Question

Please help
Simplify using boolean algebra technique and also draw a logic diagram of the given equation?

Answer 1

good luck ^^
this question is way out of my league :P

Answer 2

Have you studied Karnaugh maps?

Answer 3

we have to do it through boolean rules
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Answer 4

Did they give you a process to follow?

Answer 5

i can show you one example wait

Answer 6

No need for an example.

I think the first step is to remove any "overbars" that are over "expressions"
using De Morgan

Can you do that for
\[ A \overline{B} +A \overline{(B+C)} +B \overline{(B+C)}\]?

Answer 7

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Answer 8

B+C will become B'C'

Answer 9

ok, so you get
AB' +AB'C' + BB'C'

any ideas ?

Answer 10

yes

Answer 11

rule 8 looks applicable

Answer 12

BB'=0
SO WE HAVE
AB'+AB'C'

Answer 13

yes. but to be pedantic , we would say
AB'+AB'C' + 0 C' by rule 8
then
AB'+AB'C' + 0 by rule 3
then
AB' + (AB'C' + 0) no rule, but it's the associative rule
then
AB' + AB'C' by rule 1

Answer 14

to make it more clear, let X be short-hand for AB' and Y short for C'
then we have this pattern:
X +XY
any idea ?

Answer 15

AB'(C')

Answer 16

I would use rule 10 on
X+XY

Answer 17

to get X

and X is AB'

so the final answer is

AB'

Answer 18

ok onto the next question

Answer 19

how far did you get on 4.jpg ?

Answer 20

keep going

Answer 21

I would do
ABCD + ABCDD' +A'CD +B'CD
and eliminate the 2nd term

ABCD +A'CD +B'CD

Answer 22

HOW?

Answer 23

[ AB(C+(BD)' ) +(AB)'] CD

replace (BD)' with B'+D' and similar for (AB)'
[ AB(C + B' + D') + A' + B' ] CD
distribute the AB to get
[ ABC + ABB' +ABD' + A' + B'] CD

now distribute the CD to get
ABCCD + ABB' CD + ABC DD' + A'CD + B' CD

Answer 24

CC is just C
BB' is 0
DD' is 0
we get
ABCD + A'CD + B'CD

Answer 25

do you follow so far?

Answer 26

HAVE A LOOK AT THIS

Answer 27

?
now it's not obvious what to try next, but if we factor out the CD we get
(AB + A' + B') CD

looking at just AB + A', let X= A'
we have X + X' B
by rule 11 we simplify that to X + B which is A' + B

and we get
(A' + B + B') CD

next use rule 6 on B+B'
then rule 2
then rule 4

Answer 28

IM getting CD as answer

Answer 29

yes

Answer 30

how did you change
(AB + A' + B') to ( A'+A + B + B' ) ?

Answer 31

how did you change
(AB + A' + B') to ( A'+A + B + B' ) ?

by rule 11 (see above) you should get (A' + B + B')

Answer 32

next equation?

Answer 33

Did you fix 4.jpg ?

Answer 34

it will not be solved further

Answer 35

I meant the 3rd line from the bottom should not be
CD( (A+A') + (B+B'))
it should be
CD ( A' + B + B')
which does simplify to CD.

Answer 36

cant we move one with the sixth one ?

Answer 37

how far did you get with 6.jpg?

Answer 38

not far got stuck with it

Answer 39

we start with
ABC' + A'B'C + A'BC + A'B'C'

notice that you have A'B' in two terms, so you can factor it out.
if you do that , what do you get ?

Answer 40

yes

Answer 41

if you do that , what do you get ?

Answer 42

A'B'

Answer 43

ABC' + A'B'C + A'BC + A'B'C'

to summarize:
if we look just at the 2nd and last terms: A'B'C +A'B'C'
and factor out A'B' from each term: A'B'(C+C')
use rule 6 to get A'B' 1
use rule 4 to get A'B'

we now have
ABC' + A'BC + A'B'

the second and last terms, after we factor out A': A'(BC + B')
use rule 11 on B' + BC to get B'+C
and we now have

ABC' + A'(B' +C)
distribute the A' and we have
ABC' +A'B' + A'C

which is as simple as we can make it.