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Chapter 1 (Part a) Digital Systems and Binary Numbers

Digital Logic Design. Chapter 1 (Part a) Digital Systems and Binary Numbers . Originally by T.Tasniem Nasser Al-Yahya. Outline of Chapter 1 (Part a). 1.1 Digital Systems 1.2 Binary Numbers 1.3 Number-base Conversions 1.4 Octal and Hexadecimal Numbers.

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Chapter 1 (Part a) Digital Systems and Binary Numbers

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  1. Digital Logic Design Chapter 1 (Part a) Digital Systems and Binary Numbers Originally by T.Tasniem Nasser Al-Yahya

  2. Outline of Chapter 1 (Part a) • 1.1 Digital Systems • 1.2 Binary Numbers • 1.3 Number-base Conversions • 1.4 Octal and Hexadecimal Numbers

  3. Outline of Chapter 1 (Part a) • 1.1 Digital Systems • 1.2 Binary Numbers • 1.3 Number-base Conversions • 1.4 Octal and Hexadecimal Numbers

  4. Digital Systems • Digital systems (systems that can manipulate Discrete information) • Digital Computers. • Digital camera. • Electronic calculators. • Digital TV.

  5. Analog and Digital Signal • In digital system, discrete element of information are represented by quantities called signals. • Analog Signal • The physical quantities may vary continuously over a specified range.(infinite possible values). • Ex: voltage on a wire created by microphone • Digital Signal. • a digital signal has discrete values over a specified range • Ex: button pressed on a keyboard . X (t) Digital signal Analog signal t

  6. The signals in most digital systems use just two discrete values and therefore said to be binary. • A binary digit , called a bit, has two values : 0 and 1.

  7. The two discrete values are physically represented by ranges of voltage values called HIGH and LOW. • On yes true 1 • (voltage between 4.0 and 5.0) • Off no false 0 • (voltage between 0.0 and 1.0) • Discrete elements of information • are represented with groups of bits • called binary codes • Ex: 7 : 0111 ( Off On OnOn) V(t) Logic 1 undefined Logic 0 t Binary digital signal

  8. A digital system is an interconnection of digital modules • To understand the operation of each digital module , it is necessary to have a basic knowledge of digital circuits and their logical functions.

  9. Outline of Chapter 1 • 1.1 Digital Systems • 1.2 Binary Numbers • 1.3 Number-base Conversions • 1.4 Octal and Hexadecimal Numbers

  10. 2 1 0 -1 -2 100 10 1 0.1 0.01 500 10 2 0.7 0.04 Decimal Number System • Base (also called radix) = 10 • 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } • Coefficient Position • Integer & fraction • Coefficient Weight • Weight = (Base) Position • In general Magnitude: • Sum of “Coefficient x Weight” 5 1 2 7 4 a2*r2+a1*r1+a0*r0+a-1*r-1+a-2*r-2 (512.74)10

  11. General positional Number System • Base (also called radix) = r • r is an integer and ≥ 2 • r digits { 0, ….. , r-1} • (546)4 is not correct • Coefficient Weight • Weight = (Base) Position=i • Magnitude (The value of a digit depends not only on its value but also on its position within the number. • The magnitude M of a number ( ) in base r is:

  12. 4 2 1 1/2 1/4 2 1 0 -1 -2 Binary Number System • Base = 2 • 2 digits { 0, 1 }, called binary digits or “bits” • Weights • Weight = (Base) Position • Magnitude • Sum of “Bit x Weight” • Formal Notation • Groups of bits • 4 bits = Nibble • 8 bits = Byte 1 0 1 0 1 1 *22+0 *21+1 *20+0 *2-1+1 *2-2 =(5.25)10 (101.01)2 1 0 1 1 1 1 0 0 0 1 0 1

  13. Binary Arithmetic Operations • The next section will discuss Unsigned arithmetic operations in Binary base system • Addition • Subtraction • Multiplication

  14. Addition • The sum of two binary numbers is calculated by the same rules as in decimal except that the digits of the sum in any significant position can be only zero or one • Line up the numbers from right to left. If one number is shorter extend it by adding leading zeros to the number. (the extension is different with signed number) • Work from right to left – add each pair of digits together with carry propagation. 0+0=0 0+1=1 1+0=1 1+1=01 ( carry ) 1+1+1=11 (carry)

  15. Decimal Addition: (345)10+(89)10 1 1 Carry augend 4 5 3 addend 0 + 8 9 sum 4 3 4 = 14 ≥base 10  14-10 , add carry = 13 ≥base 10  13-10 , add carry = 4 < base 10

  16. Example 2 • Binary Addition: (111101)2+(10111)2 Carry 1 1 1 1 1 1 = (61)10 = (23)10 1 1 1 1 0 1 + 1 0 1 1 1 0 = (84)10 1 0 1 0 1 0 0 Considercarry = 2 ≥base 2  2-2 , add carry

  17. Binary Addition • Practice Calculate (101101)2+(100111)2

  18. Subtraction • Line up the numbers from right to left. If one number is shorter extend it by adding leading zeros to the number. • Work from right to left –Subtract each pair of digits together with borrowing where needed 0-0=0 0-1=1 (after borrowing) 1-0=1 1-1=0

  19. Example 1 • Decimal subtraction: (345)10-(89)10 2 3 × × minuend 4 5 3 subtrahend 0 - 8 9 difference 2 5 6 = 5<9, borrow from 4 Decrease 4 to 3 5+10 = 15 15-9 = 6 = 3<8 , borrow from 3 Decrease 3 to 2 3+10=13 13-8 = 5 = 2>0 2-0=2

  20. Example 2 • Binary subtraction : (111101)2-(10111)2 = (61)10 = (23)10 1 1 1 1 0 1 - 1 0 1 1 1 0 = (38)10 1 0 0 1 1 0

  21. Binary Subtraction • Practice Calculate (101101)2-(100111)2

  22. Binary Subtraction • In many cases binary subtraction is done in a special way by binary addition. Why? • It is much more simple to do it that way. 􀂄One simple building block called adder can be implemented and used for both binary addition and subtraction. • This subject will be discussed later

  23. Multiplication • Multiplication is Simple . • The multiplier digits are always 1 or 0,therefore , the partial products are equal either to the multiplicand or to 0 0*0=0 0*1=0 1*0=0 1*1=1

  24. Example 1 • (10111)2×(1010)2 = (23)10 = (10)10 1 0 1 1 1 x 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 1 1 + = (230)10 1 1 1 0 0 1 1 0

  25. Binary Multiplication • Practice Calculate (1011)2 * (101)2

  26. 2 1 0 -1 -2 100 10 1 0.1 0.01 500 10 2 0.7 0.04 Recall : Decimal Number System • Base (also called radix) = 10 • 10 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } • Coefficient Position • Integer & fraction • Coefficient Weight • Weight = (Base) Position • In general Magnitude: • Sum of “Coefficient x Weight” 5 1 2 7 4 a2*r2+a1*r1+a0*r0+a-1*r-1+a-2*r-2 (512.74)10

  27. 64 8 1 1/8 1/64 2 1 0 -1 -2 Octal Number System • Base = 8 • 8 digits { 0, 1, 2, 3, 4, 5, 6, 7 } • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” (Note that the digits 8 and 9 cannot appear in an octal number) 5 1 2 7 4 5 *82+1 *81+2 *80+7 *8-1+4 *8-2 =(330.9375)10 (512.74)8

  28. 256 16 1 1/16 1/256 2 1 0 -1 -2 Hexadecimal Number System • Base = 16 • 16 digits { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F } • Where A=10 B=11 C=12 D=13 E= 14 F=15 • Weights • Weight = (Base) Position • Magnitude • Sum of “Digit x Weight” 1 E 5 7 A 1 *162+14 *161+5 *160+7 *16-1+10 *16-2 =(485.4765625)10 (1E5.7A)16

  29. The Power of 2 Kilo Mega Giga Tera

  30. Outline of Chapter 1 • 1.1 Digital Systems • 1.2 Binary Numbers • 1.3 Number-base Conversions • 1.4 Octal and Hexadecimal Numbers

  31. Outline of Chapter 1 • 1.4 Octal and Hexadecimal Numbers • 1.3 Number-base Conversions

  32. Octal and Hexadecimal Numbers • The conversion from and to binary , octal , and hexadecimal plays an important role in digital computers • Digital computers use binary numbers .However , they are difficult to work with as they are long. • By using octal or hexadecimal conversion ,the human operator thinks in terms of octal and hex numbers and performs the required conversion when direct information with the machine is necessary.

  33. Binary − Octal Conversion • 8 = 23 • Each octal digit corresponds to three binary digits. Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 2 6 . 2 )8 Start from the point and proceed to the left and to the right Works both ways (Binary to Octal & Octal to Binary)

  34. Binary − Octal Conversion • Practice Convert (101011.1111 ) 2 to Octal ( 1 0 1 0 1 1 . 1 1 1 1 )2 ( . )8 Start from the point and proceed to the left and to the right

  35. Binary − Hexadecimal Conversion • 16 = 24 • Each hexadecimal digit corresponds to four binary digits Example: Assume Zeros ( 1 0 1 1 0 . 0 1 )2 ( 1 6 . 4 )16 Works both ways (Binary to Hex & Hex to Binary)

  36. Binary − Hexadecimal Conversion • Practice Convert (1101011.1111001 ) 2 to Hexadecimal ( 1 1 0 1 0 1 1 . 1 1 1 1 0 0 1 )2 ( . )16 Start from the point and proceed to the left and to the right

  37. Octal − Hexadecimal Conversion • Convert to Binary as an intermediate step Example: ( 2 6 . 2 )8 Assume Zeros Assume Zeros ( 01 0 1 1 0 . 0 1 0 )2 ( 1 6 . 4 )16 Works both ways (Octal to Hex & Hex to Octal)

  38. Decimal, Binary, Octal and Hexadecimal

  39. Outline of Chapter 1 • 1.4 Octal and Hexadecimal Numbers • 1.3 Number-base Conversions

  40. Number Base Conversions Evaluate Magnitude Octal (Base 8) Divideby 8 Evaluate Magnitude Evaluate Magnitude Decimal (Base 10) Binary (Base 2) Any system (Base r) Divide by r Divide by 2 Hexadecimal (Base 16) Divide by 16 Evaluate Magnitude

  41. Conversion form any base to Decimal (Evaluate magnitude) Decimal (Base 10) Any system (Base r) Evaluate Magnitude • We can use the evaluate magnitude method to convert any base number to decimal. (see slide 9)

  42. Conversion form any base to Decimal (Evaluate magnitude) • Example: convert 1101.01 in binary to decimal: • The decimal value is:

  43. Conversion form Decimal to any base system (Divide by base) Decimal (Base 10) Any system (Base r) Divide by r • To convert a decimal integer into any base, keep dividing by r until the quotient is 0. Collect the remainders in reverse order. • To convert a fraction, keep multiplying the fractional part by r until it becomes 0 or until we reach the required accuracy. Collect the integer parts in forward order.

  44. Decimal (Integer) to Binary Conversion • Divide the number by the ‘Base’ (=2) • Take the remainder (either 0 or 1) as a coefficient • Take the quotient and repeat the division until the quotient reaches zero. Example: (13)10 Coefficient Quotient Remainder 13 / 2 =6 1a0 =1 LSB MSB 6 / 2 =3 0a1 =0 3 / 2 =1 1a2 =1 1 / 2 =0 1a3 =1 Answer: (13)10= (a3 a2 a1 a0)2= (1101)2 MSB LSB

  45. Decimal (Fraction) to Binary Conversion • Multiply the number by the ‘Base’ (=2) • Take the integer (either 0 or 1) as a coefficient • Take the resultant fraction and repeat the division till you reach a zero or the required accuracy is attained. Example: (0.625)10 Coefficient Integer Fraction a-1 =1 0.625 * 2 =1 . 25 MSB LSB 0.25 * 2 =0 . 5a-2 =0 0.5 * 2 =1 . 0a-3 =1 Answer: (0.625)10= (0.a-1 a-2 a-3)2= (0.101)2 MSB LSB

  46. Decimal to Binary Conversion • Practice Convert (41.6875 ) 10 to binary number • it is necessary to separate the number into an integer part and a fraction part, since each part must be converted differently.

  47. Decimal to Octal Conversion Example: (175)10 Coefficient Quotient Remainder 175 / 8 =21 7a0 =7 21 / 8 =2 5a1 =5 2 / 8 =0 2a2 =2 Answer: (175)10= (a2 a1 a0)8= (257)8 Example: (0.3125)10 Coefficient Integer Fraction a-1 =2 0.3125 * 8 =2 . 5 0.5 * 8 =4 . 0a-2 =4 Answer: (0.3125)10= (0.a-1 a-2 a-3)8= (0.24)8

  48. Decimal to Octal Conversion • Practice Convert (153.513 ) 10 to Octal number • it is necessary to separate the number into an integer part and a fraction part, since each part must be converted differently.

  49. Decimal to Hexadecimal Conversion Example: (175)10 Coefficient Quotient Remainder 175 / 16 = 10 15a0 = F 10 / 16 = 0 10a1 = A Answer: (175)10= (a1 a0)16= (AF)16 Example: (0.3125)10 Coefficient Integer Fraction a-1 =2 0.3125 * 16 = 5. 0 Answer: (0.3125)10= (0.a-1)16= (0.2)16

  50. The following sections from chapter 1 have been given on the lecture but without power point slides • 1.5 ,1.6, 1.7

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