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Engin112-F08-L06-08-boolean

Course: ENGIN 112, Fall 2008
School: UMass (Amherst)
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Lectures Engin112 6-8 Binary Logic and Boolean Algebra Maciej Ciesielski Department of Electrical and Computer Engineering 09/15/2008 Recap from last lecture Codes Assignment of bit sequence to represent discrete elements BCD code and addition Weighted code Error detecting codes Today's lecture Binary storage and registers Register transfer operations Binary logic Boolean algebra Engin112 09/15/2008 2...

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Lectures Engin112 6-8 Binary Logic and Boolean Algebra Maciej Ciesielski Department of Electrical and Computer Engineering 09/15/2008 Recap from last lecture Codes Assignment of bit sequence to represent discrete elements BCD code and addition Weighted code Error detecting codes Today's lecture Binary storage and registers Register transfer operations Binary logic Boolean algebra Engin112 09/15/2008 2 Error-Detecting Code Communication between systems can be "noisy" Environmental conditions can cause bit flips Error-detecting codes If one bit is flipped, the code becomes invalid Communication system can detect that and retransmit "Parity bit" is added to binary data "Even parity": choose parity bit such that # of 1's is even "Odd parity": choose parity bit such that # of 1's is odd Example: Data=1000001, even parity=0, odd parity=1 Data=1010100, even parity=1, odd parity=0 Only one parity bit is used Engin112 09/15/2008 3 Binary Storage and Registers How is information stored on a digital system? Bits are stored in "binary cells" Binary cell can have two stable states: `0' and `1' Binary cells are grouped into registers n cells make up n-bit register Size of registers is typically predefined Simple microcontroller: 8 bits = 1 byte Pentium: 32 bits = 4 bytes Mac G5: 64 bits = 8 bytes Digital system can usually process entire registers "Register transfer" operation specify processing Engin112 09/15/2008 4 Register Example Reading from keyboard: Engin112 09/15/2008 5 Register Transfer Operations Example: addition Designing digital logic is focus of rest of course Engin112 09/15/2008 6 Binary Logic Binary logic uses two possible values `1' and `0' `yes' and `no' `true' and `false' Can be represented by variables: A, B, C, x, y, z, ... Logic functions modify input values What is the minimum number of inputs? 1 input -> pretty uninteresting, inverter or wire only 2 inputs -> basic logic functions What are possible logic functions? NOT, AND, OR Others can be derived from those (NAND, XOR, etc.) Engin112 09/15/2008 7 George Boole Father of Boolean algebra He came up with a type of linguistic algebra, the three most basic operations of which were (and still are) AND, OR and NOT. It was these three functions that formed the basis of his premise, and were the only operations necessary to perform comparisons or basic mathematical functions. Boole's system (detailed in his 'An Investigation of the Laws of Thought, on Which Are Founded the Mathematical Theories of Logic and Probabilities', 1854) was based on a binary approach, processing only two objects - the yes-no, true-false, on-off, zero-one approach. Surprisingly, given his standing in the academic community, George Boole (1815 - 1864) Boole's idea was either criticized or completely ignored by the majority of his peers. Eventually, one bright student, Claude Shannon (19162001), picked up the idea and ran with it Engin112 09/15/2008 8 NOT Function Complement operation (inversion) Single input Inverts value of input Symbolized by prime x' or overbar x Truth table Input combinations on the left Output of function on the right input output x 0 1 x' 1 0 Graphic symbol NOT gate (inverter) Little circle indicates inversion Engin112 09/15/2008 9 AND Function Operation to check if two conditions are met Two inputs Output = 1 if and only if both inputs are 1 Symbolized by dot or absence of operator xy (also written as xy ) inputs output Truth table Needs to consider 22 = 4 input combinations Graphic symbol AND gate x 0 0 1 1 y 0 1 0 1 z 0 0 0 1 Engin112 09/15/2008 10 OR Function Operation to check if at least one condition is met Two inputs Output = 1 if any one or both inputs are 1 Symbolized by "plus" sign: x + y Truth table Graphic symbol OR gate x y 0 0 0 1 1 0 1 1 z 0 1 1 1 Engin112 09/15/2008 11 Evaluation of Logic Functions Timing diagrams (waveforms) Horizontal axis is time ( t ) Vertical axis shows signals, each with two different voltage levels t Engin112 09/15/2008 12 Multiple Inputs Two inputs might not be enough 3-input AND gate: 4-input OR gate: What are the rules for aggregating functions? Boolean algebra Engin112 09/15/2008 13 Reading Assignment Read Mano 2-4 2-6 Reminder: Homework #1 due Wednesday BEFORE class ! Make sure to provide on the first/cover page the following: ENGIN 112 Fall 2008 Homework 1, date submitted Your Name (no ID needed) Please paginate + staple your homework Engin112 09/15/2008 14 Algebras What is an algebra? Mathematical system consisting of Set of elements Set of operators Axioms or postulates Why is it important? Defines rules of "calculations" Example: arithmetic on natural numbers Set of elements: N = {1,2,3,4,...} Operator: +, , * Axioms: associativity, distributivity, closure, identity elements, etc. Note: operators with two inputs are called binary Does not mean they are restricted to binary numbers! Operator(s) with one input are called unary Engin112 09/15/2008 15 Axioms of Algebraic Structures Common axioms (or postulates): 1. Closure: A set S is closed with respect to a binary operator if the operator specifies a rule for obtaining an element of S. Example: S = N and = +, * (not ) 2. Associativity: A binary operator on a set S is said to be associative if (x y) z = x (y z) Example: S = N and = +, * (not ) 3. Commutativity: A binary operator on a set S is said to be commutative if xy=yx Example: S = N and = +, * (not ) Engin112 09/15/2008 16 Axioms of Algebraic Structures Common axioms (continued): 4. Identity element: Set S is said to have an identity element with respect to binary operation on S if there exists an element e S such that e x = x e = x for every x S Example: S = Z, = +, e = 0 5. Inverse: A set S having the identity element e with respect to a binary operator is said to have an inverse whenever, for every xS, there exists and element yS such that x y = e Example: S = Z, = +, e = 0, y = -x 6. Distributivity: If and are two binary operators on a set S, is said to be distributive over whenever x (y z) = (x y) (x z) Example: S = N, = , = + Engin112 09/15/2008 17 Example Algebra Instance of an algebraic structure is a field A field consists of Set of elements Two binary operators (each having properties 15) Both operators combined have property 6 (distributivity) Example: field for ordinary algebra on real numbers Binary operator + defines addition Additive identity is 0 Additive inverse defines subtraction Binary operator defines multiplication Multiplicative identity is 1 Multiplicative inverse defines division Distributive law: over +: a (b + c) = (a b) + (a c) Engin112 09/15/2008 18 Boolean Algebra We need to define algebra for binary values Developed by George Boole in 1854 Boolean algebra: algebraic structure with Set of elements B with two operators (+, ) satisfying postulates Huntington postulates for Boolean algebra (1904): 1. Closure with respect to operator + and operator 2. Identity element 0 for operator + and 1 for operator (e x = x ) 3. Commutativity with respect to + and x+y = y+x, xy = yx 4. Distributivity of over +, and + over x(y+z) = (xy)+(xz) and x+(yz) = (x+y)(x+z) 5. Complement for every element x is x' with x+x'=1, xx'=0 6. There are at least two elements x,yB such that xy Note: + and are chosen to be intuitive Do not blindly substitute with ordinary algebra, it won't work ! Engin112 09/15/2008 19 Boolean vs. Ordinary Algebra Differences Postulates do not include associativity Can be derived for both operators Distributivity of + over holds for Boolean, but not for ordinary algebra x+(yz) = (x+y)(x+z) Boolean does not have inverse elements for + or Thus, no subtraction or division is operators Complement not defined in ordinary algebra Set of elements in Boolean algebra not yet defined But there must be at least two elements Questions What is the set of elements? What are the rules of operation for + and ? Engin112 09/15/2008 20 Two-valued Boolean Algebra Two-valued Boolean is defined as A set of elements B = {0,1} "+" operator: x 0 0 1 1 y 0 1 0 1 x+y 0 1 1 1 "" operator: x y xy 0 0 0 1 1 0 1 1 0 0 0 1 inverse x x' 0 1 1 0 Observation + is OR, is AND, Other notation: + => , Engin112 09/15/2008 ' is NOT => , NOT => 21 Check Huntington Postulates Verify that postulates hold for Boolean algebra 1. Closure? Yes, all operations with {0,1}B Yes, 0 for + and 1 for Yes, for +: 0+0=0, 1+1=1, 0+1=1+0=1 Yes, for : 00=0, 11=1, 01=10=0 Huntington postulates: Post. 1: closure Post. 2: (a) x+0=x, (b) x1=x Post. 3: (a) x+y=y+x, (b) xy=yx Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x'=1, (b) xx'=0 2. Identity? 3. Commutativity? 4. Distributivity? Yes, use truth table (p. 39 in Mano) for over + and + over x 0 0 1 1 y 0 1 0 1 x+y 0 1 1 1 5. Complement? Yes, x+x' = 1: 0+0'=0+1=1 and 1+1'=1+0=1 Yes, xx' = 0: 00'=01=0 and 11'=10=0 Yes, 01 x y xy 0 0 0 1 1 0 1 1 0 0 0 1 22 6. Two distinct values? Engin112 09/15/2008 Boolean Algebra Two-value Boolean algebra matches with Huntington postulates We can use binary logic and consider it an algebra Next: Theorems that can be derived from postulates Canonical forms (uniqueness of representation) Duality of Boolean algebra Boolean functions Engin112 09/15/2008 23 Boolean Functions Elements and operators can express a function Value of function determined by value of variables Complete function is evaluated for all possible values Function can be represented by truth table E.g., F = x + y'z x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 y' 1 1 0 0 1 1 0 0 y'z x+y'z 0 1 0 0 0 1 0 0 0 1 0 0 1 1 1 1 Evaluate in steps E.g., first y' , then y'z, etc. Engin112 09/15/2008 24 Logic Circuit Diagram of Function Logic function can be expressed as circuit diagram Direct translation from algebraic expression Example: F = x + y'z Problem: possibly multiple equivalent algebraic expressions Difficult to compare Boolean functions Engin112 09/15/2008 25 Reading Assignment Read Mano 2-4 2-6 Engin112 09/15/2008 26 Recap from Last Lecture Boolean algebra Algebra axioms, postulates Huntington's postulates Boolean functions Algebraic expression Today's lecture Simple design example Boolean theorems Comparison of Boolean functions Canonical and standard forms Engin112 09/15/2008 27 Design Problem Design the control logic for a fan in a greenhouse. The logic must sense three environmental conditions: 1. It is raining outside (variable x) 2. It is humid inside (variable y) 3. It is hot inside (variable z) F The fan should turn on when It is humid AND it is not raining AND it is hot yx'z OR It is not humid AND {(it is not raining AND it is hot) OR it is raining} y' (x'z + x) Create a Boolean function that controls the fan. F = yx'z + y'(x'z+x) Other possibilities: F = x'y'z+x'yz+xy' or F = x'z+xy' Question: How to evaluate these solutions (equivalent functions)? Engin112 09/15/2008 28 Boolean Function Representations Greenhouse example function can be expressed as: Analytically, as sum of minterms: yx'z + y'(x'z+x) = yx'z+y'x'z+y'x As a truth table x y z 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 Engin112 09/15/2008 = x'yz+x'y'z+xy'(z+z') = x'yz+x'y'z+xy'z+xy'z' F 0 1 0 1 1 1 0 0 How to derive different solutions ? How to evaluate these solutions ? 29 As a logic network Boolean Theorems Huntington's postulates define some rules Post. Post. Post. Post. closure (a) x+0=x, (b) x1=x (a) x+y=y+x, (b) xy=yx (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x'=1, (b) xx'=0 1: 2: 3: 4: Need more rules to modify algebraic expressions What is a theorem? Theorems that are derived from postulates A formula or statement that is derived from postulates (or other proven theorems) Basic theorems of Boolean algebra Theorem 1 (a): x + x = x (b): x x = x Looks straightforward, but needs to be proven ! Engin112 09/15/2008 30 Proof of x+x=x We can only use Huntington postulates: Huntington postulates: Post. 2: (a) x+0=x, (b) x1=x Post. 3: (a) x+y=y+x, (b) xy=yx Post. 4: (a) x(y+z) = xy+xz, (b) x+yz = (x+y)(x+z) Post. 5: (a) x+x'=1, (b) xx'=0 Show that x+x=x. x+x = (x+x)1 = (x+x)(x+x') = x+xx' = x+0 =x Q.E.D. by 2(b) by 5(a) by 4(b) by 5(b) by 2(a) We can now use Theorem 1(a) in future proofs Engin112 09/15/2008 31 Proof of xx=x Similar to previous proof Huntington postulates: Post. 2: (a) x+0=x, ...

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chi^2/nu= 994.01754 / 868The fit is rejectable at 99.816511 % Confidence -62.2200 -57.5400 249.91190 -57.5400 -38.0400 254.47097 -38.0400 -28.6800 258.05386 -28.6800 -9.9600
Berkeley - ASTRO - 00336489
89.219 89.399 271.606 89.719989.399 89.643 225.781 72.914389.643 89.837 278.23 87.278389.837 89.99 346.295 110.85789.99 90.125 362.141 119.62690.125 90.255 376.069 124.22790.255 90.411 313.391 103.52390.411 90.536 431.813 135.45690.536 90.634
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Source Contamination: 3.66E-06 +/- 4.5E-07 cts/s
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# t1 t2 dt rad_min rad_max cts err scl bg bg_rat wt 0.089219 0.089271 0.000052 0. 16. 11.00 3.32 0.878525 0.000000 0.281406 1 0.089271 0.089312 0.000041 0. 16. 10.44
Berkeley - ASTRO - 00336489
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Berkeley - ASTRO - 00336489
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Berkeley - ASTRO - 00336489
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Berkeley - ASTRO - 00336489
SIMPLE = T / file does conform to FITS standardBITPIX = 8 / number of bits per data pixelNAXIS = 0 / number of data axesEXTEND = T / FITS dataset may contain extensio
Berkeley - ASTRO - 00336489
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Berkeley - ASTRO - 00336489
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Berkeley - ASTRO - 00336489
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UAB - CS - 624
An Operating System Exampleclass Kerneld thread while true do skip end Kerneld class Ftpd thread while true do skip end Ftpd class Syslogd thread while true do skip end Syslogd class Lpd thread while true do skip end Lpd class Httpd thread while tru
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The Test Classclass Test instance variables atm: ATMMachine := new ATMMachine(); operations Init: () => () Init () = start(atm) end Test
UAB - CS - 624
A Sorting Example Illustrating Statementsclass St values v15 = selection_sort([3,2,9,1,3]); st1 = mk_(1,6,[3,2,-9,11,5,3]) instance variables x:nat; y:nat; l:seq1 of nat;functions min_index : seq1 of nat -> nat min_index(l) = if len l = 1 then 1 e