Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 34)
34.
For the SOP expression 
, how many 0s are in the truth table's output column?
, how many 0s are in the truth table's output column?Discussion:
5 comments Page 1 of 1.
Mahesh mangal said:
1 decade ago
000 0
001 1
010 0
011 0
100 0
101 1
110 1
111 1
001 1
010 0
011 0
100 0
101 1
110 1
111 1
(1)
Abhisek Chowdhury said:
1 decade ago
AB+B'C
= AB.1+B'C.1
= AB(C+C')+B'C(A+A') {Complement element}
= ABC+ABC'+AB'C+A'B'C {Distributive law}
As 4 numbers of output is present so 4 "1"s are present in the output. As variables are A,B,C, therefore n=3. So total outputs are 8 (2^n). number of "0"s are 8-4 = 4.
= AB.1+B'C.1
= AB(C+C')+B'C(A+A') {Complement element}
= ABC+ABC'+AB'C+A'B'C {Distributive law}
As 4 numbers of output is present so 4 "1"s are present in the output. As variables are A,B,C, therefore n=3. So total outputs are 8 (2^n). number of "0"s are 8-4 = 4.
(1)
Vidushi SInghal said:
1 decade ago
Because we can write it like (AB (C+C BAR) + B BAR +C (A+A bar) ).
IN EXPANDED FORM IT WILL BE (ABC + ABC BAR + A B BAR C BAR + A B BAR C).
SO WE NEED 4 ONE'S TO REPRESENT IT & 4 ZERO'S TO REPRESENT REST.
IN EXPANDED FORM IT WILL BE (ABC + ABC BAR + A B BAR C BAR + A B BAR C).
SO WE NEED 4 ONE'S TO REPRESENT IT & 4 ZERO'S TO REPRESENT REST.
Arpita said:
1 decade ago
We have 3 input, So 2^3=8; So we need 4 zero's and 4 One's.
PREMKUMAR said:
1 decade ago
A B C AB B' B'C AB+B'C
0 0 0 0 1 0 0
0 0 1 0 1 1 1
0 1 0 0 0 0 0
0 1 1 0 0 0 0
1 0 0 0 1 0 0
1 0 1 0 1 1 1
1 1 0 1 0 0 1
1 1 1 1 0 0 1
ANSWER CONTAIN 4 ZEROS.
0 0 0 0 1 0 0
0 0 1 0 1 1 1
0 1 0 0 0 0 0
0 1 1 0 0 0 0
1 0 0 0 1 0 0
1 0 1 0 1 1 1
1 1 0 1 0 0 1
1 1 1 1 0 0 1
ANSWER CONTAIN 4 ZEROS.
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