chapter 26 boolean algebra and logic...

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CHAPTER 26 BOOLEAN ALGEBRA AND LOGIC CIRCUITS

EXERCISE 107 Page 239

1. Determine the Boolean expression and construct a truth table for the switching circuit shown

below.

For the circuit to function, Z = C AND [(A AND B) OR ( A AND B)]

i.e. Z = ( ). . .C A B A B+

The truth table is shown below: A B C A.B A A .B A.B + A .B Z = C.( A.B + A .B) 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 0 0 0 0 0 1 1

1 1 1 1 0 0 0 0

0 0 1 1 0 0 0 0

0 0 1 1 0 0 1 1

0 0 0 1 0 0 0 1


below.

For the circuit to function, Z = C AND [(A AND B ) OR ( A )] i.e. Z = ( ). .C A B A+

The truth table is shown below: A B C A B .A B .A B + A Z = C.( A. B + A ) 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 1 1 1 0 0 0 0

1 1 0 0 1 1 0 0

0 0 0 0 1 1 0 0

1 1 1 1 1 1 0 0

0 1 0 1 0 1 0 0

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below.

For the circuit to function, Z = A AND [(B AND C ) OR ( B AND C ) OR ( A AND B )] AND B

i.e. Z = A. ( ). . . .B B C B C A B+ +

The truth table is shown below: A B C A B C .B C

.B C

.A B

( ). . .B C B C A B+ + Z = A. ( ). . . .C B C B C A B+ +

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 1 1 1 0 0 0 0

1 1 0 0 1 1 0 0

1 0 1 0 1 0 1 0

0 0 1 0 0 0 1 0

0 1 0 0 0 1 0 0

0 0 1 1 0 0 0 0

0 1 1 1 0 1 1 0

0 0 0 0 0 1 0 0

4. Determine the Boolean expression and construct a truth table for the switching circuit shown below.

For the circuit to function, Z = C AND [(B AND C AND A ) OR (A AND (B ORC )]

i.e. Z = ( ). . . .( )C B C A A B C+ +

The truth table is shown below:

A B C A

C B.C. A B + C A.( B + C ) B.C. A + A.( B +C ) Z

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 1 1 1 0 0 0 0

1 0 1 0 1 0 1 0

0 0 0 1 0 0 0 0

1 0 1 1 1 0 1 1

0 0 0 0 1 0 1 1

0 0 0 1 1 0 1 1

0 0 0 1 0 0 0 1

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5. Construct a switching circuit to meet the requirements of the Boolean expression:

A.C + A. B .C + A.B

The switching circuit for the Boolean expression A.C + A. B .C + A.B is shown below:


A.B.C.(A + B + C)

The switching circuit for the Boolean expression A.B.C.(A + B + C) is shown below:


A.(A. B .C + B.(A + C ))

The switching circuit for the Boolean expression A.(A. B .C + B.(A + C )) is shown below:

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8. Derive the Boolean expression and construct the switching circuit for column 4 in the truth table

shown below.

The Boolean expression for column 4 is: _A .

_B .C + A.B.

_C

The switching circuit is as shown below:

9. Derive the Boolean expression and construct the switching circuit for column 5 in the truth table

shown in Problem 8.


_B .

_C +

_A .B.C + A.

_B .

_C


10. Derive the Boolean expression and construct the switching circuit for column 6 in the truth

table shown in Problem 8.

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_B .

_C +

_A .B.

_C + A.

_B .

_C + A.

_B .C


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1. Simplify: P .Q + P .Q

P . Q + P .Q = ( ) ( ). . 1P Q Q P+ = from 10, Table 26.7

= P

2. Simplify: _P .Q + P.Q +

_P .

_Q

P . Q + P .Q + .P Q = ( ) ( ). . . 1 .P Q Q P Q P P Q+ + = + from 10, Table 26.7

= .P P Q+

3. Simplify: ( ). . .F G F G G F F+ + +

( ). . .F G F G G F F+ + + = . . (1) . .F G F G G F G F G G+ + = + + from 10, Table 26.7

= ( ).G F F G+ + = .1G G G G+ = + from 10, Table 26.7

= G from 9, Table 26.7

4. Simplify: ( ). . .F G F G G F G+ + +

( ). . .F G F G G F G+ + + = . .1 . . .F G F F G F G F F G+ + = + +

= ( ). .1F G G F F F+ + = +

= F + F = F from 9 and 10, Table 26.7 5. Simplify: (P + P.Q).(Q + Q.P)

(P + P.Q).(Q + Q.P) = P.Q from 15, Table 26.7

6. Simplify: . . . . . .F G H F G H F G H+ +

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. . . . . .F G H F G H F G H+ + = ( ). . . .F H G G F G H+ +

= . . .F H F G H+ from 10, Table 26.7

= ( ). .H F F G+

7. Simplify: F._G .

_H + F.G.H +

_F .G.H

F._G .

_H + F.G.H +

_F .G.H = ( ). . . .F G H G H F F+ + = ( ). . . . 1F G H G H+ from 10, Table 26.7

= . . .F G H G H+

8. Simplify: . . . . . .P Q R P Q R P Q R+ +

. . . . . .P Q R P Q R P Q R+ + = ( ). . . .Q R P P P Q R+ +

= . . .Q R P Q R+ from 10, Table 26.7 9. Simplify: . . . . . . . .F G H F G H F G H F G H+ + +

. . . . . . . .F G H F G H F G H F G H+ + + = ( ) ( ). . .F G H H F G H H+ + +

= . .F G F G+ from 10, Table 26.7,

= ( ).G F F+

= G from 10, Table 26.7

10. Simplify: F._G .H + F.G.H + F.G.

_H +

_F .G.

_H

( ) ( ). . . . . . . . . . . .F G H F G H F G H F G H F H G G G H F F+ + + = + + +

= ( ) ( ). . 1 . . 1F H G H+

= . .F H G H+

11. Simplify: R.(P.Q + P._Q ) +

_R .(

_P .

_Q +

_P .Q)

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R.(P.Q + P._Q ) +

_R .(

_P .

_Q +

_P .Q) = ( ) ( ). . . .P R Q Q P R Q Q+ + +

= ( ) ( ). . 1 . . 1P R P R+ from 10, Table 26.7

= . .P R P R+

12. Simplify: ( ) ( ). . . . . . .R P Q P Q P Q P Q R Q R+ + + +

( ) ( ). . . . . . .R P Q P Q P Q P Q R Q R+ + + + = ( ) ( ). . . . . .R P Q R P Q Q P R Q Q+ + + +

= . . . .R P Q R P P R+ + from 10, Table 26.7

= ( ). . .R P Q P R R+ +

= . .R P Q P+ from 10, Table 26.7

= .P R Q+ from 17, Table 26.7

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1. Simplify: ( ) ( ). . .A B A B

( ) ( ) ( ) ( ) ( ) ( ). . . . . . .A B A B A B A B A B A B= + = + by de Morgan’s law,

= . . . .A B A A B B+ = .(0) .B A B+ from 14 and 13, Table 26.7

= .A B

2. Simplify: ( . ) ( . )A B C A B C+ + +

( . ) ( . )A B C A B C+ + + = ( ) ( ). .A B C A B C+ + + by de Morgan’s law

= . .A B C A B C+ + +

= .( . 1)C A B A B+ + +

= C A B+ + or A B C+ + from 8, Table 26.7

3. Simplify: ( . . ). .A B B C A B+

( . . ). .A B B C A B+ = ( ) ( ) ( ). . . .A B B C A B+ by de Morgan’s law

= ( ) ( ) ( ). .A B B C A B+ + +

= ( ) ( ) ( ). .A B B C A B+ + +

= ( ) ( ). . . . .A B AC B B B C A B+ + + +

= ( ) ( ). . . .A B AC B B C A B+ + + + from 13, Table 26.7

= ( ) ( ). . .A B AC B A B+ + + from 15, Table 26.7

= ( ) ( ). 1 . .B A AC A B + + +

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= ( ) ( ). .B AC A B+ + from 8, Table 26.7

= . . . . . .A B B B AC A A B C+ + +

= . 0 0 . .A B A B C+ + + from 14, Table 26.7

= . . .A B A B C+

4. Simplify: ( ) ( ). . .A B B C A B+ +

( ) ( ) ( ). . .A B B C A B A B B C A B+ + = + + + + + by de Morgan’s law

= A B B C A B+ + + + +

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

= (1) (1) C B+ + + from 10, Table 26.7

= 1 + C + 1 + B

= 1 from 8, Table 26.7

5. Simplify: ( . . ).( . . )P Q P R P Q R+

( . . ).( . . )P Q P R P Q R+ = ( ) ( ) ( ). .P Q P R P Q R + + + + by de Morgan’s law

= ( )( ). .P Q P R P Q P R+ + + +

= ( ) ( )1 . . .Q R P Q P R+ + +

= . . . . . . . . . .P Q P R Q P Q Q P R R P Q R P R+ + + + +

= . . 0 . . . . 0P Q P R Q P R R P Q+ + + + + from 14 and 9, Table 26.7

= ( ). . .P Q R Q R R Q+ + +

= ( )( ).P Q R R Q Q+ + +

= ( ).P Q R R+ + from 10, Table 26.7

= ( ).P Q R+ from 9, Table 26.7

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l . Use Karnaugh map techniques to simplify: _X .Y + X.Y

.X Y corresponds to X = 0, Y = 1, i.e. the bottom left hand cell of the Karnaugh map, shown as a 1.

.X Y corresponds to X = 1, Y = 1, i.e. the bottom right hand cell.

The Karnaugh map is shown below.

The only variable common to both 1s is Y

Hence, _X .Y + X.Y = Y

2. Use Karnaugh map techniques to simplify: _X .

_Y +

_X .Y + X.Y


The vertical two-cell couple corresponds to: X

The horizontal two-cell couple corresponds to: Y

Hence, _X .

_Y +

_X .Y + X.Y

simplifies to: X Y+

3. Use Karnaugh map techniques to simplify: ( . ).( . )P Q P Q

.P Q corresponds to P = 0, Q = 0, i.e. the top left-hand cell of the Karnaugh map, shown as a 1.

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.P Q corresponds to P = 0, Q = 1, i.e. the bottom left-hand cell, hence .P Q corresponds to each of

the other three cells, shown as 2s.

Only one cell has both a 1 and a 2 in it, i.e. P = 0, Q = 0

Hence, ( . ).( . )P Q P Q = .P Q 4. Use Karnaugh map techniques to simplify: . .( ) . .( )AC A B C A B C B+ + + +

If a Boolean expression contains brackets it is often easier to remove them, using the laws and rules

of Boolean algebra, before plotting the function on a Karnaugh map.

Thus, . .( ) . .( )AC A B C A B C B+ + + + = . . . . . . .AC A B AC A B C A B B+ + + +

= . . . . . .0AC A B AC A B C A+ + + +

= . . . . .AC A B AC A B C+ + +

.AC corresponds to A = 1, C = 0, shown as 1s in the two right-hand cells in the top row

.A B corresponds to A = 0, B = 1, shown as 1s in the two cells in the second column

.AC corresponds to A = 0, C = 1, shown as 1s in the two left-hand cells in the bottom row

. .A B C corresponds to A = 1, B = 1, C = 1, shown as a 1 in the cell in the third column, bottom row

A four-cell couple and two two-cell couples are formed as shown by the broken lines

The only variable common to the four-cell couple is B = 1, i.e. B

The variable common to the two-cell couple on the top right of the map is A = 1 and C = 0, i.e. .AC

The variables common to the two-cell couple on the bottom left of the map is A = 0 and C = 1, i.e.

.A C

Thus, . .( ) . .( )AC A B C A B C B+ + + + simplifies to . .B AC AC+ +

5. Use Karnaugh map techniques to simplify: _P .

_Q .

_R +

_P .Q.

_R + P.Q.

_R

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_P .

_Q .

_R , i.e. P = 0, Q = 0 and R = 0, is shown on the Karnaugh map as a 1 in the top left cell

_P .Q.

_R , i.e. P = 0, Q = 1 and R = 0, is shown on the Karnaugh map as a 1 in the second left cell on

the top line

P.Q._R , i.e. P = 1, Q = 1 and R = 0, is shown on the Karnaugh map as a 1 in the third left cell on the

top line

Two two-cell couples are formed as shown

For the left-hand cell, the variables common are P = 0, R = 0, i.e. .P R

For the right-hand cell, the variables common are Q = 1, R = 0, i.e. .Q R

Hence, _P .

_Q .

_R +

_P .Q.

_R + P.Q.

_R may be simplified to: .P R + .Q R

i.e. _P .

_Q .

_R +

_P .Q.

_R + P.Q.

_R simplifies to: .P R + .Q R or ( ).R P Q+

6. Use Karnaugh map techniques to simplify: . . . . . . . .P Q R P Q R P Q R P Q R+ + +

. .P Q R , i.e. P = 0, Q = 0 and R = 0, is shown on the Karnaugh map as a 1




Two two-cell couples are formed as shown

For the cell containing the 2 and the 3, the variables common are P = 1, Q = 1, i.e. P.Q

For the cell containing the 3 and the 4, the variables common are P = 1, R = 1, i.e. P.R

Hence, . . . . . . . .P Q R P Q R P Q R P Q R+ + + may be simplified to: P.Q + P.R + . .P Q R

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i.e. . . . . . . . .P Q R P Q R P Q R P Q R+ + + simplifies to: P.(Q + R) + . .P Q R 7. Use Karnaugh map techniques to simplify: . . . . . . . . .A B C D A B C D A B C D+ +

. . .A B C D , i.e. A = 0, B = 0, C = 0 and D = 0, is shown as a 1 in the top left cell

. . .A B C D , i.e. A = 0, B = 1, C = 0 and D = 0, is shown as a 1 in the top, second left cell

. . .A B C D , i.e. A = 0, B = 1, C = 0 and D = 1, is shown as a 1 in the second line, second left cell

Two two-cell couples are formed as shown.

The variables common to the vertical couple is A = 0, B = 1, C = 0, i.e. . .A B C

The variables common to the horizontal couple is A = 0, C = 0, D = 0, i.e. . .A C D

Hence, . . . . . . . . .A B C D A B C D A B C D+ + simplifies to: . .A B C + . .A C D or ( ). .A C B D+

8. Use Karnaugh map techniques to simplify: . . . . . . . . .A B C D A B C D A B C D+ +

. . .A B C D , i.e. A = 0, B = 0, C = 1 and D = 1, is shown as a 1 on the four variable matrix

. . .A B C D , i.e. A = 0, B = 0, C = 1 and D = 0, is shown as a 2


Two two-cell couples are formed as shown.

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The variables common to the vertical couple is A = 0, B = 0, C = 1, i.e. . .A B C

The variables common to the horizontal couple is B = 0, C = 1, D = 0, i.e. . .B C D

Hence, . . . . . . . . .A B C D A B C D A B C D+ + simplifies to: . .A B C + . .B C D or ( ). .B C A D+

9. Use Karnaugh map techniques to simplify:

. . . . . . . . . . . . . . .A B C D A B C D A B C D A B C D A B C D+ + + +

. . .A B C D , i.e. A = 0, B = 1, C = 0 and D = 1, is shown as a 1 on the four-variable matrix below





A two-cell couple is formed as shown and the variables common to it are B = 1, C = 0 and D = 1,

i.e. . .B C D

A four-cell couple is formed as shown and the variables common to it are A = 1, D = 1, i.e. .A D

Hence, . . . . . . . . . . . . . . .A B C D A B C D A B C D A B C D A B C D+ + + + simplifies to:

. .B C D + .A D or ( ). .D A B C+

10. Use Karnaugh map techniques to simplify: . . . . . . . . . . . . . . .A B C D A B C D A B C D A B C D A B C D+ + + +

. . .A B C D , i.e. A = 0, B = 0, C = 0 and D = 1, is shown as a 1 on the four variable matrix below


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A four-cell couple is formed as shown and the variables common to it are A = 1, D = 0, i.e. .A D

Hence, . . . . . . . . . . . . . . .A B C D A B C D A B C D A B C D A B C D+ + + + simplifies to:

. . . .A D A B C D+

11. Use Karnaugh map techniques to simplify: . . . . . . . . . . . . . . . . . . . . .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+ + + + + +

The Karnaugh map for the given expression is shown below. A four-cell couple and three two-cell

couples are formed as shown.

The variables common to the four-cell couple are A = 0 and C = 1, i.e. .A C

The variables common to the two-cell couple on the far right of the top row are A = 1, C = 0 and

D = 0, i.e. . .A C D

The variables common to the two-cell couple on the far left and far right of the top row are B = 0,

C = 0 and D = 0, i.e. . .B C D

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The variables common to the two-cell couple at the top and bottom of the first column are A = 0,

B = 0 and D = 0, i.e. . .A B D

Hence, . . . . . . . . . . . . . . . . . . . . .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+ + + + + +

simplifies to: .A C + . .A C D + . .B C D + . .A B D

i.e. ( ). . . . .A C A C D B D A C+ + +

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1. Devise a logic system to meet the requirements of the Boolean expression: Z = _A + B.C

The logic system for the Boolean expression Z = _A + B.C is as shown below:

2. Devise a logic system to meet the requirements of the Boolean expression: Z = A._B + B.

_C

The logic system for the Boolean expression Z = A._B + B.

_C is as shown below:

3. Devise a logic system to meet the requirements of the Boolean expression:

Z = A.B._C +

_A .

_B .C

The logic system for the Boolean expression Z = A.B._C +

_A .

_B .C is as shown below:

4. Devise a logic system to meet the requirements of the Boolean expression:

Z = (_A + B).(

_C + D)

The logic system for the Boolean expression Z = (_A + B).(

_C + D) is as shown below:

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5. Simplify the expression given in column 4 of the truth table below and devise a logic circuit to meet the requirements of the simplified expression.

From column 4, 1 . . . . . . . . . .Z A B C A B C A B C A B C A B C= + + + +


The vertical two-cell couple corresponds to: A.B

The horizontal four-cell couple corresponds to: C

Hence, 1 . . . . . . . . . .Z A B C A B C A B C A B C A B C= + + + +

simplifies to: 1 .Z A B C= +

A logic circuit to meet these requirements is shown below.

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6. Simplify the expression given in column 5 of the truth table in Problem 5 and devise a logic circuit to meet the requirements of the simplified expression. From column 5, 2 . . . . . . . .Z A B C A B C A B C A B C= + + +


The vertical two-cell couple corresponds to: .A B

The horizontal two-cell couple corresponds to: B.C

Hence, 2 . . . . . . . .Z A B C A B C A B C A B C= + + +

simplifies to: 2 . .Z A B B C= +


7. Simplify the expression given in column 6 of the truth table of Problem 5 and devise a logic circuit to meet the requirements of the simplified expression. From column 6, 3 . . . . . . . . . .Z A B C A B C A B C A B C A B C= + + + +


The horizontal two-cell couple corresponds to: A.C

The four-cell couple corresponds to: B

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Hence, 3 . . . . . . . . . .Z A B C A B C A B C A B C A B C= + + + +

simplifies to: 3 .Z AC B= +


8. Simplify the Boolean expression: _P .

_Q +

_P .Q + P.Q and devise a logic circuit to meet the

requirements of the simplified expression.

Let Z =_P .

_Q +

_P .Q + P.Q


The vertical two-cell couple corresponds to: P

The horizontal two-cell couple corresponds to: Q

Hence, Z =_P .

_Q +

_P .Q + P.Q

simplifies to: Z P Q= +


9. Simplify the Boolean expression: . . . . . .P Q R P Q R P Q R+ + and devise a logic circuit to meet the


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The Karnaugh map for the Boolean expression: . . . . . .P Q R P Q R P Q R+ + is shown below.

The two-cell couple on the far right of the map corresponds to: .P R

The other two-cell couple corresponds to: .Q R

Hence, . . . . . .P Q R P Q R P Q R+ +

simplifies to: ( ). . .P R Q R or R P Q+ +


10. Simplify the Boolean expression: P._Q .R + P.

_Q .

_R +

_P .

_Q .

_R and devise a logic circuit to

meet the requirements of the simplified expression.

The Karnaugh map for the Boolean expression: P._Q .R + P.

_Q .

_R +

_P .

_Q .

_R is shown below.

The vertical two-cell couple on the far right corresponds to: .P Q

The horizontal two-cell couple corresponds to: .Q R

Hence, P._Q .R + P.

_Q .

_R +

_P .

_Q .

_R

simplifies to: ( ). . .P Q Q R or Q P R+ +


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11. Simplify the Boolean expression:


and devise a logic circuit to meet the requirements of the simplified expression.

The Karnaugh map for the Boolean expression:


is shown below.

The two-cell couple on the bottom row of the map corresponds to: . .A C D

The four-cell couple corresponds to: .B D

Hence, . . . . . . . . . . . . . . .A B C D A B C D A B C D A B C D A B C D+ + + +

simplifies to: ( ). . . . .A C D B D or D A C B+ +


12. Simplify the Boolean expression: ( ) ( ). . . .P Q R P Q R+ and devise a logic circuit to meet the


The Karnaugh map for the Boolean expression: ( ) ( ). . . .P Q R P Q R+ is shown below.

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. .P Q R is shown with a 1

. .P Q R is shown with 2s P + Q.R is shown with 3s

.P Q R+ is shown with 4s

Hence, ( ) ( ). . . .P Q R P Q R+ is represented by the cells containing both 2s and 4s

The two-cell vertical couple of the map corresponds to: .P Q

The two-cell horizontal couple corresponds to: .P R

Hence, ( ) ( ). . .P Q R P Q R+ +

simplifies to: . .P Q P R+ or ( ).P Q R+


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EXERCISE 112 3

1. Use nand-gates only to devise the logic system: Z = A + B.C

The logic system to meet the requirements of Z = A + B.C is shown below:

2. Use nand-gates only to devise the logic system: Z = A._B + B.

_C

The logic system to meet the requirements of Z = A._B + B.

_C is shown below:

3. Use nand-gates only to devise the logic system: Z = A.B. _C +

_A .

_B .C

The logic system to meet the requirements of Z = A.B. _C +

_A .

_B .C is shown below:

4. Use nor-gates only to devise the logic system: Z = (_A + B).(

_C + D)

The logic system to meet the requirements of Z = (_A + B).(

_C + D) is shown below:

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5. Use nor-gates only to devise the logic system: Z = A._B + B.

_C + C.

_D

The logic system to meet the requirements of Z = A._B + B.

_C + C.

_D is shown below:

6. Use nor-gates only to devise the logic system: Z = _P .Q + P.(Q + R)

The logic system to meet the requirements of Z = _P .Q + P.(Q + R) is shown below:

7. In a chemical process, three of the transducers used are P, Q and R, giving output signals of

either 0 or 1. Devise a logic system to give a 1 output when:

(a) P and Q and R all have 0 outputs, or when

(b) P is 0 and (Q is 1 or R is 0).

The Boolean expression to meet the requirements is:

( ). . .P Q R P Q R+ + = . . . .P Q R P Q P R+ +

= ( ). . 1 .P R Q P Q+ +

= . .P R P Q+

= ( ).P Q R+

A logic circuit to satisfy this Boolean expression is shown below:

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8. Lift doors should close (Z) if:

(a) the master switch (A) is on and either

(b) a call (B) is received from any other floor, or

(c) the doors (C) have been open for more than 10 seconds, or

(d) the selector push within the lift (D) is pressed for another floor.

Devise a logic circuit to meet these requirements.


Z = A.(B + C + D)


9. A water tank feeds three separate processes. When any two of the processes are in operation at

the same time, a signal is required to start a pump to maintain the head of water in the tank.

Devise a logic circuit using nor-gates only to give the required signal.


Z = A.(B + C) + B.(A + C) + C.(A + B)

= A.B + A.C + A.B + B.C + A.C + B.C

= A.B + A.C + B.C

i.e. Z = A.(B + C) + B.C


© 2014, John Bird

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10. A logic signal is required to give an indication when:

(a) the supply to an oven is on, and

(b) the temperature of the oven exceeds 210°C, or

(c) the temperature of the oven is less than 190°C.

Devise a logic circuit using nand-gates only to meet these requirements.


Z = A.B +A.C

i.e. Z = A.(B + C)


chapter 26 boolean algebra and logic...

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