DIGITAL ELECTRONICS IED 12303

BASIC CONCEPTS (REVIEW)

NUMBER SYSTEM

Number system- Binary number

• Base 2 system• Consist ‘0’ and ‘1’• Called as bit

Number system- Octal decimal

• Base 10 systems• 0, 1, 2, 3, 4, 5, 6, 7, 8, 9• E.g: 238• 200 + 30 + 8 = 238

Converting binary to decimal

• Convert 10011 into decimal

• (1 x 1) + (1 x 2) + (0 x 4) + (0 x 8) + (1 x 16)• = 1 + 2 + 0 + 0 + 16• = 19

Power of 2 2 2 2 2 2

1 0 0 1 1

0 1 2 34

Converting decimal to binary• Convert 87 to binary number• 87 ÷ 2 = 43• 43 ÷ 2 = 21• 21 ÷ 2 = 10 • 10 ÷ 2 = 5• 5 ÷ 2 = 2• 2 ÷ 2 = 1• 1 ÷ 2 = 0

1 0 1 0 1 1 187 =

Number system- Hexadecimal

• Base 16 system• Uses symbol of 0-9, A, B, C, D, E and F• E.g: Hexadecimal to decimal (2B6)

Powers of 16 16² 16¹ 16

Place value 256 16 1

0

256 16 12 B 6

2 11 6

512 176 6

x x x= 694 10

Number system- HexadecimalDecimal Binary Hexadecimal Decimal Binary Hexadecimal

0 0000 0 8 1000 8

1 0001 1 9 1001 9

2 0010 2 10 1010 A

3 0011 3 11 1011 B

4 0100 4 12 1100 C

5 0101 5 13 1101 D

6 0110 6 14 1110 E

7 0111 7 15 1111 F

LOGIC GATES

• An elementary building block of a digital circuit

• All digital systems are constructed by using the AND gate, the OR gate and the NOT gate

• The gate used only HIGH or LOW voltage

AND Gate Truth- TableLogic symbol Boolean Expression Truth Table

X = A.B@

X=AB

Input OutputA B X0 0 00 1 01 0 01 1 1

AX

B

OR Gate Truth- TableLogic symbol Boolean Expression Truth Table

X= A+B Input OutputA B X0 0 00 1 11 0 11 1 1

AX

B

NOT Gate Truth- Table

Logic symbol Boolean Expression Truth Table

X= AInput Output

A X

0 0 1

0 1 0

A X

XOR Gate Truth- TableLogic symbol Boolean Expression Truth Table

X= A B

(X.Y)+ (X.Y)

Input OutputA B X0 0 00 1 11 0 11 1 0

A XB

+

Derivation of Boolean expression from truth table using SOP method

• Boolean expression in a specific format• Comes from the expression’s form: a sum (OR)

of one or more products (AND)

• RULE 1: identify the rows with ones as the output, and come up with then unique product to put a one in that row

Derivation of Boolean expression from truth table using SOP method

A B C X0 0 0 00 0 1 10 1 0 00 1 1 11 0 0 11 0 1 01 1 0 11 1 1 0

• RULES 1: identify the rows with ones as the output

• A = 0, B = 0, and C= 1

• A = 0, B = 1, and C= 1• A = 1, B = 0, and C= 0

• A = 1, B = 1, and C= 0

A B C X

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

RULES 2:• Invert the input so that

the product will have 1.1.1 = 1

• A = 0, B = 0, and C= 1• A.B.C

• A = 0, B = 1, and C= 1• A.B.C

A B C X

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

• RULES 3: OR’ing all these product to give SOP expression

• A.B.C + A.B.C +A.B.C + A.B.C

RULES 3:• Invert the input so that

the product will have 1.1.1 = 1

• A = 1, B = 0, and C= 0• A.B.C

• A = 1, B = 1, and C= 0• A.B.C

A B C X0 0 0 00 0 1 10 1 0 00 1 1 11 0 0 11 0 1 01 1 0 11 1 1 0

Derivation of Boolean expression from truth table using POS method

• RULE 1: Identify the rows with ‘0’ as the output

• A = 0, B = 0, C = 0• A = 0, B = 1, C = 0 • A = 1, B = 0, C = 1• A = 1, B = 1, C = 1

A B C X

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

Derivation of Boolean expression from truth table using POS method

• RULES 2: make sum for each of these rows

• A + B + C• A + B + C • A + B + C• A + B + C

A B C X

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

Derivation of Boolean expression from truth table using POS method

• RULE 3: AND’ing all of these together gives us our POS expression

• (A+B+C)(A+B+C)(A+B+C) (A+B+C)

A B C X

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 0

REFERENCES

• Thomas L. Floyd, Digital Fundamentals, 10th Edition, Pearson Education Inc, 2009

• Roger Tokheim, Digital Electronics, 8th Edition, McGraw- Hill