logic circuits equations

Discussion in 'Homework Help' started by SHuzero, Aug 9, 2011.

  1. SHuzero

    Thread Starter New Member

    Feb 22, 2011
    14
    0
    can you help me to simplify this equation (x'y'+z)'+z+xy+wz ? please help me to simplify that equation.

    also what is the essential prime implicants in this equation f(w,x,y,z)=summation of m(0,2,4,5,6,7,8,10,13,15)?

    kindly help me with this problems.....
     
  2. Zazoo

    Member

    Jul 27, 2011
    114
    43
    Hello,
    Can you show us your attempt at a solution for each problem?
     
  3. Anestis88

    New Member

    Aug 9, 2011
    19
    2
    The equation is simplified as :

    (x'y'+z)'+z+xy because z+wz=z
    (x'y')'*z'+z +xy
    (x+y)*z'+z +xy
    (x+y+z)(z+z')+xy because x+yz=(x+y)(x+z)
    x+y+z+xy
    so x+y+z
     
  4. SHuzero

    Thread Starter New Member

    Feb 22, 2011
    14
    0
    guys help me
    what is the essential prime implicants in this equation f(w,x,y,z)=summation of m(0,2,4,5,6,7,8,10,13,15)?

    I wanted to know what is the essential prime implicants in this equation when it is in k-map

    kindly help me with this problems.....
     
  5. SHuzero

    Thread Starter New Member

    Feb 22, 2011
    14
    0
    hello guys
    thanks for the answer......i have some clarification about this part
    (x+y)*z'+z +xy
    (x+y+z)(z+z')+xy because x+yz=(x+y)(x+z)

    how come this part
    (x+y)*z'+z
    have been reduced to this
    (x+y)*z'+z
    (x+y+z)(z+z') because x+yz=(x+y)(x+z) - where is this equation....i can't see it....
    i need some explanation
     
  6. SHuzero

    Thread Starter New Member

    Feb 22, 2011
    14
    0
    this is my solution to essential prime implicants is

    f= w'z'(y+y')(x+x') + xz(w'+w)(y+y') + wx'z' (y+y')
    f= w'z' + x'z + wx'z'

    I just short cut my solution because it can be seen in k-map...

    what it the essential prime implicant
     
  7. Anestis88

    New Member

    Aug 9, 2011
    19
    2
    A+B*C=(A+B)(A+C)

    so in the (x+y)*z'+z
    if A=z , B=z' and C=x+y
    A+B*C=(A+B)(A+C)
    z+z'(x+y)=(z+z')[z+(x+y)]
     
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