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《數(shù)據(jù)與計算機通信》
(第七版)
課后習(xí)題參考答案
第2章的參考答案
2.1答案:
2.1Theguesteffectivelyplacestheorderwiththecook.Thehostcommunicatesthis
ordertotheclerk,whoplacestheorderwiththecook.Thephonesystemprovides
thephysicalmeansfortheordertobetransportedfromhosttoclerk.Thecook
givesthepizzatotheclerkwiththeorderform(actingasaheader'tothepizza).
Theclerkboxesthepizzawiththedeliveryaddress,andthedeliveryvanencloses
alloftheorderstobedelivered.Theroadprovidesthephysicalpathfordelivery.
2.2答案
a.
TblephoxeLine
ThePMsspeakasiftheyarespeakingdirectlytoeachother.Forexample,when
theFrenchPMspeaks,headdresseshisremarksdirectlytotheChinesePM.
However,themessageisactuallypassedthroughtwotranslatorsviathephone
system.TheFrenchPM'stranslatortranslateshisremarksintoEnglishand
telephonesthesetotheChinesePMstranslator,whotranslatestheseremarksinto
Chinese.
b.
b.
Anintermediatenodeservestotranslatethemessagebeforepassingiton.
2.4答案:
2.4No.Thereisnowaytobeassuredthatthelastmessagegetsthrough,exceptby
acknowledgingit.Thus,eithertheacknowledgmentprocesscontinuesforever,or
onearmyhastosendthelastmessageandthenactwithuncertainty.
2.7答案:
2.7a.No.Thiswouldviolatetheprincipleofseparationoflayers.Tolayer(N-1),
theN-levelPDUissimplydata.The(N-1)entitydoesnotknowaboutthe
internalformatoftheN-levelPDU.ItbreaksthatPDUintofragmentsand
reassemblesthemintheproperorder.
b.EachN-levelPDUmustretainitsownheader,forthesamereasongivenin(a).
第3章的參考答案
3.13答案
3.13a.(30pictures/s)(480x500pixels/picture)=7,2x106pixels/s
Eachpixelcantakeononeof32valuesandcanthereforeberepresentedby5
bits:
R=7.2x106pixels/sx5bits/pixel=36Mbps
b.Weusetheformula:C=Blog2(1+SNR)
B=4.5x106MHz=bandwidth,and
SNRdB=35=10log10(SNR),hence
SNR=1()35/10=io3-5,andtherefore
C=4.5x106log2(1+IO35)=4.5x106xIog2(3163)
C=(4.5x106x11.63)-52.335x106bps
c.Alloweachpixeltohaveoneoftenintensitylevelsandleteachpixelbeoneof
threecolors(red,blue,green)foratotalof10x3=30levelsforeachpixel
element.
3.15答案
3.15UsingShannon'sequation:C=Blog2(1+SNR)
WehaveW=300Hz(SNR)dB=3
Therefore,SNR=IO03
03
C=300log2(1+IO)=300log2(2.995)=474bps
316答案
3.16UsingNyquistsequation:C=2Blog2M
WehaveC=9600bps
a.log7M=4,becauseasignalelementencodesa4-bitword
Therefore,C=9600=2Bx4,andB=1200Hz
b.9600=2Bx8,andB=600Hz
3.19答案
已知C=20Mbps,B=3Mbps.
根據(jù)香農(nóng)定理直=6log2(1+SNR),
C=20x1。6<=3x1O'xlog?(1+SNR)
則
log2"SNR)>=6.67
SNR>=101
另解:
3.19C=Blog2(1+SNR)
66
20x10=3x10xlog2(l+SNR)
log2(l+SNR)=6.67
1+SNR=102
SNR=101
▲補充作業(yè):
設(shè)采用異步傳輸,1位起始位,2位終止位,1位奇偶位,每一個信號碼源2
位,對下述速率,分別求出相應(yīng)的有效數(shù)據(jù)速率(b/s):
(l)300Baud(2)600Baud(3)I200Baud(4)4800baud
答:
異步傳輸?shù)臄?shù)據(jù)效率為7/11,而每一個信號碼源2位,
714
R=81og2M,所以/?=打乂28=H8
714
(1)R=—x2B=——B=38l.8b/s
1111
714
(2)R=—x28=—8=763.66/s
1111
714
(3)R=—x2B=—B=\52n.3bls
1111
714
(4)R=—x2B=—B=6109.1b/s
1111
第5章的參考答案
5.6答
NRZ-L
NRZI
Bipolar。、”
?~~???~~?????
Pseudoternary|J??|||?;
Manchester[EHjHHE1
?=;UTLTLhJmrLil
B8ZS3
HDB3
5.7答:
5.7WithManchester,thereisalwaysatransitioninthemiddleofabitperiod.
n_n_nj_i_n_ru_i_i_LT
|I|I|I|O|O|I|I|O|I|O|
5.8答:
11111011111
心________________________________________________n______
,-bJ-Ln"LTLTL
c.__________________|~|_____________________
5.9答:
5.9Theerrorisatbitposition7,wherethereisanegativepulse.ForAMI,positiveand
negativepulsesareusedalternatelyforbinary1.Thepulseinposition1represents
thethirdbinary1inthedatastreamandshouldhaveapositivevalue.
第6章的參考答案
6.1答:
6.1a.Eachcharacterhas25%overhead.For10,000characters,thereare20,000extra
bits.Thiswouldtakeanextra20,000/2400=8.33seconds.
b.Thefiletakes10framesor480additionalbits.Thetransmissiontimeforthe
additionalbitsis480/2400=0.2seconds.
c.Tentimesasmanyextrabitsandtentimesaslongforboth.
d.Thenumberofoverheadbitswouldbethesame,andthetimewouldbe
decreasedbyafactorof4=9600/2400.
另答:
(a)
1+1八八~
------=20%
8+1+1
.?.額外開銷率為20%
傳輸速率:2400b/s=240w/s
傳輸時間為3S=41.67s
240
(b)
48
0.59%
8000+48
額夕卜開銷:48x10=4808〃
傳輸一-幀:建£=3.35s
2400
共擔(dān)也=1。
1000
總耗時:10x3.35=33.5s
(c)
異步、同步額外開銷不變。
耗時:異步:41.67x10=416.7s
同步:100x3.35=335s
(d)
10000
耗時:異步:------=104.2s
9600/10
同步:傳輸幀耗時:色生=0.8383s
9600
???共耗時:100x0.8383=83.83s
6.5答案
6.5LetthebitdurationbeT.Thenaframeis12Tlong.Letaclockperiodbe「.Thelast
bit(bit12)issampledat11.5T'.Forafastrunningclock,theconditiontosatisfyis
T11s
H-5F>117^-<-^=1.045=fclock<1.045加
Foraslowrunningclock,theconditiontosatisfyis
T115
11.57'<127=無>宣=0.958nfclock>^^fbit
Therefore,theoverallcondition:0.958<fclock<1.045fbit
另解:
不發(fā)生幀差錯,則8+1比特總誤差不超過50%,即小于50%/9=5.6%,精
確率在95%以上。
另解:設(shè)能夠容忍的時鐘精確率的百分比為X%,
.??(8+1+2)(100—幻=5%
?"=95?5
.?.能夠容忍的時鐘精確百分比為95.5%
6.10答:
(a)
1Q00100010011000
10001000000100001/10000000000000000000000000000000
10001000000100001::::;;;;;;;::::
looooooiooooioooo;:;::;;:!!:
10001000000100001!;;;?;;;;;;
10010001100010000?::::::
..........................................................
11001100110001666!!!!
loooioooooaiogo”;!!!
10001001101010010;;;
10001000000100001!!;
remainder=0001101110011000
(b)
||5||4||J||2
IXs
laput
Shift|ShiftRegister|Input
0000000000000000
100010000001000011
200100000010000100
301000000100001000
410000001000010000
500010010001100010
600100100011000100
/0100I000110001000
s10010001100010000
900110011001100010
1001100110011000100
1111001100110001000
1210001001101010010
1300000011011100110
1400000110111001100
1500001101110011000
16001101110110010
CRC
6.12答:
6.12
10110110
110011/1110001100000
110011:::::::
-101111:::::
::::;
111000:::;
110011-III
"IOIIOO::
noon::
111110;
110011!
CRC="11010
中文答案:
R=11010
T=1110001111010
6.13答:
6.13a.
c3<————c2<—a<—c0
Input
b.Data=10011011100
M(X)=1+X3+X4+X6+X7+X8
X4M(X)-X12+x11+X10+X8+X7+X4
P(X)P(X)
R(X)-X2
T(X)=X4M(X)+R(X)=X12+Xn+X10+X8+X7+X4+X2
Code-001010011011100
c.Code-001010001011100
T(X)
;
—R—x)y>ieldsanonzeroremainder
第7章的參考答案
7.2答案
v=------->50%
1+2。
Jpmp_20ms
ci——
tfweU4k
,:a<0.5
A/>160
,幀長度應(yīng)大于160加
7.2LetLbethenumberofbitsinaframe.Then,usingEquation7.5ofAppendix7A:
PropagationDelay20x10-380
TransmissionTime一103VL
UsingEquation7.4ofAppendix7A:
U=>0.5
l+2a-l+(160/£)
£>160
Therefore,anefficiencyofatleast50%requiresaframesizeofatleast160bits.
7.3答案
270ms
a==270
1000/IMpbs
(a)u0.0018
l+2a
w7
⑸u==0.0126
1+2。一-1+540
w127
(c)u0.2286
1+2。-'1+540
w255
(d)u0.459
1+2/-1+540
PropagationDelay270x10-J)
ZM=1O3/1O6=270
a.U=1/(1+2a)=1/541=0.002
b.UsingEquation7.6:U=W/(l+2a)=7/541=0.013
c.U=127/541=0.23
d.U=255/541=0.47
7.5答案
7.5Roundtrippropagationdelayofthelink=2xLxt
Timetotransmitaframe=B/R
Toreach100%utilization,thetransmittershouldbeabletotransmitframes
continuouslyduringaroundtrippropagationtime.Thus,thetotalnumberof
framestransmittedwithoutanACKis:
N=-a1R~+1,whereisthesmallestintegergreaterthanorequaltoX
ThisnumbercanbeaccommodatedbyanM-bitsequencenumberwith:
M=「log2(N)]
中文答案:
當(dāng)窗口大小w?2a+l時,信道得利用率為100%
,應(yīng)取w=2a+l
'"<>pL,t1.p
a=———=-------=—LRt
“eB/RB
2
:.w=-LR-t+l=2"-I
B
o
???幀號字段的長度應(yīng)為n=[log(-A-7?-r+l)]
2B
7.7答案:
7.7Assumea2-bitsequencenumber:
1.StationAsendsframes0z1,2tostationB.
2.StationBreceivesallthreeframesandcumulativelyacknowledgeswithRR3.
3.Becauseofanoiseburst,theRR3islost.
4.Atimesoutandretransmitsframe0.
5.Bhasalreadyadvanceditsreceivewindowtoacceptframes3,0,1,2.Thusit
assumesthatframe3hasbeenlostandthatthisisanewframe0,whichit
accepts.
7.8答案:
7.8Usethefollowingformulas:
a0.11.10100
S&W(1-P)/1.2(1-P)/3(1-P)/21(1-P)/201
CBN⑺(l-P)/(l-?-0.2P)(1-P)/(1+2P)7(1-7(1-P)/201(l+6P)
P)/21(1+6P)
GBN(127)(l-P)/(l+0.2P)(1-P)/(1+2P)(1-P)/(l+20P)127(l-P)/201(l+126P)
SREJ⑺1-P1-P7(1-P)/217(1-P)/201
SREJ(127)1-P1-P1-P127(1-P)/201
Foragivenvalueofa,theutilizationvalueschangeverylittleasafunctionofP
overareasonablerange(say10'3to10-12).Wehavethefollowingapproximate
valuesforP=10"6:
a0.11.010100
Stop-and-wait0.830330.050.005
GBN⑺1.01.00.330.035
GBN(127)1.01.01.00.63
SREJ(7)1.01.00330.035
SREJ(127)1.01.01.00.63
7.9答案:
7.9a.
節(jié)點A節(jié)點B
7.10答案:
7.10AlostSREJframecancauseproblems.Thesenderneverknowsthattheframewas
notreceived,unlessthereceivertimesoutandretransmitstheSREJ.
▲問題在于接收方無法通知發(fā)送方是否收到了其補發(fā)的幀。
7.11答案
7.11Fromthestandard:'ASREJframeshallnotbetransmittedifanearlierREJ
exceptionconditionhasnotbeencleared(Todosowouldrequestretransmission
ofadataframethatwouldberetransmittedbytheREJoperation.)"Inother
words,sincetheREJrequiresthestationreceivingtheREJtoretransmitthe
rejectedframeandallsubsequentframes,itisredundanttoperformaSREJona
framethatisalreadyscheduledforretransmission.
Alsofromthestandard:"Likewise,aREJframeshallnotbetransmittedifoneor
moreearlierSREJexceptionconditionshavenotbeencleared.TheREJframe
indicatestheacceptanceofallframespriortotheframerejectedbytheREJframe.
ThiswouldcontradicttheintentoftheSREJframeorframes.
REJ:發(fā)送方重發(fā)第N(R)幀及其后的各幀,接收方丟棄N(R)及其以后的各幀;
SREJ:發(fā)送方重發(fā)第N(R)幀,接受方繼續(xù)接收并保存已收到的幀。
7.12答案
7.12Leth=timetotransmitasingleframe
1024bits
"-心bps=1.024msec
Thetransmittingstationcansend7frameswithoutanacknowledgment.Fromthe
beginningofthetransmissionofthefirstframe,thetimetoreceivethe
acknowledgmentofthatframeis:
t2=270++270=541.024msec
Duringthetimet2,7framesaresent.
Dataperframe=1024-48=976
.7x976bits,..
ihrouehput=-------------------------=126zkbps
541.024x10-3sec
▲另解:
假設(shè)控制字段長8bit,FCS長16bit,則在一幀中數(shù)據(jù)比例為
1024-4x8.6=95.3%
1024
假設(shè)該鏈路可用GO-BACK-N差錯控制,則窗口尺寸可達(dá)7
“=—^—=1.3%
1+2。
數(shù)據(jù)的比特吞吐量為:
lMx95.3%xl.3%=12389M/s
7.13答案
7.13No,becausethefieldisofknownfixedlength.
7.14答案
7.14Thefollowingenhancementsarepossibilities:
?Alwaystransmitanintegralnumberofoctets
?Includealengthfield
?Donotusethesameflagtocloseoneframeandopenanother
?Ignoreanyframecontainingfewerthan32bits
?Ignoreanyframeendinginsevenormoreones
Thelengthfieldisusedtocomparethenumberofoctetsreceivedwiththenumber
transmitted.Anydiscrepanciesresultindiscardingtheframe.Thelastthree
enhancementsallowtherejectionofframeswhentheclosingflaghasbeen
destroyed.
7.16答案
7.16N(R)=2.Thisisthenumberofthenextframethatthesecondarystationexpects
toreceive.
根據(jù)題意知窗口序號為3比特,以8為模。因可以連續(xù)發(fā)送6幀,可斷定采
用回退N幀ARQ而不是選擇拒絕ARQo
因是無差錯操作,當(dāng)發(fā)送的第6幀信息幀的輪詢位置1,從站將給予RR或
RNR應(yīng)答,由于發(fā)送6幀信息前主站的N(S)為3,之后發(fā)送的信息幀的N(S)從
4開始,因此,從站返回的N(R)計數(shù)值為2,表示已接收到了4、5、6、7、0和
1幀,可以接收的下一幀的序號是2。
N(R)=010
7.17答案
7.17Oneexampleofsuchaschemeisthemultilinkprocedure(MLP)definedaspartof
layer2ofX.25.ThesameframeformatasforLAPBisused,withoneadditional
field:
FlagAddressControl|MLC一Packet|FCSFlag
Themultilinkcontrol(MLC)fieldisa16-bitfieldthatcontainsa12-bitsequence
numberthatisuniqueacrossalllinks.TheMLCandpacketfieldsformamultilink
protocol(MLP)frame.OnceanMLPframeisconstructed,itisassignedtoa
particularlinkandfurtherencapsulatedinaLAPBframe,asshownabove.The
LAPBcontrolfieldincludes,asusual,asequencenumberuniquetothatlink.
TheMLCfieldperformstwofunctions.First,LAPBframessentoutover
differentlinksmayarriveinadifferentorderfromthatinwhichtheywerefirst
constructedbythesendingMLP.ThedestinationMLPwillbufferincoming
framesandreorderthemaccordingtoMLPsequencenumber.Second,ifrepeated
attemptstotransmitaframeoveronelinkfails,theDTEorDCEwillsendthe
frameoveroneormoreotherlinks.TheMLPsequencenumberisneededfor
duplicatedetectioninthiscase.
7.18答案
7.18Theselective-rejectapproachwouldburdentheserverwiththetaskofmanaging
andmaintaininglargeamountsofinformationaboutwhathasandhasnotbeen
successfullytransmittedtotheclients;thego-back-Napproachwouldbelessofa
burdenontheserver.
該題實際上是問該B/S模式應(yīng)用究竟是回退N幀ARQ還是選擇拒絕ARQ
效率更高?
回退N幀ARQ會增加網(wǎng)上流量和服務(wù)器重傳的信息量,尤其當(dāng)線路質(zhì)量不
好時,服務(wù)器發(fā)送的數(shù)據(jù)量會劇增。
選擇拒絕ARQ會使接收和發(fā)送邏輯更復(fù)雜一些,尤其會加重服務(wù)器接收緩
沖的負(fù)擔(dān)。
WEB服務(wù)器因接收信息量小,發(fā)送信息量大,選擇拒絕ARQ的缺點對其影
響相對較小,倒是重傳信息量是主要問題。因此選擇拒絕ARQ對減輕WEB服
務(wù)器負(fù)擔(dān)可能更好一些。
第七章補充作業(yè):
1.若數(shù)據(jù)鏈路的發(fā)送窗口限度(尺寸)為4,在發(fā)送3號幀,并接受2號幀的確
認(rèn)幀后,發(fā)送方還可連續(xù)發(fā)幾幀?請給出可發(fā)幀的序號?
解答:
可以連續(xù)發(fā)送4幀,序號為3,4,5,6。
2.兩個相鄰的節(jié)點(A和B)通過后退N幀ARQ協(xié)議通信,幀順序為3位,窗口
大小為4。假定A正在發(fā)送,B正在接收,對下面兩種情況說明窗口的
位置:
①A開始發(fā)送之前
②A發(fā)送了0,1,2三個幀,而B應(yīng)答了0,1兩個幀
③A發(fā)送了3,4,5三個幀,而B應(yīng)答了第4幀
解答:
①窗口的位置在0
②窗口的位置在2
③窗口的位置在5
3.知數(shù)據(jù)幀長1024比特,其中幀頭24比特,數(shù)據(jù)速率500kb/s,線路的傳播延遲為
5ms,試計算:
①采用停等協(xié)議,求最大的信道利用率。
②設(shè)滑動窗口W>=2a+1,求窗口至少有多大?
解答:
①
5x10-3
a=----------=2.44
1024/5xlO5
?=—1―=0.17
1+2。
當(dāng)窗口大小卬22a+l時,信道得利用率為100%
應(yīng)取卬=2。+1
trop5x10-3
a二———=2.44
5
frame1024/5xlO
:.w=T-\=2a+\
???幀號字段的長度應(yīng)為n=[log2(2a+2)]=3
第8章的參考答案
8.1.請描述T1載波的原理。
答:
Bell系統(tǒng)的T1載波利用脈碼調(diào)制PCM和時分TDM技術(shù),使24路采樣聲
音信號復(fù)用一個通道。每一個幀包含193位,每一幀用125us時間傳送。T1系
統(tǒng)的數(shù)據(jù)傳輸速率為1.544Mbps。
數(shù)據(jù)位控制信號幀編碼
82在T1載波中,由非用戶數(shù)據(jù)引入的開銷占百分比是多少?
答:
T1載波是把24個話音信道多路復(fù)用在一條高速信道上,每個信道包含7位的數(shù)
據(jù)和1位的控制信令位,此外加入一位幀同步位組成基本幀。其中,用戶的開
銷為24X1(控制位)+1(基本幀)=25b總開銷為:(7+1)X24+1=193b因此,
用戶的開銷所占的百分比為:25/193X100%^13%
第9章的參考答案
本章無作業(yè)
第10章的參考答案
10.4答案
10.4a.CircuitSwitching
T=Cj+C2where
C[=CallSetupTime
C2=MessageDeliveryTime
J=S=0.2
C2=PropagationDelay+TransmissionTime
=NxD+L/B
=4x0.001+3200/9600=0.337
T=0.2+0.337=0.537sec
DatagramPacketSwitching
T=D1+D2+D3+D4where
D]=TimetoTransmitandDeliverallpacketsthroughfirsthop
D->=TimetoDeliverlastpacketacrosssecondhop
D3=TimetoDeliverlastpacketacrossthirdhop
D4=TimetoDeliverlastpacketacrossforthhop
ThereareP-H=1024-16=1008databitsperpacket.Amessageof3200bits
requirefourpackets(3200bits/1008bits/packet=3.17packetswhichweroundup
to4packets).
D[=4xt+pwhere
t=transmissiontimeforonepacket
p=propagationdelavforonehop
D]=4x(P/B)+D
=4x(1024/9600)+0.001
=0.428
D)=D3=D4=t+p
=(P/B)+D
=(1024/9600)+0.001=0.108
T=0.428+0.108+0.108+0.108
=0.752sec
VirtualCircuitPacketSwitching
T=Vj+V2where
V]=CallSetupTime
V2=DatagramPacketSwitchingTime
T=S+0.752=0.2+0.752=0.952sec
b.CircuitSwitchingvs.DiagramPacketSwitching
Tc=End-to-EndDelay,CircuitSwitching
Tc=S+NxD+L/B
Td=End-to-EndDelay,DatagramPacketSwitching
rL-
Np=Numberofpackets=———
Td-D1+(N-1)D2
D]=TimetoTransmitandDeliverallpacketsthroughfirsthop
D2=TimetoDeliverlastpacketthroughahop
D]=Np(P/B)+D
D2=P/B+D
T=(Np+N-l)(P/B)+NxD
T=Td
S+L/B=(%+N-1)(P/B)
CircuitSwitchingvs.VirtualCircuitPacketSwitching
Tv=End-to-EndDelay,VirtualCircuitPacketSwitching
Tv=S+Td
TC=TV
L/B=(Np+N-1)(P/B)
Datagramvs.VirtualCircuitPacketSwitching
Td=TV-S
10.7答案
10.7Thelayer2flowcontrolmechanismregulatesthetotalflowofdatabetweenDTE
andDCE.Eithercanpreventtheotherfromoverwhelmingit.Thelayer3flow
controlmechanismregulatestheflowoverasinglevirtualcircuit.Thus,resources
ineithertheDTEorDCEthatarededicatedtoaparticularvirtualcircuitcanbe
protectedfromoverflow.
另解:
兩者都是必要的。
因為在第三層分組中采用的流控和差錯控制雖然在格式與處理上與HDLC
相似,但因其分組中具有的D字段可以實現(xiàn)對于本地的或者是端對端的流控。
而第二層的鏈路層則采用LAPB(HDLC的子集)來實現(xiàn)大多數(shù)的鏈路控制與數(shù)
據(jù)傳輸,但不提供分組層中D字段具有的功能。
10.8答案
10.8Transmission,processing,andqueuingdelays.
X.25的分組格式中確實沒有FCS字段,但它作為PDU被傳遞到鏈路層是
由鏈路層協(xié)議將其封裝為LAPB幀,從而加上了FSC字段,這樣可以確保傳輸
LAPB幀中的數(shù)據(jù)域,從而保證X.25分組被正確地傳遞了。
10.9答案
10.9Framerelayismuchsimplified,comparedtoX.25.Themajordifferencesarethat
framerelayusesout-of-channelsignalingwhileX.25usesallin-channelcontrol.In
framerelaythereisno"hop-by-hop"flowcontrolorerrorcontrolasthereisinX.25.
Ifaframeerrorisdetecteditisjustdroppedratherthanbeingretransmitted.
Similarly,onanend-to-endbasis,thereisnoerrorcontrolorflowcontrolexcept
whatisprovidedbyhigherlevelprotocolsoutsideofframerelay.Finally,frame
relayisatwolevel(physicalandlink)layerprotocolandthemultiplexingof
logicalchannelstakesplaceatLevel2ratherthanintheLevel3PacketLayerasin
X.25.
另解:
因為X.25允許一個DTE與另一個DTE之間通過一條物理DTE-DCE鏈路,
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