Electronics and Communication Engineering - Exam Questions Papers - Discussion
Discussion Forum : Exam Questions Papers - Exam Paper 2 (Q.No. 14)
14.
Consider circuit with 4 : 16 Demux below: Now:
- f1 =
(2, 6, 11, 12) - f2 = (2, 5, 14)
= p(2, 3, 4, 5, 9, 13, 14)
Answer: Option
Explanation:
f1 = y2 + y6 + y11 + y12
and f2 = y1 + yl0 + yl3
f1(m, n, o, p) = (3, 4, 9, 13)
f2(m, n, o, p) = (2, 5, 14)
∴ = Π(2, 3, 4, 5, 9, 13, 14).
Discussion:
2 comments Page 1 of 1.
Muthumari said:
6 years ago
m,n,o,p are inverting input terminals of 4-16 demultiplexer.
f1 be the nand operation of inverting outputs of y2, y6, y11 and y12.
If any of inputs(y2, y6, y11 and y12) is high f1 also high.
y2 will be high when (m,n,o,p) be 1101.
Like that we can get for,
y6 - 1001
y11 - 0101
y12 - 0011
Hence, f1 - (3,4,9,13)
f2 also like f1 but y1, y10, y13 instead.
f1 be the nand operation of inverting outputs of y2, y6, y11 and y12.
If any of inputs(y2, y6, y11 and y12) is high f1 also high.
y2 will be high when (m,n,o,p) be 1101.
Like that we can get for,
y6 - 1001
y11 - 0101
y12 - 0011
Hence, f1 - (3,4,9,13)
f2 also like f1 but y1, y10, y13 instead.
Gopal said:
9 years ago
Please, someone explain how to got this answer?
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