Upload File

176 string recognizer example n recognize the string: 1011 n input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n...

  • Home
  • Documents
  • 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:
30
String Recognizer Example Recognize the string: 1011 Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 Output: State Machine:

Upload: lee-bennett

Post on 24-Dec-2015

223 views

Category:

Documents


3 download

Report

Tags:

TRANSCRIPT

Page 1: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

1

String Recognizer Example

Recognize the string: 1011

Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 Output:

State Machine:

Page 2: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

2

String Recognizer Example (cont.)

State Table:

Page 3: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

3

String Recognizer Example (cont.)

Implementation:

Page 4: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

4

FSMs & Flip Flops

Sometimes may be more efficient to implement FSM with other Flip Flops that Dff

Approach: Given state table & Flip Flop Excitation table,

determine how to use Flip Flop for this circuit

Present State Input Next State J K Output0 0 0 00 1 1 11 0 1 11 1 0 0

J-K Excitation Table:

Page 5: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

5

Parity with J-K (cont.)

Implementation:

Page 6: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

6

Parity with T

T Excitation Table:

Present State Input Next State T Output0 0 0 00 1 1 11 0 1 11 1 0 0

Page 7: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

7

Vending Machine with J-K

J-K Excitation Table:

Page 8: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

8

FSM Word Problem: Traffic Light ControllerA busy highway is intersected by a little used farmroad. DetectorsC sense the presence of cars waiting on the farmroad. With no caron farmroad, light remain green in highway direction. If vehicle on farmroad, highway lights go from Green to Yellow to Red, allowing the farmroad lights to become green. These stay green only as long as a farmroad car is detected but never longer than a set interval. When these are met, farm lights transition from Green to Yellow to Red, allowing highway to return to green. Even if farmroad vehicles are waiting, highway gets at least a set interval as green.

Assume you have an interval timer that generates a short time pulse(TS) and a long time pulse (TL) in response to a set (ST) signal. TSis to be used for timing yellow lights and TL for green lights.

Page 9: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

9

Traffic Light Controller (cont.)Picture of Highway/Farmroad Intersection:

Highway

Highway

Farmroad

Farmroad

HL

HL

FL

FL

C

C

Page 10: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

10

Traffic Light Controller (cont.)

• Tabulation of Inputs and Outputs:

Input SignalresetCTSTL

Output SignalHG, HY, HRFG, FY, FRST

Descriptionplace FSM in initial statedetect vehicle on farmroadshort time interval expiredlong time interval expired

Descriptionassert green/yellow/red highway lightsassert green/yellow/red farmroad lightsstart timing a short or long interval

• Tabulation of Unique States: Some light configuration imply others

StateS0S1S2S3

DescriptionHighway green (farmroad red)Highway yellow (farmroad red)Farmroad green (highway red)Farmroad yellow (highway red)

Page 11: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

11

Traffic Light Controller (cont.)

State Diagram

Page 12: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

12

Traffic Light Controller (cont.)

State Table using J-K Flip Flops

Page 13: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

13

Registers

Storage unit. Can hold an n-bit value Composed of a group of n flip-flops

Each flip-flop stores 1 bit of information

Normally use D flip-flopsD Q

Dffclk

D QDffclk

D QDffclk

D QDffclk

Page 14: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

14

Controlled RegisterReset Load Action0 0 Q = old Q1 0 Q = 00 1 Q = D

D QDffclk

D QDffclk

D QDffclk

D QDffclk

Page 15: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

15

Shift Register

Register that shifts the binary values in one or both directions

D QDffclk

D QDffclk

D QDffclk

D QDffclk

In Out

Clock

Page 16: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

16

Transfer of Data

2 modes of communication: Parallel vs. Serial Parallel: all bits transferred at the same time Serial: one bit transferred at a time

Shift register can be used for serial transfer

D QDffclk

D QDffclk

Page 17: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

17

Shift Register w/Parallel LoadShift Load Action0 0 Q = old Q1 0 ShiftX 1 Parallel Load

D QDffclk

D QDffclk

D QDffclk

D QDffclk

Page 18: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

18

Bidirectional Shift Register w/Parallel LoadShift Load Action0 0 Q = old Q0 1 Parallel Load1 0 Shift Up (V*2)1 1 Shift Down (V/2)

D QDffclk

D QDffclk

D QDffclk

D QDffclk

Page 19: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

19

Conversion between Parallel & Serial

D3 Q34-bit

D2 Q2Shift

D1 Q1Reg.

D0 Q0Shift Load Clk

D3 Q34-bit

D2 Q2Shift

D1 Q1Reg.

D0 Q0Shift Load Clk

LSI LSI

Page 20: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

20

Counters

A reg. that goes through a specific state sequence n-bit Binary Counter: counts from 0 to 2N-1 in binary Up Counter: Binary value increases by 1 Down Counter: Binary value decreases by 1 3-bit binary up counter state diagram:

Page 21: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

21

Binary Up-Counter Imp. (D Flip-Flop)

Page 22: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

22

Binary Up-Counter Imp. (T Flip-Flop)

Page 23: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

23

Complex Binary CounterLoad Count Action0 0 Q = old Q0 1 Up Count1 0 Reset1 1 Load Parallel

Page 24: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

24

Arbitrary Sequence Counters

Design a 3-bit count that goes through the sequence 000->010->100->101->111->110->001->011->000->...

Page 25: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

25

Memory

Need method for storing large amounts of data Computer programs, data, pictures, etc.

RAM: Random Access Memory, Read/Write ROM: Read-only Memory

Address Data000000 00111110000001 01101011000010 01011101000011 01100011000100 00111110000101 00000000000110 11111111000111 01010101001000 10101010001001 00100001001010 11011010

64x8 RAM

A3 D7A2 D6A1 D5A0 D4 D3 D2 D1Write D0

Page 26: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

26

8x4 RAM

000

001

010

011

100

101

110

111

3:8Decoder

Enable

S2 S1 S0

A2A1A0

In3 In2 In1 In0

Out3 Out2 Out1 Out0

Write

Page 27: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

27

RAM Cell

Requirements: Store one bit of data Change data based on input when row is selected

S QR

Input

Row Select

Page 28: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

28

RAM Expansion

Implement a big RAM from multiple small RAMSD15 D14 D13 D12 D11 D10 D9 D8 D7 D6 D5 D4 D3 D2 D1 D0Address

0000000

0010000

0100000

0110000

1000000

1010000

1100000

1110000

Page 29: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

29

RAM Expansion (cont)

Build a 16x16 RAM from 16x4 RAMs

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

Page 30: 176 String Recognizer Example n Recognize the string: 1011 n Input: 1 0 0 1 0 1 1 0 1 1 0 0 1 0 n Output: n State Machine:

30

RAM Expansion (cont)

Build a 32x16 RAM from 16x4 RAMs

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write

16x4 RAMA3 DinA2 A1 DoutA0

Write


1 1 . . 0 0 n oo n i i t t c c e SS e s s n oo n i i t t u ......1 1 . . 0 0 n oo n i i t t c c e SS e s s n oo n i i t t u u l SS oo l 1 1 . . 0 0 n oo n i i t t c c e SS e 1. 2(4

Martin Ric - University of Cambridge · JK Flip Flop jkff(0, J, K, Q, Q). jkff(1, 0, Q, Q). jkff(1, 0, 1, Q, 0). jkff(1, 1, 0, Q, 1). jkff(1, 1, 0, 1). jkff(1, 1, 0). list(N, N, [N])

2##*.0-n*#1 - ZOLLERN · 2##*.0-n*#1 Optimal forming precision ,'21+-"#0,AusAsss+#20#R1/3 0# .0-"3!2'-,$ !'*'27A + )#1 &'%&R/3 *'27A#!-,-+'!12##*.0-n*#1$-0

The Original Gold Standard Framing Coil - AngioCalc · 2020. 10. 4. · 1 1. 2 5 i n 1 . 1. 2 5 i n. 8. 8 1 2 5 i n. 0. 2 1 8 7 5 i n. 0. 7 5 i n. 0. 2 1 8 7 5 i n. 0. 7 5 i n. 0

Royal Society of Chemistry · 2016. 6. 24. · C l N N N S N * 8 0 0 k ( s S i m u l a t i o n E x p e r i m e n t a l - 1) 5 2 0 4 7 0 4 0 0 1 9 0 1 6 5 2 7 T ( ° C ) 2 1 2 0 1

TO - Defense Technical Information Center · in Vc 0 0n 0 n 0 n 0 0 H H4 H ... (n N' i--4 0D 0% CO QC r- 0o No 0% %0 MO " Ot t- %tf 0% "1 0 1- M f MN0 J 0 CO in% If"Ml 0 00 0 0%0

avivre/Algorithms.docx · Web viewLCS (Longest Common Subsequence) abcddeb. bdabaed. Opt n,m = A n = B m 1+Opt n-1,m-1 A n ≠ B m max Opt n,m-1 Opt n-1,m . d 0 b 0 a 0 c 0 1 2 b

IF Malmö Redhawks - Växjö Lakers HC · 63 VOLEJNICEK Dominik 1 0 0 0 0 0 0 - 0 N/A 71 EKLIND Oscar 0 0 2 2 1 0 0 - 1 0,00 88 SYLVEGÅRD Marcus 2 1 2 3 1 4 0 - 0 N/A 90 OLSSON Oliver

- 5- r.. / 5. · h 1 0 h n c n1 r n b. or y vir u 0 u tion 0 t by etio 0 h or n1 tion. 11 h i n c 1 to P h t n 0 t oc s n or c 0 ho on hieh t cc r 1 n , 1 t 1n 0 co c n • I 1 01

Mathematical Analysismajhu/Math203/book.pdfn!1(p n+ 1 n) = 0. 4. lim n!1 1 n2=3 n1=2 = 0. 5. lim n!1 cosn n = 0. 6. lim n!1 p n cosn p n+ sinn = 1. Exercise 1.1.2. Let a positive real

Biyomane - Somalia Water Catchment - 2016...2 ' 4 0 " N 6 2 ' 4 0 " N 6 2 ' 2 0 " N 6 2 ' 2 0 " N 6 2 ' 0 " N 6 2 ' 0 " N 6 1 ' 4 0 " N 6 1 ' 4 0 " N Cadaado Gidhays Qaradhi BIYOMANE

Glasgow 1970 - Speedway Researcher · 3. Clive Hitch 2 2 2 1 7 0 4. Pete Munday 3 2 0 1 6 0 5. Bob Kilby N EF 3 0 3 0 6. John Bishop 2* 3 N 3 N 8 1 7. Barry Duke 1 Exc 1* 0 N 1* 3

The Alpha Investor · 0 1 -Ja n -2 0 0 1 -F e b -2 0 0 1 -M a r-2 0 0 1 -A p r-2 0 0 1 -M a y -2 0 0 1 -Ju n -2 0 0 1 -Ju l-2 0 Bh a rt o n dI e x ( A p il 2 03 ) h ar t B o n d 2

BCRI - Rivers Inlet MouthBCRI 127 48'0"W 127 42'0"W 127 36'0"W 127 30'0"W 127 24'0"W 5 1 3 6 ' 0 " N 5 1 3 3 ' 0 " N 5 1 3 0 ' 0 " N 5 1 2 7 ' 0 " N 5 1 2 4 ' 0 " N BCRI - Rivers Inlet

 · 1 O 0 0 0 0 0 o N o E . O 0 N o o o o O N o o N . CD

VC areas are - Maricopa County Parks & Recreation | Maricopa … · 2021. 1. 21. · n y o n C a n yo n B l a c k (1. 1) 0. 7 2 4 0 0 2836 2 0 0 1 6 0 0 2 0 0 0 6 0 0 2658 2 8 0 0

3 -1 Chapter 3 Recursion. 3 -2 Iterative algorithm for n factorial n factorial n! = 1 if n = 0 n! = n*(n-1)*(n-2)*...*1 if n>0 Iterative algorithm prod

General Disclaimer One or more of the Following Statements ... · 1 0 0 cos n tU " (tul tu2 tu3 tu4) ' 0 1 0 cos n (1) 0 0 1 cos n The right-moat equality is true only for the ortho-skew

Coop Norrbotten Arena · 2017. 5. 15. · 51 ZANETTI Marc-Anthony 2 0 0 0 2 0 0 - 0 N/A 64 PUDAS Jonathan 2 0 0 0 -1 0 0 - 0 N/A 79 MANNBERG Daniel 2 1 0 1 1 0 0 - 0 N/A 96 LINDBLOM

n 3 n 2 n 0, 1, 2, 3, n ::: n 1, n 2,n 3,

N O … · 2 - . 1 0 / " % $ , 6 * 5 ( 4 0 3 , 3 0 + 0 0 0 0 0 0 + 0 + + 0 0 + + + 0 + + + + + + + + 0 "+ + 0 0 0 0+ 1 1 1 1 1 1 1 1 1

PIERGIORGIO BERUTO ANTONIO ORZELLI - IEEE 802 0 In a nutshell LINE BEACON IDLE DATA 0 SILENCE 0 1 … N 0 0 1 2 3 3 MIN BUS cycle MIN Time Slot 0 0 1 1 … … N N 0 MAX BUS cycle

E < 2 7 0 . / 0 > 3 1 O < . . < Y 1 N · < = 0 c W 1 0 O O < N > 7 M 0 3 1 0 2 0 7 Y 0 . < 0 2 1 N 0 2 2 7 N 0 3 1 0 b < 0 O 1 > > < 0 c @ ? `

M1 · M N 1 0 3 M N 1 0 2 M N 1 0 4 M N 1 0 2 M N 1 0 1 I III ´ 0 50 100 200 300 400 500 Metres EIS B 05 Title: Gníomhaireacht Um Fháil Iarnród P a rkg t eBu si nC , Parkgate

 · /0). //0 ; * h ;-$%i 3 " 11) , 1 * -$% ? < $ * n

34 N 27 0 13 N 51 0 1/ 1 cm:5N 2/ 1 cm:2N. 1 cm:4 N 20 0 6.5 cm 26N 10 N 1 cm:10 N 4.2 cm 42N 42N 46 0 1 cm:8 N 1 cm:3 N F x = 40 N F y = 0 F x = 0 F

Theory and conceptual design of stripline kickers Program/student... · 1 mng mg n A mn b n= 2 n sinn 0 and A nn= 0 + 1 2n sin2n 0 A mn= sin(m n) 0 m n + sin(m+ n) 0 m+ n This matrix

N%.'/$0%('1%21'345#%6$0789%:$/$1,% … · 2020-03-11 · l%.%%$'/$%0%'1""*&' +k=^=n%

IEEE TRANSACTIONS ON MAGNETICS, VOL., NO., MONTH ? 1 … · 2012. 7. 9. · 2 4 1 1+˜N 1 0 0 0 1 1+˜N 2 0 0 0 1 1+˜N 3 3 5(˜H a+ H r) (1) where all vectors are expressed relative

Fremont County Snow Load Zones...700 n 800 n 900 n 600 n 3 8 0 0 4 e3 9 0 0 2 e 110 0 n 3 6 0 0 0 e 4 1 0 0 6 e 3 5 0 0 e 4 0 0 0 1 e 3 7 0 e 1000 n 1200 n 3 0 0 e 4 2 0 e 4 3 4 4

Notas de Aula · n], 0 ≤ n ≤ M = 2 n d = N − 1 0, ... 1, 0 ≤ n ≤ M = N − 1 0, caso contrario Filtros Digitais tipo FIR 4 Resposta impulsiva - ... somat´oria do meio,

MassaM Martana Todi - abtevere.it · Todi MassaM Martana Acquasparta N A _ 0 0 3 5 N A _ 0 0 4 4 N A _ 0 0 5 6 N _ 0 0 5 5 N A _ 0 0 3 3 N A _ 0 0 4 3 N A _ 0 0 4 5 N A _ 0 0 4 1

Morbidity and Mortality Weekly Report - Centers …231 — 1 6 63 48 Missouri N 0 0 N N — 535 623 18,409 19,542 — 0 1 6 14 Nebraska — 0 0 — 1 86 149 253 5,442

• v = (v 0 , v 1 , ..., v n-1 ) : code vector v (1) = (v n-1 , v 0 , v 1 , …, v n-2 )

Mineral Properties Av a ilb e n British Columbia 7 1 5 K 1 1 1 6 0 1 1 2 0 W O 60O N 1 3 0 W O 1 4 0 O W 60 O N 55 1O N 55O N 50 O N HIGHLAND VALLEY COPPER 71 65 50 182 1 96 90 8 0