EE 261 – Introduction to Logic Circuits Module #2 – Number Systems Topics Number System Formation Base Conversions Binary Arithmetic Signed Numbers Signed Arithmetic Binary Codes Textbook Reading Assignments 2.1–2.16 Practice Problems 2.1, 2.5, 2.6, 2.7, 2.8, 2.11, 2.12 Graded Components of this Module 2 homeworks, 2 discussions, 1 quiz (all online)

EE 261 – Introduction to Logic Circuits Module #2 – Number Systems What you should be able to do after this module Convert numbers between bases (decimal, binary, octal, hexadecimal) Perform arithmetic (both signed & unsigned) using other bases Create and Decode various binary codes (BCD, Gray, ASCII, Parity)

Number Systems Number System - a system that contains a set of numbers/symbols/characters and at least one operation (+, -, …) BASE or RADIX - the number of symbols in the number system - humans use Base10 (why?) - computers use Base2 (why?) Positional Number System - we can represent 10 unique quantities w/ our Base10 system - what if we want more? - we can add a "leading" digit to our number that has increased "weight” ex) 0 1 : 9 10 11 : 99 100 101

Number Systems Radix Point - place in the string of digits at which numbers represent either whole or fractional. ex) 125.178 p - Number of digits to the LEFT of the radix point. n - Number of digits to the RIGHT of the radix point. d - Digits in the system, described with a positional subscript (NOTE: if the radix point is missing, we assume it is to the right of the #) r - Radix or Base (Base10, r=10) i - Position - starting at 0 - increasing to the left of the radix point - negative to the right of the radix point Weight - each digit has a weight based on its position = ri - we multiply the digit by its weight to find how much value that digit represents

Number Systems Value - the value of a number is the sum of each digit multiplied by its corresponding weight. Example: Given the BASE 10 (r=10) number 327.2 What is p? p=3 What is n? n=1 Write the digit notation? d2 d1 d0 . d-1 What is the weight for each position? 102 101 100 .10-1 Show the expanded decimal equivalent? = 3(102) + 2(101 ) +7(100 ) + 2(10-1 ) = 327.2 Why all this framework? Because this generic format works for all Bases.

Number Systems Binary - A number system with 2 symbols (BASE=2 or r=2). - The symbols are 0 and 1 - each symbol is called a "bit" - we’ll give a subscript to the number to indicate its base ex) 101010 (decimal) 10102 (binary) - 4 bits are called a "Nibble" - 8 bits are called a "Byte" - the leftmost bit in a string is called the Most Significant (MSB) or High Order - the rightmost bit in a string is called the Least Significant (LSB) or Low Order

Number Systems Binary Systems Example: Given the Binary number 10011 What is p? p=5 What is n? n=0 What is r? r=2 Write the bit notation? b4 b3 b2 b1 b0 What is the weight for each position? 24 23 22 21 20 Show the expanded decimal equivalent? B = 1(24) + 0(23) +0(22 ) + 1(21 ) + 1(20 ) = 1(16) + 0(8) + 0(4) + 1(2) +1(1) = 1910

Base Conversions Base - the number of symbols in a number system - we instinctively know decimal - we've talked about Binary (two symbol) - there are other bases of interest in digital systems - the bases we typically care about are powers of 2 (or associated with Binary) - these bases are typically used to represent a lot of "bits" ex) it's hard to describe the value of a 64-bit bus in 1's and 0's Octal - A number system with 8 symbols - 0,1,2,3,4,5,6,7 - each digit in this system is equivalent to 3-bits - it is a positional number system ex) 0 1 : 7 10 11 : 17 20

Base Conversions Hexadecimal - A number system with 16 symbols - we use alphabetic characters as symbols in the set above 9 - 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F - each digit in this system is equivalent to 4-bits - it is a positional number system ex) 0 1 : 9 A B C D E F 10 11 : 1F 20 :

Base Conversions How the Bases Relate to Each Other Decimal Binary Octal Hexadecimal Base10 Base2 Base8 Base16 0 0 0 0 1 1 1 1 2 10 2 2 3 11 3 3 4 100 4 4 5 101 5 5 6 110 6 6 7 111 7 7 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 15 1111 17 F - these are the commonly used bases in digital systems, we'd like to be able to convert between them

Base Conversions Converting to Decimal - we sum the products of each digit value with its positional weight - this works for whole and fractional digits - this works for Binary to Decimal - this works for Octal to Decimal - this works for Hex to Decimal

Base Conversions Base Conversion – Binary to Decimal - each digit has a “weight” of ri that depends on the position of the digit - multiply each digit by its “weight” - sum the resultant products ex) Convert 1011 to decimal 23 22 21 20 (weight) 1 0 1 1 = 1•(23) + 0• (22) + 1• (21) + 1• (20) = 1•(8) + 0• (4) + 1• (2) + 1• (1) = 8 + 0 + 2 + 1 = 11 Decimal

Base Conversions Base Conversion – Binary to Decimal with Fractions - the weight of the binary digits have negative positions ex) Convert 1011.101 to decimal 23 22 21 20 2-1 2-2 2-3 1 0 1 1 . 1 0 1 = 1•(23) + 0• (22) + 1• (21) + 1• (20) + 1• (2-1) + 0• (2-2) + 1• (2-3) = 1•(8) + 0• (4) + 1• (2) + 1• (1) + 1• (0.5) + 0• (0.25) + 1• (0.125) = 8 + 0 + 2 + 1 + 0.5 + 0 + 0.125 = 11.625 Decimal

Base Conversions Base Conversion – Hex to Decimal - the same process as “binary to decimal” except the weights are now BASE 16 - NOTE (A=10, B=11, C=12, D=13, E=14, F=15) ex) Convert 2BC from Hex to decimal 162 161 160 (weight) 2 B C = 2• (162) + B• (161) + C• (160) = 2•(256) + 11• (16) + 12• (1) = 512 + 176 + 12 = 700 Decimal

Base Conversions Base Conversion – Hex to Decimal with Fractions - the fractional digits have negative weights (BASE 16) - NOTE (A=10, B=11, C=12, D=13, E=14, F=15) ex) Convert 2BC.F to decimal 162 161 160 16-1 (weight) 2 B C . F = 2• (162) + B• (161) + C• (160) + F• (16-1) = 2•(256) + 11• (16) + 12• (1) + 15• (0.0625) = 512 + 176 + 12 + 0.938 = 700.938 Decimal

Base Conversions Base Conversion – Decimal to Binary - the decimal number is divided by 2, the remainder is recorded - the quotient is then divided by 2, the remainder is recorded - the process is repeated until the quotient is zero ex) Convert 11 decimal to binary Quotient Remainder 2 11 5 1 LSB 2 5 2 1 2 2 1 0 2 1 0 1 MSB = 1011 binary

Base Conversions Base Conversion – Decimal to Binary with Fractions - the fraction is converted to binary separately - the fraction is multiplied by 2, the 0th position digit is recorded - the remaining fraction is multiplied by 2, the 0th digit is recorded - the process is repeated until the fractional part is zero ex) Convert 0.375 decimal to binary Product 0th Digit 0.375•2 0.75 0 MSB 0.75•2 1.50 1 0.5•2 1.00 1 LSB 0.375 decimal = 0.011 binary finished

Base Conversions Base Conversion – Decimal to Hex - the same procedure is used as before but with BASE 16 as the divisor/multiplier ex) Convert 420.625 decimal to hex 1st, convert the integer part… Quotient Remainder 16 420 26 4 LSB 16 26 1 10 16 1 0 1 MSB = 1A4 2nd, convert the fractional part… Product 0th Digit 0.625•16 10.00 10 MSB = 0.A 420.625 decimal = 1A4.A hexadecimal

Base Conversions Base Conversion – Octal to Decimal / Decimal to Octal - the same procedure is used as before but with BASE 8 as the divisor/multiplier

Base Conversions Base Conversion of "Powers of 2" - converting bases that are powers of 2 are simple due to straight forward mapping - An Octal digital represents 3 bits - A Hex digit represents 4 bits

Base Conversions Converting Binary to Hexadecimal - every 4 binary bits represents one HEX digit - begin the groups of four at the LSB - if necessary, fill with leading 0’s to form the groups of four ex) Convert 111010100 from Binary to Hex 0001 1101 0100 1 D 4 Hex, notice that we had to fill with MSB 0's to get groups of four… - This works for fractions too. The only difference is that the grouping starts at the Radix Point ex) Convert 0.11 from Binary to Octal 0.1100 0.C

Base Conversions Converting Binary to Octal - every 3 binary bits represents one OCTAL digit - begin the groups of three at the LSB - if necessary, fill with leading 0’s to form the groups of three ex) Convert 111010100 from Binary to Octal 111 010 100 7 2 4 Octal ex) Convert 10100 from Binary to Octal 010 100 1 4 Octal, note we had to fill with 0's on the MSB side

Base Conversions Converting Hex to Binary - each HEX digit is made up of four binary bits ex) Convert ABC Hex to Binary A B C = 1010 1011 1100 = 101010111100

Base Conversions Converting Octal to Binary - each Octal digit is made up of three binary bits ex) Convert 567 Octal to Binary 5 6 7 = 101 110 111 = 101110111

Base Conversions Terminology NIBBLE = 4 bits BYTE = 8 bits - you should be familiar with converting Binary Nibbles to Hex & Dec - there is a table on page 28 of your textbook which lists the basic conversions

Binary Arithmetic Addition and Subtraction - same as BASE 10 math, remember borrows and carries Addition Table 0 0 1 1 + 0 + 1 + 0 + 1 0 1 1 10 Carry Subtraction Table 10 0 0 1 1 - 0 - 1 - 0 - 1 0 1 1 0 Need to "Borrow" from a more significant bit

Binary Arithmetic Binary Addition - same as BASE 10 addition - need to keep track of the carry bit for a given system size (n), i.e., 4-bit, 8-bit, 16-bit,… ex) Add the binary numbers 1011 and 1001 1 1 1 0 1 1 1 0 0 1 +_____ 1 0 1 0 0 Carry Bit

Binary Arithmetic Binary Subtraction - same as BASE 10 subtraction - need to keep track of the borrow bit for a given system size (n), i.e., 4-bit, 8-bit, 16-bit,… ex) Subtract the binary numbers 1010 and 0011 1 10 0 10 0 10 1 0 1 0 0 0 1 1 -________ 0 1 1 1 Borrow bits : if necessary, we could assume a borrow from an even higher significant bit

Signed Numbers Negative Numbers - So far, we've dealt with Positive numbers - The real world has Negative numbers - We need a method to represent negatives in binary - However, our number system doesn't have a "-", just 0's and 1's Sign Bit - We will use the MSB to represent a "+" or "-" 0 = Positive 1 = Negative

Signed Numbers Signed Magnitude Representation - a negative number system (there is more than one) - uses the MSB as the sign bit - the remaining LSB's represent the number ex) 85 dec = 1010101 +85 dec = 01010101 (additional sign bit adds one bit to the number) -85 dec = 11010101 Is this Efficient? - The number of unique combinations that a number can represents is given by: N = rx N = # of unique combinations x = # of digits in the number (p + n) r = base ex) 2-bits: N=22 = 4 (00, 01, 10, 11) 7-bits: N=27 = 128 8-bits: N=28 = 256

Signed Numbers Signed Magnitude Range - The Range of Signed Magnitude is given by: -(2x-1 - 1) < NSM < (2x-1 -1) ex) if we use 8-bits, the range is: -(28-1 - 1) < NSM < (28-1 -1) -127 < NSM < +127 - This is only 255 unique numbers - But, if we are using 8-bits, there should be 28 unique numbers, or 256? - This is because in Signed Magnitude, there are two representations for Zero 0 0000000 = +0 1 0000000 = -0

Signed Numbers Advantages of Signed Magnitude - Very easy to understand and use - the binary number is simply represented in binary, then a sign bit is added Disadvantages of Signed Magnitude Representation - We loose a possible number by having +0, and -0. - This also creates a gap in our number list, which makes simple math harder - Addition/Subtraction in general are more difficult Algorithm : if ( signs are the same) - add and give same sign else if (signs are different) - compare magnitudes - subtract smaller from larger - give the result the sign of larger number

Signed Numbers Complement Numbers - Complementing a Binary Number means: 0 changes to 1 1 changes to 0 - using this technique, number can be a "complemented" to find the negative representation, then the arithmetic becomes much simpler - the ones we care about are 1) "Radix Complement" (2's complement for binary) 2) "1's Complement

Signed Numbers Radix Complement - i.e., 10's complement, 2's complement - Technique to get the Radix Complement: - subtract current number from rx Dr-comp = rn - Dorig - Rules: - The MSB is still the sign bit (0="+", 1="-") - Now that we're subtracting, we can't have arbitrary number of bits… x is predetermined and fixed - A simpler method is : 1) Complement all digits in Dorig 2) Add 1 to the result, ignore any carry-out

Signed Numbers 2's Complement - this is Radix Complement on a BASE2 number system - straight forward to use the "complement and add 1" technique ex) Give the 8-bit, 2's complement representation of -1710 1710 = 000100012 (Notice that the MSB is the sign bit, 0=+) Step 1: 11101110 (we first complement all bits) Step 2: + 1 (we then add 1) 11101111 (Notice the Sign Bit is negative) - these 8-bits represent -1710 in 2's complement

Signed Numbers 2's Complement Checking - We can convert back to find Original magnitude and do checking ex) Give the 8-bit, 2's complement representation of -1710 -1710 = 11101111 2's Comp (Notice that the MSB is the sign bit, 1="-") Step 1: 00010000 (we first complement all bits) Step 2: + 1 (we then add 1) 00010001 (Notice the Sign Bit is positive) - these 8-bits represent +1710,Which is what we originally started with,

Signed Numbers 2's Complement Range - We need to know how many numbers we can represent using this system - Notice that we are still using a bit for the sign… - BUT, we don't duplicate Zero in this system ex) 00000000 = 010 10000000 = -12810 - The Range of 2's Comp is given by: -(2x-1) < N2Comp < (2x-1 -1) ex) if we use 8-bits, the range is: -(28-1) < N2Comp < (28-1 -1) -128 < N2Comp < +127 - There are now 256 unique numbers, this is a more efficient use of bits

Signed Numbers 1's Complement - In 2's complement, the codes are asymmetric Lowest 2's Comp # = 1000 00002Comp = -12810 Highest 2's Comp # = 0111 11112Comp = +12710 - This is good because we have all 256 possible numbers that 8-bits can give us - 1's Comp is similar, but it gives us symmetry around Zero - To find the 1's Comp, we subtract current number from (rx -1) D1Comp = (rn -1) - Dorig

Signed Numbers 1's Complement - Or we can use the simple way 1) Complement all the #'s (Don't add 1) ex) Give the 8-bit, 1's complement representation of -1710 1710 = 000100012 (Notice that the MSB is the sign bit, 0=+) Step 1: 111011101Comp (complement all bits)

Signed Numbers 1's Complement Range - The Range of 1's Complement is given by: -(2x-1 - 1) < NSM < (2x-1 -1) - Once again, we have two values for Zero 00000000 = +0 11111111 = -0

Signed Numbers Negative Representation - We have covered 3 different "signed" codes - You need to KNOW THE CODE you are using ex) Represent -2210 1) Signed Magnitude : 10010110SM 2) 2's Complement : 111010102Comp 3) 1's Complement : 111010011Comp

Signed Arithmetic Two’s Compliment Arithmetic - Two's complement has advantages when going into Hardware - Two's complement addition is straight forward because the numbers are in sequential order (+1) from their least significant (1000 0000 ) to their most significant (0111 1111) - There is only one value for Zero, so "avoiding the gap" isn't necessary

Signed Arithmetic Two’s Compliment Addition - Addition of two’s compliment numbers is performed just like standard binary addition. - However, the carry bit is ignored

Signed Arithmetic Two’s Compliment Subtraction - We can build a subtraction circuit out of the same hardware as an adder - 2's Comp inherently adds negative numbers so - to subtract, we can just complement one number, and add it. ex) Subtract 8dec from 15dec 15dec = 0000 1111 8dec = 0000 1000 -> two’s compliment -> invert 1111 0111 add1 + 1 ----------------- 1111 1000 = -8dec Now Add: 15 + (-8) = %0000 1111 %1111 1000 +_________ 1 0000 0111 = 7dec Disregard Carry

Signed Arithmetic Two’s Compliment Overflow - If a two’s compliment arithmetic operation results in a number that is outside the range of representation (i.e., 8bits : -128 < N < +127), an “overflow” has occurred. ex) -100dec – 100dec- = -200dec (can’t represent) - There are three cases when overflow occurs 1) Sum of like signs results in answer with opposite sign 2) Negative – Positive = Positive 3) Positive – Negative = Negative - Boolean logic can be used to detect these situations.

Signed Arithmetic Remember 2's Comp Range - The Range of 2's Comp is given by: -(2x-1) < N2Comp < (2x-1 -1) ex) if we use 8-bits, the range is: -(28-1) < N2Comp < (28-1 -1) -128 < N2Comp < +127 What goes on in real Hardware? - A generic, Binary adder circuit is created - the user must be aware of when 2's complement is being used - additional circuitry checks for "Overflow"

Binary Codes Codes - a string of x-bits that represent information - we've seen codes already using the binary # system 1) Signed Magnitude 2) 2's Complement 3) 1's Complement - the same information can be encoded differently - it's up to the engineer to KNOW THE CODE that is being used Code Word - the term to represent the discrete string of bits that make up the information ex) 4-bit code words in a stream of information 1111 0100 0100 word

Binary Codes Binary Coded Decimal (BCD) - Sometimes we wish to represent an individual decimal digit as a binary representation (i.e., 7-segment display to eliminate a decoder) - We do this by using 4 binary digits. Decimal BCD 0 0000 1 0001 ex) Represent 17dec 2 0010 3 0011 Binary = 10111 4 0100 BCD = 0001 0111 5 0101 6 0110 7 0111 8 1000 9 1001

Binary Codes BCD Addition - If we are using BCD, traditional addition doesn’t work. 12 = correct answer in BCD is 20. Traditional addition only considers + 08 8-bit addition carry. In BCD we need to consider 4-bit addition carry. 1A - The solution is to add 6 to every sum greater than 9 12 = When a nibble addition results in “A” or greater, we add 6 08 1A + 6 20

Binary Codes Gray Code - There are applications where we want to count, but we only want one bit to transition each time we increment counts - Reduce Power & Noise in Digital Electronics - Electromechanical devices - Such a is called a "Gray Code" ex) 3-bit Gray Code Decimal Binary Gray 0 000 000 1 001 001 2 010 011 3 011 010 4 100 110 5 101 111 6 110 101 7 111 100

Binary Codes ASCII - American Standard Code for Information Interchange - English Alphanumeric characters are represented with a 7-bit code - See Table 2-11 in text ex) ‘A’ = $40 ‘a’ = $61

Binary Codes Parity Codes - there are times when information can be corrupted during transmission - we can include an "Error Checking" bit along with the original data - this "Error Checking" bit contains additional information about the original data - this can be used by the receiver to monitor whether an error in the data occurred PARITY - the number of 1's in the information are counted, - the parity bit represents whether there are an EVEN or ODD number of 1's EVEN PARITY = 0, if there are an EVEN number of 1's in the information = 1, otherwise ODD PARITY = 0, if there are an ODD number of 1's in the information = 1, otherwise

Binary Codes Parity Example Information EVEN Parity ODD Parity 000 0 1 001 1 0 010 1 0 011 0 1 100 1 0 101 0 1 110 0 1 111 1 0

Module Overview Topics - # systems & bases - # system conversions - to & from decimal - to and from binary/oct/hex - binary arithmetic - addition (carries) - subtraction (borrows) - negative #'s in binary - Signed Magnitude - 2's Complement - 1's Complement - 2's complement arithmetic - complement & add - range & overflow - Codes - BCD, Gray, ASCII, Parity