6. A logic circuit consists of four gates as shown A В out В С What is the output? (A) 0 (B) A CB. C (C) A B C+A B (D) A.BA B+B. C 6. The Boolean expression for the circuit is A BB+C out Applying De Morgan's first theorem gives (А Ф В): (В+ C) out (А Ө В). (В+ С) The result of the EXCLUSIVE-OR is true when the inputs are different A BA B АӨВ = A BA B The negation of the EXCLUSIVE-OR is A B = A B+AB Applying De Morgan's first theorem gives (A B) (A B) (А + B) (A + В) A B - (A B) (AB) The distributive law gives AAA.B B. AB.B АФВ = 0 + A BB. A 0 = A. BA. B Substitute this result into the equation for the output and use the distributive law = (A B A B) (B C) = A B.B+ A B C+ A B. BA. B. C = 0 + A BC+ A B+ A B. C = A. B C+ A B+A B. C out Use the distributive law to factor A Bout of the second and third terms. = A B. C+ (A B) (1 C) out A BC+ A B.1 =A B. C+ A B The answer is (C).

Question

Hi,

the attached image shows a question and its answer but still, I couldn't understand it please simply explain it 

6. A logic circuit consists of four gates as shown
A
В
out
В
С
What is the output?
(A) 0
(B) A CB. C
(C) A B C+A B
(D) A.BA B+B. C
6. The Boolean expression for the circuit is
A BB+C
out
Applying De Morgan's first theorem gives
(А Ф В): (В+ C)
out
(А Ө В). (В+ С)
The result of the EXCLUSIVE-OR is true when the
inputs are different
A BA B
АӨВ
= A BA B
The negation of the EXCLUSIVE-OR is
A B = A B+AB
Applying De Morgan's first theorem gives
(A B) (A B)
(А + B) (A + В)
A B
- (A B) (AB)
The distributive law gives
AAA.B B. AB.B
АФВ
= 0 + A BB. A 0
= A. BA. B
Substitute this result into the equation for the output
and use the distributive law
= (A B A B) (B C)
= A B.B+ A B C+ A B. BA. B. C
= 0 + A BC+ A B+ A B. C
= A. B C+ A B+A B. C
out
Use the distributive law to factor A Bout of the second
and third terms.
= A B. C+ (A B) (1 C)
out
A BC+ A B.1
=A B. C+ A B
The answer is (C).

Image Transcription

6. A logic circuit consists of four gates as shown A В out В С What is the output? (A) 0 (B) A CB. C (C) A B C+A B (D) A.BA B+B. C 6. The Boolean expression for the circuit is A BB+C out Applying De Morgan's first theorem gives (А Ф В): (В+ C) out (А Ө В). (В+ С) The result of the EXCLUSIVE-OR is true when the inputs are different A BA B АӨВ = A BA B The negation of the EXCLUSIVE-OR is A B = A B+AB Applying De Morgan's first theorem gives (A B) (A B) (А + B) (A + В) A B - (A B) (AB) The distributive law gives AAA.B B. AB.B АФВ = 0 + A BB. A 0 = A. BA. B Substitute this result into the equation for the output and use the distributive law = (A B A B) (B C) = A B.B+ A B C+ A B. BA. B. C = 0 + A BC+ A B+ A B. C = A. B C+ A B+A B. C out Use the distributive law to factor A Bout of the second and third terms. = A B. C+ (A B) (1 C) out A BC+ A B.1 =A B. C+ A B The answer is (C).

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