is y1.
02:34
Then y2 is find out by using the inputs are b is inverted in the first b bar f and or gate is used say c plus d plus e.
02:49
This will be the input then it is given to and gate...