CSE 370 Solution Set #2
Spring 1999
Total Points 20

1) 2 Points:

    a. Here's a possible schematic using only NAND gates:
   

    b.
   

2) 3 Points:
    2.7 a.  (X+Y)(X+Y')
            = (X+Y)X + (X+Y)Y'
            = XX + XY + XY' + YY'
            = X + XY + XY'
            = X(1+Y+Y')
            = X
    2.7 b.   X(X+Y)
            = XX + XY
            = X + XY
            = X(1+Y)
            = X
    2.7 c.   (X+Y')Y
            = XY + YY'
            = XY + 0
            = XY
    2.7 d.  (X+Y)(X'+Z)
            = (X+Y)X' + (X+Y)Z
            = X'X + X'Y + XZ + YZ
            = 0 + X'Y + XZ + (X+X')YZ
            = X'Y + XZ + XYZ + X'YZ
            = X'Y + X'YZ + XZ + XYZ
            = X'Y(1 + Z) + XZ(1 + Y)
            = X'Y + XZ

3) 1 Points:
    a.

    b.    XY + YZ + X'Z
        = XY + (X+X')YZ + X'Z
        = XY + XYZ + X'YZ + X'Z
        = XY(1 + Z) + X'Z(1 + Y)
        = XY + X'Z

4) 0 Points:
    b. F' = ( ABC + B(C' + D') )'
            = (ABC)'( B(C' + D') )'
            = (A'+B'+C')(B' + (C' + D')')
            = (A'+B'+C')(B' + CD)
    g. F' = ( X(Y + ZW' + V'S) )'
            = X' + (Y + ZW' + V'S)'
            = X' + Y'(ZW')'(V'S)'
            = X' + Y'(Z'+W)(V+S')

5) 0 Points:
    You just needed to show the circuit in DesignWorks with probes to prove this circuit is the XOR function...
   

6) 2 Points:
    a. One function you can't make with only AND/OR gates is the NOT function
    b. One function you can't make with only XOR gates is the AND function

7) 4 Points:
    a. One way to do this problem is to use boolean algebra to simplify.   Another is to build the truth table for F, complement it's value and read the simplest form of the complement from there...

      From the table you should be able to see that F' = XY' + Z'
    b.

     From the table you should be able to see that G' = X'Y'+X'Z'+Y'Z' = X'Y'+Z'(X'+Y').  You also need to show the circuits in original & minimized form in DesignWorks.

8) 4 Points:
    a. f(A,B,C,D) = A'B'C'D'+A'B'C'D+A'B'CD'+A'BC'D+A'BCD'+AB'C'D'+AB'CD'+ABCD'
    b. f(A,B,C,D) = (A+B+C'+D')(A+B'+C+D)(A+B'+C'+D')(A'+B+C+D')(A'+B+C'+D')(A'+B'+C+D)(A'+B'+C+D')(A'+B'+C'+D')
    c. f'(A,B,C,D) = m(3,4,7,9,11,12,13,15) = A'B'CD+A'BC'D'+A'BCD+AB'C'D+AB'CD+ABC'D'+ABC'D+ABCD
    d. f'(A,B,C,D) = M(0,1,2,5,6,8,10,14) =
          (A+B+C+D)(A+B+C+D')(A+B+C'+D)(A+B'+C+D')(A+B'+C'+D)(A'+B+C+D)(A'+B+C'+D)(A'+B'+C'+D)
    To check your work: if you complement the equation from part a. then you should get the equation from part d. AND if you complement the equation from part b. then you should get the equation from part c.

9) 4 Points:
    a.  Here's the truth table for f(W,X,Y,Z) = YZ' + W'Y'Z + X'Z'

    From the table you can read f = m(0,1,2,5,6,8,10,14)

    b. To the the big M notation for the complement, you simply use the same terms in the little m notation for the function.
        f' = M(0,1,2,5,6,8,10,14)