Lecture #1 Page 1 ECE 4110– Digital SystemDesign

Lecture #1 Agenda 1. Course Logistics 2. Course Content 3. Digital Review Announcements 1. Welcome 2. Homework #1 assigned Lecture #1 Page 2

Course Overview Instructor: Omar Elkeelany Office : 332 Brown Hall Phone :(931) Web : Lecture #1 Page 3

Course Overview Textbook: “Digital Design: Principles and Practices", 4th Addition John F. Wakerly, Prentice Hall, 2006 Website: all handouts, homework assignments are ONLINE it is your responsibility to download assignments Lecture #1 Page 4

Course Overview Pre-requisites: ECE2110 / ECE3160 Grading: Homework/VHDL and Quizzes 20% Combo System project 20% Exam #1 20% Exam #2 20% Final Exam 20% - Homework Assignments are due at the beginning of class. - No Late homework will be accepted. - No make up exams will be given, unless pre-excuesd before the test. Plan on being available on the exam dates. - Term paper assignment will be given for 5110 graduate level. Lecture #1 Page 5

Course Overview Where does this course fit into the Electrical Engineering curriculum? Lecture #1 Page 6

Course Overview Where does this course fit into the Computer Engineering curriculum? Lecture #1 Page 7

Course Content What is this course? - In ECE2110 you learned: - basic combinational logic design - basic sequential logic design - In ECE3160 you learned: - how to implement logic circuits using off-the-shelf parts - ECE4110 is a follow-on course that looks at: - Large scale digital designs - Performance of digital circuitry - Programmable Logic Lecture #1 Page 8

Course Content What does "Large" mean? - Large means that you can't do it by hand. We need a way to design and simulate Millions of gates - K-maps for a Pentium would take too much paper Lecture #1 Page 9

Course Content We will learn VHDL in order to describe large digital designs - VHDL is a text based Hardware Description Language - We can simulate our digital designs created in VHDL Lecture #1 Page 10

Course Content We can also prototype our designs using an FPGA - FPGA = Field Programmable Gate Array - An FPGA is a programmable logic device - In this course, we will implement our designs and test them in FPGA hardware Lecture #1 Page 11

Course Content What topics will be covered? 1) VHDL (Exam #1 Topics) 2)Medium Scale Combinational Logic Devices 3)More Complex Finite State Machines (Exam #2 Topics) 4)Computer Systems 5)FPGA Timing and Implementation For the 5110 level, a special assignment is to: Write an original research paper on a topic related to those in this course, such as: Modern programmable logic devices, survey, features, comparisons, usage, etc.. Hardware description languages: survey, comparisons, usage, IP, etc. Sequential Logic Design methods Lecture #1 Page 12

Lecture #1 Page 13 Digital Review Combinational Logic Combinational Logic Gates : - Output depends on the logic value of the inputs - no storage

Lecture #1 Page 14 Digital Review NOTout = in’ = inf(in) = in’ = in OR out = a+bf(a,b) = a+b ANDout = a·bf(a,b) = a·b

Lecture #1 Page 15 Digital Review XORout = a  b f(a,b) = a  b NOR out = a+bf(a,b) = a+b NANDout = a·bf(a,b) = a·b

Lecture #1 Page 16 Digital Review XNORout = a  b f(a,b) = a  b Also remember about XOR Gates: f(a,b) = a  b = (a’b + b’a) Also remember the priority of logic operations (without parenthesis) is: NOT, AND, OR

Lecture #1 Page 17 Digital Review DeMorgan’s Theorems -Inverting the output of any gate results in the same function as the opposite gate (AND/OR) with inverted inputs

Lecture #1 Page 18 Digital Review DeMorgan’s Theorems -Graphically : breaking the bar changes the logic function (AND-OR) under the break out = a+b out = a·b 1) Break bar 2) Change + to · under break

Lecture #1 Page 19 Digital Review Boolean Expressions Using SOP -Logic functions can be described using a Sum of Products techniques -Sum of Products (SOP) is the summation of all minterms resulting in the truth table -A minterm is the expression for an input configuration which yields a TRUE output -A minterm expression is the AND’ing of the input "1" signal configuration SOP Expression : f(a,b) = a’·b + a·b’ Note : un-minimized Boolean expression Truth Table a bout minterm m 1 = a’·b 1 01minterm m 2 = a·b’ 1 10

Lecture #1 Page 20 Digital Review Boolean Expressions Using POS -Logic functions can be described using a Product of Sums techniques -Product of Sums (POS) is the multiplication of all maxterms resulting in the truth table -A maxterm is the expression for an input configuration which yields a FALSE output -A maxterm expression is the OR’ing of the input "0" signal configuration POS Expression : f(a,b) = (a+b) · (a'+b') Truth Table a bout 0 0 0maxterm m 0 = a+b (input configuration of 0's) maxterm m 3 = a'+b' (input configuration of 0's)

Lecture #1 Page 21 Digital Review Boolean Expressions Using SOP & POS -SOP and POS functions are equivalent SOP Expression : f(a,b) = a’·b + a·b’ is equal to POS Expression : f(a,b) = (a+b) · (a'+b')

Lecture #1 Page 22 Digital Review Karnaugh Maps -K-maps provide a graphical method to find SOP/POS expressions -K-maps also provide a graphical method to perform logic minimization K-map SOP Process 1) Circle minterms to create SOP 2) Circle in Horizontal & Vertical manner 3) Circle in groups with powers of 2 (1,2,4,8,…) Truth Table a bout a 0 1 b 0 1 No dependency on b,minterm = a No dependency on a,minterm = b SOP expression : f(a,b) = a + b

Lecture #1 Page 23 Digital Review Karnaugh Maps -K-maps provide a graphical method to find SOP/POS expressions -K-maps also provide a graphical method to perform logic minimization K-map POS Process 1) Circle maxterms to create SOP 2) Circle in Horizontal & Vertical manner 3) Circle in groups with powers of 2 (1,2,4,8,…) Truth Table a bout a 0 1 b 0 1 Dependency on a' and b',maxterm = a+b POS expression : f(a,b) = a + b

Lecture #1 Page 24 Digital Review Sequential Logic - Concept of “Storage Element” - With Storage, logic functions can depend on current & past values of inputs - Sequential State Machines can be created D-Flip-Flop - on timing event (i.e., edge of clock input), D input goes to Q output DQ Q D Q Q CLK t c2q

Lecture #1 Page 25 Digital Review State Machines - Moore :Outputs depend on present state - Mealy : Outputs depend on present state and current inputs

Lecture #1 Page 26 Digital Review State Machine Example : Design a 2-bit Gray Code Counter ) Number of States?: 4 2) Number of bits to encode states?: 2 n =4, n=2 3) Moore or Mealy?: Moore For this counter, we can make the outputs be the state codes

Lecture #1 Page 27 Digital Review State Machine Example : Design a 2-bit Gray Code Counter STATE CurrentNext A cur B cur A nxt B nxt B cur 0 1 A cur 0 1 A nxt Logic A nxt = B cur B cur 0 1 A cur 0 1 B nxt Logic B nxt = A cur ’ DQ Q DQ Q A B CLK A B counter output