Please show your steps. Thank you very much. (X +ZY XZ Y using the truth table a
ID: 3668692 • Letter: P
Question
Please show your steps. Thank you very much.
(X +ZY XZ Y using the truth table as below Prob. 6 (8pts) Prove that X.Z X Y X Z Y 0 0 0 0 0 1 010 011 1 0 0 1 0 1 1 1 0 1 1 1 Prob. 7A logic function FCA, B,C,D) takes the values of one if there are even number of one in the input A, B, C, and D. 7.a) (4pts Fill the truth table 7.b) (2pts Derive the sum-of-products logic expression for FCA,B,CD) 0000 0001 0010 0011 0100 7.c (3pts) Using Boolean algebra to convert the SOP expression into 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110Explanation / Answer
Q6:
X
Y
Z
X’
Y’
Z’
X’+Z
XZ’
(X’+Z)Y
((X’+Z)Y)’
XZ’+Y’
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
1
0
1
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
0
1
1
1
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
1
1
1
0
1
0
1
0
1
0
0
1
1
1
1
0
0
0
1
1
1
0
1
1
1
1
1
0
0
0
1
0
1
0
0
Please See above truth table you required.
Please note down below points:
be true for Y and Z also.
7.a:
ABCD
F(A,B,C,D)
0000
0
0001
0
0010
0
0011
1
0100
0
0101
1
0110
1
0111
0
1000
0
1001
1
1010
1
1011
0
1100
1
1101
0
1110
0
1111
1
Explanation:
A logic function F(A,B,C,D) takes value 1 if there are even number of 1 in input.
On the basis of above definition we will decide value of F(A,B,C,D) on various combinations of given input.
Say ABCD=0001 here there is only single 1 in Inputs so F(A,B,C,D) will be 0
Say ABCD=0011 here there double 1 in inputs so F(A,B,C,D) will be 1
Similar pattern will follow for all inputs combination.
7.b: Sum of product expression means we add the product of inputs combination which give output 1.
So in 7.a truth table for F(A,B,C,D) we will mark input combinations for which we have output=1
ABCD
F(A,B,C,D)
0000
0
0001
0
0010
0
0011
1
0100
0
0101
1
0110
1
0111
0
1000
0
1001
1
1010
1
1011
0
1100
1
1101
0
1110
0
1111
1
I have mark those combinations in Bold and itialic.
Expression = A’.B’.C.D
Because A=0, B=0, C=1,D=1 and there is only one case in product where output can be 1 and that is when all input variable has value 1 so that product of them will give value 1
Now in this case if we product input variables A.B.C.D then result will be 0 because 0.0.1.1=0
SO we have to do inverse of input whose value is 0
So final expression will be A’.B’.C.D= 1.1.1.1=1
Similar process of other input combination:
2.ABCD=0101
So final expression will be A’.B.C’.D= 1.1.1.1=1
3.ABCD=0110
So final expression will be A’.B.C.D’= 1.1.1.1=1
4.ABCD=1001
So final expression will be A.B’.C’.D= 1.1.1.1=1
5.ABCD=1010
So final expression will be A.B’.C.D’= 1.1.1.1=1
6.ABCD=1100
So final expression will be A.B.C’.D’= 1.1.1.1=1
7.ABCD=1111
So final expression will be A.B.C.D= 1.1.1.1=1
Now adding all these 7 expressions:
A’.B’.C.D+ A’.B.C’.D+ A’.B.C.D’+ A.B’.C’.D+ A.B’.C.D’+ A.B.C’.D’+ A.B.C.Dà
AB’CD+A’BC’D+AB’CD’+AB’C’D+AB’CD’+ABC’D’+ABCD-àSOPàAnswer
X
Y
Z
X’
Y’
Z’
X’+Z
XZ’
(X’+Z)Y
((X’+Z)Y)’
XZ’+Y’
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
1
0
1
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
0
1
1
1
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
1
1
1
0
1
0
1
0
1
0
0
1
1
1
1
0
0
0
1
1
1
0
1
1
1
1
1
0
0
0
1
0
1
0
0