Question

How can I use Demorgins law to express:

AB'C'D' + ACB' + CD' = Y

In terms of NAND Operators

Answer 1

you can represent it using 4 NAND gates after using demorgan laws

Answer 2

So Demorgins law is

X + Y = XY
and
(XY)' = x' + y'

I Expand it to

A + B' + C' + D' + A + C + B' + C + D'
=
(A + A) + (B' + B') + (C + C') + (D' + D')
=
A + B' + C' + D'

What am I doing wrong? I think I might just be confused about how to properly utilize the law

I'm aware a NAND Gate is just a Not And Gate

Answer 3

Demorgan laws :-

(X + Y)' = X'Y'
and
(XY)' = x' + y'

Answer 4

So,

X + Y =/= XY

Answer 5

yes. they are not same.

Answer 6

X+Y is OR gate

XY is AND gate

Answer 7

My Profs notes are kind of messed up, so my list of rules are pretty sketchy

Answer 8

yeah that makes sense blah must not be thinking

Answer 9

ok look at the two demorgan laws again, before we begin

Answer 10

you cannot expand or do anything with the given expression as it is.

Answer 11

Given function :

Y = AB'C'D' + ACB' + CD'

Answer 12

So I have to use boolean algebra to manipulate it right?

Answer 13

Ima show you how to sovle..

Answer 14

Ok thank you, just learned this stuff an hour ago and any help is appreciated greatly

Answer 15

\(Y = AB'C'D' + ACB' + CD' \)

\(Y = \overline{\overline{AB'C'D' + ACB' + CD'}} \)

Answer 16

you understand the bars right ?

Answer 17

Yes bars just mean the inverse, my latex isnt working give me a second going to restart browser

Answer 18

bar and a tick mean the same thing : NOT

Answer 19

oh ok.. take ur tiem :)

Answer 20

ha that statement is confusing, well not really but you know

Answer 21

ok back latex is working for me

Answer 22

yea.. NOT is a NOT gate

Answer 23

ok lets continue

Answer 24

Alright

Answer 25

thanks again for the help

Answer 26

\(Y = AB'C'D' + ACB' + CD' \)

\(Y = \overline{\overline{AB'C'D' + ACB' + CD'}}\)

\(Y = \overline{\overline{\color{green}{AB'C'D'} + \color{purple}{ACB' + CD'}}}\)

Answer 27

apply demorgan for green and purple terms

Answer 28

How, I'm sorry I'm still confused on how to manipulate these can you give me an example?

Answer 29

\(Y = AB'C'D' + ACB' + CD' \)

\(Y = \overline{\overline{AB'C'D' + ACB' + CD'}}\)

\(Y = \overline{\overline{\color{green}{AB'C'D'} + \color{purple}{ACB' + CD'}}}\)

\(Y = \overline{ \overline{\color{green}{(AB'C'D')}} . \overline{ \color{purple}{(ACB' + CD')}} } \)

Answer 30

have a look i used this :-

\(\overline{\color{green}{X}+\color{purple}{Y}} = \overline{\color{green}{X}} . \overline{\color{purple}{Y}\)

Answer 31

Oh Oh I see, so

( (AB'C'D')(ACB')(CD') )')'

how does this bring us any close to the solution

Answer 32

contune and see hw we get all NAND gates

Answer 33

\(Y = AB'C'D' + ACB' + CD' \)

\(Y = \overline{\overline{AB'C'D' + ACB' + CD'}}\)

\(Y = \overline{\overline{\color{green}{AB'C'D'} + \color{purple}{ACB' + CD'}}}\)

\(Y = \overline{ \overline{\color{green}{(AB'C'D')}} . \overline{ \color{purple}{(ACB' + CD')}} } \)

\(Y = \overline{ \overline{\color{green}{AB'C'D'}} . \color{purple}{\overline{ACB'} . \overline{CD'}} } \)

Answer 34

we're done

Answer 35

oh I see you just rearrange the variables

Answer 36

and you use double inverse because ((x)')'

Thank you so much

Answer 37

yes i think u see now, those are 3 NAND gates for minterms and one more outer NAND gate combining those

Answer 38

:) you have been so much help this problem has been driving me crazy :)

Answer 39

np :) digital electronics is easy to learn... 2-3 days more im sure u wil feel confident :)