ISRO CSE 2014 | Question: 15

Question

Consider the logic circuit given below:

$\text{Q =}$ __________?

• A. $\overline{\text{A}} \text{C} + \text{B} \overline{\text{C}} +\text{CD}$
• B. $\text{ABC} + \overline{\text{C}} \text{D}$
• C. $\text{AB + B} \overline{\text{C}} + \text{B} \overline{\text{D}}$
• D. $\text{A} \overline{\text{B}} + \text{A} \overline{\text{C}} + \overline{\text{C}} \text{D}$

Comments

write the output terms after the gate(and dont evaluate for each gate) evaluate only the final output

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replace last nand with bubbled or.

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whenever 2 sequential NAND gates or NOR gates are there then we can replace those NAND / NOR gates by AND - OR / OR - AND gates respectively and do simplification after replacement.

Answer 1

Option C

Q =((D'+C')B+AB)

  =AB+BC'+BD'

Therefore option C

Comments

Thank you for such a nice explanation..

Answer 2

Answer: AB + BC' + BD'

Output of  NAND gate which has input As D and C is (DC)'

Output of NAND gate which has input As  A and B is (AB)'

Output of NAND gate which has input As (DC)' and B is ((DC)' . B)'

Output of NAND gate which has input As  ((DC)' . B)' and (AB)' is (((DC)' . B)' . (AB)')'

If we solve it (((DC)' . B)' . (AB)')' then

(DC)' . B + AB 

(D' + C' )B + AB

BD' + BC' +AB

Answer 3

Hence,Option(C)AB+BC'+BD' is the correct choice.

Answer 4

option C seems correct,

eq = (((cd)' b )' ( ab )' )'

      = ( cd )' b + ab

       = ( c' + d' ) b  + ab      -----------D morgans law (ab)' = a' + b'

Answer 5

option (c) is the right answer

Answer 6