ISRO CSE 2020 | Question: 77

Question

Consider the following circuit

The function by the network above is

• A. $\overline{AB}E+EF+\overline{CD}F$
• B. $(\overline{E}+AB\overline{F})(C+D+\overline{F})$
• C. $(\overline{AB}+E)(\overline{E}+\overline{F})(C+D+\overline{F})$
• D. $(A+B)\overline{E} +\overline{EF}+CD\overline{F}$

Comments

Answer should be $(b).$

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Got it, thanks.

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Why is (C) not correct?

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Please post your solution or see my solution, you will get the difference in option (b) and (c).

Answer 1

Level $1:$

• NAND gate Output  are $: \overline{AB}$
• NOR gate Output  are $: \overline{C+D}$

Level $2:$

• $1^{st}$ AND gate output are $: (\overline{AB})E$
• $2^{nd}$ AND gate output are $: EF$
• $3^{rd}$ AND gate output are $: F(\overline{C+D})$

Level $3:$

NOR gate Output are $:\overline{(\overline{AB})E+EF+F(\overline{C+D})}$

$\implies (AB + \overline{E}) \cdot (\overline{EF})\cdot (\overline{F} + C +D)\:\:\:[\because \text{De Morgan's law:} \:\:\overline{(A+B)} = \overline{A} \cdot \overline{B}\:\text{and}\: \overline{(AB)} = \overline{A} + \overline{B}\:]$

$\implies (AB + \overline{E}) \cdot (\overline{E}+ \overline{F})\cdot (\overline{F} + C +D)$

$\implies (AB\:\overline{E} + AB\:\overline{F} + \overline{E}\:\overline{E}+ \overline{E}\:\overline{F})\cdot (\overline{F} + C +D)$

$\implies (AB\:\overline{E} + AB\:\overline{F} +\overline{E} + \overline{E}\:\overline{F})\cdot (\overline{F} + C +D)$

$\implies [\overline{E}(AB +1 + \overline{F}) +  AB\:\overline{F}] \cdot (\overline{F} + C +D)$

$\implies (\overline{E} +  AB\:\overline{F}) \cdot (C + D + \overline{F} )\:\:\:\:[\because 1+X = 1]$   

So, the correct answer is $(B).$

Answer 2

$\overline{[(\overline{A.B}).E+E.F+(\overline{C.D}).F]}$

$\rightarrow \overline{[(\bar{A}+\bar{B}).E+E.F+(\bar{C}+\bar{D}).F]}$ = $\overline{[(\bar{A}.E+\bar{B}.E+E.F+\bar{C}.F+\bar{D}.F]}$

$\rightarrow[(\overline{\bar A.E}).(\overline{\bar B.E}).(\overline{E.F}).(\overline{\bar C.F}).(\overline{\bar C.F})]$

$\rightarrow[(A+\bar E).(B+\bar E).(\bar E+\bar F).(C+\bar F).(D+\bar F)]$

$\rightarrow [(AB+A\bar E+B\bar E+\bar E)(\bar E+\bar F)(CD+C\bar F+D\bar F+\bar F)]$

$\rightarrow[(AB+\bar E)(\bar E+\bar F)(CD+\bar F)]$

Looks like they mistyped option C

Comments

Now it is edited!

Answer 3

[ ((AB)’ E) + (EF) + (F(C+D)’) ]’
= ((AB)’ E)’ (EF)’ (F(C+D)’)’
= (AB+E’)(E’+F’)(F’+C+D)
= (ABE’ + ABF’ + E’ + E’F’)(F’+C+D)
= (ABF’+ E’(AB+1+F’))(F’+C+D)
= (ABF’+E’) (F’+C+D)