Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

how to solve the following equation

Status
Not open for further replies.

faiza nasir

Junior Member level 3
Joined
May 1, 2011
Messages
26
Helped
4
Reputation
8
Reaction score
4
Trophy points
1,283
Location
Islamabad, Pakistan, Pakistan
Activity points
1,438
IF A=0 in logic expression Z= [A+EF+BC+D][A+DE+BC+DE],then
A) Z=0
B) Z=1
C) Z=BC
D) Z=BC
Where the bold characters are complements.

plz solve with complete steps.
 

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.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top