Question

#	O	W	L	S	f(O,W,L,S)
0	0	0	0	0	0
1	0	0	0	1	0
2	0	0	1	0	1
3	0	0	1	1	1
4	0	1	0	0	0
5	0	1	0	1	1
6	0	1	1	0	1
7	0	1	1	1	X
8	1	0	0	0	0
9	1	0	0	1	0
10	1	0	1	0	0
11	1	0	1	1	1
12	1	1	0	0	0
13	1	1	0	1	1
14	1	1	1	0	1
15	1	1	1	1	X

A)Show the sum-of-product, simplified (normalized) equation, using the Don’t Care cases. This time, there should only be one equation.

1)Draw the sum-of-products version of the circuit for part A

2)Draw the NAND version of the circuit from Problem A

3)Fill in the Truth Table from the equation in problem

#	O	W	L	S	EQ3.6
0	0	0	0	0
1	0	0	0	1
2	0	0	1	0
3	0	0	1	1
4	0	1	0	0
5	0	1	0	1
6	0	1	1	0
7	0	1	1	1
8	1	0	0	0
9	1	0	0	1
10	1	0	1	0
11	1	0	1	1
12	1	1	0	0
13	1	1	0	1
14	1	1	1	0
15	1	1	1	1

Answer 1

1) Given Function is

f (O, W, L, S) = ​ m (2, 3, 5, 6, 11, 13, 14) + d(7, 15)

Above Function in K-map as follows

Simplified K-map as follows

The Simplified SOP of F (O, W, L, S) = O’L+LS+WL+WS

2) Given f = O’L+LS+WL+WS

[f ']' = [(O’L+LS+WL+WS)']'  { We know that (P')'= P }

f = [(O’L)'(LS)'(WL)'(WS)']' { We know that (P+Q)'= P' Q' }

f = [(O’L)'(LS)'(WL)'(WS)']'

Simplified Circuit: Using NAND gates only

Explanation:
NAND gate is used to find the Product of Two literals P NAND Gate Q Output is (PQ)'

3)Truth Table: Given f (O, W, L, S) = O’L+LS+WL+WS

S.NO	O	W	L	S	O’	O’L	LS	WL	WS	f = O’L+LS+WL+WS
F    0	0	0	0	0	1	0	0	0	0	0
1	0	0	0	1	1	0	0	0	0	0
2	0	0	1	0	1	1	0	0	0	1
3	0	0	1	1	1	1	1	0	0	1
4	0	1	0	0	1	0	0	0	0	0
5	0	1	0	1	1	0	0	0	1	1
6	0	1	1	0	1	1	0	1	0	1
7	0	1	1	1	1	1	1	1	1	X
8	1	0	0	0	0	0	0	0	0	0
9	1	0	0	1	0	0	0	0	0	0
10	1	0	1	0	0	0	0	0	0	0
11	1	0	1	1	0	0	1	0	0	1
12	1	1	0	0	0	0	0	0	0	0
13	1	1	0	1	0	0	0	0	1	1
14	1	1	1	0	0	0	0	1	0	1
15	1	1	1	1	0	0	1	1	1	X