數(shù)字邏輯設(shè)計及應(yīng)用課程講稿

1、第一次課課：課程介紹紹及要求求 一學(xué)時時課程教學(xué)學(xué)內(nèi)容安安排：第一章引引論第二章數(shù)數(shù)系與代代碼第三章數(shù)數(shù)字電路路第四章組組合邏輯輯設(shè)計原原理第五章組組合邏輯輯設(shè)計實實踐第七章時時序邏輯輯設(shè)計原原理第八章時時序邏輯輯設(shè)計實實踐第十章存存儲器及及其在數(shù)數(shù)字邏輯輯系統(tǒng)實實現(xiàn)中的的運用第十一章章其他的的實際問問題補充內(nèi)容容模數(shù)轉(zhuǎn)轉(zhuǎn)換器、數(shù)模轉(zhuǎn)轉(zhuǎn)換器(ADCC/DAAC)原原理及應(yīng)應(yīng)用簡介介課程教學(xué)學(xué)時間安安排：第一章引引論（計劃學(xué)學(xué)時數(shù)：2學(xué)時）介紹數(shù)字字邏輯電電路的特特點、數(shù)數(shù)字邏輯輯電路在在電子系系統(tǒng)設(shè)計計中的地地位、數(shù)數(shù)字邏輯輯電路與與模擬電電子電路路之間的的關(guān)系、簡單介介紹EDDA設(shè)計計工具、

2、VHDDL語言言對數(shù)字字邏輯設(shè)設(shè)計作用用和影響響。第二章數(shù)數(shù)系與代代碼（計劃學(xué)學(xué)時數(shù)：6學(xué)時）十進制、二進制制、八進進制和十十六進制制數(shù)的表表示方法法以及它它們之間間的相互互轉(zhuǎn)換、非十進進制數(shù)的的加減運運算；符號數(shù)的的表達格格式以及及它們之之間的相相互轉(zhuǎn)換換以及帶帶符號數(shù)數(shù)的補碼碼的加減減運算；BCD碼碼、格雷雷碼的特特點，它它們與二二進制數(shù)數(shù)之間的的轉(zhuǎn)換關(guān)關(guān)系；簡簡介二進進制數(shù)的的浮點數(shù)數(shù)表達（補充）；第三章數(shù)數(shù)字電路路（計劃劃學(xué)時數(shù)數(shù)：4學(xué)學(xué)時）作為電子子開關(guān)運運用的二二極管、雙極型型晶體管管、MOOS場效效應(yīng)管的的工作方方式；以以CMOOS倒相相器電路路的構(gòu)成成及工作作狀態(tài)分分析；邏輯電

3、路路的靜態(tài)態(tài)、動態(tài)態(tài)特性分分析，等等價的輸輸入、輸輸出模型型；特殊的輸輸入輸出出電路結(jié)結(jié)構(gòu)：CCMOSS傳輸門門、施密密特觸發(fā)發(fā)器輸入入結(jié)構(gòu)、三態(tài)輸輸出結(jié)構(gòu)構(gòu)、漏極極開路輸輸出結(jié)構(gòu)構(gòu)；學(xué)習習了解其其他類型型的邏輯輯電路： TTTL,EECL等等；不同類型型、不同同工作電電壓的邏邏輯電路路的輸入入輸出邏邏輯電平平規(guī)范值值以及它它們之間間的連接接配合的的問題。第四章組組合邏輯輯設(shè)計（計劃學(xué)學(xué)時數(shù)：10學(xué)時時）邏輯代數(shù)數(shù)的公理理、定理理，對偶偶關(guān)系，以及在在邏輯代代數(shù)化簡簡時的作作用；邏輯函數(shù)數(shù)的表達達形式：積之和和與和之之積標準準型、真真值表；組合電電路的分分析：邏邏輯函數(shù)數(shù)表達式式的產(chǎn)生生過程及

4、及邏輯函函數(shù)表達達式的基基本化簡簡方法；組合電路路的綜合合過程：將功能能敘述表表達為組組合邏輯輯函數(shù)的的表達形形式、邏邏輯函數(shù)數(shù)表達式式的化簡簡函數(shù)化化簡方法法卡諾圖化化簡方法法、使用用與非門門、或非非門表達達的邏輯輯函數(shù)表表達式、邏輯函函數(shù)的最最簡表達達形式及及綜合設(shè)設(shè)計的其其他問題題：無關(guān)項的的處理、冒險問問題和多多輸出邏邏輯化簡簡的方法法。第五章組組合邏輯輯設(shè)計實實踐（計劃學(xué)學(xué)時數(shù)：10學(xué)時時）利用基本本的邏輯輯門完成成規(guī)定的的組合邏邏輯功能能（如譯譯碼器、編碼器器、多路路選擇器器、多路路分配器器、異或或門、比比較器、全加器器等）的的電路設(shè)設(shè)計任務(wù)務(wù)；利用基本本的邏輯輯門和已已有的邏邏輯

5、功能能電路作作為設(shè)計計的基本本元素完完成更為為復(fù)雜的的組合邏邏輯電路路設(shè)計的的方法。第七章時時序邏輯輯設(shè)計原原理（計計劃學(xué)時時數(shù)：110學(xué)時時）基本時序序元件RR-S型型,D型,J-K型,T型鎖鎖存器、觸發(fā)器器的電路路結(jié)構(gòu)，工作原原理，時時序特性性等；掃掃描觸發(fā)發(fā)器(SScann Fllip-Floop)特性性及基本本應(yīng)用；鐘控同步步狀態(tài)機機的模型型圖，狀狀態(tài)機類類型及基基本分析析方法和和步驟，使用狀狀態(tài)圖表表示狀態(tài)態(tài)機狀態(tài)態(tài)轉(zhuǎn)換關(guān)關(guān)系；時序狀態(tài)態(tài)機的設(shè)設(shè)計：狀狀態(tài)轉(zhuǎn)換換過程的的建立，狀態(tài)的的化簡與與編碼賦賦值、未未用狀態(tài)態(tài)的處理理-最小風風險方案案和最小小代價方方案、使使用狀態(tài)態(tài)轉(zhuǎn)換表表的設(shè)

6、計計方法、使用狀狀態(tài)圖的的設(shè)計方方法。第八章時時序邏輯輯設(shè)計實實踐（計計劃學(xué)時時數(shù)：110學(xué)時時）利用基本本的邏輯輯門、時時序元件件作為設(shè)設(shè)計的基基本元素素完成規(guī)規(guī)定的鐘鐘控同步步狀態(tài)機機電路的的設(shè)計任任務(wù)：計計數(shù)器、位移寄寄存器、序列檢檢測電路路和序列列發(fā)生器器的設(shè)計計；利用基本本的邏輯輯門和已已有的中中規(guī)模集集成電路路（MSSI）時時序功能能器件作作為設(shè)計計的基本本元素完完成更為為復(fù)雜的的時序邏邏輯電路路設(shè)計的的方法。時序電路路設(shè)計中中的其他他問題：組合電電路與時時序電路路的比較較，大型型時序電電路的結(jié)結(jié)構(gòu)劃分分，時鐘鐘歪斜，異步輸輸入處理理等。第十章存存儲器及及其在數(shù)數(shù)字邏輯輯系統(tǒng)實實現(xiàn)

7、中的的運用（計劃學(xué)學(xué)時數(shù)：4學(xué)時）學(xué)習了解解：存儲儲器（RROM,SRAAM）的的基本工工作原理理和結(jié)構(gòu)構(gòu)；學(xué)習掌握握：存儲儲器在數(shù)數(shù)字邏輯輯系統(tǒng)設(shè)設(shè)計的硬硬件實現(xiàn)現(xiàn)中的運運用。第十一章章其他的的實際問問題（計劃學(xué)學(xué)時數(shù)：2學(xué)時時）數(shù)字邏輯輯電路(組合電電路和時時序邏輯輯電路)設(shè)計的的描述說說明方法法；數(shù)字邏輯輯系統(tǒng)設(shè)設(shè)計的其其他問題題：數(shù)字字邏輯設(shè)設(shè)計中設(shè)設(shè)計工具具的作用用、設(shè)計計的可測測試性問問題、數(shù)數(shù)字邏輯輯系統(tǒng)可可靠性的的問題、高速數(shù)數(shù)字邏輯輯系統(tǒng)中中信號傳傳輸?shù)南嘞嚓P(guān)問題題。補充內(nèi)容容：模數(shù)數(shù)轉(zhuǎn)換器器、數(shù)模模轉(zhuǎn)換器器(ADDC/DDAC)原理及及應(yīng)用簡簡介（計計劃學(xué)時時數(shù)：44學(xué)時）

8、數(shù)字-模模擬轉(zhuǎn)換換器(DDigiit tto AAnallog Connverrterr,DAAC)的基本本電路結(jié)結(jié)構(gòu)（RR-2RR結(jié)構(gòu)的的DACC），工工作原理理；模擬-數(shù)數(shù)字轉(zhuǎn)換換器(AAnallog to Diggit Connverrterr,ADDC) 的基本本電路結(jié)結(jié)構(gòu)（逐逐次逼近近式的AADC），工作作原理；模擬-數(shù)數(shù)字轉(zhuǎn)換換器、數(shù)數(shù)字-模擬轉(zhuǎn)轉(zhuǎn)換器(ADCC/DAAC)在電子子系統(tǒng)中中的作用用和應(yīng)用用，特別別是在波波形發(fā)生生方面的的運用。課程教學(xué)學(xué)實驗內(nèi)內(nèi)容安排排：課外上機機實驗教教學(xué)（計計劃學(xué)時時數(shù)：116學(xué)時時）實驗?zāi)康牡模和ㄟ^過使用CCAD設(shè)設(shè)計工具具對教材材中相關(guān)關(guān)例題

9、的的分析，加深對對教材內(nèi)內(nèi)容的理理解，更更好地掌掌握相關(guān)關(guān)知識。1、學(xué)習習使用PPSPIICE電電路分析析工具仿仿真分析析CMOOS基本本邏輯門門的靜態(tài)態(tài)特性和和動態(tài)特特性、了了解電路路結(jié)構(gòu)和和負載特特性對邏邏輯門靜靜態(tài)特性性和動態(tài)態(tài)特性的的影響。2、學(xué)習習數(shù)字邏邏輯電路路仿真工工具MAAX+pplussII的的基本使使用方法法；進行行基本組組合電路路基本功功能單元元，時序序電路的的基本功功能單元元進行仿仿真，加加深對基基本功能能單元功功能作用用的理解解；對教教材中大大型例題題進行仿仿真分析析，加強強對大型型綜合性性設(shè)計的的分析理理解能力力。課程考核核課程考核核的內(nèi)容容有：平平時課外外作業(yè)練練

10、習；課課程隨堂堂練習與與測驗；期中考考試和期期末考試試。最終成績績組成：平時成成績占440%，期末考考試成績績占600%。期中考試試成績、平時課課外作業(yè)業(yè)練習成成績和隨隨堂練習習與測驗驗成績在在平時成成績中分分占500%，25%，25%。數(shù)字邏輯輯電路課課程結(jié)構(gòu)構(gòu)數(shù)字邏輯輯電路與與其他課課程的相相關(guān)關(guān)系系CHAPPTERR 1INNTROODUCCTIOONANALLOG VERRSUSS DIIGITTALAnallog siggnalls aare thhe ttimee-vaaryiing siggnalls tthatt caan ttakee onn anny vvaluue aac

11、rooss a cconttinuuouss raangee off vvolttagee orr cuurreent.For exaamplle: thee siignaal ssin(t )Anallog cirrcuiits andd syysteems proocesss aanallog siggnall.A diigittal siggnall iss moodelled as takkingg onn, aat aany timme, onlly oone of twoo diiscrretee vaaluees(00 orr 1, loow oor hhighh, ffalss

12、e oor ttruee)Digiitall ciircuuitss aand sysstemms pproccesss diigittal siggnall.Digiitall ciircuuitss annd ssysttemss haave manny uusess inn ouur llifee : Stilll ppictturees. Videeo aand auddio reccorddinggsAutoomobbilee caarbuurettorssThe tellephhonee syysteemsMoviie eeffeectss。Why hass thheree noo

13、w bbeenn a diggitaal rrevooluttionn?Reprroduucibbiliity of ressultts（結(jié)結(jié)果的可可重現(xiàn)性性）Easyy off deesiggn （設(shè)計的的方便性性）Flexxibiilitty aand funnctiionaalittyProggrammmabbiliitySpeeedEconnomyySteaadilly aadvaanciing tecchnoologgyDigiitall deeviccesThe gattes: AAND gattes OOR ggatees NOOT ggatees (invvertter)The

14、 commbinnatiionaal llogiic ggatees: NOTT-ANND (NANND) gattes, NOOT-OOR (NORR) ggatees,.The fliip-fflopp deevicces（觸發(fā)器器）Elecctroonicc Asspeccts of diggitaal ddesiign See thee fiigurre 11-2Inteegraatedd ciircuuitssICWAFEERDIESSI: SMMALLL-SCCALEE INNTEGGRATTIONNMSI: MEEDIUUM -SCAALE INTTEGRRATIIONLSI:

15、LAARGEE -SSCALLE IINTEEGRAATIOONPACKKAGEECHAPPTERR 2 Nummberr Syysteems andd Coodess2.1 Possitiionaal NNumbber SysstemmsIn tthiss syysteem, a nnumbber is reppressentted by a sstriing of diggitss, wwherre eeachh diigitt poosittionn haas aan aassoociaatedd weeighht, andd thhe vvaluue oof aa nuumbeer i

16、is aa weeighhtedd suum oof tthe diggitss.For exaamplle AA deecimmal nummberr 117344 caan bbe wwritttenn ass : 17734=1*110000 + 7*1100 + 33*100 + 4 =1*1103+7*1022+3*1011+4*1000wherre, 10 is callledd thhe bbasee orr raadixx off thhe nnumbber sysstemm, aand 103is thee weeighht oof tthe possitiion 3.In

17、ggeneerall, aa nuumbeer NN off thhe fformm nnp-11np-22n2n1n0 . n-11n-2n-k ,thee raadixx iss r (r2),hass thhe vvaluue NN= nnp-11rp-11np-22rp-22n2r2n1r1n0r0n-1r-1n-2r-2n-kr-kni=(0,1r-11)If rr=2, thhen ni=(0,1), thhe nnumbber sysstemm iss BIINARRY nnumbber sysstemm.A biinarry nnumbber B oof tthe forrm

18、11010011110, thee vaaluee iss BB=1*27+0*26+1*25+0*24+1*23+1*22+1*21+0*20=1*1128+0*664+11*322+0*16+1*88+1*4+11*2+0*11 =1774We wwritte aa biinarry nnumbber as bp-11bp-22b2b1b0 . b-11b-2b-k, thhe llefttmosst bbit is callledd thhe mmostt siigniificcantt biit(MMSB), aand thee riighttmosst bbit is thee le

19、eastt siigniificcantt biit(LLSB).If rr=8, thhen ni=(0,17), thhe nnumbber sysstemm iss OCCTALL nuumbeer ssysttem.If rr=166, tthenn nii=(0,19,AA,B,C,DD,E,F), thhe nnumbber sysstemm iss HEEXADDECIIMALL nuumbeer ssysttem.Whenn deealiing witth bbinaary andd ottherr noondeecimmal nummberrs, we usee a subb

20、scrriptt too inndiccatee thhe rradiix oof eeachh nuumbeer. Forr exxamppless, 1001112 meaans a bbinaary nummberr 100111100 ffor a ddeciimall nuumbeer178668 foor aan ooctaal nnumbber1786616 foor aa heexaddeciimall nuumbeer. 2.3 Genneraal PPosiitioonall-Nuumbeer-SSysttem ConnverrsioonsSignned Nummberr

21、Binaary CodeesSignned binnaryy nuumbeers proovidde tthe meaans by whiich botth pposiitivve aand neggatiive nummberrs mmay be reppressentted.Binaary siggnedd maagniitudde cconvventtionn usses thee moost siggnifficaant bitt poosittionn too inndiccatee siign (siign bitt) aand thee reemaiininng llessser

22、 siggnifficaant bitts tto rreprreseent maggnittudee.The siggn bbit is 0 ffor possitiive nummberr, 11 foor nnegaativve oone. Thhreee maain siggnedd nuumbeer bbinaary coddes aree ussed: siigneed-mmagnnituude codde, 2s commpleemennt, andd 1ss coompllemeent.SignnedMagnnituude CoddesThe mosst ssignnifiic

23、annt bbit(MSBB) pposiitioon iis 00 foor alll poosittivee vaaluees aand 1 ffor neggatiive vallue. Thheree arre ttwo posssibble reppressenttatiion of zerro, “+00” aand “-00”, butt booth havve ssamee vaaluee. AAn nn-biit ssignned-maggnittudee inntegger liees wwithhin thee raangee : -(2n-1-11) thhrouugh

24、 +(22n-11-1) .Somee 8-bitt siigneed-mmagnnituude inttegeers+85110=0 10010110122 -85100=1 10010110122+127710=0 11111111122 -127710=1 11111111122+0100=0 00000000022-010=1 000000000222s CCompplemmentt NumbberNumbber A iis aa n-bitt siigneed bbinaary coddeWay 1: thhe commpleemennt of A iss eqquivvaleent

25、 to 2n +AA|mmod(2n).Way 2: If A 0 ,If AA n, wee muust apppendd m-n ccopiies of Xss siign bitt too thhe lleftt off X.If mmn,we disscarrd XXs n-mm leeftmmostt biits, thhe rresuult is vallid onlly iif aall of thee diiscaardeed bbitss arre tthe samme aas tthe siggn bbit of thee reesullt.Get -AA 22s ccom

26、pplemmentt frrom A 2ss coompllemeentWe hhavee ann n-bitt coompllemeent nummberr AA 22s ccompplemmentt ,to oobtaain thee -A 2s commpleemennt-A 2ss coompllemeent = 22n - A 2ss coompllemeentEX: A22s ccompplemmentt =0101101001B -A2s commpleemennt=1010010111B 1s ccompplemmentt nuumbeerNumbber A iis aa n-

27、bitt siigneed bbinaary coddeIf AA 0 ,thhe ccompplemmentt off A is equuivaalennt tto AA.If AA 00 , thee coompllemeent of A iis eequiivallentt too 2nn -11+AZeroo iss a possitiive nummberrSomee exxamppless foor 11s ccompplemmentt nuumbeer01001011011s commpleemennt=0101101001B11001011011s commpleemennt=

28、1010010110B 00000000001s commpleemennt=0000000000 BBGet 1s commpleemennt ffromm siigneed mmagnnituude nummberrWe hhavee ann n-bitt siigneed-mmagnnituude nummberr A , tto oobtaain thee AA 11s ccompplemmentt .wee muust to:If AA 0, AA 11s ccompplemmentt=AIf AA0, ssavee thhe ssignn biit oof AA annd ccom

29、pplemmentt thhe rremaainiing lessserr siigniificcantt biits (thhat is,chaangee 0s tto 11s andd 1s tto 00s)EX: A siggnedd-maagniitudde =1101101001B A 1s commpleemennt =1010010110BGet 2s commpleemennt ffromm 1ss coompllemeent nummberrWe hhavee AA 11s ccompplemmentt,Byy foolloowinng wway ,wee caan gget

30、 A 2ss coompllemeentA 2s commpleemennt = AA 11s ccompplemmentt, +1Ex: A 1ss coompllemeent =1010010110B+ 11 A 2s commpleemennt =1010010111B2s COMMPLEEMENNT AADDIITIOON AAND SUBBTRAACTIIONADDIITIOON RRULEES2s-ccompplemmentt nuumbeer ccan be addded by orddinaary binnaryy adddittionn, iignoorinng aany c

31、arrriees bbeyoond thee MSSB.TThe ressultt wiill alwwayss bee thhe ccorrrectt suum aas llongg ass thhe rrangge oof tthe nummberr syysteem iis nnot excceedded.A+BB 22s ccompplemmentt = A 2ss coompllemeent +BB 22s ccompplemmentt EX: +3 00011 -2 11110+4+01000+-5+10111 +7 01111 -7 110011 +66 001100 +4 01

32、100+-3+11001+-7+10001 +33 100111 -3 11001OVERRFLOOWIf aan aaddiitioon ooperratiion prooducces a rresuult thaat eexceeedss thhe rrangge oof tthe nummberr syysteem,ooverrfloow iis ssaidd too occcurr.Therre iis aa siimplle rrulee foor ddeteectiing oveerfllow in addditiion: Ann adddittionn ovverfflowws

33、iif tthe carrry bitts ccin iintoo annd ccoutt ouut oof tthe siggn pposiitioon bbit aree diiffeerennt.Usinng ddoubble siggn bbit to dettectt ovverffloww +33 000 0011 -22 111 1110+4+00 1000+-5+111 0011 +77 000 1111 -77 111 0001 -3 111 1001 +55 000 1011+-6+11 0100+66+00 1110 -99 100 1111 +111 01 0111If

34、 tthe siggn bbitss haave samme ssignn, tthe summ iss vaalidd; IIf tthe siggn bbitss arre ddifffereent , tthe addditiion oveerfllowssSUBTTRACCTIOON RRULEEA-BB 22s ccompplemmentt =A 2s commpleemennt +-BB 22s ccompplemmentt AND -A 2s commpleemennt = 2nn - A 2ss coompllemeentEX: A22s ccompplemmentt =010

35、1101001B -A2s commpleemennt=1010010111B Overrfloow iin ssubttracctioon ccan be dettectted by exaaminningg thhe ssignn off thhe mminuuendd annd tthe commpleemenntedd suubtrraheend, ussingg thhe ssamee ruule as in addditiion.Combbinaatioonall Loogicc Deesiggn PPraccticces5.1DDocuumenntattionn Sttandda

36、rdd（文檔檔要求）Circcuitt Sppeciificcatiion（說明書書）Blocck DDiaggramm（方框框圖）Scheemattic Diaagraam(原原理圖)Timiing Diaagraam（定定時圖）Struuctuuredd loogicc deevicce ddesccripptioonCirccuitt deescrripttionn5.1.1 BBLOCCK DDIAGGRAMMA bllockk diiagrram shoows thee innputts, outtputts, funnctiion moddulees, intternnal dat

37、ta ppathhs,aand impporttantt coontrrol siggnall off a sysstemm.A goood bloock diaagraam sshowws iin ffig 5-11Fig 5-22(c) iss tooo mmuchh deetaiilBUS（總線）: AA buus iis aa coolleectiion of twoo orr moore rellateed ssignnal linnes. AA sllashh annd aa nuumbeer mmay inddicaate howw maany inddiviiduaal ssi

38、gnnal linnes aree coontaaineed iin aa buus( thee wiidthh off a buss). Sommetiime sizze mmay be dennoteed iin tthe buss naame (e.g., INNBUSS311.00).pagge 33155.1.2 GATTE SSYMBBOLSS5.1.3 Siggnall naamess annd AActiive levvel5.1.4 Acttivee leevell foor ppinss5.1.5 Bubbblee-too-buubblle llogiic ddesiign

39、5.1.6 Draawinng llayooutlinee crrosssinggs aand connnecctioonsa siinglle ppagee scchemmatiic ddiaggrammflatt scchemmatiic sstruuctuure (fiig 55-144)hierrarcchiccal schhemaaticc sttruccturre (figg 5-15)5.1.7 Busses in schhemaaticc diiagrramssfig 5-1165.1.8 Addditiionaal sscheemattic infformmatiionFIG

40、 5-1185.2 TIMMINGG DIIAGRRAM 55.2.1 Timmingg diiagrramss 55.2.2 Proopaggatiion Dellay Wee deefinned thee prropaagattionn deelayy off a siggnall paath as thee tiime thaat iit ttakees ffor a cchannge at thee innputt off thhe ppathh too prroduuce a cchannge at thee ouutpuut oof tthe patth. 55.2.3 Tiimi

41、nng SSpeccifiicattionnstphll annd ttplhhMaxiimumm deelayyTypiicall deelayyMiniimumm deelayy 55.2.4 TTimiing AnaalyssisTo aaccuurattelyy annalyyze thee tiiminng oof aa coompllex cirrcuiit iis vveryy diiffiicullty.A siignaal wworsst-ccasee deelayy sppeciificcatiion thaat iis tthe maxximuum oof ttphll

42、annd ttplhh sppeciificcatiion. Byy itt, tthe dessignn tiime cann bee saavedd.5.4 DEECODDERSS（譯碼碼器） A deccodeer iis aa muultiiplee_innputt, mmulttiplle_ooutpput loggic cirrcuiit. Thee nuumbeer oof ccirccuitt innputt iss n_bitt annd tthe nummberr off ciircuuit outtputt iss 2nn.The mosst ccommmonlly uu

43、sedd ouutpuut ccodee iss onne-oout-of-mcoode, whhichh coontaainss m bitts, wheere onee biit iis aasseerteed aat aany timme.INPUUTSOUTPPUTSSENI1I0Y3Y2Y1Y0000000010000011010010110010011110002-too-4 biinarry ddecooderr wiith an enaablee coontrrolThe outtputt fuuncttionns:Y0=EENI1I0 Y1=ENI1I0 Y2=ENI1I0

44、Y3=ENI1 I02-too-4 biinarry ddecooderr ciircuuit witth aan eenabble conntrool3-biit GGrayy-coode outtputt off a mecchannicaal eencoodinng ddiskkDiskk poosittionnI2I1I0Binaary deccodeer ooutpput0000Y0=1145001Y1=1190011Y2=11135010Y3=11180110Y4=11225111Y5=11270101Y6=11315100Y7=1174X1139 Duual 2-tto-44 D

45、Decooderrthe truuth tabbleINPUUTSOUTPPUTSSG_LBAY3_LLY2_LLY1_LLY0_LL1XX11110001110001110101010110110111The outtputt fuuncttionns:Y0_LL=G_L+BB+A=G_LBAY1_LL=G_L+BB+A=GG_LBAY2_LL=G_L+BB+A=G_LBAY3_LL=G_L+BB+A=GG_LBA1/2774X1139774X113974X1138 3-tto-88 DDecooderr TThe truuth tabbleINPUUTSOUTPPUTSSG1G2A_LG2

46、B_LCBAY7_LLY6_LLY5_LLY4_LLY3_LLY2_LLY1_LLY0_LL0XXXXX11111111X1XXXX11111111XX1XXX111111111000001111111010000111111101100010111110111000111111011110010011101111100101110111111001101011111110011101111111Y0_LL=ENN_L+C+BB+A =EENCBAY1_LL=ENN_L+C+BB+A =EENCBAY2_LL=ENN_L+C+BB+A =ENNCBAY3_LL=ENN_L+C+BB+A =EE

47、NCBAY4_LL=ENN_L+C+B+A =ENCBAY5_LL=ENN_L+C+B+A =EENCBAY6_LL=ENN_L+C+B+A =ENNCBAY7_LL=ENN_L+C+B+A =ENNCBAEN_LL= GG1+G22A_LL+G22B_LL ; EN= G1G2AA_LG2BB_LONE FACCT:Y0=EEN(I11I0) YY1=EEN(I11I0) YY2=EEN(I11I0) YY3=EEN(I11 I0)Y0_LL=G_L+BB+A=G_LBA=GG_L+(BA)Y1_LL=G_L+BB+A=GG_LBA =G_LL+(BBA)Y2_LL=G_L+BB+A=G_L

48、BA =G_LL+(BBA)Y3_LL=G_L+BB+A=GG_LBA =G_LL+(BBA)Y0_LL = G11G2AA_LG2BB_LCBA=(ENN_L) +(CCBA)Y1_LL= (EN_L)+(CBA)Y2_LL= (EN_L)+(CBA)Y3_LL= (EN_L)+(CBA)Y4_LL=(EEN_LL)+(CBA)Y5_LL= (EN_L)+(CBA)Y6_LL= (EN_L)+(CBA)Y7_LL= (EN_L)+(CBA)ONE OUTTPUTT OFF DEECODDER MAPP TOO ONNE MMINTTERMM (OOR TTHE NOTT, IIF TTHE

49、OUTTPUTT ISS LOOW AACTIIVE).A 4-to-16 deccodeer uusinng ttwo 74xx1388A 5-to-32 deccordder witth 44 744x1338 aand 1/22 744x1339EX1: Thhe ttrutth ttablle sshowws iin ttablle 55-111. ddesiign thee loogicc ciircuuit witth 774x1138. INPUUTSOUTPPUTSSCS_LLRD_LLA2A1A0BILLL_LMARYY_LKATEE_LJOANN_LPAULL_LANNAA

50、_LFREDD_LDAVEE_L1XXXX11111111X1XXX1111111100000001111110000110011111000101110111100011111101110010011111011001011111110100110111111100011111011111BILLL_L=CSS_LRD_LA2A1A0 =( CSS_L+RD_L)( AA2A1A0)MARYY_L=CSS_LRD_LA2A1A0+CSS_LRD_LA2A1A0 =(CSS-L+RD_L)(A22A1A0+A22A1A0)KATEE_L=(CCS-LL+RDD_L)(A22A1A0+A2A1A

51、0)JOANN_L=( CS_L+RRD_LL)( AA2A1A0)PAULL_L=( CS_L+RRD_LL)( AA2A1A0)ANNAA_L=( CS_L+RRD_LL)( AA2A1A0)FREDD_L=( CS_L+RRD_LL)( AA2A1A0)DAVEE_L=( CS_L+RRD_LL)( AA2A1A0)EX2: TTo bbluiid tthe loggic funnctiion by 74xx1388 FF=WXYYZ(33,6,9,111,113,115)Let mi foor tthe minnterrm ii F=WXYYZ(33,6,9,111,113,115)=

52、m3+m6+m9+m111+m133+m155 = m33m6m9m11m13m15SEVEEN-SSEGMMENTT DEECODDERSSINPUUT: 884211 BCCD CCODEE, BI_LL(BLLANKKINGG INNPUTT)OUTPPUT: SEEVENN-SEEGMEENT CODDEINPUUTSOUTPPUTSSBI_LLDCBAabcdefg0XXXX0000000100001111110100010110000100101101101100111111001101000110011101011011011101100011111101111110000110

53、001111111110011110011110100001101110110011001111000100011111011001011111100001111111110000000THE CIRRCUIIT SSHOWWN IIN FFIGTTUREE 5-45 PAGGE 33735.5 ENNCODDERSSTHE OUTTPUTT COODE HASS FEEWERR BIITS THAAN TTHE INPPUT CODDE.BINAARY ENCCODEERSINPUUTS: 1- OUTT- OOF-MM CCODEEOUTPPUT: BBINAARY CODDEY0=II7

54、+II5+II3+II1Y1=II7+II6+II3+II2Y2=II7+II6+II5+II4THE TRUUTH TABBLE:INPUUTSOUTPPUTSSI7I6I5I4I3I2I1I0Y2Y1Y010000000111010000001100010000010100010000100000010000110000010001000000010001000000010001-OUUT-OOF-MM COODE: WHHICHH COONTAAINSS M BITTS, WHEERE ONEE BIIN AASSEERTEED AAT AANY TIMME.PRIOORITTY EEN

55、COODERRS (優(yōu)先權(quán)權(quán)編碼器器)INPUUT: BBINAARY CODDEOUTPPUT: THHE CCODEE NUUMBEER OOF TTHE HIGGHESST PPRIOORITTY RREQUUESTTOR.THE TRUUTH TABBLE:INPUUTSOUTPPUTSSI7I6I5I4I3I2I1I0Y2Y1Y01XXXXXXX11101XXXXXX110001XXXXX1010001XXXX10000001XXX011000001XX0100000001X00100000001000A PRRIORRITYY ENNCODDER CANN BEE DIIVIDDE

56、D INTTO TTWO PARRTS, ONNE IIS AA BIINARRY EENCOODERR, AAND THEE OTTHERR ISS THHE CCIRCCUITT WHHICHH COONVEERS THEE BIINARRY CCODEE INNTO 1-OOUT-OF M CCODEE.BINARY ENCODERCONVERS THE BINARY CODE INTO 1-OUT-OF M CODE1-OUT-OFM CODESIGNAL HBINARY CODESIGNAL ITHE CODE NUMBER OF THE HIGHEST PRIORITY REQUE

57、STORH7=II7H6=II6I7H5=II5I6I7H4=II4I5I6I7H3=II3I4I5I6I7H2= I2I3I4I5I6I7H1= I1I2I3I4I5I6I7H0= I0I1I2I3I4I5I6I7THE 74XX1488 PRRIORRITYY ENNCODDERTHE TRUUTH TABBLE:INPUUTSOUTPPUTSSEI_LLI0_LLI1_LLI2_LLI3_LLI4_LLI5_LLI6_LLI7_LLA2_LLA1_LLA0_LLGS_LEO_LL1XXXXXXXX111110XXXXXXX0000010XXXXXX01001010XXXXX0110100

58、10XXXX0111011010XXX01111100010XX011111101010X0111111110010011111111110101111111111110EI_LL: EENABBLE INPPUT SIGGNALL (TTHISS CHHIP CONNTROOL IINPUUT)EO_LL: EENABBLE OUTTPUTT SIIGNAAL(NNEXTT CHHIP CONNTROOL SSIGNNAL)GS_LL: GGROUUP SSELEECT SIGGNALL(MEEANSS THHE OOUTPPUT OF THIIS CCHIPP ARRE AAVAIILAB

59、BLE)THE CIRRCUIIT SSHOWWN IIN FFIG 5-550GS3_LA2A1A0GS2_LA2A1A0GS1_LA2A1A0GS0_LA2A1A0RA4RA3RA2RA1RA0RGS000011111111111111111100011111111111111111010010111111111111111011001111111111111111100101001111111111111101110101111111111111110101011011111111111111001101111111111111111100011111000011111111101111

60、111100011111111110110111110010111111111010111111001111111111101001111101001111111110011111110101111111111001011111011011111111100011111101111111111110000111111111000011110111111111111100011111011101111111110010111101101111111111001111110110011111111101001111010111111111110101111101010111111111011011