1. Chapter 4: The Building Blocks: Binary Numbers, Boolean Logic, and Gates Invitation to Computer Science,

2. Invitation to Computer Science, C++ Version, Third Edition 2 Objectives In this chapter, you will learn about: The binary numbering system Boolean logic and gates Building computer circuits Control circuits

3. Invitation to Computer Science, C++ Version, Third Edition 3 Introduction Chapter 4 focuses on hardware design (also called logic design)  How to represent and store information inside a computer  How to use the principles of symbolic logic to design gates  How to use gates to construct circuits that perform operations such as adding and comparing numbers, and fetching instructions

4. Invitation to Computer Science, C++ Version, Third Edition 4 The Binary Numbering System A computer’s internal storage techniques are different from the way people represent information in daily lives Information inside a digital computer is stored as a collection of binary data  Bit = BInary digiT

5. Invitation to Computer Science, C++ Version, Third Edition 5 Binary numbering system  Base-2  Built from ones and zeros  Each position is a power of 2 1101 = 1 * 2 3 + 1 * 2 2 + 0 * 2 1 + 1 * 2 0 Decimal numbering system  Base-10  Each position is a power of 10 3052 = 3 * 10 3 + 0 * 10 2 + 5 * 10 1 + 2 * 10 0 Binary Representation of Numbers

6. Invitation to Computer Science, C++ Version, Third Edition 6 Figure 4.2 Binary-to-Decimal Conversion Table

7. Invitation to Computer Science, C++ Version, Third Edition 7 Conversion from Base 10 to Base 2 To convert a number n from base 10 to base 2 you must: 1. Divide n by 2, then its quote by 2 and continue this way until the quote is 0 2. Collect all the remainders produced by the divisions performed in Step 1 3. Write the remainders in the opposite order they have been produced

8. Invitation to Computer Science, C++ Version, Third Edition 8 Conversion from Base 10 to Base 2 (cont.) Example: What is 178 10 in base 2? 178 10 = 1011 0010 2 178 0 89 1 44 0 22 0 11 1 5 1 2 0 1 1 0 2 1*2 7 + 1*2 5 + 1*2 4 +1*2 1 = 128 + 32 + 16 + 2 = 178 10 quote remainder................

9. Invitation to Computer Science, C++ Version, Third Edition 9 Conversion from Base 10 to Base 2 To convert a number n from base 2 to base 10 you must: 1. Multiply each digit by 2 p where p is the position of the digit. Of course the digits that are 0 will produce a product equal to zero, so you can limit you attention only to the digits equal to 1 2. Add all the products obtained in Step 1.

10. Invitation to Computer Science, C++ Version, Third Edition 10 Conversion from Base 2 to Base 10 Example: What is 1011011 2 in base 10? 101 1011 2 = = 1*2 6 + 1*2 4 + 1*2 3 + 1*2 1 +1*2 0 = = 64 + 16 + 8 + 2 + 1 = = 91 10 101 1011 2 = 91 10

11. Invitation to Computer Science, C++ Version, Third Edition 11 Conversion from Base 10 to Base k General Rule: To convert a number n from base 10 to base k you must: 1. Divide n by k, then its quote by k and continue this way until the quote is 0 2. Collect all the remainders produced by the divisions performed in Step 1 3. Write the remainders in the opposite order they have been produced

12. Invitation to Computer Science, C++ Version, Third Edition 12 Conversion from Base 10 to Base k (cont.) Example: What is 178 10 in base 5? 178 10 =1203 5 178 3 35 0 7 2 1 1 0 5 1*5 3 + 2*5 2 + 0*5 1 + 3*5 0 = 125 + 50 + 0 + 3 = 178 10 quote remainder................ Conversion from Base 5 to base 10

13. Invitation to Computer Science, C++ Version, Third Edition 13 Conversion from Base k to Base 10 General Rule: To convert a number n from base k to base 10 you must: 1. Multiply each digit by k p where p is the position of the digit. Since the digits that are 0 will produce a product equal to zero, you can skip them. 2. Add all the products obtained in Step 1.

14. Invitation to Computer Science, C++ Version, Third Edition 14 Conversion from Base k to Base t with k and t different from 2 or 10 To convert a number n from base k to base t you must perform two conversions: 1. Convert n from base k to base 10 2. Convert the result from base 10 to base t

15. Invitation to Computer Science, C++ Version, Third Edition 15 Conversion from Base k to Base t with k and t different from 2 or 10 Example: What is 153 6 in base 4? 153 6 =69 10 1*6 2 + 5*6 1 +3*6 0 = 36 + 30 + 3 = 69 10 69 1 17 1 4 0 1 1 0 4 First: Convert from Base 6 to base 10 Second: Convert the result from Base 10 to base 4 69 10 = 1011 4 153 6 = 1011 4 Answer

16. Invitation to Computer Science, C++ Version, Third Edition 16 Conversion Algorithms Base 10 is convenient for us while reading and writing numbers….however for the machine:  The algorithm to convert to base 10 (to display results - output) and the algorithm to convert from base 10 (to insert data - input) require time. Can we display or insert data in a base different from 10?  Base 2 – Binary numbers are too long. Better a base closer to 10 What about 8 or 16?

17. Invitation to Computer Science, C++ Version, Third Edition 17 Fast conversion between bases that are power of 2 The power of 2 are special bases (ex. 4, 8, 16,..)  the conversion between these bases is immediate Example: Convert 3762 8 to base 2 Since 8 = 2 3 > 2 we expand each digit in 3 binary digits  This is possible because with 3 bits we can represent 8 distinct numbers in binary 3 7 6 2 8 0 1 1 1 1 1 1 1 0 0 1 0 2 3762 8 = 111 1111 0010 2 Answer

18. Invitation to Computer Science, C++ Version, Third Edition 18 Fast conversion between bases that are power of 2 Example: Convert 3B0 16 to base 2 Since 16 = 2 4 > 2 we expand each digit in 4 binary digits  This is possible because with 4 bits we can represent 16 distinct numbers in binary  In Base 16 there are 16 distinct symbols:  0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F 3 B 0 16 3B0 16 = 11 1011 0000 2 Answer 0 0 1 1 1 0 1 1 0 0 0 0 2

19. Invitation to Computer Science, C++ Version, Third Edition 19 Example 2: Convert 1110 1010 2 to base 16  Since 2< 16 =2 4 we GROUP 4 binary digit in 1 hexadecimal digits E A 16 Fast conversion between bases that are power of 2 Example 1: Convert 1110 1010 2 to base 8 Since 2 < 8=2 3 we GROUP 3 binary digit in 1 octal digits 3 5 2 8 1110 1010 2 = 352 8 0 1 1 1 0 1 0 1 0 2 1 1 1 0 1 0 1 0 2 1110 1010 2 = EA 16

20. Invitation to Computer Science, C++ Version, Third Edition 20 Converting decimal numbers in binary To convert a decimal number n= int.dec in base 2 you must: 1. Convert the integer part int as shown so far 2. Convert the decimal part in this way: 1. Multiply the decimal part dec by 2, then the decimal part of the product just obtained by 2 and continue this way until the product obtained already occurred or is 0. 2. Collect all the integer parts produced during the multiplication in Step 2.1 3. Write the integers in the same order they have been produced

21. Invitation to Computer Science, C++ Version, Third Edition 21 Converting decimal numbers in binary Example: What is 178.25 10 in base 2? Step 1:Step 2: 178.25 10 = 1011 0010. 01 2 178 0 89 1 44 0 22 0 11 1 5 1 2 0 1 1 0 2 0.25  2 = 0.5 0 0.5  2 = 1.0 1 0 2

22. Invitation to Computer Science, C++ Version, Third Edition 22 Converting decimal numbers in binary Example: What is 0.325 10 in base 2? 0.325 10 = 0. 0101001 2 0.325  2 = 0.65 0 0.65  2 = 1.30 1 0.30  2 = 0.60 0 0.60  2 = 1.20 1 0.20  2 = 0.40 0 0.40  2 = 0.80 0 0.80  2 = 1.60 1 0.60 2 period

23. Invitation to Computer Science, C++ Version, Third Edition 23 Summary Conversion of an integer from Base 10 to Base 2 and vice versa Conversion of an integer from Base 10 to Base k and vice versa Conversion of an integer from Base k to Base t with k and t different from 2 or 10 Fast conversion between bases that are power of 2 Converting decimal numbers in binary

24. Invitation to Computer Science, C++ Version, Third Edition 24 End of Lesson

25. Invitation to Computer Science, C++ Version, Third Edition 25 Representing integers  Decimal integers are converted to binary integers Given k bits, the largest unsigned integer is 2 k – 1; the smallest is 0  with 4 bits, the largest unsigned number is 2 4 -1 = 15  Signed integers must also represent the sign (positive or negative). 1 bit is used for the sign. Given k bits, the largest signed integer is 2 k-1 - 1; the smallest is –(2 k-1 – 1)  with 4 bits, the largest signed number is 2 3 -1 = 7 Binary Representation of Numbers

26. Invitation to Computer Science, C++ Version, Third Edition 26 Representation of Numbers Signed integer representation  Sign/magnitude notation 1 bit to represent the sign N bits to represent the integer  Example: if n=8 + 33 is represented as 0 0100001 - 33 is represented as 1 0100001 + 1 is represented as 0 0000001 - 1 is represented as 1 0000001 What about 0? + 0 is represented as 0 0000000 - 0 is represented as 1 0000000 Big Problem!!! There are two representations for 0!!! Sign Magnitude 1 111 -7 1 110 -6 1 101 -5 1 100 -4 1 011 -3 1 010 -2 1 001 -1 1 000 -0 0 000 +0 0 001 +1 0 010 +2 0 011 +3 0 100 +4 0 101 +5 0 110 +6 0 111 +7

27. Invitation to Computer Science, C++ Version, Third Edition 27 Representation of Numbers Signed integer representation  Two’s complement notation a) a positive number follows the sign/magnitude notation b) a negative number is obtained from the corresponding positive number in two steps: 1. Flip all the bits (0 becomes 1 and 1 becomes 0) 2. Add one to the result

28. Invitation to Computer Science, C++ Version, Third Edition 28 Representation of Numbers Two’s complement notation Example: assume we use 8 bits precision Q. What is +33 10 in two’s complement? A. + 33 10 is represented as 0 0100001 Q. What is -33 10 in two’s complement? A. - 33 is represented as 1 1011111 Step 1: take +33 10 (i.e. 0 0100001) Step 2: flip it (i.e. 1 1011110) Step 3: add 1 to it 1 1011110 + 1 1 1011111 = - 33 10

29. Invitation to Computer Science, C++ Version, Third Edition 29 Representation of Numbers Advantages of the two’s complement representation  One representation for the zero  Easy subtractions Subtractions are simply additions  The sign is embedded in the number! Example: Suppose we use 8 bits precision 15 0000 1111 - 5+ 1111 1011 + 10 1 0000 1010 Two’s complement 1 1111 -1 1 1110 -2 1 1101 -3 1 1100 -4 1 1011 -5 1 1010 -6 1 1001 -7 1 1000 -8 1 0111 -9 1 0110 -10 1 0101 -11 1 0100 -12 1 0011 -13 1 0010 -14 1 0001 -15 1 0000 -16 00000 +0 0 0001 +1 0 0010 +2 0 0011 +3 0 0100 +4 0 0101 +5 0 0110 +6 0 0111 +7 0 1000 +8 0 1001 +9 0 1010 +10 0 1011 +11 0 1100 +12 0 1101 +13 0 1110 +14 0 1111 +15 - 9 111 10111 - 5+ 111 11011 - 14 1111 10010 overflow

30. Invitation to Computer Science, C++ Version, Third Edition 30 Representation of Numbers Q: Given n bits precision, what is the interval of numbers that can be represented? A : In sign/magnitude the interval is (-2 n-1, 2 n-1 )  Note: the interval is open on both sides!  There are 2 distinct representations of the 0! In two’s complement the interval is [-2 n-1, 2 n-1 )  Note: the interval is open on one side! Q: With n bits, how many distinct numbers can be represented? A : 2 n distinct numbers

31. Invitation to Computer Science, C++ Version, Third Edition 31 Representation of Numbers Representing real numbers  Real numbers may be put into binary Scientific Notation  Mantissa  Bases  Exponent  Number then are normalized so that first significant digit is immediately to the right of the binary point  Then only Mantissa and Exponent are stored. Example: Consider Binary number: +6.25 10 = +110.01 2 is Normalized as: +.11001 2  2 3

32. Invitation to Computer Science, C++ Version, Third Edition 32 Physical Representation of Numbers How to store numbers in memory: Assume that a number is stored in 16 bits storage location  1 bit for the sign  9 bits for the mantissa  1 bit for the sign of the exponent  5 bits for the exponent +6.25 10 = +110.01 2 = +.11001 2  2 3 has  0 for the sign  1101 for the mantissa  0 for the sign of the exponent  3 for the exponent and in a 16 bits storage location is represented as 0110100000000011 Decimal point 01 1 0 0 1 0 0 0 000 0 0 1 1 Sign Mantissa Sign Exponent

33. Invitation to Computer Science, C++ Version, Third Edition 33 Physical Representation of Numbers: Examples 3/16 10 = 1/8+1/16 10 = +0.0011 2 = +.11 2  2 -2 has  0 for the sign  11 for the mantissa  1 for the sign of the exponent  2 for the exponent and in a 16 bits storage location is represented as 0110000000100010 -19/64 10 = 1/4+1/32+1/64 10 = -0.010011 2 = -.10011 2  2 -1 has  1 for the sign  10011 for the mantissa  1 for the sign of the exponent  1 for the exponent and in a 16 bits storage location is represented as 1100110000100001 11 0 0 1 1 0 0 0 010 0 0 0 1 Sign Mantissa Sign Exponent 01 1 0 0 0 0 0 0 010 0 0 1 0 Sign Mantissa Sign Exponent

34. Invitation to Computer Science, C++ Version, Third Edition 34 Characters are mapped onto binary numbers  ASCII code set (look for the code in your book) 8 bits per character; 256 character codes  UNICODE code set 16 bits per character; 65,536 character codes Text strings are sequences of characters in some encoding http://ascii-table.com/ Binary Representation of Numeric and Textual Information (continued)

35. Invitation to Computer Science, C++ Version, Third Edition 35 Binary Representation of Sound Multimedia data is sampled to store a digital form, with or without detectable differences Representing sound data  Sound data must be digitized for storage in a computer  Digitizing means periodic sampling of amplitude values

36. Invitation to Computer Science, C++ Version, Third Edition 36 Binary Representation of Sound (cont.)  From samples, original sound may be approximated  To improve the approximation: Sample more frequently Use more bits for each sample value

37. Invitation to Computer Science, C++ Version, Third Edition 37 Figure 4.5 Digitization of an Analog Signal (a) Sampling the Original Signal (b) Recreating the Signal from the Sampled Values

38. Invitation to Computer Science, C++ Version, Third Edition 38 Representing image data  Images are sampled by reading color and intensity values at even intervals across the image  Each sampled point is a pixel  Image quality depends on number of bits at each pixel Binary Representation of Images

39. Invitation to Computer Science, C++ Version, Third Edition 39 Binary Representation of Images (cont.) RGB – Encoding scheme  8 bits per color [0, 255] distinct values  A pixel in RGB uses 24 bits  Black – absence of colors 0, 0, 0  White – all colors together 255, 255, 255 Example: representation of the above image in RGB 0,0,0 0,0,0 0,0,0 0,0,0 0,0,0 0,0,0 0,0,0 255,255,255 0,0,0 0,0,0 0,0,0 255,255,255 255,255,255 255,255,255 0,0,0 0,0,0 0,0,0 255,255,255 0,0,0 0,0,0 0,0,0 0,0,0 0,0,0 0,0,0 0,0,0

40. Invitation to Computer Science, C++ Version, Third Edition 40 Binary Representation of Images (cont.) 00000000 11111111 00000000 11111111 00000000 11111111 00000000

41. Invitation to Computer Science, C++ Version, Third Edition 41 Storage size of Sound, Images, Movies Q. What is the total number of bits required to store 3 minutes of standard recording in MP3 over two channels? 44.100 samples/sec * 16 bits/sec * 180 sec = 127,008,000 bits = 15,876,000 bytes = 15,504 KB = 15.1MB per channel 15.1MB * 2 channels = 30.2 MB Q. What is the total number of bits required to store a picture taken from a digital camera with 3,500,000 pixels? 3,500,000 pixels/image * 24 bits/pixel * 1 image = 84,000,000 bits = 10,500,000 bytes = 10,254 KB = 10MB Q. What is the total number of bits required to store 5 minutes a movie that shows 30 frames per second and has 3 million of pixels per frame? 3,000,000 pixels/frame * 24 bits/pixel * 30 frame/sec * 300 sec = 648,000,000,000 bits = 81,000,000,000 bytes = 79,101,562.5 KB = 77,247.6 MB = 75.4 GB

42. Invitation to Computer Science, C++ Version, Third Edition 42 End of Lesson

43. Invitation to Computer Science, C++ Version, Third Edition 43 The Reliability of Binary Representation Electronic devices are most reliable in a bistable environment Bistable environment  Distinguishing only two electronic states Current flowing or not Direction of flow Computers are bistable: hence binary representations

44. Invitation to Computer Science, C++ Version, Third Edition 44 Magnetic core  Historic device for computer memory  Tiny magnetized rings: flow of current sets the direction of magnetic field  Binary values 0 and 1 are represented using the direction of the magnetic field Binary Storage Devices

45. Invitation to Computer Science, C++ Version, Third Edition 45 Figure 4.9 Using Magnetic Cores to Represent Binary Values

46. Invitation to Computer Science, C++ Version, Third Edition 46 Transistors  Solid-state switches: either permits or blocks current flow  A control input causes state change  Constructed from semiconductors Binary Storage Devices (continued)

47. Invitation to Computer Science, C++ Version, Third Edition 47 Figure 4.11 Simplified Model of a Transistor

48. Invitation to Computer Science, C++ Version, Third Edition 48 Boolean Logic and Gates: Boolean Logic Boolean logic describes operations on true/false values True/false maps easily onto bistable environment Boolean logic operations on electronic signals may be built out of transistors and other electronic devices

49. Invitation to Computer Science, C++ Version, Third Edition 49 Boolean Logic (continued) Boolean operations  a AND b True only when a is true and b is true  a OR b True when either a is true or b is true, or both are true  NOT a True when a is false, and vice versa

50. Invitation to Computer Science, C++ Version, Third Edition 50 Boolean expressions  Constructed by combining together Boolean operations Example: (a AND b) OR ((NOT b) AND (NOT a)) Truth tables capture the output/value of a Boolean expression  A column for each input plus the output  A row for each combination of input values Boolean Logic (continued)

51. Invitation to Computer Science, C++ Version, Third Edition 51 abValue 001 010 100 111 Example: (a AND b) OR ((NOT b) and (NOT a)) From Boolean Expression to Truth Table

52. Invitation to Computer Science, C++ Version, Third Edition 52 Gates  Hardware devices built from transistors to mimic Boolean logic AND gate  Two input lines, one output line  Outputs a 1 when both inputs are 1

53. Invitation to Computer Science, C++ Version, Third Edition 53 Gates (continued) OR gate  Two input lines, one output line  Outputs a 1 when either input is 1 NOT gate  One input line, one output line  Outputs a 1 when input is 0 and vice versa

54. Invitation to Computer Science, C++ Version, Third Edition 54 Figure 4.15 The Three Basic Gates and Their Symbols

55. Invitation to Computer Science, C++ Version, Third Edition 55 Abstraction in hardware design  Map hardware devices to Boolean logic  Design more complex devices in terms of logic, not electronics  Conversion from logic to hardware design may be automated Gates (continued)

56. Invitation to Computer Science, C++ Version, Third Edition 56 Building Computer Circuits: Introduction A circuit is a collection of logic gates:  Transforms a set of binary inputs into a set of binary outputs  Values of the outputs depend only on the current values of the inputs Combinational circuits have no cycles in them (no outputs feed back into their own inputs)

57. Invitation to Computer Science, C++ Version, Third Edition 57 Figure 4.19 Diagram of a Typical Computer Circuit

58. Invitation to Computer Science, C++ Version, Third Edition 58 A Circuit Construction Algorithm Sum-of-products algorithm is one way to design circuits:  Step1: From Truth Table to Boolean Expression  Step 2: From Boolean Expression to Gate Layout

59. Invitation to Computer Science, C++ Version, Third Edition 59 Figure 4.21 The Sum-of-Products Circuit Construction Algorithm

60. Invitation to Computer Science, C++ Version, Third Edition 60 Sum-of-products algorithm  Truth table captures every input/output possible for circuit  Repeat process for each output line Build a Boolean expression using AND and NOT for each 1 of the output line Combine together all the expressions with ORs Build circuit from whole Boolean expression A Circuit Construction Algorithm (continued)

61. Invitation to Computer Science, C++ Version, Third Edition 61 An application of Step 1 of the SOP algorithm: From Truth Table to Boolean Expression 1.Ignore rows where the output is 0. 2.Create the product of the selections in this way: for each row with output 1: - if the input is 1, select the variable name. - if the input is 0, select the NOT of the variable name. 3.Create the Boolean expression for the circuit by summing each of the products created in step 2. ==> This is the Boolean expression for the circuit to be constructed. a b a XOR b 0 0 0 1 0 1 0 1 1 1 1 0 ab’+a’b Note: a’=NOT a b’=NOT b

62. Invitation to Computer Science, C++ Version, Third Edition 62 Another example a b c output 0 0 0 1 0 0 1 1 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 0 1 1 1 a'b'c' + a'b’c + ab'c' + abc' + abc or output = a'b'c' + a'b’c +ab'c' + abc' + abc

63. Invitation to Computer Science, C++ Version, Third Edition 63 Examples Of Circuit Design And Construction Compare-for-equality circuit Addition circuit Both circuits can be built using the sum-of- products algorithm

64. Invitation to Computer Science, C++ Version, Third Edition 64 Compare-for-equality circuit  CE compares two unsigned binary integers for equality Given two n-bit numbers, a = a 0 a 1 a 2 …a n-1 and b = b 0 b 1 b 2 …b n-1 output 1 if the numbers are the same and 0 if they are not  Built by combining together 1-bit comparison circuits (1- CE)  Integers are equal if corresponding bits are equal (AND together 1-CD circuits for each pair of bits) A Compare-for-equality Circuit

65. Invitation to Computer Science, C++ Version, Third Edition 65 1-CE circuit truth table Or output = a’b’+ab abOutput 001 010 100 111 A Compare-for-equality Circuit (continued)

66. Invitation to Computer Science, C++ Version, Third Edition 66 Figure 4.22 One-Bit Compare for Equality Circuit

67. Invitation to Computer Science, C++ Version, Third Edition 67 1-CE Boolean expression  First case: (NOT a) AND (NOT b)  Second case: a AND b  Combined: ((NOT a) AND (NOT b)) OR (a AND b) A Compare-for-equality Circuit (continued)

68. Invitation to Computer Science, C++ Version, Third Edition 68 A Compare-for-equality Circuit (cont.) 1-CE Generalize the process by AND-ing together 1-CD circuits for each pair of bits The 1-CE circuit is represented by the box o o o o a 0 b 0 a 1 b 1 a 2 b 2 a n-1 b n-1 out 1-CE

69. Invitation to Computer Science, C++ Version, Third Edition 69 An Addition Circuit Addition circuit  Adds two unsigned binary integers, setting output bits and an overflow  Built from 1-bit adders (1-ADD)  Starting with rightmost bits, each pair produces A value for that order (i.e. s i =a i +b i ) A carry bit for next place to the left (i.e. c i )

70. Invitation to Computer Science, C++ Version, Third Edition 70 1-ADD truth table  Input One bit from each input integer One carry bit (always zero for rightmost bit)  Output One bit for output place value One “carry” bit An Addition Circuit (continued)

71. Invitation to Computer Science, C++ Version, Third Edition 71 Figure 4.24 The 1-ADD Circuit and Truth Table

72. Invitation to Computer Science, C++ Version, Third Edition 72 The 1-bit adder

73. Invitation to Computer Science, C++ Version, Third Edition 73 Put rightmost bits into 1-ADD, with zero for the input carry Send 1-ADD’s output value to output, and put its carry value as input to 1-ADD for next bits to left. Repeat process for all bits. Building the full adder s 0 =a 0 +b 0 s 1 = a 1 +b 1 s 2 = a 2 +b 2 s n-1 = a n-1 +b n-1 s n = c n c 0 =0 a 0 b 0 a 1 b 1 a 2 b 2 a n-1 b n-1 o o o o 1-ADD c1c1 c2c2 c n-1 c3c3

74. Invitation to Computer Science, C++ Version, Third Edition 74 Control Circuits Do not perform computations Choose order of operations or select among data values Major types of controls circuits  Decoders Sends a 1 on one output line, based on what input line indicates  Multiplexors Select one of inputs to send to output We will use these control circuits while studying next chapter.

75. Invitation to Computer Science, C++ Version, Third Edition 75 Decoder  Form N input lines 2 N output lines  N input lines indicate a binary number, which is used to select one of the output lines  Selected output sends a 1, all others send 0 Control Circuits (Decoder)

76. Invitation to Computer Science, C++ Version, Third Edition 76 Decoder purpose  Given a number code for some operation, trigger just that operation to take place  Numbers might be codes for arithmetic: add, subtract, etc.  Decoder signals which operation takes place next Control Circuits (continued)

77. Invitation to Computer Science, C++ Version, Third Edition 77 Figure 4.29 A 2-to-4 Decoder Circuit

78. Invitation to Computer Science, C++ Version, Third Edition 78 Decoder Circuit A decoder is a control circuit which has  N input and 2 N output A 3 to 2 3 example of decoder a b c o0 o1 o2 o3 o4 o5 o6 o7 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 1 1 1 0 0 0 0 0 0 0 1 a b c 0123456701234567

79. Invitation to Computer Science, C++ Version, Third Edition 79 Multiplexor form  2 N regular input lines  N selector input lines  1 output line Multiplexor purpose  Given a code number for some input, selects that input to pass along to its output  Used to choose the right input value to send to a computational circuit Control Circuits (Multiplexor)

80. Invitation to Computer Science, C++ Version, Third Edition 80 Figure 4.28 A Two-Input Multiplexor Circuit

81. Invitation to Computer Science, C++ Version, Third Edition 81 Multiplexor Circuit with N input multiplexor circuit 2 N input lines 0 1 2 2 N -1 N selector lines output Interpret the selector lines as a binary number. Suppose that the selector line write the number 00….01 which is equal to 1. For our example, the output is the value on the line numbered 1.....

82. Invitation to Computer Science, C++ Version, Third Edition 82 Summary Digital computers use binary representations of data: numbers, text, multimedia Binary values create a bistable environment, making computers reliable Boolean logic maps easily onto electronic hardware Circuits are constructed using Boolean expressions as an abstraction Computational and control circuits may be built from Boolean gates