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Dr.Khaled Kh. Sharaf

Faculty Of Computers And Information

Technology

Second Term 2018 - 2019

DIGITAL DESIGN

Discussion

Chapter 2

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Chapter 2

Chapter 2

Question 1Demonstrate the validity of the following identities by means of truth tables:(a)DeMorgan’s theorem for three variables:

(x + y + z)’ = x’y’z’

x’y’z’z’y’x’(x+y+z)’x+y+zxyz

111110000

001101001

010101010

000101011

011001100

001001101

010001110

000001111

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Chapter 2

Question 1Demonstrate the validity of the following identities by means of truth tables:(a)DeMorgan’s theorem for three variables:

(xyz)’ = x’ + y’ + z’

X +’y +’z’z’y’x’(x+y+z)’x+y+zxyz

111110000

001101001

010101010

000101011

011001100

001001101

010001110

000001111

Chapter 2

Question 1(b) The distributive law: x + yz = ( x + y )( x + z )

(x+y)(x+z)(x+z)(x+y)x+yzxyz

0000000

0100001

0010010

1111011

1111100

1111101

1111110

1111111

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Chapter 2

Question 1Demonstrate the validity of the following identities by means of truth tables:(c) The distributive law: x(y + z) = xy + xz

xy+xzxzxyx(y+z)xyz

0000000

0000001

0000010

0000011

0000100

1101101

1011110

1111111

Chapter 2

Question 1Demonstrate the validity of the following identities by means of truth tables:(d) The associative law: x + (y + z) = (x + y) + z

(x + y) + z(x + y) x+(y+z)(y+z)xxyz

00000000

10110001

11110010

11110011

11101100

11111101

11111110

11111111

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Chapter 2

Question 1Demonstrate the validity of the following identities by means of truth tables:(e) The associative law and x(yz) = (xy)z

(xy)z xyx(yz)yzxyz

0000000

0000001

0000010

0001011

0000100

0000101

0100110

1111111

Chapter 2

Question 2Simplify the following Boolean expressions to a minimum number of literals and draw the original and simplified expressions

xy + xy’ = x(x+y’)= x

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Chapter 2

Question 2

(b)(x + y)(x + y’)= x+yy’=x(x+y’) + y(x+y’)

= xx + xy’ + xy + yy’= x + x(y’+ y) + 0= x + x

= x

remembre xx = xyy’=0y+y’=1X+x=x

Chapter 2

Question 2

(c) xyz + x’y + xyz’ = xy(z+z’) + x’y= xy + x’y= y(x + x’)= y

remembre z+z’=1X+x’=1

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Chapter 2

Question 2

(d) (A + B)’(A’ + B’)’= (A’B’)(AB)=(A’B’)(BA)= A’(B’B)A=0

remembre BB’=0

Chapter 2

Question 2

(e) (a + b + c’)(a’b’ + c)= aa’b’ + ac + ba’b’ + bc + c’a’b’ + c’c= 0 + ac + 0 + bc + c’a’b’ + 0 = ac + bc + a’b’c’

??remembre aa’=0bb’=0

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Chapter 2

Question 2

(f) a’bc + abc’ + abc + a’bc’= a’b(c+c’) + ab(c+c’) = a’b + ab = (a’+ a)b= b

??

remembre a’+ a=1

Chapter 2

Question 3

Find the complement of the following expressions:(a) F = xy’+ x’y

F’ = (xy’+ x’y)’ = (x’ + y)(x + y’)= x’x + x’y’ + xy + yy’= 0 + x’y’ + xy + 0= x’y’ + xy

(b) F = (a + c)(a + b’)(a’ + b + c’)F’ = [(a + c)(a + b’)(a’ + b + c’)]’

= (a+c)’+ (a + b’)’ + (a’+b+c’)’= a’c’ + a’b + ab’c=

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Chapter 2

Question 4

Answer

Chapter 2

Question 5

List the truth table of the function:(a) F = xy + xy’ + y’z (b) F= bc + a’c’

11 10 01 11 00

Fxyzm

00000

10011

00102

00113

11004

11015

11106

11117

FabcM

10000

00011

10102

10113

01004

01015

01106

11117

(a) F(x,y,z)= ∑(1,4,5,6,7) (b) F(a,b,c)= ∑(0,2,3,7)

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Chapter 2

Question 6

Chapter 2

Answer Q6

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Chapter 2

Answer Q6

Chapter 2

Answer Q6

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Chapter 2

Question 7

T2T1CBA

01000

01100

01010

10110

10001

10101

10011

10111

Simplify the following Boolean function T1 and T2 to minimum number of literals:

T1= A’B’C’ + A’B’C + A’BC’

= A’B’(C’+C) + A’C’(B’+B)

= A’B’ + A’C’

= A’(B’ + C’)

T2=T1’

T2= [A’(B’ + C’)]’

= A + BC

Chapter 2

Question 8

Express the following function as sum of minterms and as product of maxterms:

F(A,B,C,D) = B’D + A’D + BD

01 01 11

F=∑m(1,3,5,7,9,11,13,15)

F=∑M(0,2,4,6,8,10,12,14)

M A B C D B’D A’D BD0 0 0 0 01 0 0 0 1 x X2 0 0 1 03 0 0 1 1 x X4 0 1 0 05 0 1 0 1 X X6 0 1 1 07 0 1 1 1 x X8 1 0 0 09 1 0 0 1 x10 1 0 1 011 1 0 1 1 x12 1 1 0 013 1 1 0 1 X14 1 1 1 015 1 1 1 1 x

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Chapter 2

Question 9

Answer

Chapter 2

Question 10

Answer

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Chapter 2

Question 11

Convert each of the following expression into sum of products and product of sum:

(a)(u + xw)(x + u’v)

= ux + uu’v + xwx + xwu’v

= ux + 0 + xw + xwu’v

= ux + xw( 1 + u’v)

= ux + xw (SOP form)

= x(u + w) (POS form)

Remember

(1 + u’v)=1

Chapter 2

Question 12

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Chapter 2

Answer 12

Chapter 2

Question 13

Answer