Logic Circuits I ECE 1411 Thursday 4:45pm-7:20pm. Nathan Pihlstrom.

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1 Logic Circuits I ECE 1411 Thursday 4:45pm-7:20pm Nathan Pihlstrom

My Background B.S.E.E. from Colorado State University M.S.E.E. from Colorado State University M.B.A.

Covidien 2011-2013 LSI/Avago 2013 - present UCCS Fall Semester

2 My Background B.S.E.E. from Colorado State University M.S.E.E. from Colorado State University M.B.A. from UCCS Ford Microelectronics, Inc./Visteon Intel Corp Marvell Hittite Microwave Covidien LSI/Avago present UCCS Fall Semester 2013 ECE4280/5280 UCCS Spring Semester 2014 ECE4211/5211 UCCS Fall Semester 2014 ECE4280/2411

3 Chapter 1

4 1.2 Binary Numbers In general, a decimal number with a decimal point is represented by a series of coefficients: a 5 a 4 a 3 a 2 a 1 a 0.a -1 a -2 a -3 where the coefficients a j are any of the 10 digits (0, 1, 2, 3, 9) and the subscript value j give the place value. Thus for a base (or radix) 10 number: 10 5 a a a a a a a a a -3 For example: 7, = 7x x x x x x10-2

5 1.2 Binary Numbers The modern binary system was discovered by the German mathematician and philosopher Gottfried Liebniz in The binary number system has only two possible values: 0 and 1. Each coefficient is multiplied by a power of the radix and the result added to obtain the decimal equivalent of the number. Thus for a base (or radix) 2 number: 2 5 a a a a a a a a a -3 For example: = 1x x x x x x x2-2 = = 26.75

6 1.2 Binary Numbers In general, a number expressed in a base-r system has coefficients (a j ) multiplied by powers of r: a n r n + a n-1 r n a 2 r 2 + a 1 r 1 + a 0 r 0 + a -1 r -1 + a -2 r a -m r -m To distinguish between numbers of different bases, the coefficients are enclosed in parentheses with a subscript equal to the based used: (4021.2) 5 = 4x x x x x5-1 = (511.4) 10 Note: The coefficients for base 5 can only be 0, 1, 2, 3, and 4.

7 1.2 Binary Numbers Octal numbers: A base-8 system that has coefficients of: 0, 1, 2, 3, 4, 5, 6, and 7. Note that 8 and 9 are not part of the set. (127.4) 8 = 1x x x x8-1 = (87.5) 10 Hexadecimal: A = 10, B = 11, C = 12, D = 13, E = 14, F = 15 (B65F) 16 = 11x x x x16 0 = (46,687) 10 For number bases greater than 10, the letters of the alphabet are used to supplement the 10 decimal digits.

8 1.2 Binary Numbers The digits in a binary number are called bits. When a bit is 0, it does not contribute to the sum during conversion. Thus, the conversion from binary to decimal can be obtained by adding only the numbers with powers of two corresponding to the bits that are equal to 1: (110101) 2 = = (53) 10

9 1.2 Binary Numbers Arithmetic operations with numbers in base r follow the same rules as for decimal base numbers. Be careful to only use r-allowable digits when working in bases other than 10. 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 only be 0 or 1. The rules for binary subtraction are the same as decimal except that the borrow in any given significant position adds 2 to a minuend digit. Multiplier digits are always 1 or 0. The partial products are equal either to a shifted (left) copy of the multiplicand or to 0. augend: minuend: multiplicand: 1011 addend: subtrahend: multiplier: x 101 sum: difference: partial products: 1011 product:

10 Powers of Two: n 2n (bits) 1.2 Binary Numbers (bytes) n 2n (bits) (bytes) n 2n (bits) (bytes) 0 1 n/a n/a n/a * (1Kb) (1Mb) (1KB) (1MB) 30 1Gb 33 1GB 40 1Tb * = nibble 43 1TB

11 1.3 Number-Base Conversions As shown previously, the conversion of a number in base r to decimal is done by expanding the number in a power series and adding up all the terms. The conversion of a decimal number to a number in base r is done by dividing the number and all successive quotients by r and accumulating the remainders. If the decimal number contains a radix (decimal) point, it is necessary to separate the number into an integer part and a fraction part.

12 1.3 Number-Base Conversions Example 1.1: Convert (41) 10 to binary Integer Quotient Remainder Coefficient 41/2 = 20 + ½ a 0 = 1 20/2 = a 1 = 0 10/2 = a 2 = 0 5/2 = 2 + ½ a 3 = 1 2/2 = a 4 = 0 1/2 = 0 + 1/2 a 5 = 1 Therefore (41) 10 = (a 5 a 4 a 3 a 2 a 1 a 0 ) = (101001) 2

13 1.3 Number-Base Conversions Example 1.3: Convert (0.6875)10 to binary Integer Fraction Coefficient x 2 = a -1 = x 2 = a -2 = x 2 = a -3 = x 2 = a -4 = 1 Therefore (0.6875) 10 = (0.a -1 a -2 a -3 a -4 ) = (0.1011) 2

14 1.4 Octal and Hexadecimal Numbers Shorter patterns of hex characters are easier to recognize than log patterns of 1 s and 0 s. Conversion to binary to octal/hexadecimal is easily accomplished by partitioning the number into groups of three/four digits, starting from the binary point and proceeding from the left and right. Given: ) 2 (10_110_001_101_ _100_000_110) 2 = ( ) 8 (10_1100_0110_ _0000_0110) 2 = (2C6B.F06) 16

15 1.4 Octal and Hexadecimal Numbers

16 1.4 Octal and Hexadecimal Numbers Microsoft Windows calculator has ability to convert decimal/binary/octal/hexadecimal Calculator -> View -> Programmer Note: Grouping of bit in fours as hex is more popular and octal is rarely used

17 1.5 Complement of Numbers Number complements are used to simplify subtraction operation and logical manipulation in digital computers. Two types of complements for each base-r system: Radix complement Referred to the r s complement 2 s complement for binary 10 s complement for decimal Diminished radix complement r-1 complement 1 s complement for binary 9 s complement for decimal

18 1.5 Complement of Numbers Diminished Radix Complement: Defined as (r n 1) N where: N is a given number n is the number of 0 s following a single 1 r is the base, 10 for decimal, 2 for binary Diminished» 10 1 = 9 for decimal» 2 1 = 1 for binary Examples: 9 s complement of is (10 6 1) = s complement of is (10 6 1) = s complement of 101_1000 is (2 7 1) 101_1000 = 010_0111 Where (2 7 1) = (1000_0000 1) = 0111_ s complement of 010_1101 is (2 7 1) 010_1101 = 101_0010 Where (2 7 1) = (1000_0000 1) = 0111_1111 Or for binary, simply change all 1 s to 0 s and all 0 s to 1 s (invert).

19 1.5 Complement of Numbers Radix complement: Defined as r n 1 for N!= 0 and 0 for N = 0 n is the number of 0 s following a single 1 r is the base, 10 for decimal, 2 for binary r s complement is obtained by adding 1 to the (r-1) s complement Examples: 10 s complement of 012_398 is (10 6 1) 012_ _000_ _398 = 987_ s complement of 246_700 is (10 6 1) 012_ _000_ _700 = 753_300 2 complement of 110_ 1100 is (2 7 1) 110_ _ _1100 = 001_ complement of 011_ 0111 is (2 7 1) 011_ _ _ 0111 = 100_1001 Or for binary, leave all least significant 0 s and the first 1 unchanged and invert all other higher significant digits.

20 1.5 Complement of Numbers Subtraction with Complements 10 s complement M = 72532, N = 3250 M N: M = 72_ s Complement of N = + 96_750 Sum = 169_282 Discard end carry 10 5 = -100_000 Answer = 69_282 M = 3250, N = M N: M = 03_ s Complement of N = + 27_468 Sum = 30_718 No end carry, therefore the answer is -(10 s complement of 30_718) = -69_ _000 3_250 = 96_ _000 72_532 = 27_468

21 1.5 Complement of Numbers Subtraction with Complements 2 s complement Given X = 101_0100 and Y = 100_0011 X Y: X = 101_ s Complement of Y = + 011_1101 Sum = 1001_0001 Discard end carry 2 7 = -1000_0000 Answer = 001_0001 Y - X: Y = 100_ s Complement of X = + 010_1100 Sum = 110_1111 No end carry, therefore the answer is -(2 s complement of 110_1111) = -001_0001

22 1.5 Complement of Numbers Subtraction with Complements 1 s complement Given X = 101_0100 and Y = 100_0011 X Y: X = 101_ s Complement of Y = + 011_1100 Sum = 1001_0000 End around carry = +0000_0001 Answer = 001_0001 Y - X: Y = 100_ s Complement of X = + 010_1011 Sum = 110_1110 No end carry, therefore the answer is -(1 s complement of 110_1110) = -001_0001

23 1.6 Signed Binary Numbers In ordinary numbers, a negative number is indicated with a minus sign. Computers however, must represent everything with a 1 or a 0. It is customary to represent the sign with a bit placed in the leftmost position of the number 0 for positive 1 for negative The users determine whether the number is signed or unsigned Binary Value Signed Decimal Unsigned Decimal 0_ _

24 1.6 Signed Binary Numbers Consider -9 represented as a byte value: Signed magnitude 1_0000_1001 Signed-1 s complement 1111_0110 Signed-2 s complement 1111_0111 where the 1 s and 2 s complements are obtained using all bits, including the sign bit

25 1.6 Signed Binary Numbers Changing a positive number to a negative is easily done by taking the 2 s complement of the positive.

26 1.6 Signed Binary Numbers Arithmetic Addition The addition of two signed binary numbers with negative numbers represented in a signed-2 s complement form is obtained from the addition of the two numbers, including their sign bits. A carry out of the sign-bit position is discarded Negative results are automatically in 2 s complement form _ _ _ _ _ _ _ s complement of 6 -> _ _0011 < - 2 s complement of 13 -> _ _1001 < - 2 s complement of 7 overflow 2 s complement of 19 -> _1101

27 1.6 Signed Binary Numbers Arithmetic Subtraction Take the 2 s complement of the subtrahend (including the sign bit) and add to the minuend (including the sign bit). A carry out of the sign-bit is discarded. (+A) (+B) = (+A) + (-B) (+A) (-B) = (+A) + (+B) Remember: minuend subtrahend difference _ s complement of 6 -> _1010 -(+13) 1111_0011 < - 2 s complement of 13 -> -(+13) 1111_ _1001 < - 2 s complement of 7 overflow 2 s complement of -19 -> -19 1_1110_ _ s complement of 6 -> _1010 -(-13) 0000_1101 -(-13) 0000_ _ _0111

28 1.7 Binary Codes Binary-Coded Decimal Code (BCD) A number with k decimal digits will require 4k bits in BCD The binary combinations 1010 through 1111 are not used and have no meaning BCD numbers are decimal number and not binary numbers, although they use bits in their representation (396) 10 = ( ) BCD = (1_1000_1100) 2 (185) 10 = ( ) BCD = (1011_1001) 2 (15) 10 = ( ) BCD = (1111) 2 (10) 10 = ( ) BCD = (1010) 2

29 BCD Addition 1.7 Binary Codes Since each digit does not exceed 9, the sum cannot be greater than = 19 When binary sum is <= 1001 (w/o carry), the BCD sum is correct When binary sum is > 1001, add, 6 = (0110) 2 to correct BCD and produce carry Invalid BCD -> _0001 <- Invalid BCD Add (0110) 2 -> <- Add (0110) 2 1_0010 1_0111

30 1.7 Binary Codes BCD Addition (cont.) The addition of two n-digit unsigned BCD numbers follows the same procedure: BCD Binary Sum _ Add BCD Sum

31 Other Decimal Codes Binary codes for decimal digits require a minimum of four digits Excess-3 is an un-weighted code in which each coded combination is obtained from the corresponding binary value plus 3 The 2421 and excess-3 codes are examples of self complementing codes 9 s complement of a decimal number is obtained by complementing each bit of the code 1.7 Binary Codes

32 Gray Code 1.7 Binary Codes Only one bit in the code group changes in going from one number to the next Used for optical rotary encoders Q Binary Q Q Q Gray Q Q

33 ASCII Character Code 1.7 Binary Codes American Standard Code of Information Interchange Uses seven bits to code 128 characters As most computer manipulate bytes, ASCII characters are most often stored one per byte The extra bit is used to indicate other symbols (additional 128 available)

34 1.7 Binary Codes

35 1.7 Binary Codes Error-Detecting Code Parity bit is an extra bit included with a message to make the total number of 1 s either even or odd. Note: detects error only. Error Correcting Code (ECC) covered in Logic Circuits II With even parity With odd parity ASCII A = 100_ _ _0001 ASCII T = 101_ _ _0100

36 1.8 Binary Storage and Registers Register Most commonly device to hold data A register is group of binary cells. Nibble: 4-bits Byte: 8-bits Word: 2-Bytes/16 bits Dword: 2-Words/32-bits Qword: 2-Dwords/64-bits A 16-bit register can be in one of 2 16 (64k) states and can store any binary number from 0 to

37 1.9 Binary Logic Definition of Binary Logic AND: Represented by a dot or by the absence of an operator x y = z or xy = z OR: Represented by a plus sign x+y = z NOT: Represented by a prime _ or sometimes an over-bar x = z or x = z

38 Logic Gates 1.9 Binary Logic

39 1.9 Binary Logic Maximum output high voltage = V OH Minimum output low voltage = V OL Input voltages <= V IL are considered low Input voltages >= V IH are considered high N MH = V OH V IH N ML = V IL -V OL

40 Homework 10 points each:

Digital Systems and Binary Numbers

Digital Systems and Binary Numbers Prof. Wangrok Oh Dept. of Information Communications Eng. Chungnam National University Prof. Wangrok Oh(CNU) 1 / 51 Overview 1 Course Summary 2 Binary Numbers 3 Number-Base