how to solve the following equation

Status
Not open for further replies.
There are no bold characters in your expression. I think it's better to use "!" in front of the character you want to complement; f.i.; !A means "not A".
Of course with BC I think you mean "B and C". Is it correct ? In this case !BC will mean "not B and C" while B!C will mean "B and not C"
 

ok sorry for incorrect equation here's the correct equation

IF A=0 in logic expression Z= [A+EF+!BC+D][A+!D!E+!BC+!D!E],then
A) Z=0
B) Z=1
C) Z=BC
D) Z=BC

plz solve with complete steps.
 

Z= [A+EF+!BC+D][A+!D!E+!BC+!D!E]
if A = 0 ==> Z= [EF+!BC+D][!D!E+!BC+!D!E]
take !BC as common term,
Z = !BC[(EF+1+D)(!D!E+1+!D!E)]
since 1+A = A,
Z = !BC[(EF+D)(!D!E+!D!E)]
since A+A = A
Z = !BC[(EF+D)(!D!E)]
now
Z = !BC[EF!D!E + D!D!E]
since A!A = 0
Z = !BC[0+0]
==> Z = 0
 

you have first to expand the equation, like a normal expession remembering that:
x+x=x*x=x
x+!x=1
x*!x=0

using these rules and rearranging I obtained (hope I didn't do errors):

Z=a*(1+!be+!bc+ef+d)+!bc*(ef +!d!e+1+d)

so in the brackets there is always 1.

Z=a+!bc

then Z=1 if a=1 or !bc=1 that is b=0 and c=1
z=0 if a=0 and !bc=0
z=bc never
z=!bc if a=0

I'm from a mobile phone and is difficult to explain better. on saturday I'll have access to my PC.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…