Computer Number Systems and Boolean Algebra Test A3 Solutions PDF

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Course/Section:_____________ ____

Start Time:____________

Professor’s Name:__________________

Computer Number Systems & Boolean Algebra Test A3 Name: __SOLUTIONS_____________ Time: 75 Minutes _____ Student Number: ______________________ 50 1) Complete the table below by converting the given in its current number system to the missing ones. (1 mark each = 6 marks) Binary Octal Decimal Hexadecimal 101111002

2748

18810

฀฀฀฀16

101000112

2438

16310

฀฀316

2) Perform the indicated operation in the given number systems. You must show your calculations. (2 marks each = 8 marks) a) 17118 + 64348 b) 22338 − 12538 1 89 1 7 11 6 4 34 7 (11) 4 5 8 10 3 4 5

c) C4A16 + 17E16 = ฀฀฀฀816 1 12 4 10 1 7 14 13 12 (24) 16 13 12 8

1 (1)1 2 2 (1)3 3 1 2 5 3 7 6 0

d) FADE16 − 2C3F16 = ฀฀฀฀9฀฀16 14 12 15 (1)10 13 (1)14 2 12 3 15 12 14 9 15

3) Subtract the given binary numbers by the requested method. Show your calculations. (2 marks each = 4 marks) a) Using one’s complement 11101100 − 10011011 10011011 → 01100100 (1′s complement) 11101100 + 01100100 1 01010000 → 01010000 + 1 (carry) = 01010001 b) Using two’s complement 1100011 − 0011001 0011001 → 1100110 �1′ s comp. � → 1100110 + 1 = 1100111 (2 ′ s comp. ) 1100011 + 1100111 1 1001010 → 1001010 (ignore the carry)

4) Draw the logic diagrams that represent the Boolean expressions below.

(3 marks each)

������������ a) ฀฀฀฀ ⨁ �฀฀฀฀

A B

F

C ��������  + ฀฀� b) ฀฀฀฀ + �฀฀

B C

F

D

5) Write the Boolean expressions for the given circuits.

(3 marks)

a)

฀ ฀ = ฀฀(฀ ฀ + ฀฀ )

A B

F

C

b) Write the corresponding truth table for questions 5(a) above. ฀฀ 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 0 0 1 1 1 1

฀฀ 0 1 0 1 0 1 0 1

฀ ฀ + ฀฀ 1 1 0 1 1 1 0 1

(3 marks) ฀฀(฀ ฀ + ฀฀ ) 0 0 0 0 1 1 0 1

6) Write the Boolean expressions for the given circuits.

(3 marks)

a)

 � + ฀฀฀฀ ฀ ฀ =������� ฀฀⨁฀฀

A B

F

C

b) Write the corresponding truth table for question 6(a) above ฀฀

฀฀

฀฀

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 0 0 1 1 1 1

� ฀฀

� ฀฀⨁฀฀

0 0 1 1 0 0 1 1

0 0 1 1 1 1 0 0

������� � ฀฀⨁฀฀ 1 1 0 0 0 0 1 1

(3 marks) ฀฀฀฀ 1 0 0 0 1 0 0 0

�������  � + ฀฀฀฀ ฀฀⨁฀฀ 1 1 0 0 1 0 1 1

7) Simplify the following expression and justify each step (see attached formula sheet). (3 marks) ������������ � + ฀฀ � ฀฀ ฀฀฀฀

������������ � (฀ ฀ + ฀฀)------------------8a = ฀฀ � + ���������� (฀ ฀ + ฀฀) = ฀฀

------------------14b

(฀ ฀ + ฀฀) = ฀ ฀ +����������

------------------13a

8) a) Draw the logic diagram for the Boolean expression ฀���������� ฀ + ฀฀฀฀

A B

(3 marks)

F

C

b) Simplify the expression in 8(a) and justify each step

(3 marks)

�฀��������� ฀฀฀฀ )------------------14a ฀ + ฀฀฀฀= ฀฀�(������ � + ฀฀ ) ------------------14b = ฀฀(฀฀ � + ฀฀) ------------------13a = ฀฀(฀฀

c) Draw the logic diagram of the simplified expression.

A B

(2 marks)

F

C

d) Construct a truth table to prove that the expressions are equal ฀฀ 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 0 0 1 1 1 1

� ฀฀ 0 0 1 1 0 0 1 1

฀฀ 0 1 0 1 0 1 0 1

� + ฀฀ ฀฀ 1 0 1 1 1 0 1 1

� + ฀฀) ฀฀(฀฀ 0 0 0 0 1 0 1 1

(3 marks) ฀฀฀฀ 0 1 0 0 0 1 0 0 Equal

฀ ฀ + ฀฀฀฀ 1 1 1 1 0 1 0 0

฀��� ฀������� + ฀฀฀฀ 0 0 0 0 1 0 1 1

Formula Sheet BASE 16 VALUES: A = 10, B = 11, C = 12, D = 13, E = 14, F = 15 Boolean Algebra Rules (from text plus two more) ฀ ฀ = 1 if ฀฀ ≠ 0 ฀ ฀ = 0 if ฀฀ ≠ 1 1a 1b 0∙0 =0 2a 2b 0+0=0 1+1=1 1∙1 =1 3a 3b 1∙0 =0 4a 4b 1+0=1 5a 5b 1=0 0=1 ฀฀฀฀ = ฀฀฀฀ (Commutative) ฀ ฀ + ฀ ฀ = ฀ ฀ + ฀฀ (Commutative) 6a 6b ฀฀(฀฀฀฀) = ฀฀฀฀(฀฀) (Associative) 7a 7b ฀ ฀ + (฀ ฀ + ฀฀ ) = (฀ ฀ + ฀฀) + ฀฀ (Associativ ฀฀(฀ ฀ + ฀฀) = ฀฀฀฀ + ฀฀฀฀ 8a 8b ฀ ฀ + ฀฀฀฀ = (฀ ฀ + ฀฀)(฀ ฀ + ฀฀) (Distributiv (Distributive) ฀฀ ∙ 0 = 0 ฀ ฀ + 0 = ฀฀ 9a 9b ฀฀ ∙ 1 = ฀฀ ฀฀+1=1 10a 10b ฀฀ ∙ ฀฀ = ฀฀ ฀ ฀ + ฀ ฀ = ฀฀ 11a 11b 12a 12b ฀฀ ∙ ฀฀ = 0 ฀฀+฀ ฀ = 1 13a 13b ฀ ฀ = ฀฀ ฀ ฀ = ฀฀ 14a 15a 16a 17 18a 18b

฀฀ ฀฀ ฀ ฀ = ฀ ฀ + ฀ ฀ + ฀฀ (De Morgan 14b ฀ ฀ + ฀ ฀ + ฀ ฀ = ฀฀฀฀฀฀ (De Morgan) ฀฀(฀ ฀ + ฀฀) = ฀฀ (Absorption) ฀ ฀ + ฀฀฀฀ = ฀฀ (Absorption) 15b 16b ฀฀�฀฀ + ฀฀� = ฀฀฀฀ (Absorption) ฀ ฀ + ฀฀฀฀ = ฀ ฀ + ฀฀ (Absorption) (฀ ฀ + ฀฀)�฀฀ + ฀฀� = ฀฀฀฀ + ฀฀฀฀ (Multiplying Out) ฀฀฀฀ + ฀฀฀฀ + ฀฀฀฀ = ฀฀฀฀ + ฀฀฀฀ (Consensus Theorem) (฀ ฀ + ฀฀ )�฀฀ + ฀฀�(฀ ฀ + ฀฀) = (฀ ฀ + ฀฀)�฀฀ + ฀฀� (Consensus Theorem)

Logic Gates AND

Symbol:

OR

A• B

NAND

Symbol:

Symbol:

NOT

A+ B

NOR

A• B

Symbol:

Symbol:

A

XOR

A+ B

Symbol:

A⊕ B...