Thi Nghiem Vien Thong

8/2/2019 Thi Nghiem Vien Thong

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TRNG AI HOC TON C THANG

KHOA IEN IEN T----------

MON HOCTH NGHIEM VIEN THONG

GIANG VIEN: ThS. HOANG MANH HA


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TRNG I HC TN C THNG

KHOA IN - IN TB MN IN T- VIN THNG

TI LIU HNG DN TH NGHIM

VIN THNG

Tp.H Ch Minh, thng 4 - 2010


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NI QUYPHNG TH NGHIM IN-IN T

IU I. TRC KHI N PHNG TH NGHIM SINH VIN PHI:1. Nm vng quy nh an ton ca phng th nghim.2. Nm vng l thuyt v c k ti liu hng dn bi thc nghim.3. Lm bi chun b trc mi bui th nghim. Sinh vin khng lm bi chun b theo ng

yu cu s khng c vo lm th nghim v xem nh vng bui th nghim .4. n phng th nghim ng giquy nh v gi trt t chung. Tr 15 pht khng c vo

th nghim v xem nh vng bui th nghim .5. Mang theo th sinh vin v gn bng tn trn o.6. Tt in thoi di dng trc khi vo phng th nghim.

IU II. VO PHNG TH NGHIM SINH VIN PHI:

1. Ct cp, ti xch vo ni quy nh, khng mang dng c nhn vo phng th nghim.2. Khng mang thc n, ung vo phng th nghim.3. Ngi ng ch quy nh ca nhm mnh, khng i li ln xn.4. Khng ht thuc l, khng khc nh v vt rc ba bi.5. Khng tho lun ln ting trong nhm.6. Khng t di chuyn cc thit b th nghim

IU III. KHI TIN HNH TH NGHIM SINH VIN PHI:1. Nghim tc tun theo s hng dn ca cn b ph trch.2. K nhn thit b, dng c v ti liu km theo lm bi th nghim.3. c k ni dung, yu cu ca th nghim trc khi thao tc.4. Khi my c s c phi bo ngay cho cn b ph trch, khng t tin sa cha.5. Thn trng, chu o trong mi thao tc, c thc trch nhim gi gn tt thit b.6. Sinh vin lm h hng my mc, dng c th nghim th phi bi thng cho Nh trng v

s b trim th nghim.7. Sau khi hon thnh bi th nghim phi tt my, ct in v lau sch bn my, sp xp thit

b trv v tr ban u v bn giao cho cn b ph trch.IU IV.

1. Mi sinh vin phi lm bo co th nghim bng chnh s liu ca mnh thu thp c vnp cho cn b hng dn ng hn nh, cha np bo co bi trc th khng c lm

bi k tip.2. Sinh vin vng qu 01 bui th nghim hoc vng khng xin php s b cm thi.3.

Sinh vin cha hon thnh mn th nghim th phi hc li theo quy nh ca phng o to.4. Sinh vin hon thnh ton b cc bi th nghim theo quy nh sc thi nhn im ktthc mn hc.

IU V.1. Cc sinh vin c trch nhim nghim chnh chp hnh bn ni quy ny.2. Sinh vin no vi phm, cn b ph trch th nghim c quyn cnh bo, trim thi.

Trng hp vi phm lp li hoc phm li nghim trng, sinh vin s bnh ch lm thnghim v s ba ra hi ng k lut nh trng.

Tp.HCM, Ngy 20 thng 09 nm 2009

KHOA IN-IN T

( k)

PGS TS PHM HNG LIN

TRNG H TN C THNGKHOA IN IN T

---------------------

CNG HA X HI CH NGHA VIT NAMc lp - Tdo - Hnh phc

**************


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Tai Liu Hng Dn Thi Nghim Vin Thng Bai 1

1-2

Hnh 1.3

Ch siu ch (m) l t s ca bin nh tn hiu tin tc trn bin nh ca tn hiusng mang. Phng php tnh xc nh ch siu ch (m) t tn hiu AM da vo hnh1-3 v cng thc di y:

M =BA

BA

+

Phn trm iu ch l ch siu chc biu din theo phn trm (m x 100).

Hnh 1.4

Dng sng AM trnh by trong hnh 1-4 l iu ch 100% (ch siu ch bng 1); cc imtrng tip xc vi ng chun zero. Cng sut v hiu sut trong truyn AM lin quan trctip n ch siu ch; iu ch 100% l nhm t cng sut bin ln nht.

Hnh 1.5.

Khi qu iu ch xy ra (ln hn 100%), c hai bin ca hnh bao iu ch bng qua ngchun zero, nh hnh 1-5. Trong truyn thng AM, qu iu ch gy ra cc tn s bin gi gil bin splatter. Splatter ny gy ra mo trong my thu v nhiu vi cc trm radio khc.

Khi ch siu ch tng, mc cng sut ca cc bin (PSB) tng khi cng sut sng mang (PC)gi khng i. Bi v thng tin c ch c cha trong tn hiu tn s radio (RF) nm trongcc bin, do c tht cc i cng sut bin bng cch tng ch siu ch mong mun.Tuy nhin, trong AM, ch siu ch phi khng c ln hn 1, nu khng mo v nhiu sxy ra.

Cng sut tng (PT) trong tn hiu AM l tng ca cng sut sng mang (PC) v cng sutbin di v bin trn (PSB).

PT = PC + PSB


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1-3

Hiu sut truyn () l phn s ca cng sut tng mc cha trong cc bin.

=T

SB

P

P

Hiu sut truyn cng lin quan n ch siu ch.

= m

2

/ (2 + m

2

)2. My pht AM

Hnh 1.6.

C nhiu cch to ra mt tn hiu AM. Nh hnh 1-7, mch tch hp biu ch cn bng(IC), mt phn t phi tuyn, c th c chc nng iu ch bin . Cc tn hiu tin tc vsng mang l cc tn hiu vo mch iu ch bin .

Hnh 1.7

Mch iu ch bin trn tn hiu tin tc v sng mang dch tn s ca tn hiu tin tc:n dch tn hiu tin tc n tn s ca tn hiu sng mang. Hnh bao ca tn hiu sng mang l

bn copy ca tn hiu tin tc v thay i ti cc tn s ging vi tn s tn hiu tin tc.

B khuch i cng sut RF l phn cui cng trc khi n anten pht. N cung cp skhuch i cng sut cn thit cho anten bc x cc tn hiu RF trn cc khong cch di.

Hnh 1-8 trnh by s ca mch khuch i cng sut trong khi mch AM/SSBTRANSMITTER. Vng ny gm c RF POWER AMPLIFIER (Q1) v ANTENNAMATCHING NETWORK c chc nng to trkhng ng ra Q1 cn thit truyn cngsut yu cu n anten trkhng thp (c m phng bi R5).


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1-4

Hnh 1.8.

in dung C1 trong b chn cch ly in p cc nn DC bi in trR1. Cc in trR2v R3 lm thnh mch chia p cung cp in p DC khng i cc nn khong 4.8V.

B khuch i cng sut RF (Q1) l mt b khuch i E chung. Q1 lun c phn ccthun bi v in p DC cc nn Q1 khng i. Kt qu l Q1 hot ng nh mt bkhuch i lp A.

Hnh 1.9.

Trong b khuch i lp A, dng cc thu dn 3600 cho tn hiu ng vo, v tn hiu ng rakhng b mo. Trong b khuch i lp B, dng cc thu dn 1800 cho tn hiu ng vo, vtn hiu ng ra b mo. Trong b khuch i lp C, dng cc thu dn t hn 1800 cho tn hiung vo, v tn hiu ng ra b mo ng k (xem hnh 1-9).

i vi trng hp pht qung b, hiu sut l quan trng bc nht trong cc b khuch icng sut bi v bt k suy hao cng sut u dn n t cng sut hn cho qu trnh bc xv tn hiu khng truyn c xa. Cc b khuch i lp C cung cp hiu sut tt hn lp B,v cc b khuch i lp B cung cp hiu sut tt hn lp A.

3. My thu AM

Hnh 1-10 trnh by s khi n gin ca my thu AM qung b. Tng RF v b dao ngni c iu chnh ng thi, tng RF s khuch i tn hiu AM. Bi tn chuyn tn sAM thnh tn s IF (trung tn) 455 kHz, v tng IF lc v khuch i tn hiu trung tn IF.B tch sng khi phc li tn hiu tin tc t tn hiu IF, v cui cng tng audio khuch itn hiu tin tc n loa.


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1-5

Hnh 1.10.

Xem hnh 1-11. Tn hiu AM t anten my pht AM kt ni n in tr1M (R8) gimcng sut ca tn hiu AM c nhn ti my thu. Trkhng ng vo b khuch i RF xpx 3 k. R8 ni n mt bin p t ngu L4 c t s vng l 7.7. Vi t s ny, trkhng ngvo ca b lc RF s xp x 50 khi b lc cng hng tn s 1000 kHz.

CH : Ti tn s cng hng ca RF FILTER, XL4 = XC7 v mch c trkhng l in trthun.

Hnh 1.11.

Nhc trnh by trong hnh 1-12, mch RF FILTER gm mt in cm v mt in dungcnh mc song song. l mch lc thng di LC song song. Tn s cng hng(fr) xyra khi in khng v in dung bng nhau (XL = XC). Ti fr, trkhng RF FILTER l in trthun v tn hiu ng ra l ln nht.

Hnh 1.12.

iu chnh RF FILTER cho tn s cng hng l 1000 kHz: iu chnh cun in cm (L4)sao cho tn hiu ng ra RF AMPLIFIER ln nht.

RF AMPLIFIER khuch i tn hiu AM 1000 kHz n RF FILTER v tng mc cng sutca n khong 72 dB ( li khong 16,000,000).

Sinh vin iu chnh L5 cho mch RLC cc thu c cng hng 1000 kHz cho li cc i. Khi mch RLC cc thu c cng hng 1000 kHz, in dung v in khngtrit tiu, v mch ch cn li in trthun.

Cc mc cng sut vo v ra ca RF STAGE v cc thnh phn khc ca mt my thu AMthng decibel (dB) lin quan n mc cng sut tham chiu. Thng thng s dng


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1-6

miliwatt (mW). Biu thc di y cho quan h gia cng sut tnh bng dBm vi cng sut(P) tnh bng mW.

dBm = 10 x [log10 (P/1 mW)]

Mt dBm l mt lng cng sut tht s, khc vi dB i din cho t s cng sut. Cch dngca dBm th thun li khi x l vi cc mch nhiu tng. Hiu s gia mc dBm ti ng ra v

ng vo mt tng chnh l li cng sut tnh bng dB ca tng (hnh 1-13).

Hnh 1.13.

Loi IC dng cho biu ch cn bng cng c dng cho bi tn (xem hnh 1-14).Bi tn c hai ng vo: M (tin tc) v C (sng mang). Ng vo tn hiu RF ca bi tn(M) ly t ng ra b khuch i RF. Ng vo tn hiu dao ng ni ca bi tn (C) l mt

tn hiu 1455 kHz c to ra t khi mch VCO-HI.

Hnh 1.14.

Bi v tn hiu ng vo RF n bi tn c ba thnh phn tn s, tn hiu ng ra bi tncn bng s gm cc tn s tng v hiu ca ba thnh phn RF. Cc tn s ng vo RF khng

c bin ng k trong ng ra.B lc IF l mt b lc s c bng thng 20 kHz, n loi b tt c cc tn s di 455 kHz vtrn 465 kHz. cho bi tn to ra mt tn hiu vi tn s hiu l 455 kHz n b lc IF,tn s VCO-HI phi c chnh chnh xc 1455 kHz.

B tch sng ng bao (xem hnh 1-15 v 1-16): trn na chu k dng ca tn hiu ng ra,tin np n in p nh ng vo. V th, in p qua R12 v C10 s bng in p ca tnhiu ng vo (tr st p diode) bi v diode (CR1) c phn cc thun. Khi tn hiu ng ranh hn gi tr ny, diode (CR1) tt v tin (C10) bt u x t t qua in tr(R12) vitc c xc nh bi thi hng RC.

Hnh 1.15.


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1-7

Hnh 1.16.

Trn na chu k dng k tip ca tn hiu ng vo, CR1 dn v C10 c np in n gitr mi c xc nh bi tn hiu ng vo. Qu trnh lp li mt cch tng v lin tc.

C mt gi tr ti u cho thi hng x ca RC. Nu thi hng x qu ln hoc qu nh, ng raca b tch sng s khng cng dng vi hnh bao ca tn hiu AM ng vo. nh hngca thi hng qu ln hoc qu nhc trnh by trong hnh 1-17.

Hnh 1.17.

iu kh khn chnh ca b tch sng hnh bao l phi c mt in p xp x t 0.4 V n 0.6V ri trn diode khi diode dn. Vn trnn nghim trng i vi cc tn hiu nh hay cc

tn hiu iu ch 100%. nh hng ca st p phn cc thun ca diode c trnh by tronghnh 1-18.

Hnh 1.18.

III. YU CU THIT B B chn . Board mch ANALOG COMMUNICATIONS Ngun cung cp 15 Vdc Dao ng k hai knh My pht sng sine/vung

IV. TIN HNH TH NGHIM


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1-8

1. My pht AM

Trnh t tin hnh A - Cc tn hiu AM

Trong phn TRNH TTIN HNH ny, bn s quan st tc ng ca tn hiu tin tc trntn hiu AM.

1. Xc nh v tr cc khi mch AM/SSB TRANSMITTER v VCO-LO v kt ni mchnh trn hnh 1-19. Gn jumperv tr1000 kHz trn khi mch VCO-LO. t cc cng tcS1, S2, v S3 OFF.

Hnh 1.19.

2. Kt ni u d dao ng k knh 1 ti ng vo tn hiu tin tc (M) ca MODULATOR.Khi quan st tn hiu trn knh 1, iu chnh my pht tn hiu c mt tn hiu sng sine0.2 Vpk-pk, 2 kHz ti M.

3. Kt ni u d knh 2 n ng vo tn hiu sng mang (C) ca MODULATOR. Khi quanst tn hiu trn knh 2, iu chnh VCO-LO cho mt tn hiu 0.2 Vpk-pk, 1000 kHz ti C. iuchnh tn s sng mang vi nm NEGATIVE SUPPLY trn b chn , v iu chnh bin sng mang vi nm vn trn khi mch VCO-LO.

4. Kt ni u d knh 2 n ng ra ca MODULATOR. t VERT MODE ca dao ngk v tr ALT, v trigger trn knh 1 (tn hiu tin tc).

5. iu chnh nm in th MODULATOR cho dng sng AM trn knh 2 dao ng kl 2.0 V gia cc nh trn v di, nh hnh 1-20.

Hnh 1.20.

6. Hnh bao tn hiu AM (knh 2) c tn s v dng ging nh tn hiu tin tc (knh 1)? C Khng

7. Chnh tn s tn hiu sng mang (fc) bng 1000 kHz v tn s tn hiu tin tc (fm) bng 2kHz. Cc tn s g hin din trn ph tn s ca tn hiu AM?

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1-9

8. Thay i chc nng tn hiu my pht t mt sng sine thnh mt sng hnh vung. Hnhbao ca tn hiu ng ra AM c thay i t mt sng sine thnh mt sng vung?

C Khng

9. Chnh li chc nng my pht tn hiu cho mt sng sine. Khi quan st tn hiu ng raAM trn knh 2, thay i nm iu khin AF FREQUENCY ca my pht tn hiu thay

i tn s tn hiu tin tc. Tn s ca hnh bao tn hiu AM c thay i tng ng vi tn sca tn hiu tin tc?

C Khng

10.Chnh li tn s tn hiu tin tc n 2 kHz. Khi quan st tn hiu ng ra AM, thay i nmiu khin AF LEVEL ca my pht tn hiu thay i bin ca tn hiu tin tc. Bin ca hnh bao tn hiu AM c thay i tng ng vi bin ca tn hiu tin tc?

C Khng

Trnh t tin hnh B - Ch siu ch v phn trm iu ch

Trong phn TRNH TTIN HNH ny, bn s thc hin cc php o dao ng k ca tnhiu AM v tnh ch siu ch (m) v phn trm iu ch (%Mod.).

Hnh 1.21.

11.Kim tra cc cng tc S1, S2, v S3: cc cng tc ny phi v tr OFF.12.Trn knh 1 dao ng k, iu chnh in p nh-nh ca tn hiu tin tc n 0.2Vpk-pk.

Nu cn, iu chnh nm in th MODULATOR sao cho dng sng AM trnh by trn knh2 l 2.0 V gia cc nh trn v di (xem hnh 1-20). 2.0 V l so A trong hnh 1-21.

Trn knh 2 dao ng k, o cao dc (volt) gia cc im trng trn v di (so Btrong hnh 1-21) ca dng sng c iu ch: B = ____________ V.

13.Tnh ch siu ch: m = _________________________.14.Tnh phn trm iu ch %Mod = _____________________.

Hnh 1.22.


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1-11

1. Kt ni mch my pht AM, nh trnh by trong hnh 1-24.

Hnh 1.24.

2. Trn khi mch VCO-LO, ni jumper vo v tr 1000 kHz.3. t cc cng tt S1 v S2 v tr OFF, S3 v tr ON. Khi S3 ON, ANTENNAMATCHING IMPEDANCE s tng chnh n 330.

4. Kt ni u d knh 1 dao ng k n ng vo tn hiu sng mang ca MODULATOR(C).

5. Khi quan st tn hiu trn knh 1, chnh bin tn hiu sng mang n 0.1Vpk-pk bngcch iu chnh nm trn khi mch VCO-LO v iu chnh tn s tn hiu sng mang n1000 kHz bng cch chnh nm NEGATIVE SUPPLY trn b chn .

6. Kt ni u d knh 2 dao ng k n ng vo tn hiu tin tc MODULATOR (M).7. Khi quan st tn hiu trn knh 2 dao ng k, chnh tn hiu my pht tn hiu c tnhiu sng sine 0.1Vpk-pk, 2 kHz ti ng vo tin tc ca MODULATOR.

Kt ni u d knh 1 dao ng k n ng ra ca anten (R5). t sweep n 0.1 ms/DIV, vtrigger trn knh 2. iu chnh nm MODULATOR dng sng AM c iu ch 100%,nh trnh by trong hnh 1-22.

Trnh t tin hnh B - B lc RF

Trong phn TRNH TTIN HNH ny, bn iu chnh b lc RF cho tn s cng hng

l 1000 kHz, l tn s ca tn hiu c pht.8. Dng jumper kt ni TRANSMITTERn in tr 1M (R8) ti ng vo khi mchAM/SSB RECEIVER (hnh 1-25). Kt ni u d knh 1 dao ng k n ng vo R8. Tnhiu AM c c pht n ng vo R8?

C Khng

Hnh 1.25.

16.B nhng kt ni khng cn thit trong cc phn RF FILTER v RF AMPLIFIER camch my thu ngoi tr jumper kt ni mch TRANSMITTER, hnh 1-26 trnh by s RFFILTER v RF AMPLIFIER.


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1-12

Hnh 1.26.

17.Kt ni u d knh 1 dao ng k n ng ra RF AMPLIFIER. iu chnh L5 imgia cho mt tn hiu xut hin trn knh 1.

18.iu chnh in cm L4 cho tn hiu nh nh cc i ti ng ra RF AMPLIFIER. Tns cng hng (fr) ca RF FILTER l bao nhiu? ____________ kHz.

19.Vi tn hiu sng mang 1000 kHz v tn hiu tin tc 2 kHz, LSB m b lc RF phi choqua l tn hiu g? Tn s bao nhiu?

...................................................................................................................................................

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20.Vi tn hiu sng mang 1000 kHz v tn hiu tin tc 2 kHz, bng thng (BW) ti thiuca RF FILTER cn thit cho qua tn hiu AM thu c?

...................................................................................................................................................

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Trnh t tin hnh C - B khuch i RF: li cng sut cc i

Trong phn TRNH TTIN HNH ny, sinh vin siu hng mch RLC trong mchkhuch i cc thu RF sao cho li RF AMPLIFIER ln nht. Sau o tn hiu ng ra tnh cng sut ng ra rms v li cng sut ca tng RF.

21.Khi quan st tn hiu trn knh 1, iu chnh in cm R5 trong mch cc thu RFAMPLIFIER cho tn hiu sng mang nh nh ln nht ti ng ra RF AMPLIFIER.

22.Trn knh 1, o in p nh nh ca tn hiu sng mang ti ng ra RF AMPLIFIER(VRF(0)). Ghi li kt qu: ______________ V.

23.Chuyn i gi tr VRF(o)pk-pk tnh bc 22 thnh gi tr rms (VRF(o)rms = VRF(o)pk-pkx0.3535). Dng kt qu ny thay vo biu thc di y tnh cng sut rms ca tn hiusng mang ti ng ra RF AMPLIFIER. Trkhng ng ra RF AMPLIFIER l 2 k. Ghi ktqu theo n v microwatt. PRF(o) = VRF (o)

2 / 2 k.

...................................................................................................................................................

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24.Cng sut ng vo v ng ra ca tn hiu sng mang n v i khi tng RF (RF FILTERv RF AMPLIFIER) c trnh by. Tnh v ghi cng sut ng vo theo dBm.

dBmRF(i) = 10 x [log10 (PRF(i) / 1 mW)] = ______________ dBm.

25.Tnh cng sut ng ra theo dBm.dBmRF(o) = 10 x [log10 (PRF(o) / 1mW)] = ______________ dBm.

26.T cng sut ng vo v ng ra theo dBm, tnh li cng sut ca tng RF theo dB.


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1-13

APRF = dBmRF(O) dBmRF(i) = _____________ dB.

27.Kt ni u d knh 1 dao ng k n im M ti khi MIXER. Chnh L5 vim gia mt tn hiu xut hin trn knh 1.

28.Chnh in cm L4 cho tn hiu AM peak-peak ln nht ti ng ra RF AMPLIFIER.29.Chnh in cm thay i (L5) trong mch cc thu RF AMPLIFIER sao cho tn hiu AMc bin nh -nh ln nht ti ng ra RF AMPLIFIER.Tin trnh thc hnh D - Bi tn

Trong phn TI N TRNH THC HNH ny, sinh vin s kho st nh hng ca bMIXER trn tn hiu AM bng cch quan st cc tn hiu ng ra v ng vo ca MIXER.

15.Kt ni ng ra ca khi mch VCO-HI 1455 kHz n ng vo dao ng ni (C) caMIXER nh hnh 1-27. Chnh nm in th VCO-HI xoay mt vng theo chiu kim ng h.Kt ni ng ra MIXERn ng vo IF FILTER bng mt jumper.

Hnh 1.27.

16.Kt ni u d knh 2 dao ng k n ng ra ca MIXER. Vn nm iu chnh caMIXER cho ti khi tn hiu ng ra xut hin nh trong hnh 1-28. Siu chnh ny lm triti tn s VCO-LO 1455 kHz trong tn hiu ng ra.

Hnh 1.28.

17.Kt ni u d knh 2 dao ng k n ng ra ca IF FILTER. Khi quan st ng ra IFFILTER, chnh tn s VCO-HI 1455 kHz bng cch iu chnh nm POSITIVE SUPPLYtrn b chn sao cho tn hiu c in p nh - nh ln nht. Nu vic iu chnh nykhng chnh xc, tn hiu AM s khng xut hin. Kt ni u d knh 2 dao ng k nng ra ca MIXER, v kt ni u d knh 1 n ng vo M ca MIXER.

18.Chnh in th ca MIXER cho tn hiu ng ra c hnh dng r. So snh tn hiu ng ratrn knh 2 vi tn hiu ng vo trn knh 1 ca MIXER. C mt hnh bao iu ch khc

ging nh tn hiu trnh by trong hnh 1-29, trong tn hiu AM ti ng ra MIXER khng?

C Khng


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1-14

Hnh 1.29.

19.Chnh sweep dao ng k n 1s/DIV v trigger trn knh 2. Tn hiu ng ra caMIXER s ging vi tn hiu phc tp trnh by trn hnh 1-30. o thigian gia cc nhca dng sng phc tp, l so xp x ca chu k (T). Mi vch chia ngang l 1 s.

T = _____________s

Hnh 1.30.

20.T chu k (T), tnh ton tn s ca dng sng phc tp (f =1/T) theo n v kilohertz.f = __________ kHz

21. chng t cc thnh phn tn s trong khong tn s 2455 kHz c hin din, chnhsweep dao ng k n 0.2 /DIV. Tn hiu trn knh 2 s xut hin. o v ghi tn s (f).Mi vch chia l 0.2 s.

f = ______________ kHz.

Tin trnh thc hnh G - B lc IF

Trong phn TIN TRNH THC HNH ny, sinh vin s quan st tn s ca b dao ngni nh hng n tn hiu ng ra bng cch so snh cc tn hiu ng vo v ng ra ca blc IF .

23.Kt ni u d knh mt dao ng k n ng ra IF FILTER. Knh 2 s kt ni n ngvo IF FILTER. Chnh sweep dao ng k n 0.2 ms/DIV. Trigger trn knh 2.24.So snh ng ra IF FILTER trn knh 1 vi ng vo IF FILTER trn knh 2. C mt hnh

bao iu ch khc xut hin trong tn hiu ti ng ra IF FILTER khng ?

C Khng

25.Khi quan st tn hiu ng ra IF FILTER trn knh 1, thay i mt t tn s ca tn hiu1455kHz n b MIXER bng cch iu chnh vng chnh tinh ca nm POSITIVESUPPLY trn b chn .


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1-15

26.Ti sao tn hiu ng ra IF FILTER bin mt khi tn s ca tn hiu 1455 kHz n MIXERb tng hoc gim mt t?

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Tin trnh thc hnh H - B tch sng hnh baoTrong phn TI N TRNH THC HNH ny, sinh vin s quan st tn hiu ng raENVELOPE DETECTOR ca my thu v so snh n vi tn hiu tin tc c gi n my

pht kim tra ng ra ca my thu c tht s l tn hiu tin tc hay khng.

27.Kt ni u d knh 1 dao ng k n ng vo tn hiu tin tc MODULATOR trn khimch AM/SBB TRANSMITTER, v kt ni u d knh 2 n ng ra ENVELOPEDETECTOR.

28.Tn hiu ti ng ra ENVELOPE DETECTOR c tn s ging nh tn hiu tin tc khng? C Khng

29.Ti my pht tn hiu, thay i tn s ca tn hiu tin tc 2 kHz. Tn s ng raENVELOPE DETECTOR c thay i theo tn s ca tn hiu tin tc?

C Khng

30.Cc nh m b xn phng do mo ca tn hiu tin tc c gy ra do tn hiu iu ch100% n ENVELOPE DETECTOR. Gim ch siu ch ca tn hiu AM pht bng cchxoay t t nm iu chnh trn khi MODULATOR theo ngc chiu kim ng h, ngthi quan st cc nh m ca tn hiu tin tc c khi phc trli bnh thng.

31.Ti my pht tn hiu, thay i bin ca tn hiu tin tc 2 kHz. Bin ng raENVELOPE DETECTOR c thay i theo bin ca tn hiu tin tc?

C Khng

Bt cng tc CM 9 (chuyn sang v tr ON) thay i thi hng x RC ca mchENVELOPE DETECTOR. Quan st tn hiu tin tc c khi phc. S tng hay gim cngtt CM 9 c thay i thi hng x RC? (sinh vin c th bt cng tc CM 9 ng v mxem s khc nhau.)

VI. KT LUN

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BI 2

IU CH V GII IU CH GC (FM V PM)

I. MC CH

Hon thnh bi th nghim ny, sinh vin c th m t qu trnh iu ch v gii iu ch gcbng cch dng cc khi mch VCO-LO, PHASE MODULATOR, v QUADRATUREDETECTOR trn board mch ANALOG COMMUNICATIONS.

II. C S L THUYT

1. iu ch

Qu trnh iu ch gc gm hai dng: iu ch tn s (FM) v iu ch pha (PM). FM v PMc lin h vi nhau; khi ci ny thay i th ci kia cng thay i theo. Nh hnh 2-1(a) minhha, khi bin tn hiu tin tc l zero, khng c thay i tn s sng mang FM; tn hiuc tn sti tn s trung tm ca n, tc l tn s sng mang cha iu ch. Cc thay idng hay m trong tn s sng mang so vi tn s trung tm ca n c gi l lch tn

shay di tn.Sau y l ba khi nim iu ch tn s (FM) m sinh vin cn nh:

1. Tn s tn hiu sng mang ch thay i t l vi bin tn hiu tin tc.2. Tn s ca tn hiu tin tc khng nh hng n lch tn s ca tn hiu sng mangnhng nh hng n tc lch.

3. Bt k cc thay i bin ca sng mang FM u khng cha thng tin ca tn hiutin tc; ch cc lch v tn s mi cha thng tin.

Nh minh ha trong hnh 2-1, lch tn s sng mang (cng hoc tr) trong tn hiu FM lln nht khi bin ca tn hiu tin tc c gi trnh ln nht hay nh nht. Khi tn hiu tin

tc l zero, lch tn s sng mang l zero, bi v sng mang ti tn s trung tm ca n.

Hnh 2.1.

Quan st hnh 2-2. Bi v cc thay i bin FM khng cha bt k thng tin ca tn hiu

tin tc, bin ca sng mang FM c thc gii hn trong khong gi tr mong mun. Ktqu l cc gai bin nhiu c thc gim bi cc mch gii hn. Cc b khuch i lpC hiu sut cao, chnh hng bin nhng khng nh hng tn s, c thc dngtrong thit b FM.

Truyn FM c kh nng loi nhiu ngu nhin tt hn truyn AM. Bi v s loi nhiu tthn my thu, FM c th truyn trn khong cch ln hn AM vi cng mt tn s v cngsut my pht


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2-2

Hnh 2.2.

di tn ca tn hiu FM l lch ti a ca tn s tn hiu FM so vi tn s sng mang.

Hnh 2.3.

Chsiu chFM (MI) l t s ca di tn trn tn s tn hiu tin tc (fm)

= / fm

V d, nu tn hiu tin tc 5 kHz (fm) gy ra lch tn s sng mang l 10 kHz (), MI s

l 2. = 10/5 = 2

Phn trm iu ch(% Mod) l t s gia di tn vi tn s sng mang. Tn s trung tmca tn hiu sng mang l tn s sng mang khng c iu ch: tn s khng c tn hiu tintc (bin l zero).

Khi lch sng mang l t hn 75kHz qui nh bi FCC cho iu ch 100% (hnh 2-4),phn trm iu ch cng gim. V d, lch ca 56.25 kHz (3/4 ca 75 kHz) l iu ch75%. lch ca 37.5 kHz (1/2 ca 75 kHz) l iu ch 50%.

Hnh 2.4.


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2-3

Ging nh AM, cc bin tn hiu sng mang FM cha thng tin tn hiu tin tc (hnh 2-5).Ph ca tn hiu FM gm mt vch phtn s cf ng vi thnh phn tn s sng mang v

cc cp vch ph bn cc tn s mc ff , mc ff 2 , . . . ln ca cc vch ph ny c

tnh da vo hm Bessel.

Hnh 2.5.

Bng thng ca tn hiu FM l di tn s m trong tp trung 99% cng sut tn hiu FM.Bng thng ny ph thuc vo hai thng s : bin v tn s tn hiu tin tc. Carsson a ra cng thc tnh gn ng bng thng tn hiu FM, gi l quy tc Carsson, nh sau:

BW 2( + mf ) (3)

vi mf l tn s tn hiu tin tc.

Bng sau y ch ra s lng cc vch ph (k c vch ph trung tm) nm trong bng thngtn hiu FM ng vi gi tr cho trc ca ch siu ch:

Ch siu ch 1 2 3 4 5 6 7 8 9 10

S vch ph 7 9 13 15 17 19 23 25 27 29

Bng 2.1.

Trong PM, tn hiu sng mang thay i pha ca n (v tn s) theo thay i trong bin vtn s tn hiu tin tc (hnh 2-6). dch pha t l vi bin tn hiu tin tc. Khi pha casng mang thay i, cng xy ra s lch tn s. lch tn s ( di tn) t l vi tc vlng dch pha. Tc dch pha t l vi tn s ca tn hiu tin tc. V th lch tn strong PM i xng trc tip vi bin v tn s ca tn hiu tin tc.

Hnh 2.6.

i vi FM, s thay i tn s tn hiu tin tc khng gy ra cc lch tn s trong tn hiusng mang. Ch khi bin tn hiu tin tc thay i mi gy ra s lch tn s sng mang(xem hnh 2-7).


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Hnh 2.7.

c tstn hiu trn nhiu tt trong iu ch pha v tn s, ngi ta thng dng cc bkhuch i lp C hiu sut cao bi v mo bin khng nh hng cht lng tn hiu. Ccnhc im ca pht thanh FM v PM l bng thng rng v s cn thit phi thit lp ngtruyn thng (line-of-sight) cho cc tn hiu FM.

2. Gii iu ch

Cc b gii iu ch FM cn gi l cc b tch sng hay cc b tch pha. B tch sng cu

phng l mt trong cc mch gii iu ch FM. Cc mch gii iu ch FM khc gm btch sng Foster-Seeley, b tch sng t s, b tch sng m xung, v b tch sng dngvng kho pha. Tt c cc mch ny chuyn s thay i tn s FM thnh bin v tn sca tn hiu tin tc.

Phn u ca my thu FM rt ging vi my thu AM hay SSB. Nh hnh 2-8 trnh by, phnu ca my thu FM gm mt tng RF, mt bi tn, b dao ng ni, v mt tng IF.

Ngoi ra c th c thm mch tng iu chnh khuch i (AGC). Tn s chun chotng IF l 10.7 MHz, thay v 455 kHz nhiu ch AM. Tng RF FM, bi tn, v tng IFca my thu FM rt ging vi cc mch thu AM tng ng. Bi v cc khi ny ckho st trong phn AM nn chng khng hin din trong my thu FM trn board th nghim(qu trnh gii iu ch).

Hnh 2.8.

Hnh 2-8 trnh by my thu FM gm mt mch hn bin v mt b tch sng. B tch sng lmt thit b lc ra tn hiu tin tc t cc thay i tn s hay pha ca tn hiu sng mang.Thut ng mch tch sng thng c dng thay cho thut ng b gii iu ch FM. Btch sng to ra tn hiu tin tc c ln ph thuc vo lch tn s sng mang: tc thay i ca tn s sng mang xc nh tn s ca tn hiu tin tc, v lch tn s khngi xc nh bin ca tn hiu tin tc.

Mch deemphasis mang phn tn s cao ca tn hiu tin tc tr li thnh cc tn s thp hnvi chnh xc cao. Tng audio khuch i v iu khin m lng cho tn hiu tin tcc khi phc.

Thut ngcu phngdin t quan h v pha gia hai tn hiu c tn s bng nhau v vungpha 900. Trn board mch th nghim, khi mch QUADRATURE DETECTOR thc hinchc nng b tch sng: gii hn v gii iu ch tn hiu FM. B tch sng FM chuyn


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2-5

lch v tc thay i tn s thnh bin v tn s ca tn hiu tin tc. Sn gin cab tch sng cu phng c trnh by trong hnh 2-9.

Hnh 2.9.

Ti ng vo ca b tch sng cu phng, tn hiu FM chia thnh hai ng. Mt tn hiuFM i n b gii hn / b dch pha, c tc dng chuyn di tn thnh di pha vi gi trlch pha khong 900. Cc tn hiu c gi l cu phng khi chng c pha lch nhau 900.Tn hiu FM ban u v tn hiu FM dch pha l hai ng vo ca b tch sng pha, tc biu ch cn bng. B tch sng pha to ra mt tn hiu vi tn s bng hai ln tn s FM vmt in p DC thay i theo s lch pha gia hai ng vo.

Bi v s lch pha gia cc tn hiu ng vo b tch sng pha thay i theo lch tn s catn hiu FM nn in p ng ra DC ca b tch sng pha thay i theo bin v tn s catn hiu tin tc FM. Mt b lc thng thp ti ng ra ca b tch sng pha loi b tn hiu tns cao v cho qua in p ng ra DC thay i, chnh l tn hiu tin tc c khi phc.

III. YU CU THIT B

B chn F.A.C.E.T. Board mch DIGITAL COMMUNICATIONS 2 Ngun cung cp 15 Vdc Dao ng k hai knh My pht sng sine V.O.M

IV. TRNHT TH NGHIM

1. iu ch FM v PM

Tin trnh thc hnh A iu ch tn s (FM)

Trong phn TI N TRNH THC HNH ny, sinh vin siu ch tn s tn hiu sngmang, o cc thng s ca n, v quan st cc c tnh ca n.

Tn hiu FM c pht bi khi mch VCO-LO. Trong bi ny, sinh vin st jumperv

tr 452 kHz. (cha thc hin lc ny). Nm iu chnh trn khi mch VCO-LO dng iu chnh bin ng ra. iu chnh tn s ng ra VCO-LO, sinh vin siu chnh nmNEGATIVE SUPPLY trn phn bn tri ca b chn .

Sn gin ca khi mch VCO-LO c trnh by trong hnh 2-10. Tn s b dao ngc xc nh bng cch iu hng mch LC. Sinh vin c thiu hng mch LC bngcch thay i gi trin th NEGATIVE SUPPLY ti anode ca diode varactorCR2. Gi trca in p m tc ng n in dung ca CR2, do tc ng n tn s cng hng camch LC. Khi in p NEGATIVE SUPPLY trnn m hn th tn s ng ra ca VCO-LO


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tng. Ti 0V DC, tn s ng ra khong 310 kHz. Ti 10 V DC, tn s ng ra khong 510kHz.

Hnh 2.10.

1. Trn khi mch VCO- LO, ni jumperv tr 452 kHz (hnh 2-11). Xoay nm chnh bin mt vng theo chiu kim ng h.

2. Kt ni u d knh 2 dao ng k n (FM)OUT trn VCO-LO. Chnh knh 2 n 200

mV/DIV v TIME/DIV n 0.5 s/DIV.3. Chnh voltmeter cho in p DC. Ktni voltmeter DC n T trn khi mch VCO-LO.

4. thay i tn s VCO-LO, chnh NEGATIVE SUPPLY n gi tr 4.0 V DC tiim T.

5. Kt ni u d knh 1 dao ng k n T.Chnh knh 1 n 1.0V/DIV v Vertical Mode nv tr DC.

6. Chnh knh 1 v knh 2 dao ng k cc dng sng xut hin nh trong hnh 2-12. Tnhiu DC knh 1 (ti T - ch th4.0V) s nm trn vch chia th hai tnh ca mn hnhdao ng k.

Hnh 2.12.

7. o chnh xc chu k (T) gia cc nh ca tn hiu sng mang FM khng c iu chtrn knh 2. Ghi kt qu theo n v ms: T = _________ ms.

8. T chu k (T), tnh ton tn s trung tm (f) ca tn hiu sng mang FM khng c iuch. Ghi kt qu theo n v kHz: f = 1/T = ___________ Hz.

9. Sinh vin thay i in p m phng s thay i bin tn hiu tin tc (mt tn hiusine). Quan st mn hnh dao ng k, chnh nm NEGATIVE SUPPLY theo chiu kim

Hnh 2-11


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2-7

ng h v sau theo ngc chiu kim ng h cho in p DC trn knh 1 thay ikhong 1V DC. Trn knh 2, tn s sng mang FM c thay i khi cc in p DC trn knh1 thay i khng?

C Khng

10.Xc nh lch tn s ca sng mang FM khi bin tn hiu tin tc thay i 1V DC.Chnh nm NEGATIVE SUPPLY theo chiu kim ng h thay i in p ti T trn khimch VCO-LO n 5.0 V DC.

11.o chnh xc chu k (T) gia cc nh ca tn hiu sng mang FM c iu ch trnknh 2. Ghi kt qu theo n vs: T = _________s.

12.T T, tnh ton tn s ca tn hiu sng mangFM c iu ch. Ghi kt qu bng kHz.

f = ____________ kHz

13.Tnh lch tn s FM khi bin ca tnhiu tin tc thay i 1V. Tn s trung tm FM lgi tr sinh vin tnh c trong bc 8 v tn s

ca tn hiu tin tc -1 V DC l gi tr sinh vin vatnh ton trong bc 12: = _________ kHz

14. chnh tn s sng mang trli gi tr tn strung tm, chnh nm NEGATIVE SUPPLY theochiu kim ng h thay i in p ti T trn khi mch VCO-LO t -5.0 V DC n -4.0V DC.

15.By gisinh vin s quan st nh hng ca tn hiu tin tc 2 Vpk-pk, 5 kHz i vi tn ssng mang FM. Kt ni SIGNAL GENERATOR n (M) trn khi mch VCO-LO, nhtrong hnh 5-15.

16.Chnh SIGNAL GENERATOR to mt sng sine 2.0Vpk-pk, 5 kHz ti T. iu chnh nytng ng vi thay i in p ti T l 1V.

17.Chnh TIME/DIV n 0.5 s/DIV, v trigger trn knh 2. Quan st trn knh 2 thy tnhiu FM. Khi bin tn hiu tin tc l 1V, in p ti T gim n 5V v tn s tng t tns trung tm n gi tr sinh vin tnh ton c trong bc 12. Khi bin tn hiu tin tc l1V, tn s s gim so vi tn s trung tm.

18.Tnh ton ch siu ch (MI) cho mt tn hiu FM vi lch tn s (fcd) sinh vin xcnh bc 13 v vi mt tn hiu tin tc 5 kHz (fm) (MI = fcd /fm).

.........................................................................................................................................

19.Dng gi tr MI tnh ton c bc 18, tnh s vch ph c trong bng thng tn hiuFM (xem bng 2-1). Nu MI khng phi l s nguyn, dng MI cao nht k tip tnh slng cc vch ph. S vch ph = ________ .

20.Vi mt tn hiu tin tc 2.0 Vpk-pk, 5 kHz, cc cp ph bin chim khong 5 kHz trn mibin ca tn s trung tm. Tnh ton bng thng (BW) ca tn hiu tin tc. Ghi kt qu theon v kilohertz. (BW = SSB x 5 kHz x2 = _______ kHz)

Tin trnh thc hnh B Siu ch pha (PM)

Trong phn TIN TRNH THC HNH ny, sinh vin s thc hin iu ch pha mt tnhiu sng mang, o s thay i pha, v quan st cc c tnh ca n.

Hnh 2-13


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Sinh vin dng khi mch PHASE MODULATOR, nh trong hnh 2-14(a), to ra mt tnhiu PM. Sn gin ca khi mch PHASE MODULATORc trnh by trong hnh2-14(b).

Hnh 2.14(a) .Hnh 2-14(b).

Vic iu hng mch LC xc nh s dch phaca tn hiu sng mang. Sinh vin iu hngmch LC bng cch thay i gi tr ca in pPOSITIVE SUPPLY ti cathode ca diode

varator CR5. Gi tr ca in p POSITIVESUPPLY xc nh in dung CR5, t nhhng n tn s cng hng ca mch LC. Sinhvin s chnh in p POSITIVE SUPPLY ng ra ca MODULATOR (ng vo LIMITER)l cng pha vi tn hiu sng mang.

Khi in p POSITIVE SUPPLY tng, ng ra ca MODULATOR sm pha hn ng vo. Khiin p POSITIVE SUPPLYgim, ng ra ca MODULATOR chm pha hn ng vo (hnh2-15).

Khi mt tn hiu tin tc sng sine c a n M, ln ca n lm cho pha ca ng raMODULATOR thay i.

V bin ca tn hiu PM khng cha bt k thng tin g ca tn hiu tin tc, cc gai nhntn hiu b gy ra do nhiu c th b ct b ci thin t s tn hiu trn nhiu ca tn hiuPM. Mch LIMITTER gi bin ca tn hiu PM trong phm vi mong mun (hnh 2-16).Mch LIMTTER gm mt opamp vi li l 1.0 v hai diode Schottky (CR6 v CR7) cni t ng ra n ng vo. Chiu phn cc ca cc diode b ngc nhau: cc anode ni ncc cathode. Cc diode phn cc ngc gii hn cc nh m v dng ca tn hiu n gitrin p thun ca diode. Khi nh tn hiu PM tin n khong 0.2 V, diode vi anode can c ni n ng ra dn v duy tr in p nh dng PM ti gi tr khong 0.2 V. Tngt, diode cn li duy tr in p nh m PM ti 0.2 V. Cc in p khng mong mun trn vdi 0.2 V b loi b bi mch LIMTTER.

CH : Diode Shottky c st p thun thp trong khong t 0.2 V n 0.4 V.

Hnh 2.16.

Hnh 2-15


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21.Trn khi mch VCO-LO, ni jumpervo v tr 452 kHz.

22.Kt ni (FM) OUT trn khi mchVCO-LO n im C ca khiMODULATOR trn khi mch PHASEMODULATOR (hnh 2-17).

23.Trong phn THC HNH ny, ccu d dao ng k nn chnh n v trX10 v tn hiu b suy hao nhiu. Kt ni u d knh 1 dao ng k n im C trn khimch PHASE MODULATOR. Chnh knh 1 ti 200mV/DIV v TIME/DIV bng 1.0s/DIV.

24.Vn nm iu chnh trn khi mch VCO-LO chnh bin tn hiu ti C bng600mVpk-pk.

25.Chnh volmetercho in p DC. Ktni volmeter DC n im T trn khi mchVCO-LO.

26.Chnh nm NEGATIVE SUPPLY gc trnbn tri b chn c in p -4.5V DC ti T,tc l gi tr tn s tn hiu ti C khong 475 kHz.

27.Kt ni u d knh 2 dao ng k n imgia MODULATOR v LIMITER. Chnh knh 2n 200 mV/DIV.

28. thay i tn s VCO-HI, chnh nm POSITIVE SUPPLY gc trn bn phi ca bchn tn hiu ng ra MODULATOR (knh 2) cng pha vi tn hiu VCO-LO (knh 1),nh trong hnh 2-18.

29.Khi quan st mn hnh dao ng k vn t t nm POSITIVE SUPPLY trn b chn theo chiu kim ng h v sau theo ngc chiu kim ng h. Thay i in pPOSITIVE SUPPLY tc l thay i bin tn hiu tin tc.

30.Khi in p POSITIVE SUPPLY thay i, quan h pha gia tn hiu ng raMODULATOR trn knh 2 vi tn hiu ng vo knh 1 c thay i khng?

C Khng

31.Chnh in p POSITIVE SUPPLY cc tnhiu trn knh 1 v knh 2 cng pha.

32.Quan st nh hng ca mt tn hiu tin tc3Vpk-pk, 5 kHz i vi tn s sng mang PM. Ktni SIGNAL GENERATOR n im M trn

MODULATOR. Kt ni u d knh 1 dao ngk n M, nh trong hnh 2-19. chnh knh 1 n1 V/DIV v TIME/DIV bng 0.1 ms/DIV.

33.Chnh SIGNAL GENERATOR to mtsng sine 3Vpk-pk, 5kHz ti M trn khiMODULATOR.

34.Kt ni u d knh 1 dao ng k n C, vkt ni u d knh 2 dao ng k n im gia MODULATOR v DETECTOR. Cc u

Hnh 2.18

Hnh 2-19

Hnh 2-17


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d nn v tr X10. Chnh knh 1 v knh 2 n 200mV/DIV, v chnh TIME/DIV n1s/DIV. Trigger trn knh 1 v chnh VERT MODE v tr ALT.

35.Knh 1 hin th sng mang khng c iu ch, v knh 2 hin th tn hiu iu ch pha.Cc tn hiu ny nh th no?

.........................................................................................................................................

36.Kt ni u d knh 1 dao ng k n ng ra LIMITER. So snh cc tn hiu ng vo vng ra ca LIMITER. LIMITER c lm gim bin ca tn hiu PM khng?

C Khng

2. Gii iu ch

Tin trnh thc hnh A - B dch pha v mch hn binTrong phn TIN TRNH THC HNH ny, sinh vin s quan st b dch pha lm thay i

pha ca tn hiu sng mang FM nh th no v mch hn bin lm gim bin ca tn hiusau khi dch pha nh th no.

Mch gii hn/dch pha gm mt b khuch i, mt tin, mt mch LC, v mt mch hnbin. Tn hiu FM chia thnh hai ng ti ng vo ca b tch sng cu phng. Mt tnhiu FM l ng vo n b khuch i Opamp o vi li bng 2. Tin c tc dngdch pha tn hiu FM i 900. Bi v tn s cng hng (fr) ca mch LC bng tn s trung tmFM nn tn s trung tm FM, mch LC trthnh in trthun. Do s dch pha 900 catn s trung tm khng bnh hng bi mch LC. Tuy nhin, cc tn s ln hn hay nhhn fru b dch pha t hay nhiu hn 90

0 mt cch tng ng, so vi tn hiu FM ban u.

Tn hiu FM sau khi dch pha c a n b gii hn. Nh hnh 2-20 trnh by, mch hnbin gm c hai diode c ni t ng ra xung t vi cc cc ca n c b tr ngcnhau: cc anode ni vi cc cathode. Cc diode c cc ngc nhau lm gii hn bin ngra v ti thiu ho bt k siu ch AM no m b dch pha c th gy ra.

Hnh 2.20.

Hnh 2.21.

1. Trn khi mch VCO-LO, ni jumperv tr 452 kHz (hnh 2-21).


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2. Kt ni (FM) OUT trn khi mch VCO-LO n ng vo FM trn khi mchQUADRATURE DETECTOR.

3. Kt ni u d knh 1 dao ng k n ng vo FM trn khi mch QUADRATUREDETECTOR. Chnh knh 1 n 100 mV/DIV vTIME/DIV bng 1s/DIV. Vi nm iu chnhtrn VCO-LO, chnh tn hiu sng mang FM cha

iu ch ti ng vo FM n 300 mVpk-pk.4. Ko nm vn TIME VARIABLE dao ng kra, v iu chnh tn hiu knh 1 cho mt chuk (3600) chim 8 vch chia ngang, nh tronghnh 2-22. Mi vch chia ngang dao ng k bng450.

5. Kt ni u d knh 2 dao ng k n ng raca PHASE SHIFTER/LIMITER trn khi mchQUADRATURE DETECTOR. Chnh knh 2 n200mV/DIV. Chnh tn s FM bng cch iu chnh nm NEGATIVE SUPPLY gc trn

bn tri ca b chn cho n khi tn hiu trn knh 2 c bin ln nht. Nu cn thit,chnh li tn hiu knh 1 cho mt chu k (3600) chim 8 vch chia ngang.

6. Khi bin ng ra ca PHASE SHIFTER/LIMITER l ln nht, tn s trung tm FMbng vi tn s g:

Tn s cng hng mch LC (fr) Mt na tn s ca tn hiu FM gc?

7. lch pha gia tn hiu sng mang FM cha iu ch trn knh 1 v tn hiu ng raPHASE SHIFTER/LIMITER trn knh 2 l bao nhiu?

= ____________

8. Trong mch PHASE SHIFTER/LIMITER, thnh phn no gy ra dch pha 900 gia cctn hiu ng vo v ng ra:

B khuch i in dung Mch LC Mch hn bin?9. Chnh nm NEGATIVE SUPPLY trn b chn mt vng theo chiu kim ng h vsau theo ngc chiu kim ng h thay i tn s FM. Ti sao s lch pha gia cc tnhiu ng ra v ng vo tng v gim trong khi bin lun lun gim: do mch LC gy radch pha khi tn s FM thay i hay do dch pha 900 gy ra bi tin thay i?

....................................................................................................................................................

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10.Nu mt tn hiu tin tc iu ch sng mang FM, ng ra PHASE SHIFTER/LIMITER sthay i pha tu theo lch tn s ca tn hiu FM c iu ch?

..................................................................................................................................................

..................................................................................................................................................

11.Ti sao tn hiu ng ra PHASE SHIFTER/LIMITER trn knh 2 c cc nh v trngbng phng?

Mch hn bin hn ch bin ca tn hiu ng ra

Tn hiu ng ra khng cng pha vi tn hiu ng vo?

Hnh 2-22


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12.Khi quan st tn hiu ng ra PHASE SHIFTER/LIMITER (knh 2), gim bin ca tnhiu ng vo (knh 1) n khong 100 mVpk-pk v sau tr li n 300 mVpk-pk bng cchvn nm iu chnh trn khi mch VCO-LO theo chiu kim ng h v sau theo ngcchiu kim ng h.

13.Khi gim bin tn hiu ng vo n 100 mVpk-pk, tn hiu ng ra PHASESHIFTER/LIMITER (knh 2) c trthnh mt sng sine hay bng phng hn khng?

C Khng

Tin trnh thc hnh B B tch sng pha v b lc

Trong PHN TIN TRNH THC HNH ny, sinh vin s kho st b tch sng pha v blc khi phc tn hiu tin tc.

Hnh 2.23.

Tn hiu FM ban u v tn hiu FM dch pha 900 t mch hn bin c kt hp vi nhau b tch sng pha, l mt biu ch cn bng (hnh 2-23). Biu ch cn bng ny kthp cc tn s ng vo FM to ra cc thnh phn tn s tng hiu. Do cc tn hiu ng voc tn s bng nhau nn tn s tng bng 2 ln tn s FM v thnh phn tn s hiu trthnhin p DC thay i theo lch pha so vi gi tr trung tm l 900. Tn hiu tin tc gy racc lch pha tn s FM, v c chuyn thnh s khc nhau v pha, do thnh phn inp DC (hiu) ng ra b tch sng pha thay i trc tip theo tn hiu tin tc. V th, ng raca b tch sng pha cha thnh phn tn s tng v thnh phn tn hiu tin tc.

Mch RL ti ng ra ca b tch sng pha l mt b lc thng thp c tc dng loi b thnhphn tn s tng v cho qua thnh phn tn hiu tin tc.

15.Kt ni mch v cc u d dao ng k knh 1 v knh 2 nh hnh 2-24.

Hnh 2.24. Hnh 2.25

16.iu chnh cc tn hiu trn dao ng k knh 1 v knh 2, nh hnh 2-25. TN HIU KNH 1: iu chnh 300mVpk-pk. iu chnh dao ng k mt chu k(3600) chim 8 vch chia.

TN HIU KNH 2: iu chnh nm NEGATIVE SUPPLY trn b chn cho dngsng trn knh 2 vung 900 so vi pha ca tn hiu knh 1.


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2-13

17.Kt ni u d knh 2 n ng ra PHASE DETECTOR.18.Tn hiu ng ra ca PHASE DETECTOR trn knh 2 l g? Thnh phn tn s tng Thnh phn tn s hiu

19.Thnh phn tn s hiu ng ra ca PHASE DETECTOR l g? Tn s trung tm FM Mt in p DC

20.Yu t no thay i in p ng ra DC ca PHASE DETECTOR (thnh phn hiu)? Cc thay i tn s Cc thay i pha

21.Kt ni u d knh 2 n ng vo ca PHASE DETECTOR. Nu cn thit, chnh tn sFM bng nm vn NEGATIVE SUPPLY cho lch pha gia cc tn hiu ng vo trnknh 1 v 2 l 900 (xem bc 16).

22.Chnh voltmetero cc in p DC. Kt ni dy dn ca voltmetern ng ra caPHASE DETECTOR, v kt ni dy chung n t. Vi s khc pha 900 gia cc tn hiung vo, o v ghi in p DC ti ng ra ca PHASE DETECTOR (V900 = _______ V)

23.iu chnh lch pha gia cc tn hiu trn knh 1 v 2 bng 1350bng cch chnh nmvn NEGATIVE SUPPLY theo ngc chiu kim ng h (hnh 6-31(a)). Vi lchpha1350 gia cc tn hiu ng vo, o v ghi in p DC ti ng ra ca PHASE DETECTOR:

V1350 = _________ V

Hnh 2.26(a). Hnh 2.26(b).

24.iu chnh lch pha gia cc tn hiu trn knh 1 v 2 n 450 bng cch chnh nmvn NEGATIVE SUPPLY theo chiu kim ng h (xem hnh 2-26(b)). Vi lch pha 450gia cc tn hiu ng vo, o v ghi in p DC ti ng ra ca PHASE DETECTOR:

V450 = ______ V

25.Khi lch pha tng hay gim t gi tr 900, in p ng ra DC c thay i khng? C Khng

26.iu chnh s khc nhau v pha gia cc tn hiu trn knh 1 v 2 trli 900 (hnh 6-8)bng cch chnh tn s FM vi nm NEGATIVE SUPPLY.

27.Sinh vin siu ch sng mang FM bi mt tn hiu tin tc 300 mVpk-pk, 3 kHz. Kt niSIGNAL GENERATORn (M) trn khi mch VCO-LO (hnh 2-27).


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2-14

Hnh 2.27.

28.Kt ni u d knh 1 n T trn VCO-LO. y nm TIME VARIABLE trn dao ngk vo, v xoay n mt vng theo chiu kim ng h. Chnh knh 1 n 100 mV/DIV, chnhTIME/DIV n 0.1 ms/DIV, v trigger trn knh 1.

29.Chnh SIGNAL GENERATOR pht mt sng sine 300mVpk-pk, 3 kHz ti im T trnVCO-LO (knh 1).

30.Kt ni u d knh 1 n ng vo FM ca QUADRATURE DETECTOR. Kt ni ud knh 2 n ng ra PHASE SHIFTER/LIMITER. Chnh knh 1 n 100mV/DIV, v chnhknh 2 n 200mV/DIV. Chnh TIME/DIV n 0.5 s/DIV. Chnh VERT MODE v trALT, v trigger trn knh 1.

31.So snh cc tn hiu ti hai ng vo PHASE DETECTOR. Pha ca tn hiu FM trn knh2 c ang thay i theo tn hiu trn knh 1?

C Khng

32.Kt ni u d knh 2 n ng ra ca PHASE DETECTOR quan st tn hiu tn stng. Trn knh 2 mc DC, tc ng chun zero (im gia) ca tn hiu tn s tng, cthay i?

C Khng

33.Chnh TIME/DIV ca dao ng k n 0.2 ms/DIV. Kt ni u d knh 1 n im T tikhi mch VCO-LO quan st tn hiu tin tc. Trigger trn knh 1. Quan st cc thay iDC ca tn hiu ng ra PHASE DETECTOR trn knh 2.

34.So snh tn hiu tin tc trn knh 1 vi cc thay i DC ca ng ra PHASE DETECTORtrn knh 2. Cc thay i DC ca ng ra PHASE DETECTOR c tn s ging nh tn hiutin tc khng?

C Khng

35.Kt ni u d knh 2 n ng ra ca FILTER. Chnh knh 2 n 50mV/DIV. Quan sttn hiu tin tc trn knh 1 v ng ra FILTER trn knh 2. Thay i tn s v bin tn hiutin tc. Tn hiu tin tc c khi phc trn knh 2 c thay i theo bin v tn s tn hiutin tc trn knh 1 khng?

C

Khng36.FILTER c loi b tn s tng hay tn s tn hiu tin tc khng?

C Khng

V. KT LUN

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________


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3-1

BI 3

M HA NG TRUYN (Line Coding)

I. MC CH

Khi han tt bi th nghim ny, sinh vin c th m tc ba k thut m ha ngtruyn thng dng trong thng tin s l NRZ, RZ v Manchester, v gii thch c u nhcim ca tng loi. Sinh vin cng hiu c cch m ha ng truyn bng mt trong ba

phng php trn cng nh cch gii m v nhng yu cu cn thit gii m my thu.

II. C S L THUYT

KNH TRUYN

Hnh 3-1

Vic truyn d liu gia hai thit b vin thng c thc hin theo mt trong hai phngthc: truyn tn hiu di nn (cn gi l tn hiu bng gc) v truyn tn hiu bng di(passband). Trong truyn tn hiu di nn, tn hiu tin tc sc truyn trc tip trn knhtruyn m khng cn iu ch. Khi truyn tn hiu di nn th cc c tnh ca tn hiu phi

ph hp vi knh truyn.

Cc c trng ca tn hiu bao gm: thng tin nh thi, s dch mc DC do chui 0 hocchui 1 ko di v ph tn s ca tn hiu.

1. M ha

M ha l thay i cch thc truyn d liu trn ng truyn. Cc k thut m ha khcnhau u c mt s trao i gia cc yu t: p ng tn s, bng thng, vn nh thi vs dch mc DC.

Cc k thut m ha c kho st trong bi th nghimny bao gm: NRZ (non-return-to-zero), RZ (return-to-zero) v Manchester.

a) M ha NRZ:

Cc mc in p nh phn ca d liu NRZ c ginguyn trong sut mt chu k bit v khng trv 0 trongsut chu k bit.

b) M ha RZ:

Cc mc nh phn ca d liu c biu din bng mcin p tng ng trong na chu k bit, sau trv 0trong na chu k bit k tip.

c) M ha Manchester:


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3-2

Cc bit 1 v bit 0 ca d liu c biu din bngmc in p tng ng trong mt na chu k bit u,v bng mc in p ngc li na chu k bit sau.

Cs la chn k thut m ha thch hp l davo ba c trng ca tn hiu nu trn. Cc ctrng ny c chn sao cho ph hp vi knhtruyn v m bo sng b gia bn pht v

bn thu. Mt s phng php m ha c kt hpthng tin v xung clock ngay trong d liu truyn,mt s phng php khc i hi phi truyn cxung clock v d liu. Khi c s chuyn i mccc chu k bit th my thu c th da vo khi phc xung clock. Nu cc tn hiu d liutruyn i khng c s chuyn trng thi trong ccchu k bit th buc phi truyn thm xung clock ring ng b. V phng din ny, mManchester c u im nht v n m bo t nht mt ln chuyn i mc trong mt chu k

bit.

V cch biu din cc mc logic bng in p, c hai dngsng tn hiu c dng: dng n cc (5V v 0V) v dnglng cc (+5V v -5V).

C th thc hin giao tip vi cc knh truyn bng hai cch:ghp trc tip hoc ghp AC. Trong h thng ghp AC, thnh

phn DC b chn v cc thnh phn tn s thp b suy gim.

Nhiu phng tin truyn c c tnh tn s ging nh mt b lc thng di, ch cho phptruyn tn hiu trong mt di tn t 1Cf n 2Cf .

Hnh v di y th hin cc thnh phn nng lng ca cc k thut m ha nu trn.

Phng php NRZ lng cc c nhc im l huht cc thnh phn nng lng ca n t p trung khong tn s zero (DC). Trong khi , phng phpManchester di cc thnh phn nng lng ln vngtn s cao hn nn c th truyn c c trn cc

knh truyn c p ng b chn vng DC. Tuynhin, i li, phng php Manchester li lm tngbng thng ca tn hiu, do c th dn n li khi truyn trn knh c bng thng gii hn.


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3-4

Nu ng vo ca cng XOR, ta t tn hiu NRZ v tn hiu xung clock th ng ra ca ns l tn hiu m ha Manchester.

gii m tn hiu m ha Manchester v dng NRZ, ta li dng mt cng XORmy thu.

Khi truyn trn khong cch xa, s kt ni trc tip gia my pht v my thu l khng khthi. Do s khng c ng tn hiu clock chung. D liu truyn i phi c iu ch

bng cc k thut nh ASKM, FSK, v sc gii iu ch ti my thu. ng thi, my thu phi c mt b phn ng b xung clock khi phc tn hiu clock t d liu pht.

III. YU CU THIT B

B chn F.A.C.E.T. Board mch DIGITAL COMMUNICATIONS 2 Ngun cung cp 15 Vdc Dao ng k hai knh My pht sng sine V.O.M

IV. TRNHT TH NGHIM

1. M ha

Trnh t tin hnh A M ha NRZ1. Xc nh v tr khi mch ENCODER. Kt ni u d knh 1 ca dao ng k vi imCLK v knh 2 vi im SYNC.

2. Ni u d EXT vi im SYNC ng b cc tn hiu quan st vi nhau. Ni GNDca dao ng k vi im GND trn board mch.


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3-5

Hnh 3-1

3. iu chnh knh 1 v knh 2 ca dao ng k v tr 5V/DIV v time base v tr0.5ms/DIV. Trigger bi cnh ln ca ng vo EXT.

4. Chnh dao ng k c dng sng nh hnh 3-2.

Hnh 3-2

5. o chu k xung clock trn knh 1: T = ___________ ms.6. C bao nhiu chu k xung clock gia cc xung ng b? S chu k clock = __________.7. Di chuyn u d knh 2 n v tr NRZ. Khng thay i cc thng s ca dao ng k.

Hnh 3-3

8. Mi chu k xung clockng vi 1 bit d liu. Hy xc nh chui bit nh phn (8 bit) cpht? Data = _____________.

9. Quan st dng sng NRZ, tn hiu c gi nguyn mc in p (cao hoc thp) trong sutchu k bit?

C

Khng

10.Thng tin nh thi l mt c trng quan trngca cc phng php m ha ng truyn. Mt ctrng khc cng rt quan trng l tc truyn dliu. Tc baud (tc truyn d liu) c nhngha l nghch o ca khang thi gian ca mt

phn t tn hiu ngn nht.

11.Chu k xung clock c o bc 5: T = _____ms. Vy tc baud ca tn hiuNRZ quan st c l bao nhiu: baud rate = ________ baud.

Hnh 3-4


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3-6

12.Tc truyn d liu cng c nh ngha theo s bit truyn c trong 1s, gi l tc bit. Tc bit v tc baud c th bng nhau hoc khc nhau, ty vo phng php m hav cc phn t tn hiu.

13.Tc bit (tnh bng bps) ca tn hiu NRZ nu trn l bao nhiu: R = _______ bps.14.Tc bit v tc baud ca tn hiu NRZ quan st c c bng nhau khng? Ti sao?

__________________________________________________________________________.

Trnh t tin hnh B M ha RZ

15.Tho tt c cc jumper v dy ni trn board mch DIGITAL COMMUNICATIONS 2.16.Ni u d knh 1 ca dao ng k n v tr CLK v knh 2 n v tr SYNC, u dEXT cng ni vi v tr SYNC.

17.Chnh knh 1 v knh 2 ca dao ng k 5V/DIV v TIME/DIV = 0.5ms/DIV. Triggerbi cnh ln ca EXT.

Hnh 3-5

18.iu chnh dao ng k quan st c dng sng nh hnh 3-6.

Hnh 3-6

19.Di chuyn knh 2 dao ng k n v tr RZ.

Hnh 3-7

20.Da trn tn hiu RZ quan st c, hy xc nh chui bit nh phn pht i? _________.21.Quan st tn hiu RZ trn knh 2. Tn hiu ny mang thng tin v xung clock: mt cch y ch mt phn khng mang thng tin g

22.Tc baud ca tn hiu RZ trn knh 2: baud rate = ________ baud.


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3-7

23.Tc bit ca tn hiu RZ quan st c trn knh 2? R = ________ bps.24.Tc bit v tc baud ca tn hiu RZ khng bng nhau? V sao?

__________________________________________________________________________.

Trnh t tin hnh C M ha Manchester


Hnh 3-8

27.Chnh knh 1 v knh 2 ca dao ng k 5V/DIV v TIME/DIV = 0.5ms/DIV. Triggerbi cnh ln ca EXT.28.iu chnh dao ng k quan st c dng sng nh hnh 3-9.

Hnh 3-9

29.Di chuyn knh 2 dao ng k n v tr MAN.

Hnh 3-10

30.Tn hiu trn knh 2 biu din chui bit nh phn (8 bit) no? ____________.31.Quan st tn hiu RZ trn knh 2. Tn hiu ny mang thng tin v xung clock: mt cch y ch mt phn khng mang thng tin g

32.Tc baud ca tn hiu m ha Manchester trn knh 2: baud rate = ________ baud.33.Tc bit ca tn hiu m ha Manchester quan st c trn knh 2? R = ________ bps.34.Tc bit v tc baud ca tn hiu RZ khng bng nhau? V sao?

__________________________________________________________________________.


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3-8

Lu : mt khi nim lin quan n baud rate l bng thng tn hiu. Tn hiu c tc baudcng cao i hi bng thng cng rng.

35.Di chuyn knh 1 v v tr NRZ. Quan st v o tc baud ca tn hiu NRZ?baud rate = ________ baud.

36.Gia hai tn hiu NRZ v Manchester, tn hiu no c bng thng ln hn? NRZ Manchester

Trnh t tin hnh D Tn hiu n cc v lng cc



40.iu chnh dao ng k quan st c dng sng nh hnh 3-9.41.

Kt ni jumperkhi MODULATORS nh hnh v 3-11. Di chuyn u d knh 1 v vtr NRZ v u d knh 2 v v tr POLAR (hnh 3-11).

Hnh 3-11

42.Chnh vertical mode v GND cho c hai knh. Sau iu chnh v tr ng chun cahai knh nh hnh 3-12.

Hnh 3-12 Hnh 3-13

43.Chnh vertical mode trv DC. Knh 1 hin th tn hiu NRZ dng n cc cn knh 2hin th tn hiu NRZ lng cc (hnh 3-13).44.Quan st tn hiu n cc. Mc nh phn 1 c biu din bng in p dng cn mc 0c biu din bng in p zero. Dng sng tn hiu nm pha trn ng chun. o mcin p dng biu din bit 1: V1 = _______ V.

45.Quan st tn hiu lng cc. Mc 1 biu din bng in p dng v mc 0 biu dinbng mc in p ngc li. Dng sng tn hiu nm v hai pha ca ng chun. o mcin p m biu din bit 0: V0 = _______ V.


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3-9

46.Tho knh 2 ca dao ng k ra khi mch v chuyn knh 1 n v tr POLAR trn khiMODULATORS (hnh 3-14). Knh 1 s hin th chui bit 10110100 dng NRZ lng cc .

Hnh 3-14

47.Dng ng ho DC o thnh phn DC ca tn hiu NRZ lng cc ni trn.DC = ___________ mVDC

48.Hiu chnh mch chuyn d liu nh phn thnh 01000100. Dng ng ho DC o li thnh phn DC ca tn hiu NRZ lng cc mi.

DC = __________ VDC

49.Tho jumper khi v tr NRZ trn khi MODULATORS. Dng dy dn ni t v tr MANtrn khi ENCODER sang khi MODULATORS nh hnh 3-15.

Hnh 3-15

50.Dng ng ho DC o thnh phn DC ca tn hiu Manchester lng cc ni trn.DC = ___________ mVDC

51.Hiu chnh mch chuyn d liu nh phn thnh 01000100. Dng ng ho DC o li thnh phn DC ca tn hiu Manchester lng cc mi.

DC = __________ mVDC

52.T kt quo, cho bit phng php m ha no c thnh phn DC nh nht (khngquan tm n data), m NRZ hay Manchester?1. Gii m

Trnh t tin hnh A Gii m tn hiu RZ

1. Xc nh v tr khi mch ENCODER v ni u d knh 1 ca dao ng k n v trCLK, knh 2 n v tr SYNC, u d EXT cng ni vi v tr SYNC.

2. Chnh knh 1 v knh 2 ca dao ng k 5V/DIV v TIME/DIV = 0.5ms/DIV. Triggerbi cnh ln ca EXT.


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3-10

3. iu chnh dao ng k quan st c dng sng nh hnh 3-9.4. Trn khi ENCODING, gn jumper ni gia RZ v ng vo ca khi D Flip-Flop. Dichuyn knh 2 n v tr RZ (xem hnh 3-16).

Hnh 3-16

5. Quan st d liu RZ (knh 2) ng vo D Flip-Flop. Khi gii m RZ thnh NRZ, tn hiuclock phi:

truyn trn ng truyn ring c khi phc t d liu RZ

6. Di chuyn knh 2 n ng ra D Flip-Flop v knh 1 n v tr NRZ trn khi ENCODER.

Hnh 3-17

7. So snh d liu m ha NRZ (CH1) v d liu NRZ sau khi gii m (CH2). Tn hiu giim b tr so vi tn hiu m ha:

mt chu k xung clock na chu k xung clock

8. Di chuyn u d knh 1 sang v tr RZ (ng vo D). S chuyn trng thi t thp ln caoln u tin ca xung clock lm ng ra Q ca b gii m chuyn sang:

mc cao mc thp

9. D liu gii m NRZ ph thuc mc d liu RZ ti mi cnh ln ca xung clock. chuk bit th 5, d liu NRZ ng ra l bit 1 hay 0? _________.

Trnh t tin hnh B Gii m tn hiu m ha Manchester

10.Xc nh v tr khi mch ENCODER v ni u d knh 1 ca dao ng k n v trCLK, knh 2 n v tr SYNC, u d EXT cng ni vi v tr SYNC.


12.iu chnh dao ng k quan st c dng sng nh hnh 3-9.13.Trn khi ENCODING, gn jumper ni gia RZ v ng vo ca khi D Flip-Flop. Dichuyn knh 2 n v tr RZ (xem hnh 3-18).


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3-12

v tr SYNC.


21.iu chnh dao ng k quan st c dng sng nh hnh 3-9.22.Trn khi ENCODING, gn jumper ni gia MAN v ng vo ca khi MAN SYNCDECODER Flip-Flop. Vn nm iu chnh LOCK ht c theo chiu kim ng h. Chuynu d knh 2 sang v tr RCLK trn khi MAN SYNC DECODER (xem hnh 3-24).

Hnh 3-24

23.Trong khi quan st cc dng sng, vn t t nm LOCK theo ngc chiu kim ng hcho n khi tn hiu RCLK (knh 2) ging vi tn hiu CLK (knh 1). Lc ny, mch ng b bm theo tn hiu m ha Manchesterng vo. Cc thnh phn ca mch ng b baogm: mch pht hin cnh (EDGE DET), cng AND (AND), vng kha pha (PLL) v bdch pha ( SHIFT) (xem hnh 3-25).

Hnh 3-25

24.Di chuyn knh 1 ti v tr MAN v knh 2 ti v tr EDGE DET. Tn hiu MAN l tnhiu vo EDGE DET mch pht hin cnh. ng ra ca EDGE DET l mt chui xung hp.Mch EDGE DET c to t cng AND v cng XOR (hnh 3-26).

Hnh 3-26

25.Quan st cc dng sng. Mch pht hin cnh to ra mt xung hp khi no?_______________________________________________________________________.

26.Di chuyn knh 2 n v tr SHIFT v knh 1 n v tr EDGE DET.


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3-13

27.Quan st cc dng sng. Tn hiu ra cng AND cc v tr xung th 4 v th 9 ca EDGEDET smc cao hay thp? ________________.

28.Di chuyn knh 2 n v tr AND OUT. Tn hiu ng ra cng AND ging vi tn hiu b EDGE DET tr cc xung v tr:

u chu k bit gia chu k bit

29.Cc xung ng ra cng AND c a ti ng vo ca mch vng kho pha (PLL). Dichuyn u d knh 2 ti v tr VCO OUT v u d knh 1 n v tr AND OUT.

30.So snh cc tn hiu. Tn s ca tn hiu sng vung ca PLL (knh 2) bng vi: tn s ng ra cng AND hai ln tn s ng ra cng AND

31.Di chuyn knh 2 ca dao ng k n v tr RCLK. So snh cc dng sng. Tn s tnhiu ra khi b chia (knh 2) c bng vi tn hiu u ra cng AND (knh 1)? Lu rngcc xung nh thi v tr AND OUT ng b vi xung clock ca my pht.

C Khng

32.Di chuyn knh 1 n v tr SHIFT. So snh cc dng sng. Tn hiu trn knh 2 l mtsng vung ging vi tn hiu RCLK nhng b lch pha mt gc bng bao nhiu? ________.

33.Di chuyn knh 1 n v tr NRZ trn khi ENCODING. Di chuyn knh 2 n v trNRZ OUT trn khi MAN SYNC DECODER.

34.So snh tn hiu m ho NRZ (knh 1) v tn hiu NRZ sau khi gii m. D liu sau khigii m b tr so vi d liu m ho:

1 chu k xung clock chu k xung clock chu k xung clock

V. CU HI

1. Cc phng php m ho NRZ, RZ, Manchester.2. Thng tin nh thi c cha trong d liu m ho Manchester bng cch no?3. Phn bit tn hiu n cc v lng cc.4. u v nhc im ca m ho Manchester so vi m ho NRZ v RZ ?5. nh ngha tc baud.6. M ho NRZ v RZ, phng php no tt hn xt kha cnh thng tin v xung clock ?7. Phng php gii m tn hiu Manchester thnh d liu NRZ ?8. Phng php gii m tn hiu RZ thnh d liu NRZ ?9. Hot ng ca mch ng b xung clock ?


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4-1

BI 4

IU CH FSK

I. MC CH

Khi hon tt bi th nghim sinh vin c th: M tc s lin h gia tn hiu iu ch FSK vi tn hiu siu ch (baseband). M tc vic iu ch FSK s dng analog multilexer. M tc ph tn ca tn hiu FSK

II. C S L THUYT

a s knh truyn tng t khng ph hp ti tn hiu s baseband.

Hnh 4-1

Bng thng gii hn ca knh truyn lm mo dng tn hiu v tn hiu s c cc thnh phnph nm ngoi bng thng ca knh truyn. Khi , cn phi phi c siu ch tn hiu, v ti to li tn hiu cn phi c s gii iu ch. Tn hiu c iu ch bi sng mang nmtrong bng thng ca knh truyn do khng c s suy hao tn hiu.

Hnh 4-2

FSK l mt dng ca iu tn (FM), trong tn hiu iu ch (baseband) iu khin tn ssng mang. Khc vi FM , tn hiu iu ch trong FSK l tn hiu s. Tn s sng mang cchuyn i gia hai tn s, vic chuyn i ny da trn tn hiu iu ch.

C hai loi gii iu ch FSK: ng b v bt ng b. Vic gii iu ch bt ng b sdng phng php lc a v dng ASK v s dng b tch sng ng bao ti to tn


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4-2

hiu s baseband. Vic gii iu chng b cn phi c mt tn hiu tham chiu v t spht hin c s thay i ca tn s sng mang t ti to li tn hiu baseband.

Hnh 4-3

1. iu ch FSK

Hnh 4-4

Vic iu ch FSK s dng hai chuyn mch (switch) tng tc thit lp nh b dnknh tng t (analog multiplexer). Chuyn mch s ng khi p iu khin mc 5V,chuyn mch ngt (h) khi p iu khin mc -5V. in p iu khin POLAR v POLARINV th c to ra t s thay i mc logic ca tn hiu iu ch. Tn hiu sng mang ngra l tn hiu tng ng vi switch c ng.

2. Gii iu ch FSK bt ng b

Hnh 4-5

B gii iu ch FSK phc hi tn hiu s baseband bng cch pht hin s thay i tn strong tn hiu FSK.

Mt tn hiu FSK bao gm 2 tn hiu on-off keying (OOK). Hai thnh phn OOK c cng

bin .


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4-3

Hnh 4-6

Hnh 4-7

Ph tn ca tn hiu FSK l kt hp ca cc thnh phn tn sc sinh ra t 2 thnh phntn hiu OOK.

Hnh 4-8

Mt b lc thng di c s dng cho qua mt thnh phn tn hiu OOK trong khi lmsuy gim mt thnh phn OOK cn li.

Ng ra ca b lc bin tn hiu s thay i khi tn hiu FSK thay i tn s.

B tch sng bt ng b s phc hi tn hiu NRZ bng cch pht hin bin tn hiu sau

b lc. B lc thng di v b tch sng bt ng b kt hp vi nhau c c chc nngpht hin s thay i ca sng mang.

B tch sng bt ng b c 3 khi thc hin chc nng phc hi tn hiu s baseband ttn hiu c bin thay i (sau b lc thng di): b chnh lu ton k chuyn i tn hiuc bin thay i thnh tn hiu c mc DC tng ng vi cc mc bin , b lcthng thp lm trn li dng tn hiu sau khi chnh lu v mt b so snh phc hi mcin p chun thnh cc mc in p logic.


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4-4

Hnh 4-9

3. Gii iu ch FSKng b

B gii iu chng b s phc hi trng thi ca tn hin baseband NRZ t tn s ca tnhiu FSK.B gii iu chng b s dng b bin i tn s thnh in p, n s dng c tnh thayi tn s ca tn hiu FSK to ra mc in p thay i.

Hnh 4-10

B chuyn i tn s thnh in p bao gm mt vng kho pha c cu hnh pht hins thay i tn s ca tn hiu FSK. Mt b lc thng thp v mt b so snh in p s phchi tn hiu v mc logic 5V.

PLL gi cho tn hiu tham chiu v tn hiu FSKng b vi nhau bng cch hiu chnh tns b VCO.

Hnh 4-11

Mc in p ng vo b VCO iu khin tn s ca tn hiu tham chiu. Khi tn hiu thamchiu v tn hiu FSK c tn s ging nhau, mc in p trung bnh ng vo VCO bm theotn s tn hiu.

Mt b lc thng thp c s dng tch c mc trung bnh ca in p a vo bVCO. Mc in p ny i din cho trng thi tn hiu s baseband. Mc logic c phc hisau b so snh.

Tn hiu ng vo b VCO c sinh ra t b so pha v mt b lc thng thp. B so pha sosnh tn hiu ra t b VCO vi tn hiu FSK. B lc thng thp gip n nh tn hiu trckhi vo b VCO. B VCO, b so pha v b lc thng thp kt hp vi nhau to thnh vngkho pha (PLL).

III. YU CU THIT B

B chn F.A.C.E.T. Board mch DIGITAL COMMUNICATIONS 2


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4-5

Ngun cung cp 15 Vdc Dao ng k hai knh My pht sng sine V.O.M

IV. TRNHT TH NGHIM

1. iu ch FSK

1. Kt ni probe vi ng vo EXT trn oscilloscope, ni probe ny vi im SYNC trn khiENCODER . Trigger theo cnh ln ca EXT. Phng php trigger ny s xuyn sut tintrnh th nghim ny.

Hnh 4-12

2. Kt ni knh 1 ca oscilloscope vi ng ra NRZ ca khi ENCODER v chnhoscilloscope hin th t nht 2 bit ca tn hiu baseband NRZ.

3. Xc nh tc baud ca tn hiu hin th trn knh 1: NRZ = ______baud

Hnh 4-13

4. Kt ni tn hiu NRZ n biu ch FSK (s dng jump). Kt ni knh 2 n ng ra cabiu ch FSK, hiu chnh oscilloscope hin th c 2 tn hiu NRZ v FSK.

5.o tc baud trn knh 2: FSK = _________baud6. C 2 tn hiu c cng tc baud?

ng Sai

7.o bin tn hiu FSK (knh 2) khi tn hiu NRZ mc cao: _______( )p p High V = 8.o bin tn hiu FSK (knh 2) khi tn hiu NRZ mc thp: _______( )

p p Low V

=

9. Quan h gia 2 in p ny? High > Low High < Low High = Low

10. Quan st oscilloscope, xc nh pha ca sng mang trc khi c s thay i tn s ticanh ln v xung ca tn hiu NRZ.

11. Xc nh gc pha ca tn hiu FSK ngay trc khi tn hiu NRZ chuyn t cao xungthp:


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4-7

25.Ni knh 1 n HIGH TONE, knh 2 n LOW TONE. ng, mkha CM 12.26. iu g xy ra hi thay i trng thi kho CM 12?

a. LOW TONE b loi b b. HIGH TONE b loi b

c. LOW TONE v HIGH TONE u b loi b. d. Khng c g xy ra

27. Chuyn knh 2 ti ng ra FSK. ng mkho CM 12. Loi iu ch no xut hin ng ra FSK khi kha CM 12 ng?

On-Off Keying Frequency-shift keying Phase-shift keying

28. Ti tn s no trong ph tn ca tn hiu FSK c bin nh? DC v 600Hz 600Hz v 1200Hz 1200Hz v 2400Hz 2025Hz v 2225Hz

29. Kt ni knh 1 n im POLAR. ng mkho CM 6 v quan st tn hiu FSK.30. Kho CM 6 gy ra s thay i t ngt trong tn hiu FSK?

ng Sai

S gin on lm gia tng rng bng tn ca tn hiu FSK. Cc biu chc thit k

tit kim bng thng bng cch gim s gin on. C th quan st thy pha tn hiu FSKtrng thi 1800 khi thay i tn s.

31. Gc pha ca tn hiu sng mang FSK 2400Hz ti ngay trc thi im tn hiu POLARchuyn t cao xung thp?

00

450 90

0 180

0

Hnh 4-15

32. Chuyn knh 2 ti im HIGH TONE, ng ngt kho CM 6. Kho CM 6 to ra s ginon nh th no?

a. Kha CM 6 thay i tn s ca tn hiu HIGH TONE.b. Kha CM 6 thay i bin ca tn hiu HIGH TONE.c. Kha CM 6 thay i pha ca tn hiu HIGH TONE.d. Kha CM 6 thay i pha ca tn hiu POLAR.

2. Gii iu ch FSK bt ng b

1. Kt ni probe vi ng vo EXT trn oscilloscope , ni probe ny vi im SYNC trnkhi ENCODER . Trigger theo cnh ln ca EXT.


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4-8

Hnh 4-16

2. Kt ni knh 1 ca oscilloscope vi ng ra NRZ ca khi ENCODE v chnhoscilloscope hin th t nht 2 bit ca tn hiu baseband NRZ.

3. S dng jump ni tn hiu NRZ n biu ch FSK v dng dy ni ni tn hiuFSKn ng vo khi CHANNEL.

Hnh 4-17

4. t mc nhiu mc cao nht. Ni knh 2 n ng ra ca khi CHANNEL v iuchnh oscilloscope quan st c 2 tn hiu NRZ v FSK.

Khi CHANNEL m phng mt ng truyn truyn tn hiu FSK sau biu ch.

Hnh 4-18

5. Chuyn knh 2 n ng ra b lc Bandpass. Ni ng ra knh truyn v ng vo b lcBANDPASS.

6. o bin tn hiu ng ra b lc BANDPASS khi tn hiu NRZ mc cao:NRZ high = _________Vp-p

7. o bin tn hiu ng ra b lc BANDPASS khi tn hiu NRZ mc thp:NRZ low = _________Vp-p


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4-9

Hnh 4-19

B tch sng ng bao khi phc tn hiu NRZ t s thay i bin tn hiu c to tb lc BANDPASS.

Hnh 4-20

8. Di chuyn knh 2 n ng ra ca b chnh lu ton k (Full-Wave Rectifier-FWR).9. Ni ng ra b lc thng di vi ng vo b chnh lu tn k.B chnh lu ton k to ra tn hiu ng ra ca n thng qua 2 tng: tng u l b chnhlu chnh xc ton k c o, tng 2 l tng cng (c o) kt hp tn hiu c chnh luvi tn hiu ban u to ra tn hiu c chnh lu ton kng ra.

Hnh 4-21

10. Chuyn knh 2 n ng ra b lc thng thp. B lc thng thp cho tn hiu baseband tns thp qua v lm suy gim sng mang tn s cao. N lm phng tn hiu va c chnhlu v mc DC v c dng gn ging vi tn hiu baseband ban u.

11. Chui bit hin th trn knh 2? _______________ .12. Mc in p sau b lc thng thp (knh 2) c cng mc in p vi tn hiu NRZ (knh1) khng?

C Khng

13. Chuyn knh 2 n ng ra khi so snh in p. Hiu chnh POSITIVE SUPPLY mccao nht. T t chnh POSITIVE SUPPLY xung thp cho n khi ng ra b so snh gingvi tn hiu NRZ knh 1.

14. Chuyn knh 1 n ng ra b lc thng thp. B so snh c khi phc mc 5V logic ttn hiu ti ng ra b lc thng thp?

C Khng


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4-10

Ng ra b so snh mc 5V khi ng ra b lc thng thp cao hn mc ngng c to ra tPOSITIVE SUPPLY. Mc 0V tng ng khi p ng ra b lc thng thp thp hn mcngng.

15. Kha CM 7 thay i CHANNEL. ng mkho CM 7. Tn hiu NRZ khi phc ccn chnh xc khng khi CM7 ng?

C

Khng

Hnh 4-22

16. Chuyn knh 1 ti b lc thng di. Xc nh li bin tn hiu ng ra ng vi tn sthp v tn s cao. ng ngt kho CM7. Kho CM7 nh hng nh th no n ng ra blc thng di?

a. chnh lch bin tng lnb. Bin mc thp gim nhiu hn mc caoc. Bin mc cao gim nhiu hn mc thpd. C hai mc bin gim vi lng bng nhau

17. Chuyn knh 2 n ng ra khi CHANNEL. Quan st tn hiu ng ra CHANNEL khithay i kho CM7.

18. Gii thch snh hng ca CM7 n s thay i ca tn hiu ng ra b lc bandpass?___________________________________________________________________________

___________________________________________________________________________

___________________________________________________________________________

19. Chuyn knh 1 n ng vo khi CHANNEL. Quan st oscilloscope khi ng mkhoCM7.

20. CM7 nh hng n CHANNEL nh th no? Thay i bng thng Thay i li

Thay i tn hiu ng vo Tt c cc pht biu trn

21. Chng ta c th thay i mch nh th no b cho s gim bng thng ca knhtruyn?

a. truyn HIGH TONE vi bin tng ln hn so vi LOW TONE.b. gim bng thng ca b lc bandpass b thuc. s dng tn s sng mang trong khong bng thng ca knh truynd. tt c cc gii php trn

3. Gii iu ch FSKng b

1. Kt ni probe vi ng vo EXT trn oscilloscope , ni probe ny vi im SYNC trnkhi ENCODER . Trigger ca oscilloscope c t theo cnh ln ca EXT.


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4-11

2. Kt ni knh 1 ca oscilloscope vi ng ra NRZ ca khi ENCODER v chnhoscilloscope hin th t nht 2 bit ca tn hiu baseband NRZ.

Hnh 4-23

3. S dng jumper ni tn hiu NRZ n biu ch FSK v dng dy ni ni tn hiuFSKn ng vo khi CHANNEL.4. t mc nhiu mc cao nht. Ni knh 2 n ng ra CHANNEL v hiu chnhoscilloscope hin th c 2 tn hiu NRZ v FSK.

5. Gi tr nh phn ca 2 bit u tin hin th trn oscilloscope? Data bit = _________.6. S dng jumper ni ng ra knh truyn n b SYS DETECTOR v PLL (REF IN)vi ng ra knh truyn.

7. Loi b tt c cc kt ni bn trong khi PLL. Ni knh 1 n ng ra VCO.

Hnh 4-24

8. Ng ra VCO c ng b vi tn hiu FSK? C Khng

9. Ni gia A/F v ng vo b so pha (CIN). ng hi tip t VCO qua D flip flop chophp b so pha khp kn vng kho pha.

10.Ni knh 2 n ng vo b VCO (VCIN) v knh 1 n ng ra knh truyn. c th quan st ng ra b so pha (PC) b lc thng thp RC khng c kt ni vo. Bso pha to ra in p cho ng vo b VCO (VCIN) bng cch so snh 2 ng vo ca n bng

b XOR.

11. Khi no ng ra b so pha mc cao? Khi cc ng vo ging nhau Khi cc ng vo khc nhau

Khi c 2 ng vo mc cao


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4-13

Hnh 4-26

Hnh 4-27

trb thu khong 400us.

22. 2 bit u tin phc hi b thu l g? 2 bit u = __________binary23.Ni knh 1 vi tn hiu NRZ nguyn thy. Tn hiu NRZ v tn hiu phc hi c gingnhau v mc in p?

C Khng

B so snh in p phc hi mc 5V logic bng cch so snh tn hiu sau b lc thng thpvi mt ngng in p c thiu chnh c. Ngng ny c iu chnh bng nm

NEGATIVE SUPPLY.

24.Ni knh 2 ti ng ra b so snh. Hiu chnh NEGATIVE SUPPLY ng ra b so snhl tn hiu cng dng vi NRZ nhng b tr.

25. Tn hiu khi phc c cng mc logic vi tn hiu NRZ. C Khng

26. Chuyn knh 1 n ng ra CHANNEL v knh 2 n ng ra LP FILTER ca khi SYNCDETECTOR. t knh 1 mc 2V/Div, knh 2 mc 500mV/DIV v thi gian qut1ms/DIV.

27. o in p DC cc i trng thi mc logic thp ti ng ra b lc thng thp (knh 2).Maximal low = _________mV DC

Hnh 4-28


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4-14

28. o in p dc cc tiu trng thi mc logic cao ti ng ra b lc hong thp (knh 2).Minimal high = _________mV DC

Hnh 4-29

29. Bt CM7 gim bng thng knh truyn, iu ny dn n bin tn hiu FSK cngb thay i theo tn sb thu. Xc nh li mc in p ra b lc thng thp.

maximal low = _________mV DC

minimal high = _________mV DC

30. S thay i bin tn hiu FSK c nh hng n vic tch sng tn hiu NRZ? C Khng

31. Bt CM10 to s khng lin tc ca pha tn hiu FSK. Xc nh li mc in p ra b lc thng thp.

maximal low = _________mV DC

minimal high = _________mV DC

32.Nhiu pha c lm thay i tn hiu ng ra ca b lc thng thp? C Khng

V. KT LUN__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________


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5-1

BI 5

IU CH V GII IU CH PSK

I. M

C

CHSau bi th nghim, sinh vin c th gii thch c cch to ra tn hiu PSK, ng b sngmang, v tch sng ng b.

II. C S L THUYT

Hnh 5-1

PSK l mt dng iu ch pha, tn hiu sng mang b dch (shift) pha mi khi tn hiu s (tnhiu iu ch) thay i trng thi.

Hnh trn minh ha ti thi im ngay khi tn hiu NRZ chuyn t cao xung thp, tn hiuPSK chuyn v trng thi pha = 00, cng pha vi tn hiu song mang.

So snh vi sng mang, pha ca tn hiu PSK dch 00 khi tn hiu NRZ trng thi thp, v180

0khi tn hiu NRZ trng thi cao.

Hnh 5-2

Hnh 5-2 minh ha khi iu ch PSK trn board mch. Tn hiu s c cc mc logic 0V v+5V. Mt b dch mc chuyn tn hiu ny thnh mc -5V v +5V. Tn hiu ny sau cnhn vi tn hiu sng mang bng biu ch cn bng to ra tn hiu PSK.

Hnh 5-3


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5-2

iu ch PSK cng c thc s dng vi cc tn hiu m ha khc. Hnh 5-3 biu dinmt tn hiu lng cc RZ, tn hiu sng mang v tn hiu PSK sinh ra t vic nhn 2 tn hiu vi nhau bng biu ch cn bng.

Khi gii iu ch PSK, vic ti to li tn hiu sng mang ti pha b thu l cn thit. iuny c thc hin bng vic ng b sng mang pha b thu thng qua cc khi: nhn itn s, PLL , b chia 2, v mch dch pha 900.

Hnh 5-4

Tn hiu c sinh ra sau c kt hp vi tn hiu PSKb Product Detector (tch sngnhn). Tn hiu sau b Product Detectorc qua b lc thng thp, v kt qu l chui xungc sa dng bng b so snh in p ti to li tn hiu s ban u.

III. YU CU THIT B B chn F.A.C.E.T. Board mch DIGITAL COMMUNICATIONS 2 Ngun cung cp 15 Vdc Dao ng k hai knh

IV. TRNHT TH NGHIM

1. iu ch PSK

Trnh t tin hnh A To tn hiu PSK

Trong tin trnh th nghim ny, sinh vin to tn hiu PSK t tn hiu m ha RZ, NRZ,Manchester v kim tra kt qu bng oscilloscope.

1. Kt ni knh 1 osciloscope n v tr SYNC trong khi ENCODING. Hiu chnhoscilloscope hin th mt chu k ca tn hiu SYNC trong khong rng ca mn hnh.

2. Loi b kt ni knh 1 vi SYNC . Ni SYNC vi ng vo EXT trigger ca osciloscope.

Hnh 5-5

3. Trong khi iu ch (MODULATOR) , kt ni NRZ vi n ng vo khi ASK/PSKbng mt jump ni. Dng 1 jumper chn kiu iu ch PSK (hnh 5-5).

4. Chnh nm OFFSET ton thang, v nm BAL (cn bng) gia thang.5. Kt ni knh 1 n NRZ v knh 2 n im kin tra tn hiu trn khi ASK/PSK.


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5-3

Tn hiu s lng cc sau c nhn vi tn hiu sng mang hnh sin bng biu chcn bng.

Hnh 5-6

iu ny c gii thch thng qua s mch ca khi ASK/PSK nh hnh 5-6. Mt ngvo biu ch cn bng l tn hiu sng mang (CARRIER), mc DC ca n c iuchnh bi bin trBAL. Ng vo th 2 c chn thng qua cu ni tng ng vi im A(ASK), P (PSK). Khi im P c kt ni vi ng vo biu ch cn bng, tn hiu s phii qua mt b dch mc trc .

6.

Loi m ha c th c ca tn hiu sa vo? Ch l NRZ NRZ, RZ hoc Manchester

Khi im A c chn, mt in p offset DC iu khin bi bin trOFFSET c cngvi tn hiu s trc khi i vo biu ch cn bng. Ng ra ca biu ch sc a raim kim tra vi mt ng ra o v mt ng ra khng o. Ng ra + c m c thli cc khi mch khc.

7. Khi jumperv tr no th tn hiu c qua b dch mc trc khi vo biu ch cnbng?

A P

8. So snh tn hiu NRZ (knh 1) v tn hiu vo b iu ch cn bng (knh 2) trnosciloscope. Tn hiu vo biu ch cn bng khc vi tn hiu NRZ nh th no?

Tn hiu tn hiu vo biu ch cn bngbo

Mc logic ca tn hiu vo biu ch cn bng b dch v dng lng cc.

C hai trn

9. Quan st osciloscope khi ln lt di chuyn knh 2 n ng ra + v -. Ng ra no 00ngay sau thi im tn hiu NRZ chuyn sang mc thp?

+ -

10.Chuyn knh 2 n ng ra ca khi ASK/PSK. Quan st tn hiu trn osciloscope khi iuchnh nm BAL v tr cc i v cc tiu. Thng s no ca tn hiu PSK thay i khichnh nm BAL?

Tn s Pha dch offset

11.Bin trBAL b cho s bt cn bng ca mch. c kt qu tt nht, hy chnh BALcho n khi nh ca tt c cc chu k sng sin nm trn mt ng thng.


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5-4

Hnh 5-7

12.Tho jumper ni NRZ n ng vo ca khi ASK/PSK. Ni tn hiu RZ t khiENCODERn ng vo khi ASK/PSK. Chuyn knh 1 n ng vo biu ch cn bng.iu chnh osciloscope hin th dng sng nh hnh 5-8.

Hnh 5-8

13.Quan st s dch pha xy ra ti mi thi im chuyn giao ca tn hiu xung. C baonhiu gc pha khc nhau xy ra ti nhng thi im chuyn giao?

5 3 2

14.Chuyn ng ni tn hiu cp cho biu chn v tr MAN. Hiu chnh osciloscope quan st. C th thy rng s dch pha ch xy ra ti thi im chuyn trng thi ca tnhiu s.

Hnh 5-915.Chnh osciloscope vi tc qut thi gian mc 50us/DIV v quan st dng tn hiunh hnh 5-10.


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5-5

Hnh 5-10

16.C th thy tn hiu PSK c pha l 1800 ngay sau khi tn hiu xung mc thp. Hon ttmt chu k pha tn hiu trv 1800 ngay trc khi tn hiu ln mc cao, pha tc thi ca tnhiu PSK l bao nhiu sau chuyn i ny?

00 90

0 180

0

Board mch ny s dng pha 00 v 1800 cho tn hiu sin ti thi im chuyn i tn hiu s.Nhng gi tr pha ny c chn vn gi tnh lin tc ca dng sng iu ch lm gimnhiu ti a.

17.Bt kha CM6 dch pha ca tn hiu sng mang, iu ny cng thay i pha ca tnhiu PSK ti thi im tn hiu s thay i. C kt lun g v s thay i dng tn hiu PSK?

Dng sng khng lin tc Gc pha khng i ti thi im chuyn trng thi

C 2 trn

18.Tho tt c cc kt ni trn board.2. Gii iu ch PSK

Trnh t tin hnh B Tch sng ng b

Trong tin trnh ny sinh vin s s dng b tch sng ng b c c tn hiu basebandt tn hiu PSK thu c. Kt quc kim tra bng osciloscope.

Hnh 5-11

1. Trong khi mch iu ch MODULATORS, s dng 2 jumper chn NRZ lm tn hiuvo v loi iu ch l PSK.

2. Kt ni ng ra PSKn ng vo khi CHANNEL, v ni ng ra khi CHANNEL ti khiSYSC DETECTOR bng jumper.

3. S dng 3 jumper ni kt ni cc phn trong khi SYSC DETECTOR nh hnh 5-11.


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5-6

4. Kt ni u d EXT ca osciloscope n im SYNC v u d knh 1 n im NRZca khi ENCODING. Chnh osciloscope hin th 2 bit u trong chui d liu NRZ trntan b chiu rng mn hnh.

5.Ni knh 2 ti im kim tra sng mang (CARRIER) ca khi iu ch(MODULATORS). C bao nhiu chu k sng mang trong mt chu k bit?

__________________ chu k.

6. Chuyn knh 1 n im CARRIER v knh 2 n ng ra khi CHANNEL. Chnh nmvn NOISE v mc cc tiu.

7. Chnh nm BAL trong khi MODULATORS c c tn hiu PSK nh hnh 5-12.Ch : tn hiu NRZ biu din y khng xut hin trn osciloscope.

Hnh 5-12

8. Chuyn knh 1 n im ng ra RECT ca b DOUBLER (nhn i tn s). So snh tnhiu PSK v tn hiu ng ra b chnh lu. Kt qu ca vic chnh lu tn hiu PSK l:

Cc tnh ca tt cnh ca tn hiu PSK bo

nh dng bo thnh nh m.

nh m bo thnh nh dng.

9. Vic chnh lu loi khi tn hiu PSK: Thng tin dch pha Tin tc C hai yu t trn

Hnh 5-13

Nh vy mc ch ca vic tch sng mang ng b l vic ti to ch sng mang v thngtin dch pha khng c yu cu trong qu trnh ny.

10. Chuyn knh 1 n CARRIER v knh 2 n ng ra b DOUBLER. Thng s no casng mang c nhn i sau khi qua b DOUBLER?

Tn s Chu k


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5-7

Hnh 5-14

11. Di chuyn knh 2 n v tr VCO trong khi PLL. Ng ra VCO c cng tn s vi: Sng mang Tn hiu ng ra b nhn i (DOUBLER)

12. Chuyn knh 2 n VCD (VCO Divided). So snh sng mang v dng sng VCD.

Hnh 5-15

13. VCD l tn hiu t ng ra VCO c chia i tn s. Quan st thy rng 2 tn hiu ccng tn s nhng lch pha. Pha sai lch gia sng mang v chui xung vung VCD l baonhiu?

00 90

0 180

0

Vic ti to sng mang ch cn vic dch pha tn hiu VCD 900 khp vi tn hiu sng

mang.

14. Di chuyn knh 2 n v tr 900 ca b dch pha (PHASE SHIFTER). C th kt lun tvic so snh tn hiu sng mang vi tn hiu sau b dch pha 900?

Tin tc b loi khi tn hiu PSK.

Tn hiu sng mang c phc hi t tn hiu PSK.

C 2 trn.

15. Di chuyn knh 2 n ng ra b dch pha v suy gim, im ny cng chnh l mt tronghai ng vo ca b trn (MIXER). Tn hiu cn li i vo b MIXER l g?

Sng mang Tn hiu PSK VCD

16. Tng nhy knh 2 n 50mV/DIV quan st tn hiu ng ra b dch pha (PHASESHIFTER). Dch tn hiu knh 2 chng ln tn hiu sng mang knh 1 (dng nmPOSITION trn osciloscope). Kim tra li s cng tn s v pha ca 2 tn hiu. Tn hiu ngra b dch pha l:

Tn hiu lng cc (polar) Tn hiu n cc (unipolar)

17. Chuyn knh 1 kn tn hiu PSK ti ng vo b MIXER.


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5-8

Hnh 5-16 minh ha tn hiu ng vo b MIXER. B MIXER l biu ch cn bng. ng raca n l tch ca hai tn hiu ng vo. Ng ra b dch pha l tn hiu lng cc. Do , bMIXER nhn tn hin PSK vi s dng cho xung dch pha mc cao, v nhn vi s mcho xung dch pha mc thp.

Hnh 5-16

V d, nh m tn hiu PSKc nhn vi s m (xung dch pha 0), kt qu to ra mt nh dng ti ng ra MIXER.

18. Dng sng no l ng ti ng ra b MIXER theo hnh minh ha di y? A B C D

Hnh 5-17

19. Chuyn knh 2 n ng ra b MIXER. Hiu chnh osciloscope v nm BAL trn khiMIXER c c dng sng nh hnh 5-18.

Hnh 5-18


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5-9

20. Tn hiu ra b MIXER gm vi nh dng v k l vi nh m. Khi no cc tnhca cc nh thay i?

Mi ln tn hiu PSK qua im zero

Mi ln nh ca tn hiu PSK thay i cc tnh.

Khi tn hiu PSK dch pha.

21. Chuyn knh 1 n ng ra b lc thng thp (LP FILTER) v hiu chnh dao ng k mc 0,5V/DIV. C th kt lun t dng sng knh 1 (b lc thng thp)?

Lm suy gim tn s sng mang. Cho tn s sng mang qua.

22. Chuyn knh 2 n ng ra b so snh. Chnh nm NEGATVE SUPPLY n v tr tonthang. Chnh t t nm NEGATIVE SUPPLY theo chiu ngc li c c tn hiu trnknh 2 nh bn di (lu nm NEGATIVE SUPPLY t mc ngng in p so snh).

Hnh 5-19

23. Chc nng ca b so snh? Chnh dng xung Lc thng thp B pht hin tch (product detector).

24. Di chuyn knh 1 ti v tr NRZ. Hiu chnh osciloscope c c dng sng nh hnh5-20.

Hnh 5-20

Ch : Nu cn thit, chuyn v thay th jumper ti ng ra CHANNEL c c tnhiu ng trn knh 2.

25.

Dng sng hin th trn osciloscope ny cho thy b tch sng ng b c th: Gii m mt tn hiu NRZ Gii iu ch mt tn hiu PSK C hai.

26. Loi b jumperng ra CHANNEL. S dng dy ni kt ni ng ra CHANNEL nng vo khi tch sng bt ng b (ASYNC DETECTOR).


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5-10

Hnh 5-21

27. Di chuyn knh 2 n ng ra b so snh ca khi tch sng bt ng b. Quan st viu chnh nm POSITIVE SUPPLY. C th kt lun g qua cc bc trn?

Mt tn hiu PSK ch c th gii iu ch bi b tch sng ng b.

Mt tn hiu PSK ch c th gii iu ch bi b tch sng ng b hoc bt ng b.

28. Tho b tt c cc kt ni trn board mch.V. KT LUN

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6-1

BI 6

IU CH V GII IU CH ASK

I. MC CH

Khi hon tt bi th nghim ny, sinh vin c th m tc cc nguyn tc cbn cng nhcc khi mch c th thc hin iu ch v gii iu ch s theo phng php ASK.

II. C S L THUYT

Hnh 6-1

Hu ht cc knh truyn tng tu c bng thng gii hn v khng khp vi bng thngca tn hiu s bng gc. truyn tn hiu s trn knh truyn analog, trc tin cn thchin u ch bng mt phng php no . iu ch l trn bng tn gc ca tin tc vi tns sng mang chuyn vo bng tn trng khp vi bng thng ca knh truyn.

Biu ch bin (ASK) to ra nhng thay i trong bin sng mang t l vi tn hiutin tc.

Hnh 6-2

Phng php iu ch PSK c th minh ha bi hnh nh mt cng tc in tc iukhin bng d liu tin tc. Khi d liu bng 1 th sng mang bin ln c ni n ng ra,cn khi d liu bng 0 th sng mang bin nhc ni n ng ra.

Hnh 6-3

Mt trng hp iu ch ASKn gin hn c minh ha bi cng tc in t SPST. Khidata bng 1, kha ng, sng mang c ni n ng ra, khi data bng 0, kha mv sngmang khng c ni n ng ra. Dng ny c gi l kha on-off (OOK - On-Off Keying).


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6-2

S thay i t ngt t tn hiu sng mang (khi data bng 1) trv 0 (khi data bng 0) i hiphi tang them bng thong cho tn hiu sau iu ch.

Hnh 6-4

1. iu ch ASK

Nh trnh by, bin sng mang s thay i gia hai mc ty theo gi tr ca d liu tintc nh phn.

Hnh 6-5

Theo phng php iu ch ASK trn board mch th nghim, u tin d liu tin tc sc dch mc bng cch cng thm mt in p DC dng vo d liu NRZ. Sau , tnhiu c a n biu ch cn bng nhn vi sng mang. Khi data mc 1, li ca

biu ch cn bng sc iu chnh cnh.

Hnh 6-6

Hnh 6-7 th hin s mch n gin ca biu ch ASK.

Hnh 6-7

Khi in p offset ca mch dch mc c chnh v 0 th mc 0 tng ng vi trng thikhng c sng mang ti ng ra, cn mc 1 ng vi dng sng mang y . y l dng ngin ca ASK, g

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Thi Nghiem Vien Thong