Solution of map method- chapter 3 ( Morris Mano)

 MAP METHOD:

3.1: Obtain the simplified expressions in sum of products for the following boolean function:

a) F(x,y,z) =m2+m3+m6+m7

solution: 

        F=Y

b)  F(A,B,C,D)=m7+m13+m14+m15

solution:

          F=BCD+ABC+ABD

c) F(A,B,C,D) = m4+m6+m7+m15

solution: 

        F=A'BD' + BCD

d) F(w,x,y,z) = m2+m3+m12+m13+m14+m15

solution: 

            

            F=wx + w'x'y

3.2: Obtain the simplified expressions in sum of products for the following Boolean function:

a)      xy + x’y’z’ + x’yz’

            F=x’z’+xy

b)      A’B+ BC’+B’C’

            F=C’+ A’B

c)      a’b’ + bc + a’bc’

          F=a’+bc

d)      xy’z + xyz’ + x’yz + xyz

       F= yz + xz + xy

3.3: obtain the simplified expressions in sum of products for the following Boolean functions:

a) D(A’+B) + B’(C+AD)

                   F=D+ B’C

b) ABD + A’C’D’ + A’B + A’CD’ + AB’D'

             F=BD + B’D’ + A’D’

c) k’lm’ + k’m’n + klm’n’ + lmn’

                  F=ln’ + k’m’n

d)      A’B’C’D’ + AC’D’ + B’CD’ + A’BCD + BC’D

          F=B’D’ + A’BD + ABC’

e)      xz’ + w’xy’ + w(x’y + xy’)

           F= x’z + x’y’ + wx’y

3.4: obtain the simplified expressions in sum of products for the following Boolean function

a) F(A,B,C,D,E) =m0+m1+m4+m5+m16+m17+m21+m25+m29

         F=B’C’D’ + AD’E + A’B’D’

 b) BDE + B’C’D + CDE + A’B’CE + A’B’C + B’C’D’E’

 

            F= B’C’E’ + DE + A’B’C

c) A’B’CE’ + A’B’C’D’ + B’D’E’ + B’CD’  + CDE’ + BDE’

 

        F=BDE’ + CDE’ + B’CD’ + B’D’E’ + A’B’D’

 

3.5: given the following table:

x	y	z	F1	F2
0	0	0	0	0
0	0	1	1	0
0	1	0	1	0
0	1	1	0	1
1	0	0	1	0
1	0	1	0	1
1	1	0	0	1
1	1	1	1	1

 

 

a)     a) express F1 and F2 in product of maxterms

F1=π(0,3,5,6)

F2=π(0,1,2,4)

b)     b) obtain the simplified functions in sum of products

F1=m1+m2+m4+m7

F2=m3+m5+m6+m7

 

            F1=x’y’z + x’yz’ + xy’z’ + xyz

 

                   F2= yz + zx + xy

                                            

c)     c) obtain the simplified function in product of sum

 F1’=x’y’z’ + xy’z + x’yz + xyz’

  F1= (x+y+z) (x’+y+z’) (x+y’+z’) (x’+y’+z)

 

  F2’=x’z’ + y’z’ + x’y’

  F2=(x+z) (y+z) (x+y) 

                                                          

3.6: obtain the simplified expressions in product of sum:

a) F(x,y,z)=π(0,1,4,5)

                     F’=Y’

                      F=Y

                                                               

b)F(A,B,C,D)=π(0,1,2,3,4,10,11)

  F’ = A’B’+B’C + A’C’D’

  F=(A+B) (B+C’) 

c) F(w,x,y,z)= π(1,3,5,7,13,15)

 

  F’ =w’z + xz

  F= (w+z’) (x’+z’)                                                                

                                                                 

3.7: obtain the simplified expression in

   1) sum of products

  2) product of sum

a) x’z’ + y’z’ + yz’ + xyz

F=z’+ xy

F’ = x’z + y’z

F= (x+z’) (y+z’)

b) (A+B’+D) (A’+B+D) (C+D) (C’+D’)

F=C’D + A’B’CD’ + ABCD’

F’=C’D’ + CD + A’BC + AB’C

    =(C+D) (C’+D’) (A+B’+D’) (A’+B+D’)

 

c) (A’+B’+D’) (A+B’+C’)(A’+B+D’) (B+C’+D’)

F=B’D’ + A’C’ + AD’

F’=AD+CD +A’BC

F=(A’+D’) (C’+D’) (A+B’+C’)

 

d) (A’+B’+D) (A’+D’) (A+B+D’) (A+B’+C+D)

F=B’D’ + A’BD + A’CD’

F’=B’D + AB + BC’D’

F= (B+D’) (A’+B’) (B’+C+D)

e) w’yz’+ vw’z’ + vw’x + v’wz + v’w’y’z’

 

F= w’z’ + v’wz + vw’x

F’= vw + wz’ + w’x’z + v’w’z

F= (v’+w’) (w’+z) (w+x+z’) (v+w+z’)

3.8: draw the gate implementation of the simplified boolean functions in problem 3.7 using AND and OR gates

a) F=z’+ xy

F= (x+z’) (y+z’)

b) F=C’D + A’B’CD’ + ABCD’

F  =(C+D) (C’+D’) (A+B’+D’) (A’+B+D’)

c)F=B’D’ + A’C’ + AD’

F=(A’+D’) (C’+D’) (A+B’+C’)

   d)   F=B’D’ + A’BD + A’CD’

F= (B+D’) (A’+B’) (B’+C+D)

b)    e)   F= w’z’ + v’wz + vw’x

F= (v’+w’) (w’+z) (w+x+z’) (v+w+z’)

3.9: simplify each of the following functions and implement them with NAND gates. Give two alternatives.

a) F1=AC’+ACE+ACE’+A’CD’+A’D’E’

F1=D’E’+A+CD’

F1’=A’D+A’C’E

F1=(A’D+A’C’E)’

b) F2=(B’+D’)(A’+C’+D)(A+B’+C’+D)(A’+B+C’+D’)

F2’=BD+BC+AC

F2=(BD+BC+AC)’

F2=C’D’+A’B’+B’C’

3.15: simplify the Boolean function in sum of products using the don’t care conditions

a) F=y’+x’z’

     d=yz+xy

F=1

b) F=B’C’D’ + BCD’ + ABCD’

   d=B’CD’+A’BC’D

F=B’D’ + CD’