ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT, OAKLAND UNIVERSITY ECE-2700: Digital Logic Design Fall Notes - Unit 1

Similar documents
ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT, OAKLAND UNIVERSITY ECE-378: Computer Hardware Design Winter Notes - Unit 1

Boolean algebra.

Boolean Algebra. Boolean Algebra

Digital Control of Electric Drives

Solutions - Homework 1 (Due date: September 9:30 am) Presentation and clarity are very important!

CS12N: The Coming Revolution in Computer Architecture Laboratory 2 Preparation

Fachgebiet Rechnersysteme1. 1. Boolean Algebra. 1. Boolean Algebra. Verification Technology. Content. 1.1 Boolean algebra basics (recap)

Combinational Logic. Precedence. Quick Quiz 25/9/12. Schematics à Boolean Expression. 3 Representations of Logic Functions. Dr. Hayden So.

Introduction to Electrical & Electronic Engineering ENGG1203

Overview of Today s Lecture:

Engr354: Digital Logic Circuits

Control with binary code. William Sandqvist

expression simply by forming an OR of the ANDs of all input variables for which the output is

Boolean Algebra. Boolean Algebras

Unit 4. Combinational Circuits

CS 330 Formal Methods and Models

Fast Boolean Algebra

Chapter 1. Chapter 1 1

IST 4 Information and Logic

Note 12. Introduction to Digital Control Systems

Fault Modeling. EE5375 ADD II Prof. MacDonald

1 2 : 4 5. Why Digital Systems? Lesson 1: Introduction to Digital Logic Design. Numbering systems. Sample Problems 1 5 min. Lesson 1-b: Logic Gates

Chapter 1: Boolean Logic

Chapter 3 Single Random Variables and Probability Distributions (Part 2)

Basics of Digital Logic

EECS 141 Due 04/19/02, 5pm, in 558 Cory

Elements of Computing Systems, Nisan & Schocken, MIT Press. Boolean Logic

Review of Gaussian Quadrature method

CHAPTER 3 LOGIC GATES & BOOLEAN ALGEBRA

ENGR 3861 Digital Logic Boolean Algebra. Fall 2007

Lecture 3. Introduction digital logic. Notes. Notes. Notes. Representations. February Bern University of Applied Sciences.

Coalgebra, Lecture 15: Equations for Deterministic Automata

Lecture 11 Binary Decision Diagrams (BDDs)

Good Review book ( ) ( ) ( )

CS 330 Formal Methods and Models Dana Richards, George Mason University, Spring 2016 Quiz Solutions

CHAPTER 1 PROGRAM OF MATRICES

Genetic Programming. Outline. Evolutionary Strategies. Evolutionary strategies Genetic programming Summary

SIMPLIFICATION OF BOOLEAN ALGEBRA. Presented By: Ms. Poonam Anand

Convert the NFA into DFA

AUTOMATA AND LANGUAGES. Definition 1.5: Finite Automaton

Chapter 2. Random Variables and Probability Distributions

Reasoning and programming. Lecture 5: Invariants and Logic. Boolean expressions. Reasoning. Examples

Lecture 08: Feb. 08, 2019

Unit 8 Problem Solutions

CISC 4090 Theory of Computation

Chapter 6 Techniques of Integration

INF1383 -Bancos de Dados

ECE223. R eouven Elbaz Office room: DC3576

IST 4 Information and Logic

CS 310 (sec 20) - Winter Final Exam (solutions) SOLUTIONS

p-adic Egyptian Fractions

CMPSCI 250: Introduction to Computation. Lecture #31: What DFA s Can and Can t Do David Mix Barrington 9 April 2014

Some Theory of Computation Exercises Week 1

Let's start with an example:

EE 108A Lecture 2 (c) W. J. Dally and P. Levis 2

Asynchronous Sequen<al Circuits

KNOWLEDGE-BASED AGENTS INFERENCE

First Midterm Examination

Talen en Automaten Test 1, Mon 7 th Dec, h45 17h30

Bridging the gap: GCSE AS Level

CHAPTER 9 MULTIPLEXERS, DECODERS, AND PROGRAMMABLE LOGIC DEVICES

CS 301. Lecture 04 Regular Expressions. Stephen Checkoway. January 29, 2018

Things to Memorize: A Partial List. January 27, 2017

Chapter 3. Vector Spaces

Homework 3 Solutions

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions

Worked out examples Finite Automata

BİL 354 Veritabanı Sistemleri. Relational Algebra (İlişkisel Cebir)

Parse trees, ambiguity, and Chomsky normal form

4.1. Probability Density Functions

Lecture 6. Notes. Notes. Notes. Representations Z A B and A B R. BTE Electronics Fundamentals August Bern University of Applied Sciences

Chapter 4 Regular Grammar and Regular Sets. (Solutions / Hints)

Lecture 2 : Propositions DRAFT

Propositional models. Historical models of computation. Application: binary addition. Boolean functions. Implementation using switches.

Homework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama

Logic Synthesis and Verification

Continuous Random Variable X:

Lecture 3: Curves in Calculus. Table of contents

Introduction to Group Theory

Industrial Electrical Engineering and Automation

1 Nondeterministic Finite Automata

Exercise 3 Logic Control

Scanner. Specifying patterns. Specifying patterns. Operations on languages. A scanner must recognize the units of syntax Some parts are easy:

3 Regular expressions

Nondeterminism and Nodeterministic Automata

Goals for Lecture. Binary Logic and Gates (MK 2.1) Binary Variables. Notation Examples. Logical Operations

IST 4 Information and Logic

Assignment 1 Automata, Languages, and Computability. 1 Finite State Automata and Regular Languages

DATABASE DESIGN I - 1DL300

Chapter 4. Additional Variational Concepts

Worksheet A EXPONENTIALS AND LOGARITHMS PMT. 1 Express each of the following in the form log a b = c. a 10 3 = 1000 b 3 4 = 81 c 256 = 2 8 d 7 0 = 1

Vectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3-dimensional vectors:

Revision Sheet. (a) Give a regular expression for each of the following languages:

Designing finite automata II

CS103B Handout 18 Winter 2007 February 28, 2007 Finite Automata

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

CS 330 Formal Methods and Models

CARLETON UNIVERSITY. 1.0 Problems and Most Solutions, Sect B, 2005

CHAPTER 1 Regular Languages. Contents

Transcription:

INTRODUTION TO LOGI IRUITS Notes - Unit 1 OOLEN LGER This is the oundtion or designing nd nlyzing digitl systems. It dels with the cse where vriles ssume only one o two vlues: TRUE (usully represented y the symol '1'), nd LSE (usully represented y the symol '0'). This is lso clled Two-vlued oolen lger or Switching lger. circuit consisting o switches cn e represented in terms o oolen lgeric equtions. The equtions cn e then mnipulted into the orm representing the simplest circuit. The circuit my then e immeditely drwn rom the equtions. This powerul method irst ppered in: symolic nlysis o Rely nd Switching ircuits, lude E. Shnnon, Trnsctions o the IEE, vol. 57, no. 12, Dec. 198, pp. 71-721. SI OPERTIONS nd re oolen vriles. oolen vriles re used to represent the inputs or outputs o digitl circuit. OPERTION OOLEN EPRESSION OPERTION NOT ( or ) Logicl negtion ND. Logicl conjunction o two sttements OR + Logicl disjunction o two sttements TRUTH TLES ND LOGI GTES Truth Tle: tulr listing o unction vlues or ll possile comintions o vlues on its input rguments. I there re n inputs, there re 2 n possile comintions. Logic Gtes: Hrdwre components tht produce logic 1 or logic 0 depending on the stte o inputs. oolen unctions cn e implemented with logic s. NOT = ' 0 1 1 0 = ' ND =. =. OR = + = + Logic Gtes (ND, OR) cn hve multiple inputs: Z =..Z... Z = ++Z+......... IOMS 0.0 = 0 1.1 = 1 0.1 = 1.0 = 0 0 = 1 1+1=1 0+0 = 0 1+0 = 0+1 = 1 1 = 0 1 Instructor: Dniel Llmocc

THEOREMS Vrile dominnt rule ommuttive rule omplement rule Idempotency Identity Element Doule negtion ssocitive rule Distriutive rule. 1 = + 0 =. =. + = +. = 0 + = 1. = + =. 0 = 0 + 1 = 1 =. (. Z) = (. ). Z + ( + Z) = ( + ) + Z. ( + Z) =. +. Z +. Z = ( + ). ( + Z) Other Theorems sorption djcency onsensus DeMorgn Simpliiction. ( + ) =. +. = +. =. (1 + ) = +. =. (1 + ) =. +. = ( + )( + ) =. + Z + Z = + Z ( + )( + Z)( + Z) = ( + )( + Z) orollry: ( + )( + Z) = + Z. = +,.. Z = + + Z + + =., + + Z+... =.. Z. ( + ) =. + = + useul ppliction o the theorems is on the simpliiction o oolen unctions which leds to the reduction o the mount o logic s: = ( + + D + E)( + + D ) + E = ( + )( + ), = +, = D + E = ( + )( + ) = = + = ( )Z + + Z = Z + Z = Z( + ) = Z = + Z = ( + )( + ) = + + + = + ( + ) = + = = x 1 x 2 + x 1 x 2 + x 2 x 1 = x 1 x 2 + (x 1 2 + x ) 2 = x 1 x 2 + x 1 = x 1 + x 1 x 2 = (x 1 + x 1 )(x 1 + x 2 ) = x 1 + x 2 = ( + ) + = (. + ) = ( + ). + = ( ). + = 2 Instructor: Dniel Llmocc

DERIVING OOLEN UNTIONS ROM TRUTH TLES Using 1s: 0 0 1 0 1 1 1 1 = + + Using 0s: 0 0 1 0 1 1 1 1 = ( + + )( + + ) Other Logic Gtes NND NOR OR = + = NOR = + = Instructor: Dniel Llmocc

SUM O PRODUTS (SOP) ND PRODUT O SUMS (POS) USING MINTERMS ND MTERMS: MINTERMS nd MTERMS ( vrile unction) x 1 x 2 x Minterms Mxterms 0 m 0 = x 1 x 2 x M 0 = x 1 + x 2 + x 1 m 1 = x 1 x 2 M 1 = x 1 + x 2 + x 2 m 2 = x 1 x 2 x M 2 = x 1 + x 2 + x m = x 1 x 2 x M = x 1 + x 2 + x 4 m 4 = x 1 x 2 x M 4 = x 1 + x 2 + x 5 m 5 = x 1 x 2 M 5 = x 1 + x 2 + x 6 m 6 = x 1 x 2 x M 6 = x 1 + x 2 + x 7 m 7 = x 1 x 2 x M 7 = x 1 + x 2 + x unction with n vriles cn hve up to 2 n minterms (or 2 n mxterms) rom m 0 to m 2 n 1 (or rom M 0 to M 2 n 1) Note tht: m i = M i. lso, or n vriles, the totl numer o dierent unctions is 2 2n. unction cn e expressed s sum o minterms or s product o mxterms: When minterm evlutes to 1, it implies tht the unction evlutes to 1. Exmple: (x, y) = xy + x y. Here, (x, y) = 1 i xy = 00,11. Thus, the minterms re m 0 nd m. (x, y) = m 0 + m When mxterm evlutes to 0, it implies tht the unction evlutes to 0. Exmple: (x, y) = (x + y )(x + y). Here, (x, y) = 0 i xy = 10,01. Thus, the mxterms re M 1 nd M 2. (x, y) = M 1 M 2 sum o products (SOP) tht include only minterms or product o sums (POS) tht contin only mxterms re clled nonicl orms. I SOP includes terms tht re not minterms (or POS includes terms tht re not mxterms), they re clled noncnonicl orms. or exmple: (x 1, x 2, x ) = x 1 x 2 x + x 1 x 2 (x 1, x 2, x ) = (x 1 + x 2 + x )(x 1 + x ) 2 (x 1, x 2, x ) = x 1 x 2 x + x 1 x 2 + x 1 2 x + (x 1 + x 2 + x ) Exmple: Z Sum o Products 0 = Z + Z + Z + Z 0 (,, Z) = (m 1, m 4, m 5, m 6 ). 0 (,, Z) = m(1,4,5,6) lso: (,, Z) = m(0,2,,7) 0 1 Product o Sums 1 = ( + + Z)( + + Z)( + + Z )( + + Z ) 1 (,, Z) = (M 0, M 2, M, M 7 ). 0 (,, Z) = M(0,2,,7) lso: (,, Z) = M(1,4,5,6) Note how (,, Z) = m(1,4,5,6) = M(0,2,,7). TIMING DIGRMS G G 4 Instructor: Dniel Llmocc

ILIN PG IMPLEMENTTION - DESIGN LOW Design Entry: Here, the circuit is speciied vi Hrdwre Description Lnguge (HDL), Schemtic, or wveorm. The process o veriiction o the HDL syntx o schemtic connections is clled Synthesis. ehviorl Simultion: This is crucil step. our Design Entry might e 'error-ree' syntx-wise, ut it might not work s expected. Here, we provide time-vrying stimuli to the inputs o logic circuit nd veriy tht the outputs re correct. When the stimuli is written in HDL, it is clled 'test-ench'. This process is very similr to using signl genertor to crete the inputs, nd using scope to visulize the outputs over time. Physicl Mpping: Here we speciy which inputs nd outputs mp to the speciic components o the PG we selected nd the Printed ircuit ord (P) tht houses the PG. In ilinx Vivdo, this is done vi ile clled onstrints ile (.xdc) Timing Simultion: ehviorl Simultion only simultes the circuit 'logiclly', i.e., it does not tke into ccount nlog nd electricl eects. Timing simultion does consider the dely tht exist etween inputs nd outputs, nd thereore it is very useul to determine glitches, hzrds, etc. Implementtion: Here, we "progrm" the PG. In this step, we gr conigurtion ile (clled 'itstrem') nd then downlod it onto the PG conigurtion memory. PRTIE EERISES Simpliy the ollowing unctions: = Z + Z + Z + Z (,, Z) = (m 0, m 2, m 6 ) = ( + + Z)( + + Z ) = ( + + D)( + D) = ( + D ) + (,, Z) = (M, M 4, M 7 ) Using oolen lger Theorems, prove tht: The OR opertion is ssocitive: c = ( ) c = ( c) = ( c). ( c) = () (c) Provide the oolen unctions nd sketch the logic circuit. Use the two representtions: i) Sum o Products, ii) Product o Sums. lso, provide the minterms nd mxterms representtions. 1 2 4 5 6 7 1 0 0 1 1 1 0 0 1 1 1 0 1 0 1 1 0 0 1 1 1 1 1 Otin the logic unction (nd minimize i possile) o the ollowing circuits: Drw the timing digrm o the ollowing circuit: Design circuit tht veriies the logicl opertion o the OR. = '1' (LED ON) i the OR works properly. ssumption: when the OR is not working, it is generting 1's insted o 0's nd vice vers. Tip: irst, generte the truth tle. x? Security comintion: lock only opens when the 8 switches re set s in the igure. Get the unction tht opens the lock ( logicl '1' is generted) when the switches re conigured s in the igure. Ech switch represents oolen vrile. Here, n open lock is represented y n LED tht is ON. ON (1) O (0) 5 Instructor: Dniel Llmocc