# 6. A logic circuit consists of four gates as shownAВoutВСWhat is the output?(A) 0(B) A CB. C(C) A B C+A B(D) A.BA B+B. C6. The Boolean expression for the circuit isA BB+CoutApplying 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 theinputs are differentA BA BАӨВ= A BA BThe negation of the EXCLUSIVE-OR isA B = A B+ABApplying De Morgan's first theorem gives(A B) (A B)(А + B) (A + В)A B- (A B) (AB)The distributive law givesAAA.B B. AB.BАФВ= 0 + A BB. A 0= A. BA. BSubstitute this result into the equation for the outputand 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. CoutUse the distributive law to factor A Bout of the secondand third terms.= A B. C+ (A B) (1 C)outA BC+ A B.1=A B. C+ A BThe answer is (C).

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the attached image shows a question and its answer but still, I couldn't understand it please simply explain it help_outlineImage Transcriptionclose6. 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). fullscreen
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Step 1

Although, given logic circuit consist of four gates

First one is NOT gate,second is XOR gate, third and fourth is NOR gate.

Now calculating output of logic gates one by one help_outlineImage TranscriptioncloseA В out В Output of NOT gate= A although Boolean expression of XOR gate is given by AB+AB where A and B are inputs of XOR gate threfore ouput of XOR gate for given logic circuit is given by =AB+AB (inputs areA and B = AB+AB hence output of XOR is AB+AB fullscreen
Step 2

Similarly output of NOR gate which inputs are B and C is given by

Step 3

Hence outputs of XOR and first NOR gate is the input of second NOR g...

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