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Boolean Logic & Number Systems

Boolean Logic & Number Systems. Today: Some Announcements First Hour : Introduction to Logic Section 1.2, 1.3 of Katz’s Textbook In-class Activity #1 Second Hour : Number Systems Appendix A.1 and A.2 of Katz’s Textbook In-class Activity #2.

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Boolean Logic & Number Systems

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  1. Boolean Logic & Number Systems • Today: • Some Announcements • First Hour: Introduction to Logic • Section 1.2, 1.3 of Katz’s Textbook • In-class Activity #1 • Second Hour: Number Systems • Appendix A.1 and A.2 of Katz’s Textbook • In-class Activity #2

  2. An electronic copy of Katz’s textbook is on the website! Wait-listed students can now login User name = ecse2610_stu Password = ecse2610_stu Some Announcements

  3. Boolean Equations Gates Inverter X Z = X X Z X Y X AND Y AND 0 0 0 X Y Z = X •Y = X Y Z 0 1 0 1 0 0 X OR Y 1 1 1 NOT X X Y 0 1 0 0 0 OR 1 0 0 1 1 X Y Z Z = X + Y 1 0 1 1 1 1 Truth Tables

  4. Half adder adds two binary digits to form Sum and Carry Sum equals A plus B (modulo 2) Carry equals (A plus B minus Sum)/2 A B C in S um C out A B Sum Carry 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 0 1 0 1 0 1 0 Full adder adds two binary digits and Carry in (Cin) to form Sum and Carry Out (Cout) 0 1 1 0 1 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 Sum equals A plus B plus Cin (modulo 2) Carry equals (A plus B plus Cin minus Sum)/2 Example Truth Tables

  5. Sum =A B+A B A 0 0 1 1 B 0 1 0 1 Sum 0 1 1 0 Carry = A B Truth Tables to Equations Step 1: Identify each truth table row where the function value is 1 Step 2: Construct (ANDed) product terms of the inputs for these rows If input variable is 0, it appears in complemented form If 1, it appears un-complemented Step3: “OR” together these product terms Half adder Carry 0 0 0 1

  6. Sum = A B Cin + A B Cin + A B Cin + A B Cin Sum 0 1 1 0 1 0 0 1 A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 Cin 0 1 0 1 0 1 0 1 Cout =A B Cin+A B Cin+A B Cin+A B Cin Full Adder Example Cout 0 0 0 1 0 1 1 1

  7. C out 0 A B C in 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 Equations to Truth Tables Consider: Cout = A Cin+B Cin+A B Fill in rest of Cout with 0s Put a 1 in each row where product term is true Each product term in the equation covers exactly two rows in the truth table Verify the equivalence of this Boolean equation: compare this Cout with the original Cout truth table Notice This row is covered 3 times

  8. Sum = A B + A B A AB AB B Equations to Circuits Half Adder A Sum B Carry AB Carry = A B

  9. Sum = A B Cin + A B Cin + A B Cin + A B Cin Full Adder Notation: \A = A, etc. Cout = A B Cin + A B Cin + A B Cin + A B Cin

  10. Equivalent Circuits  Cout = A B Cin + A B Cin + A B Cin + A B Cin Cout = A Cin + B Cin + A B

  11. Some Practicalities Fan-out number of inputs a gate output is connected to Fan-in number of inputs available on a gate "Rules of Composition" place limits on fan-in/fan-out

  12. Reference: Section 1.2, 1.3 of Katz Textbook Do Activity #1 Now

  13. Computers and Numbers • Most modern computers use logic circuits that exhibit two states (binary digital) • Large binary numbers are unwieldy, so alternative representations are used. • Hexadecimal numbers are a compromise between efficient representation and conversion to a more natural (to most humans) decimal system. • It is important to understand these other bases when designing hardware and software.

  14. Notation • Unless otherwise noted, the default base for quantities is decimal (base 10) • Subscripts following a quantity indicate the base • Examples: • 12310 is base 10 or decimal, so is 123 • 12316 is base 16 or hexadecimal • 1012 is base 2 or binary • Other notation (used later on in this course): • %101 is binary • $123 is hexadecimal

  15. Decimal Numbers Uses 10 digits [ 0, 1, 2, . . . 9 ] Positional Number Notation Weight of digit determined by its position. • Example: • 246 = 200 + 40 + 6 • = 2 x 102 + 4 x 101 + 6 x 100 • In general: • N =  Ni x 10i where Ni [0, 1, 2, . . .8, 9], • where Ni are the weights and the base of the number • system is raised to the exponent i. • Note: decimal fractions occur when i is negative • 0.35 = 3 x 10-1 + 5 x 10-2

  16. Binary Numbers Uses 2 digits [ 0 & 1 ] Positional Number Notation • Example: • 1110012 = 1000002 + 100002 + 10002 + 0002 + 002 + 12 • = 1 x 25 + 1 x 24 + 1 x 23 + 0 x 22 + 0 x 21 + 1 x 20 • = 1 x 32 + 1 x 16 + 1 x 8 + 0 x 4 + 0 x 2 + 1 x 1 • = 5710 • In general: • N =  Ni x 2i where Ni [0, 1]. • Note: binary fractions occur when i is negative • 0.112 = 1x 2-1 + 1 x 2-2 = 0.510 + 0.2510 = 0.7510

  17. Hexadecimal (Hex) Numbers Uses 16 digits [ 0, 1, 2, . . . 9, A, B, C, D, E, F ] Positional Number Notation Note: A - F represent the decimal values of 10 - 15, respectively. • Example: • 89AB16 = 800016 + 90016 + A016 + B16 • = 8 x 163 + 9 x 162 + A x 161 + B x 160 • = 8 x 4096 + 9 x 256 + 10 x 16 + 11 x 1 = 3524310 • In general: • N =  Ni x 16i where Ni [0, 1, 2, ..., 9, A, B, C, D, E, F]. • Note: hexadecimal fractions occur when i is negative • 0.9A16 = 9 x 16-1 + 10 x 16-2 = 0.562510 + 0.039062510 = 0.601562510

  18. Conversion from any base to decimal: Use positions and weights ! Eg: Decimal value =  Ni x Bi where Ni [0, 1, 2, . . . NB-1] (  means “sum of”) Eg: 89AB16= 8 x 4096 + 9 x 256 + 10 x 16 + 11 x 1 = 3524310 Conversion among the 2n bases (binary, hexadecimal, octal, etc.): First convert to binary (base 2) & then regroup into n bits Then convert the groups of bits into the proper digits. Conversion Among Bases

  19. Example: convert 100110001110011012 hexadecimal Group by 4s, starting at the right  expand into 4 bit groups 1 0011 0001 1100 11012 1 3 1 C D16 Hexadecimal  Binary Hex Binary Decimal 0 0000 0 1 0001 1 2 0010 2 3 0011 3 4 0100 4 5 0101 5 6 0110 6 7 0111 7 8 1000 8 9 1001 9 A 1010 10 B 1011 11 C 1100 12 D 1101 13 E 1110 14 F 1111 15 Memorize this table!!!It makes conversions between hexadecimal and binary trivial

  20. Successive Division: an easy way to convert Decimal to Binary Decimal  Binary Based on the remainders after dividing the decimal number repeatedly by 2. If the decimal number is even, the corresponding binary number will end in a 0, and if it is odd, the binary number will end in a 1, and so on. Example: 57 / 2 = 28, remainder = 1 (binary number will end with 1) 28 / 2 = 14, remainder = 0 14 / 2 = 7, remainder = 0 7 / 2 = 3, remainder = 1 3 / 2 = 1, remainder = 1 1 / 2 = 0, remainder = 1 (binary number will start with 1) Therefore, collecting the remainders, 5710 = 1110012 (((((0x2 + 1)x2 + 1)x2 + 1)x2 +0)x2 +0)x2 +1 = 1x32 + 1x16 + 1x8 + 0x4 + 0x2 + 1x1 = 57 (check: 32 + 16 + 8 + 0 + 0 + 1 = 57)

  21. Decimal  Hex Successive Division: an easy way to convert Decimal to Hexadecimal Based on the remainders after dividing the decimal number by higher powers of 16. Example: 35243 / 16 = 2202, remainder = 11 B (hex number will end with B) 2202 / 16 = 137, remainder = 10 A 137 / 16 = 8, remainder = 9 8 / 16 = 0, remainder = 8 (hex number will start with 8) Therefore, collecting the remainders, 3524310 = 89AB16 (check: 8 x 4096 + 9 x 256 +10 x 16 + 11 = 35243) Notice how simple this method is

  22. Hex  Decimal Multiplication: an easy way to convert Hexadecimal to Decimal Multiply the hexadecimal weights by powers of 16. Example: BA89 B x 163 + A x 162 + 8 x 16 + 9 (CAUTION: mixed bases) = 11 x 4096 + 10 x 256 + 8 x 16 + 9 = 45056 + 2560 + 128 + 9 = 47753 We can check by converting back to hex: 47753 / 16 = 2984 + remainder, 9 2984 / 16 = 186 + remainder, 8 186 / 16 = 11 + remainder, 10  A 11 / 16 = 0 + remainder, 11  B  4775310 BA8916

  23. Multiplication: an easy way to convert Binary to Decimal Binary  Decimal We can multiple the binary weights by powers of 2. However, sometimes it is just as easy to convert the binary number to hexadecimal first and and then into decimal. Compare: 110101102 27 + 26 + 24 + 22 + 2 (exploiting the binary weights) = 128 + 64 + 16 + 4 + 2 = 21410 110101102 D616 13 x 16 + 6 = 208 + 6 = 21410

  24. Due: End of Class Today RETAIN THE LAST PAGE (#3)!! For Next Class: Bring Randy Katz Textbook Course website has an electronic copy available. Required Reading: Sec A.1, A.2, 2.1, 2.2 (up to 2.2.4) of Katz This reading is necessary for getting points in the Studio Activity! Studio Session #2 Tuesday/Wednesday Do Activity #2 Now

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