Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 23)
23.
Use Boolean algebra to find the most simplified SOP expression for F = ABD + CD + ACD + ABC + ABCD.
Discussion:
8 comments Page 1 of 1.
Aayush mehara said:
3 years ago
ABD + CD + ACD + ABC + ABCD.
= CD + ACD + ABD + ABC(D+D') + ABCD,
= CD(1+A) + ABD + ABCD + ABCD'+ ABCD,
= CD + ABD(1+C) + ABC(D'+D) {1+C=1, D+D'=1}.
= CD + ABD + ABC.
= CD + ACD + ABD + ABC(D+D') + ABCD,
= CD(1+A) + ABD + ABCD + ABCD'+ ABCD,
= CD + ABD(1+C) + ABC(D'+D) {1+C=1, D+D'=1}.
= CD + ABD + ABC.
(7)
Veo said:
4 years ago
@Svs.
Where is (ABCD) (ACD)? Please explain it clearly.
Where is (ABCD) (ACD)? Please explain it clearly.
(1)
Vaishnavi sherala said:
5 years ago
ABD + CD(1+A) + ABC(1+D).
ABD + CD + ABC ---> (1+A) ,(1+D)as per OR law.
ABD + CD + ABC ---> (1+A) ,(1+D)as per OR law.
Svs0 said:
9 years ago
The same but taking C as common instead of D
F = ABD + CD + ACD + ABC + ABCD.
= ABD(1 + C) + CD(A + 1) + ABC.
= ABD + CD + ABC.
(1 + X) = 1.
F = ABD + CD + ACD + ABC + ABCD.
= ABD(1 + C) + CD(A + 1) + ABC.
= ABD + CD + ABC.
(1 + X) = 1.
SUGU said:
1 decade ago
= ABD+(1+A)CD+(1+D)ABC.
= ABD+CD+ABC;(1+A) = 1.
= ABD+CD+ABC;(1+A) = 1.
Khan said:
1 decade ago
= ABD+CD(1+A)+ABC(1+D).
= ABD+CD(1)+ABC(1)..as(1+A=1 &1+D=1).
= ABD+CD+ABC.
Where did you get the 1 from ?
= ABD+CD(1)+ABC(1)..as(1+A=1 &1+D=1).
= ABD+CD+ABC.
Where did you get the 1 from ?
Vishal said:
1 decade ago
ABD+CD+ACD+ABC+ABCD
=ABD+CD(1+A)+ABC(1+D)
=ABD+CD(1)+ABC(1)..as(1+A=1 &1+D=1)
=ABD+CD+ABC
=ABD+ABC+CD
=ABD+CD(1+A)+ABC(1+D)
=ABD+CD(1)+ABC(1)..as(1+A=1 &1+D=1)
=ABD+CD+ABC
=ABD+ABC+CD
(2)
Pushpa said:
1 decade ago
In the expression the commom term is ABD, write that term one time i.e.
ABD + CD (1+A) + ABC (1+D) , 1+D and 1+A
which is appoximately equal to one, then ABD+ABC+CD.
ABD + CD (1+A) + ABC (1+D) , 1+D and 1+A
which is appoximately equal to one, then ABD+ABC+CD.
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