Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 1 (Q.No. 20)
20.
The circuit of the given figure realizes the function
Answer: Option
Explanation:
or
Y = (A + B)C + DE.
Discussion:
10 comments Page 1 of 1.
DILEEP said:
7 years ago
A NAND B = (A.B)'
D NAND E = (D.E)'
(D.E)' BY Applying NAND it becomes D.E.
(A.B)' NAND C =((A.B)'.C)'=(A.B)+C'.
((A.B)'.C)' NAND (D.E)'=(((A.B)'.C)'.(DE))'.
USING DeMorgan's theorem.
(A.B)'.C + (DE)'.
= (A'+B').C + (D.E)'.
D NAND E = (D.E)'
(D.E)' BY Applying NAND it becomes D.E.
(A.B)' NAND C =((A.B)'.C)'=(A.B)+C'.
((A.B)'.C)' NAND (D.E)'=(((A.B)'.C)'.(DE))'.
USING DeMorgan's theorem.
(A.B)'.C + (DE)'.
= (A'+B').C + (D.E)'.
(2)
Giribabu said:
8 years ago
Y= (((AB)' . C)' . DE)'.
Using DeMorgan's theorem.
Y=(((AB)'' + C' ). DE)'
=((AB+C') . DE)'
= (AB+C')' + (DE)'
=(AB)' . C + (DE)'
or
=(A'+B').C + (DE)'.
Using DeMorgan's theorem.
Y=(((AB)'' + C' ). DE)'
=((AB+C') . DE)'
= (AB+C')' + (DE)'
=(AB)' . C + (DE)'
or
=(A'+B').C + (DE)'.
(4)
Keerthi said:
8 years ago
Y = (((AB) '. C')'. DE)'
Using De Morgan's theorem.
= (AB)'. C+(DE)'
Y = (A'+B'). C+(DE)'.
Using De Morgan's theorem.
= (AB)'. C+(DE)'
Y = (A'+B'). C+(DE)'.
(1)
Tharun said:
7 years ago
Please explain the answer clearly to understand to me.
Kamatchi said:
8 years ago
I can't understand this. Please explain in detail.
Sujideepika said:
9 years ago
Please, Can anyone explain this?
Sai said:
7 years ago
Thanks for the answer @Giribabu.
Sana said:
9 years ago
It is using demorgans theorem.
Sandy said:
5 years ago
Nice explanation, thanks all.
Samantha said:
1 decade ago
How it is?
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