Chapter 1(Rev)
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Transcript of Chapter 1(Rev)
![Page 1: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/1.jpg)
DIGITAL ELECTRONICS IED 12303
BASIC CONCEPTS (REVIEW)
![Page 2: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/2.jpg)
NUMBER SYSTEM
![Page 3: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/3.jpg)
Number system- Binary number
• Base 2 system• Consist ‘0’ and ‘1’• Called as bit
![Page 4: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/4.jpg)
Number system- Octal decimal
• Base 10 systems• 0, 1, 2, 3, 4, 5, 6, 7, 8, 9• E.g: 238• 200 + 30 + 8 = 238
![Page 5: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/5.jpg)
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
![Page 6: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/6.jpg)
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 =
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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
![Page 8: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/8.jpg)
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
![Page 9: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/9.jpg)
LOGIC GATES
![Page 10: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/10.jpg)
• 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
![Page 11: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/11.jpg)
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
![Page 12: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/12.jpg)
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
![Page 13: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/13.jpg)
NOT Gate Truth- Table
Logic symbol Boolean Expression Truth Table
X= AInput Output
A X
0 0 1
0 1 0
A X
![Page 14: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/14.jpg)
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
+
![Page 15: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/15.jpg)
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
![Page 16: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/16.jpg)
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
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• 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
![Page 18: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/18.jpg)
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
![Page 19: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/19.jpg)
• 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
![Page 20: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/20.jpg)
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
![Page 21: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/21.jpg)
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
![Page 22: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/22.jpg)
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
![Page 23: Chapter 1(Rev)](https://reader035.fdocuments.us/reader035/viewer/2022062221/55cf8cde5503462b13903735/html5/thumbnails/23.jpg)
REFERENCES
• Thomas L. Floyd, Digital Fundamentals, 10th Edition, Pearson Education Inc, 2009
• Roger Tokheim, Digital Electronics, 8th Edition, McGraw- Hill