Question

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Question#1: Write the complement of the following function using product of maxterms
F(A,B,C,D) = ∑(0,2,4,6,8,10,12,14)






Question#2: Write the following function using sum of minterms
F(A,B,C,D) = ABC` + ABCD` + AC`







Question#3: Write the truth table for the complement of the following function
F(A,B,C,D) = ∏(0,2,4,6,8,10,12,14)




Question#4: Simplify the following function using Algebra F = Y(X + Y) + (X+Y)’Z + YZ








Question#5: What is the expression of the following function in product of maxterms
F(x,y,z) = (x+y)(x'+z)(y'+z'),

Answer 1

1)

F(A,B,C,D) = ∑(0,2,4,6,8,10,12,14)

complement of the function F'(A,B,C,D) in sum of minterm form means it contains the term that is not present in this sum of minterm expression.

so F'(A,B,C,D) = ∑(1,3,5,7,9,11,13,15)

now complement of function in product of maxterms form means it contains the term that is not present in this complement of sum of minterm expression.

so F'(A,B,C,D)=Π(0,2,4,6,8,10,12,14).

2)

Function in sum of minterms form

F(A,B,C,D) = ABC` + ABCD` + AC`

                   =ABC’(D+D’) + ABCD’ +AC’(B+B’)(D+D’) //expand 1st term by anding it with(D+D'), expand 3rd term by anding with(B+B')(D+D')

                   =ABC’D + ABC’D’ +ABCD’ + AC’BD + AC’BD’ + AC’B’D + AC’B’D’

                   = ABC’D + ABC’D’ + ABCD’+ ABC’D + ABC’D’ + AB’C’D + AB’C’D’

                   = ABC’D +ABC’D’ + ABCD’ + AB’C’D + AB’C’D’ (same terms written one time)

                   =m13,m12,m14,m9,m8

=Σ(8,9,12,13,14)

3)

complement of the function
F(A,B,C,D) = ∏(0,2,4,6,8,10,12,14)

complement of the function F'(A,B,C,D) in product of maxterm form means it contains the term that is not present in this product of maxterm expression.

F'(A,B,C,D) = ∏(1,3,5,7,9,11,13,15)

Truth table

A	B	C	D	Maxterm	F	F’
0	0	0	0	M0=A+B+C+D	0	1
0	0	0	1	M1=A+B+C+D’	1	0
0	0	1	0	M2=A+B+C’+D	0	1
0	0	1	1	M3=A+B+C’+D’	1	0
0	1	0	0	M4=A+B’+C+D	0	1
0	1	0	1	M5=A+B’+C+D’	1	0
0	1	1	0	M6=A+B’+C’+D	0	1
0	1	1	1	M7=A+B’+C’+D’	1	0
1	0	0	0	M8=A’+B+C+D	0	1
1	0	0	1	M9=A’+B+C+D’	1	0
1	0	1	0	M10=A’+B+C’+D	0	1
1	0	1	1	M11=A’+B+C’+D’	1	0
1	1	0	0	M12=A’+B’+C+D	0	1
1	1	0	1	M13=A’+B’+C+D’	1	0
1	1	1	0	M14=A’+B’+C’+D	0	1
1	1	1	1	M15=A’+B’+C’+D’	1	0

4)

F= Y(X + Y) + (X+Y)’Z + YZ

= YX+YY+X’Y’Z+YZ ( (X+Y)’= X’Y’)

= XY+Y+X’Y’Z+YZ    (YY=Y)

= Y(X+1)+X’Y’Z+YZ   

= Y + X’Y’Z +YZ   (X+1=1)

= Y+YZ +X’Y’Z

= Y(1+Z) +X’Y’Z

= Y + X’Y’Z (1+Z=1)

5)

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

             =(x+y+zz’)(x’+yy’+z)(xx’+y’+z’) //expand 1st term by oring it with zz', expand 2nd term by oring it with yy', expand 3rd term by oring it with xx'

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

            =M0.M1.M4.M6.M3.M7

            = Π(0,1,3,4,6,7)