Edge-Triggered D Flip-flop. Formal Analysis. Fundamental-Mode Sequential Circuits. D latch: How do flip-flops work?

Similar documents
MAT3707. Tutorial letter 201/1/2017 DISCRETE MATHEMATICS: COMBINATORICS. Semester 1. Department of Mathematical Sciences MAT3707/201/1/2017

Present state Next state Q + M N

Solutions to Homework 5

OpenMx Matrices and Operators

Minimum Spanning Trees

d e c b a d c b a d e c b a a c a d c c e b

Cycles and Simple Cycles. Paths and Simple Paths. Trees. Problem: There is No Completely Standard Terminology!

BASIC CAGE DETAILS D C SHOWN CLOSED TOP SPRING FINGERS INNER WALL TABS ARE COINED OVER BASE AND COVER FOR RIGIDITY

BASIC CAGE DETAILS SHOWN 3D MODEL: PSM ASY INNER WALL TABS ARE COINED OVER BASE AND COVER FOR RIGIDITY SPRING FINGERS CLOSED TOP

a g f 8 e 11 Also: Minimum degree, maximum degree, vertex of degree d 1 adjacent to vertex of degree d 2,...

Grade 7/8 Math Circles March 4/5, Graph Theory I- Solutions

ECE COMBINATIONAL BUILDING BLOCKS - INVEST 13 DECODERS AND ENCODERS

CS 103 BFS Alorithm. Mark Redekopp

, each of which is a tree, and whose roots r 1. , respectively, are children of r. Data Structures & File Management

16.unified Introduction to Computers and Programming. SOLUTIONS to Examination 4/30/04 9:05am - 10:00am

Steinberg s Conjecture is false

Priority Search Trees - Part I

VLSI Testing Assignment 2

N1.1 Homework Answers

Outline. Computer Science 331. Computation of Min-Cost Spanning Trees. Costs of Spanning Trees in Weighted Graphs

University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2014 Lecture 20: Transition State Theory. ERD: 25.14

CS September 2018

A L A BA M A L A W R E V IE W

(2) If we multiplied a row of B by λ, then the value is also multiplied by λ(here lambda could be 0). namely

Internet Algorithms. (Oblivious) Routing. Lecture 10 06/24/11. Wereferto. demands(requirements), forall vertex pairs,,

Exponential Functions

CS553 Lecture Register Allocation I 3

5/1/2018. Huffman Coding Trees. Huffman Coding Trees. Huffman Coding Trees. Huffman Coding Trees. Huffman Coding Trees. Huffman Coding Trees

1 Introduction to Modulo 7 Arithmetic

Seven-Segment Display Driver

24CKT POLARIZATION OPTIONS SHOWN BELOW ARE REPRESENTATIVE FOR 16 AND 20CKT

Design and Analysis of Algorithms (Autumn 2017)

Multipoint Alternate Marking method for passive and hybrid performance monitoring

CS150 Sp 98 R. Newton & K. Pister 1

EE1000 Project 4 Digital Volt Meter

Tangram Fractions Overview: Students will analyze standard and nonstandard

Functions and Graphs 1. (a) (b) (c) (f) (e) (d) 2. (a) (b) (c) (d)

Frequency Response & Digital Filters

Planar convex hulls (I)

The University of Sydney MATH 2009

AP Calculus Multiple-Choice Question Collection

ENGR 7181 LECTURE NOTES WEEK 5 Dr. Amir G. Aghdam Concordia University

learning objectives learn what graphs are in mathematical terms learn how to represent graphs in computers learn about typical graph algorithms

Math 166 Week in Review 2 Sections 1.1b, 1.2, 1.3, & 1.4

Case Study 1 PHA 5127 Fall 2006 Revised 9/19/06

Software Process Models there are many process model s in th e li t e ra t u re, s om e a r e prescriptions and some are descriptions you need to mode

Chapter Taylor Theorem Revisited

S i m p l i f y i n g A l g e b r a SIMPLIFYING ALGEBRA.

Having a glimpse of some of the possibilities for solutions of linear systems, we move to methods of finding these solutions. The basic idea we shall

I N A C O M P L E X W O R L D

TP A.31 The physics of squirt

ROSEMOUNT 3051SAM SCALABLE PRESSURE TRANSMITTER COPLANAR FLANGE PROCESS CONNECTION

Evans, Lipson, Wallace, Greenwood

ENGR 323 BHW 15 Van Bonn 1/7

Synthesis of asynchronous control circuits with automatically generated relative timing assumptions

Weighted Graphs. Weighted graphs may be either directed or undirected.

Outline. 1 Introduction. 2 Min-Cost Spanning Trees. 4 Example

Outlines: Graphs Part-4. Applications of Depth-First Search. Directed Acyclic Graph (DAG) Generic scheduling problem.

Designing A Uniformly Loaded Arch Or Cable

QUESTIONS BEGIN HERE!

12/3/12. Outline. Part 10. Graphs. Circuits. Euler paths/circuits. Euler s bridge problem (Bridges of Konigsberg Problem)

Online Supplement: Advance Selling in a Supply Chain under Uncertain Supply and Demand

CPS 616 W2017 MIDTERM SOLUTIONS 1

5/9/13. Part 10. Graphs. Outline. Circuits. Introduction Terminology Implementing Graphs

dy 1. If fx ( ) is continuous at x = 3, then 13. If y x ) for x 0, then f (g(x)) = g (f (x)) when x = a. ½ b. ½ c. 1 b. 4x a. 3 b. 3 c.

Graphs. Graphs. Graphs: Basic Terminology. Directed Graphs. Dr Papalaskari 1

Schematic of a mixed flow reactor (both advection and dispersion must be accounted for)

Complete Solutions for MATH 3012 Quiz 2, October 25, 2011, WTT

Physics 43 HW #9 Chapter 40 Key

CATAVASII LA NAȘTEREA DOMNULUI DUMNEZEU ȘI MÂNTUITORULUI NOSTRU, IISUS HRISTOS. CÂNTAREA I-A. Ήχος Πα. to os se e e na aș te e e slă ă ă vi i i i i

AP Calculus BC AP Exam Problems Chapters 1 3

Atomic Physics. Final Mon. May 12, 12:25-2:25, Ingraham B10 Get prepared for the Final!

y = 2xe x + x 2 e x at (0, 3). solution: Since y is implicitly related to x we have to use implicit differentiation: 3 6y = 0 y = 1 2 x ln(b) ln(b)

LR(0) Analysis. LR(0) Analysis

a b v a v b v c v = a d + bd +c d +ae r = p + a 0 s = r + b 0 4 ac + ad + bc + bd + e 5 = a + b = q 0 c + qc 0 + qc (a) s v (b)

In which direction do compass needles always align? Why?

Paths. Connectivity. Euler and Hamilton Paths. Planar graphs.

PHY 410. Final Examination, Spring May 4, 2009 (5:45-7:45 p.m.)

Problem Value Score Earned No/Wrong Rec -3 Total

( ) Differential Equations. Unit-7. Exact Differential Equations: M d x + N d y = 0. Verify the condition

3-2-1 ANN Architecture

ENGR 7181 LECTURE NOTES WEEK 3 Dr. Amir G. Aghdam Concordia University

Divided. diamonds. Mimic the look of facets in a bracelet that s deceptively deep RIGHT-ANGLE WEAVE. designed by Peggy Brinkman Matteliano

VTU NOTES QUESTION PAPERS NEWS RESULTS FORUMS

NOTE: ONLY RIGHT IDLER (CONFIGURATION A) ARM SHOWN IN VIEWS ON THIS PAGE

Chapter 3. Table of content Chapter 1: Switching Algebra Chapter 2: Logical Levels, Timing & Delays

Chapter 7 Conformance Checking

Multiple RS-232 Drivers & Receivers

Experiment # 3 Introduction to Digital Logic Simulation and Xilinx Schematic Editor

Final Exam Solutions

METHOD OF DIFFERENTIATION

1973 AP Calculus AB: Section I

is an appropriate single phase forced convection heat transfer coefficient (e.g. Weisman), and h

P a g e 5 1 of R e p o r t P B 4 / 0 9

Using the Printable Sticker Function. Using the Edit Screen. Computer. Tablet. ScanNCutCanvas

Trigonometric functions

- Prefixes 'mono', 'uni', 'bi' and 'du' - Why are there no asprins in the jungle? Because the parrots ate them all.

Outline. Heat Exchangers. Heat Exchangers. Compact Heat Exchangers. Compact Heat Exchangers II. Heat Exchangers April 18, ME 375 Heat Transfer 1

MA1506 Tutorial 2 Solutions. Question 1. (1a) 1 ) y x. e x. 1 exp (in general, Integrating factor is. ye dx. So ) (1b) e e. e c.

Note If the candidate believes that e x = 0 solves to x = 0 or gives an extra solution of x = 0, then withhold the final accuracy mark.

Transcription:

E-Trir D Flip-Flop Funamntal-Mo Squntial Ciruits PR A How o lip-lops work? How to analys aviour o lip-lops? R How to sin unamntal-mo iruits? Funamntal mo rstrition - only on input an an at a tim; iruit must stal or nxt input an an. Clk D B S How os it work? How was it sin? E-Trir D Flip-lop Intuitiv xplanation: lok=0 => R=S=1, CL=PR=1, an - ar l. itr D=1 => A=1, B=0 or D=0 => A=0, B=1 D an an ut an - ar l. Lt lok an 0->1 Eitr B=1, A=0 => R=1, S=0 => =0, -=1 or B=0, A=1 => R=0, S=1 => =1, -=0 Wat i D ans wil lok=1? a) D was 0 an ans to 1 A=0, B=1 => R=1, S=0 as S=0, an in D os not an B. ) D was 1 an ans to 0 A=1, B=0 => R=0, S=1 B ans rom 0 to 1, ut A os not an. Fallin lok as no t. Funamntal mo rstrition applis. D E Formal Analysis Y D lat: + Y To analys tis iruit, rak t ak loop. It is onvnint to prtn tat all t ats av 0 lays an tat tr is a init ut small lay in t ak loop.

Analysis o D lat Y + an tout o as t nxt stat. Y + = D.E + Y.(D + E) Transition Tal Y 0 1 DE 00 0 0 1 0 1 0 1 1 Y + Stat Tal S 00 DE S0 S0 S0 S1 S0 S1 S1 S0 S1 S1 S* Stal stats - nxt stat is t sam as prsnt stat (irl). Stat & Output Tal Stat Tal DE S 00 S0 S0, S0, S1, S0, S1 S1, S0, S1, S1, S*, = D.E + Y.E + Y.D = D.E + Y S 00 DE S0 S0 S0 S1 S0 S1 S1 S0 S1 S1 D ans to 1 E ans to 1 S* Can tra wat appns wn inputs an rom DE = 00, S= S0. D os to 1 ollow y Eto1.

Stat Tal PR E-Trir D Flip-Flop S 00 DE Y1 Y1 + S0 S0 S0 S1 S0 S1 S1 S0 S1 S1 Clk Y3 Y3 + S* Wat appns i D&E now an almost simultanously? Exat squn is unprital. Final rsult is unprital. D Y2 Y2 + Multipl Fak Loops Assum PR an CL ar at 1 - n inor. + Y1 = Y2.D + Y1.Clk + Y2 = Y2.D + Y1.Clk + Clk + Y3 = Y1.Clk + Y3.(Y2.D + Y1.Clk + Clk) = Y1.Clk + Y3.(Y2.D + Y1.Clk + Clk) = Y3 + Y1. Y2.Clk + Y1.Clk. D Transition Tal Y1 Y2 Y3 00 ClkD 0 0 0 0 0 000 000 0 0 1 1 1 Y2 000 tn 000 0 1 0 0 1 1 Y3 000 0 1 1 1 1 1 000 1 0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Y3 tn Y2 Y1* Y2* Y3*

Ras Suppos startin stat is Y1,Y2,Y3 =1, Clk,D=00 an Clk ans to 1 Exat orr o stat ans pns on t orr o an o intrnal varials Ra - Multipl intrnal varials an stat as a rsult o ONE input anin stat I inal stat os not pn on orr o intrnal stat ans: non-ritial ra Suppos t transition tal look lik: Transition Tal Y1 Y2 Y3 00 ClkD 0 0 0 0 0 000 000 0 0 1 1 1 Y2 000 tn 000 0 1 0 0 1 1 Y3 1 0 1 1 1 1 1 000 1 0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Y1* Y2* Y3* Critial Ra to avoi! Stat & Output Tal S 00 ClkD S0 S2, S2, S0, S0, S1 S3, S3, S0, S0, S2 S2, S6, S6, S0, S3 S3, S7, S7, S0, S4 S2, S2, S7, S7, S5 S3, S3, S7, S7, S6 S2, S6, S7, S7, S7 S3, S7, S7, S7, S1,S4,S5 ar unstal S2,, S3,, S6, annot S*, - ra Flow Tal S 00 ClkD S0 S2, S6, S0, S0, S2 S2, S6, -,- S0, S3 S3, S7, -,- S0, S6 S2, S6, S7, -,- S7 S3, S7, S7, S7, S*, - Eliminat unstal stats, multipl ops an unraal stats

Flow Tal S 00 ClkD SM SM, S6, SM, SM, S3 S3, S7, -,- SM, S6 SM, S6, S7, -,- S7 S3, S7, S7, S7, S*, - S0 an S2 ar ompatil an may mr 1. 2. 3. 4. 5. 6. Funamntal Mo Dsin Stat Dsin Spiiations Driv a primitiv low tal Ru t low tal Mak a ra-r stat assinmnt Otain t transition tal an output map Otain azar-r stat quations Stat Minimisation Two stats an onsir quivalnt i tir outputs an nxt stats ar t sam (or all ominations o inputs). i.. ty ar inistinuisal rom outsi. Stats&karquivalnt (t two rows ar t sam). Rpla all instans o k y (only applis to row j). Nowstats&jarquivalnt. Stat Minimisation Prsnt stat s 00 xy a,0,0 a,0 a,0,0,0 a,0 a,0,0 i,0 a,0,1,1,0,0,1,1 i,0 a,0,1,0,0 a,0 a,0,1 j,0 a,0,1,0,0,0 a,0 i i,0 i,0 a,0 a,0 j,0 k,0 a,0 a,0 k,0 i,0 a,0,1 s*,z

Stat Minimisation Prsnt stat s 00 xy a,0,0 a,0 a,0,0,0 a,0 a,0,0 i,0 a,0,1,1,0,0,1,1 i,0 a,0,1,0,0 a,0 a,0,1,0 a,0,1,0,0,0 a,0 i i,0 i,0 a,0 a,0 Stat Minimisation Not tat tis is an inormal approa. W av ru t numr o stats rom to 9. W an ru t numr o stats to 6. Tis an on y onstrutin quivaln lasss. Can also on usin an impliation tal. Impliation tal an also us or inompltly spii systms. Impliation tal ontains all possil ominations o stats: s*,z i - - - -a -i -i Impliation Tal -i i- - a I outputs ar irnt, put X. I outputs ar sam, list nxt stats tat must quivalnt. (I nxt stats ar t sam as prsnt stats, put a tik.) i - - - -a -i -i Impliation Tal -i a Now o rit to lt pass trou t tal, puttin in Xs wr impliations av alray n isount...&i annot quivalnt as -i as an X. 1st pass sown. i- -

i - - - -a -i -i Impliation Tal -i 2n pass. i- - a Impliation Tal Tr ar now tr squars lt unross. In tis as, w an u tat stats (,i), (a,), (,) ar quivalnt. Tus t stats n to implmnt t systm ar (,i), (a,), (,), (), (), (). Can av mor omplx ass: i - - - -i -i - a- a- - - -i -i - a- a- a- Impliation Tal - -i - -i - a- - a- a- Atr all passs. - - - - - - - a Equivaln Classs Start at Rit o impliation tal. In olumn, t squar (,) is not ross out so w list t pair (,). In olumn, w in (,i). In olumn, w in ot (,i) an (,), so w an a to (,i). Similarly, w an a to (,) an a to (,,i). i - - - (,) (,i) (,) (,i) (,) (,,i) (,) (,,i) (,,) a (a,,,i) (,,) Equivaln Classs: (a,,,i) (,,) () ()

Inompltly Spii Systm Suppos tat w o not ar wat t output is in rtain stats. Furtr, rtain nxt stat transitions may not in. Tis an xploit to minimis stats. W on't ar aout intrnal stats, only aout outputs! W now talk aout ompatil stats. Two stats ar ompatil i (an only i) tir outputs, i spii, ar t sam atr t sam squn o inputs... onsir a systm wit 4 inputs an 2 outputs spii y t ollowin stat an output tal. (N.B. w av not onsir all ominations o inputs - only on input is tru at any tim.) Inompltly spii systm s Input K L M N a a,- a,- a,-,00,-,-,- -,- -,-,-,- -,-,- -,-,- -,-,-,- -,- -,- -,- -,-,- a,,- -,- -,- a,,-,-,- a,00 s*,yz a- a- a- a- a- a- - - - - - - - - - - - Impliation Tal - - - - - - - a Dtrmination o maximum ompatils Sam pross as or inin quivaln lasss. Start at rit o tal an list ompatils. - - (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) (,) a (,) (,) (,) (,) (,) (,) (,) (a) Not tat, unlik a ully spii systm, (,) ar ompatil, (,) ar ompatil, ut (,) ar not ompatil. Tus w av two ompatil pairs, insta o on lass.

Maximal Compatils Not tat w now av 8 ompatil sts. W start wit 8 stats, so tis osn't appar to av ain us anytin! But w o not n all t sts. W ar tryin to in ompatil stats, tus o not n to us a st. W n t ollowin sts: (a) (,) ut all t otr stats our in two sts. Tis is analoous to runany in K-maps. Tror, w slt suiint sts to ovr all t stats.. (a) (,) (,) (,) (,) or (a) (,) (,) (,) (,) Stat Assinmnt m stats an r stat varials ivs 2r! /(2 r - m)! possil assinmnts. No way o trminin wi is st. Can assin to mak stat varials maninul. Can assin to minimis numr o its anin twn stats. Not rom t impliation tal, tat (,) implis (,) an (,) implis (,). Stat Dsin Exampl Nativ -trir T lip-lop Inputs T C Output Commnts a 1 1 0 Initial stat 1 0 1 Atr a 1 1 1 Initial output 1 0 0 Atr 0 0 0 Atror 0 1 0 Atraor 00 Nativ -trir T Flip-lop /0 a/0 /1 /1 00 00 00 00 /0 /0 /1 /1 00 0 0 1 Atror 0 1 1 Atror

Primitiv Flow Tal Stat 00 TC Output a - a 0-1 - 1 - a 0-0 a - 0-1 - 1 a- Impliation Tal a Compatil stats From Impliation tal, ompatil stats ar: (a,) (,) (,) (,) (,) (,) (,) (,) Ts an mr - on tniqu is: a Ru Flow Tal Stat 00 TC Output (a,) A D A A B 0 (,,) B B B C B 1 (,) C B C C D 1 (,,) D D D A D 0

Stat Assinmnt Ru Flow Tal A=00 B= Y1 Y2 00 TC Output 00 00 00 0 1 D= C= 1 00 0 Y1* Y2* -v -trir T lip-lop K-maps (Entris orrsponin to stal stats in rn): TC Y1 Y2 00 00 1 0 0 0 TC Y1 Y2 00 Y2*: 00 0 0 1 1 0 1 1 1 1 1 1 0 Y1*: 0 0 1 0 0 1 1 1 1 1 0 1 Y1* T.C.Y2 C.Y1.Y2 T.Y1.Y2 C.Y1.Y2 T.C.Y2 T.Y1.Y2 T.C.Y1 Tr ar 8 prout trms in Y1*. Not tat a 1 to 0 an in T or C an aus a azar, vn twn stal stats. Tror all prout trms ar n, xpt T.C.Y1 aus tat ovrs two stal stats, wit t sam inputs. Altou Y2 is 1 in on stat an 0 in t otr, t transition an t appn. 0 0 0 0 Y2* T.Y2 C.Y2 Y1.Y2 T.C.Y1 W n n Y1.Y2 or Y2* aus it ovrs a transition twn stal stats wn T ans rom 1 to 0. T rul is: inlu all prout trms xpt tos tat ovr stal pairs in t sam olumn (i.. wit t sam inputs) an tos wr a transition rom a stal stat to an unstal stat (in t sam row) an only our as a rsult o a 0 to 1 transition.