Morris Mano chapter 2- solution: boolean algebra and logic gates

 

Boolean algebra and logic gates

2.5 :  Simplify the following to minimum number of literals

a) xy +xy'

=x ( y + y' )

=x

b) ( x+y ) ( x+y' )

= xx + xy' + xy + yy'

=x + x( y+y' )

=x + x

=x

c) xyz +x'y + xyz'

= xy ( z+z' ) + x'y

=xy + x'y

=y (x+x')

d)zx+ zx'y

= zx + zy ( theorem 7)

= z(x+y)

e) (A+B)' (A'+B')'

=A'B' (AB)

= AA' BB'

=0

f) y ( wz'+wz) + xy

= wy(z'+z) + xy

=y ( x+w)

2.6 : Reduce the following boolean expressions to the required no of literals

a) ABC + A'B'C + A'BC + ABC' + A'B'C'(to five literals)

= AB ( C+C' ) + A'B'(C+C') +A'BC

=AB+A'B'+A'BC

=AB + BC + A'B' ( theorem 7)

= B(A+C) + A'B'

b) BC + AC' + AB + BCD (to four literals)

= BC (A+D) + A(B+C')

=BC + AB +AC' ( theorem 9)

= BC + AC'

c) [ (CD)' + A]' + A + CD + AB ( to three literals)

= CDA' + A + CD + AB

= CD (1+A') + A (1+B)

= CD + A

d) (A+C+D) (A+C+D') (A+C'+D) (A+B') ( to four literals)

=( A+C+DD') (A+C'+D) (A+B')

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

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

=(A+CD)(A+B')

=A + CDB'

2.7 : Find the complement of the following boolean functions and reduce them to a minimum number of literals:

b) B'D + A'BC' + ACD + A'BC

[B'D + A'BC' + ACD + A'BC]'

=(B'D)' (A'BC')' (ACD)' (A'BC)'

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

= (B+D') (A+B'+CC') (A'+C'+D')

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

=(AB+BB'+AD'+B'D') (A'+C'+D')

=(AB+AD'+B'D') (A'+C'+D')

=(AB+B'D') (A+C'+D')

=AAB + AC'B + ABD' + AB'D' + B'C'D' + B'D'D' 

=AB(1+C') + AD'(B+B') + B'D'(1+C')

=AB + AD' + B'D'

=A(B+D') + B'D'

c) [(AB)'A] [(AB)'B]

[[(AB)'A] [(AB)'B]]'

=((AB)'A)' + ((AB)'B)'

=( AB + A') + ( AB + B') 

= A'+B + B' + A

=1 + 1

= 1

d) AB' + C'D'

[ AB' + C'D']'

= (AB')' (C'D')'

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

2.8: given two boolean functions F1 and F2.

a) show that the boolean function E=F1+F2 obtained by ORing the two functions contains the sum of all the minterms in F1 and F2.

solution: 

F1= m1+m3+m5+m7

                                                                  F2=m0+m1+m4+m5

                                                                  E=m0+m1+m3+m4+m5+m7

                                                                    =F1+F2

 b) show that the boolean function E=F1.F2 obtained by AND ing  the two functions contains those minterms common to F1 and F2.

                            Now, F1=M1+M3+M5+M7

                                        F2=M0+M1+M4+M5
                                        G=M1+M5

2.9: Obtain the truth table of the function

      F=xy + xy' + y'z

 solution: 

2.11: Given the boolean function

                              F=xy + x'y' +y'z

a) Implement it with AND, OR and NOT gates

b) Implement only with OR and NOT gates

solution:

F=xy + x'y' + y'z

  = (xy)'' + (x+y)' + (y'z)''

  =(x'+y')' + (x+y)' +(y+z')'

c)Implement it only with AND and NOT gate

solution:

 F= xy + x'y' +y'z

   =(xy + x'y' + y'z)''

   =[ (xy)' (x'y')' (y'z)' ]'

   

2.12: Simplify the function T1 and T2 to a minimum number of literals

solution: 

           T1= m0 + m1 +m2

                =A'B'C' + A'B'C + A'BC'

                =A'B'(C+C') + A'BC'

                =A'B' + A'BC'

                =A'B' + A'C'

                =A'(B'+C')

            T2= T1'

                =[A'(B'+C')]'

                =A+(B'+C')'

                =A+BC

2.13: Express the following function in a sum of minterms and a product of maxterms

a) F(A,B,C,D) = D(A'+B) + B'D

F(A,B,C,D)

=A'D+BD+B'D

=A'(B+B')D + (A+A')BD + (A+A')B'D 

=A'BD + A'B'D + ABD + A'BD + AB'D + A'B'D

=A'B(C+C')D + A'B'(C+C')D + AB(C+C')D + A'B(C+C')D + AB'(C+C')D + A'B'(C+C')D

=A'BCD + A'BC'D + A'B'CD + A'B'C'D + ABCD + ABC'D + A'BCD + A'BC'D + AB'CD + AB'C'D + A'B'CD + A'B'C'D

=A'BCD + A'BC'D + A'B'CD + A'B'C'D + ABCD + ABC'D + AB'CD + AB'C'D 

=m1+m3+m5+m7+m9+m11+m13+m15

=M2+M4+M6+M8+M10+M12+M14

b) F(w,x,y,z) =y'z+wxy'+wxz'+w'x'z

F(w,x,y,z)

=(x+x')y'z + wxy'(z+z') + wx(y+y')z' + w'x'(y+y')z

=xy'z + x'y'z + wxy'z + wxy'z' + wxyz' + wxy'z' + w'x'yz + w'x'y'z

=(w+w') xy'z + (w+w')x'y'z + wxy'z + wxy'z' + wxyz' + wxy'z' + w'x'y'z 

=wxyz' + w'xyz' + wx'y'z + w'x'y'z + wxy'z + wxy'z' + wxyz' + wxy'z' + w'x'yz + w'x'y'z

=wxy'z + w'xy'z + wx'y'z + w'x'y'z + wxy'z' + wxyz' + w'x'yz

=m1+m3+m5+m9+m12+m13+m14

=M0+M2+M4+M6+M7+M8+M10+M11+M15

c) F(A,B,C,D)

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

=(A+B'+C+DD') (A+B'+CC'+DD') (A+BB'+C'+D') (A'+B+C+D') (AA'+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')(A'+B+C+D')(A+B+C'+D')(A'+B+C'+D')

=M3+M4+M5+M6+M7+M9+M11

=m0+m1+m2+m8+m10+m12+m13+m14+m15

d) F(A,B,C)

=(A'+B) (B'+C)

=(A'+B+CC')(AA'+B'+C)

=(A'+B+C)(A'+B+C')(A+B'+C)(A'+B'+C)

=M2+M4+M5+M6

=m0+m1+m3+m7

e) F(x,y,z)

=1

=x'y'z'+x'y'z+x'yz'+x'yz+xy'z'+xy'z+xyz'+xyz

=m0+m1+m2+m3+m4+m5+m6+m7

NO MAXTERM 

NOTE: If F=0, no minterm

f) F(x,y,z)

=(xy+z) (y+xz)

=(x+z)(y+z)(y+x)(y+z)

=(x+yy'+z)(xx'+y+z)(x+y+zz')

=(x+y+z)(x+y'+z)(x+y+z)(x'+y+z)(x+y+z)(x+y+z')

=M0+M1+M2+M4

=m3+m5+m6+m7