Awesome q2a theme
0 votes
17 views

  1.  $\sum m(2,3,6,7)$
  2. $\sum m(0,1,2,3)$
  3. $\sum m(0,1,4,5)$
  4. $\sum m(4,5,6,7)$
ago in Digital Logic by (613 points) | 17 views
0
Option (c) ? Please tell me it is (c)
0
Yah C
AIR 1 incoming
0
Please explain
I am unable to understand
0
Haha - one can hope.

Were you able to solve it ?
0
Nope

I got stuck in the decoder part
0
Is the last bubble just adjacent to f a NOT?
+1
I hope not. That's a very odd place to insert a negation. Even if it is, just complement the final output and see if it matches any of the options. If not, ignore that particular bubble.

1 Answer

+1 vote
Best answer
Stage 1 is a MUX - finding the output should be straightforward

$Y = \overline{S_1}.\overline{S_0}.1 + \overline{S_1}.{S_0}.0 + {S_1}.\overline{S_0}.0 + {S_1}.{S_0}.1$
substitute for $S_1, S_0$

$Y = \overline{A}\overline{C} + {A}{C}$

Move on to the decoder now. We're only bothered with the outputs $O_0$ and $O_2$, lets not bother with any others:

$O_0 = \overline{B}\overline{Y}$, think about why this is. The decoder output $O_0$ is selected when input is $00$ (converts encoded binary to decoded decimal)

$O_2 =  \overline{B}{Y}$, Output $O_2$ selected when input is $10$

And the outputs are bubbled (negated) before fanning into the NAND:

$f(A,B,C) = \overline{\overline{O_0}.\overline{O_2}} = O_0 + O_2$

Now just substitute for Y in the expressions for $O_0 + O_2$ we obtained earlier, and you will get the required minterms of $\Sigma(0, 1, 4, 5)$
ago by (1.2k points)
selected ago by
+1
This is a case where going the SOP method will take time (4 variables here, not a good approach). So try manipulating the expression.

$(\overline{A}+\overline{B})(C + D) = (\overline{A}+\overline{B}).C + (\overline{A}+\overline{B}).D$

$=\overline{AB}.C + \overline{AB}.D$

double complement

$=\overline{\overline{\overline{AB}.C + \overline{AB}.D}}$

$=\overline{\overline{\overline{AB}.C}.\overline{\overline{AB}.D}}$

Count the bars - $\overline{AB}$ can be calculated once and reused. So I'm guessing answer is 4..
0
Beautiful.. :)
0
Haha thanks. The solution I typed out seems elegant. The working is anything but. As soon as I decided to expand the given expression - things fell in place. I assume this is the kind of problem I'll skip and tackle at the end if I have enough time. If it clicks - its a very short solution. Else, like Vimal mentioned, its NP-C.
0
Though NP-C it amazes me how a human mind can apply various procedures best suited according to the given problem of that context to reach a solution quite quickly, where as any well defined algorithm fails to do so in general. There is lack of flexibility in algorithms, where a human mind has it in plenty.
0
Which is precisely why all those RSA crypto questions in ME where they ask you find $e$ or $d$ for some insanely large $n$ are to be considered bad. The basis of RSA is the discrete logarithm problem, which itself is NP-I. But alas.
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Welcome to GATE CSE Doubts, where you can ask questions and receive answers from other members of the community.
Top Users Jan 2020
  1. shashin

    1163 Points

  2. Vimal Patel

    306 Points

  3. Deepakk Poonia (Dee)

    305 Points

  4. Debapaul

    237 Points

  5. Satbir

    192 Points

  6. SuvasishDutta

    137 Points

  7. Pratyush Priyam Kuan

    118 Points

  8. tp21

    108 Points

  9. pranay562

    95 Points

  10. DukeThunders

    94 Points

Monthly Top User and those within 60% of his/her points will get a share of monthly revenue of GO subject to a minimum payout of Rs. 500. Current monthly budget for Top Users is Rs. 75.
2,983 questions
1,509 answers
8,933 comments
89,814 users