CHƯƠNG 6: TRUYỀN DẪN VÔ TUYẾN TÍN HIỆU SỐ...

Bi ging in t thng tin Bin son Ths Nguyn Hong Huy

Trang 90

CHNG 6: TRUYN DN V TUYN TN HIU S THNG DI

(15 tit) PHN 1: L THUYT (12 tit) 6.1 iu ch s 6.1.1. Kha dch bin (ASK) Kha dch bin (ASK Amplitude Shift Keying) cn c gi l kha bt tt (OOK On/Off Keying) iu bin mt sng mang hnh sin bng cch bt tt vi mt tn hiu nh phn n cc.

Hnh 6.1 Cc tn hiu nh phn c iu ch s

Tn hiu ASK c biu din nh sau:

ttAmts ccos)()( trong m(t) l tn hiu d liu bng c s n cc nh hnh 6.1. Tn hiu ASK c th c tch sng bng mt b tch sng ng bao (tch sng khng kt hp) hoc mt b tch sng tch (tch sng kt hp).


Trang 91

Hnh 6.2 Tch sng ASK

6.1.2. Kha dch tn (FSK) Kha dch tn (FSK Frequency Shift Keying) dch tn s mt sng mang hnh sin t mt tn s vt (tng ng vi vic gi i mt s nh phn 1) sang mt tn s trng tng ng vi vic gi i mt s nh phn 0) theo tn hiu bng c s. Tn hiu FSK c biu din nh sau:

)cos()cos(

)(22

11

tAtA

ts

trong 1 c gi l tn s vt v 2 c gi l tn s trng 6.1.3. Kha dch pha nh phn (BPSK) Kha dch pha nh phn (BPSK Binary Phase Shift Keying) s dch pha mt sng mang hnh sin i 00 hoc 1800 bng mt tn hiu nh phn n cc (hnh 6.1). Tn hiu PSK c biu din nh sau:

))(cos()( tmDtAts pc trong m(t) l tn hiu d liu bng c s cc nh hnh 6.1. tch sng BPSK phi s dng tch sng ng b tng t nh hnh 6.2b 6.1.4. Kha dch pha cu phng (QPSK) v kha dch pha M (MPSK) Nu my pht PM c tn hiu iu ch s vi M = 4 mc th tn hiu u ra l tn hiu kha dch pha M (MPSK). Cc mc pha ln lt c th l 00, 900, 1800, 2700 hoc 450, 1350, 2250, 3150. hai nhm tn hiu ny c bn l ging nhau tr lch trong chun pha sng mang. Trng hp MPSK trong pha ca sng mang l 450, 1350, 2250, 3150 cn c gi l QPSK.

Hnh 6.3 S o pha tn hieu MPSK


Trang 92

6.1.5. ieu che bien o cau phng (QAM) Tng t nh MPSK s dung s sai lech pha e ieu che nhng khac cho tn hieu QAM (Quadrature Amplitude Modulation) co the co nhieu mc bien o va nhieu mc pha hn. QAM 16 (co M = 16 mc) va QAM 64 (co M = 64 mc) thng c s dung trong truyen hnh so.

Hnh 6.4 S o pha tn hieu QAM 16

6.2 Phn tch tuyn truyn dn v tuyn 6.2.1. Tn hao ng truyn v cng sut tn hiu thu Cng sut thu c mt anten vi h s khuch i Gr c th biu din nh sau:

p

rr L

GEIRPP .

trong : EIRP = PtGt l cng sut pht x tng ng ca anten ng hng; Pt l cng sut pht; Gt l h s khuch i ca anten pht; Gr l h s khuch i anten thu; Lp l tn hao ng truyn. i vi anten parabol, h s khuch i anten thng c tnh theo cng thc sau:

2472.10 fDG trong : f l tn s sng mang [GHz], D l ng knh gng phn x [m] v l hiu sut mt m. Thng thng = 0.55 0.73 Trong khng gian t do tn hao ng truyn c xc nh nh sau:

2

2)4(dFSL

trong : d l khong cch gia anten pht v anten thu, l bc sng. dng dB:

2

2)4(lg10dFSL

FSL = 92,5 + 20lgf [GHz] + 20lgd [km], dB hay: FSL = 32,5 + 20lgf [MHz] + 20lgd [km], dB


Trang 93

Hp th ca kh trong kh quyn l nguyn nhn gy ra tn hao kh quyn. Cc tn hao ny thng vo khong vi phn ca dB (k hiu l AA). Tng in ly gy ra dch phn cc sng in t dn n tn hao lch phn cc (k hiu l PL). Ngoi ra, s lch nh hng v mt ng chnh hng phn cc ca anten cng gy ra tn hao k hiu l AML Tng tn hao ng truyn Lp khi tri quang ng c xc nh theo cng thc sau: Lp = FSL + RFL + AML + AA + PL [dB] trong : FSL l tn hao trong khng gian t do; RFL l tn hao phid my thu; AA l tn hao hp th kh quyn; PL l tn hao lch phn cc. Phng trnh cho cng sut thu dB nh sau: Pr = EIRP + Gr Lp trong :Pr l cng sut thu [dBW], EIRP l cng sut pht x ng hng tng ng [dBW]; Lp l tn hao ng truyn [dB]. 6.2.2. H s nhiu v nhit nhiu Nhiu nhit: Cng sut tn hiu thu trong mt ng truyn v tuyn thng rt nh, vo khong picowat. Cng sut ny s c my thu khuch i n cng sut ln. Tuy nhin do lun lun c nhiu u vo my thu nn nu tn hiu thu khng ln hn nhiu, khuch i s khng c tc dng v n khuch i c nhiu. Tnh trng ny cn tr nn ti t hn v chnh b khuch i cng b sung thm nhiu gi l nhiu nhit. Trong thit b nhiu nhit gy ra do chuyn ng nhit ca cc in t trong cc vt dn. N c to ra cc phn t ghp c tn hao gia anten vi my thu v cc tng u ca my thu. Mt ph cng sut nhiu nhit khng i tt c cc tn s thp hn 1012 Hz., v th c gi l nhiu trng. Qu trnh nhiu nhit my thu c m hnh ho bng qu trnh tp m trng Gauss cng (AWGN: additive white Gauss noise) v c biu th bng cng sut nhiu cc i c th c u vo b khuch i nh sau: N = kTf [W] trong k = 1,38.10-23 l hng s Bonzmant; T l nhit nhiu o bng Kenvin [K] v f l bng thng knh. Mt ph cng sut tp m (PSD) n bin trong trng hp ny c xc nh nh sau:

kTf

NN

0 [W/Hz]

Nhiu anten: Cc anten thu a nhiu vo cc ng truyn. C th phn chia nhiu do anten a vo thnh hai nhm: nhiu xut s t tn hao anten v nhiu bu tri. Nhiu bu tri l thut ng miu t pht x vi ba t v tr do cc phn t c lm nng trong v tr gy ra. S pht x ny trong thc t bao ph ph rng hn ph vi ba. Nhit nhiu tng ng ca bu tri nhn t anten mt t c cho hnh 6.2. th pha di dnh cho anten hng thng nh u (thin nh) cn th cao hn dnh cho anten hng ngay trn ng chn tri. S tng nhit nhiu trong trng hp th hai l do s pht x nhit ca tri t v y l l do thit lp gii hn di ca gc ngng anten bng 50 bng C v 100 bng Ku.


Trang 94

Cc th cho thy ti u tn s thp ca ph, nhiu gim khi tng tn s. Khi anten hng thin nh, nhit nhiu gim xung cn 3 K ti cc tn s nm trong khong t 1 n 10 GHz. Pha trn 10 GHz c hai nh nhit . Mi c ch tn hao hp th u to ra nhiu nhit v tn ti lin quan trc tip gia tn hao v nhiu nhit.

Hnh 6.5 Hnh 6.5 p dng cho cc anten mt t. Cc anten v tinh thng thng hng xung mt t v v th chng thu pht x nhit t mt t. Trong trng hp ny nhit nhiu nhit tng ng ca anten ngoi tr cc tn hao ca anten vo khong 2900K Cc tn hao anten cng vi nhiu thu t pht x v tng nhit nhiu anten ny l tng ca nhiu tng ng ca tt c cc ngun trn. i vi cc anten bng C mt t, thng thng tng nhit nhiu anten vo khong 60K v i vi bng Ku vo khong 80 K trong iu kin bu tri quang ng. Tt nhin khng th p dng cc gi tr ny cho cc trng hp c bit v chng c dn ra y ch cho ta mt khi nim v cc i lng c th c.


Trang 95

H s nhiu: H s nhiu c nh ngha l t s gia t s tn hiu trn nhiu u vo vi t s tn hiu trn nhiu u ra ca phn t thu:

out

in

SNRSNRNF

H s nhiu ca my thu ch yu c xc nh bi cc tng u ca my thu. hnh 6.6 nhiu gy ra do b khuch i ca my thu c quy i thnh nhiu u vo my thu v c k hiu l Nai.

Hnh 6.6

T hnh 6.6 ta c:

i

ai

aiir

ir

NN

NNAAPNPNF

1)(/

/

trong : Pr l cng sut thu, A l khuch i ca mch gy nhiu, Ni l nhiu u vo v Nai l nhiu quy i u vo ca phn t gy nhiu. c th p dng c NF ta phi s dng ngun nhiu tham kho Ni. Nh vy h s nhiu s cho thy thit b s to ra nhiu ln hn bao nhiu ln nhiu ca ngun tham kho. H s nhiu c th c xc nh i vi ngun nhiu tham kho nhit T = 290K. Khi , mt cng sut nhiu ca ngun tham kho nh sau:

HzdBWNHzWkTN

dB /204/1042901038.1

0

21230

Nhit nhiu Ta c: Nai = (NF 1)Ni Nu thay Ni = kTif v Nai = kTrf, trong : Ti l nhit ngun tham kho; Tr l nhit nhiu hiu dng ca my thu ta c: kTrf = (NF-1)kTif Tr = (NF 1)Ti V ta chn Ti = 290K, nn: Tr = (NF 1)290K Ta thy rng c th m hnh ho mt b khuch i c nhiu nh ngun nhiu b sung (hnh 6.6) hot ng nhit nhiu hiu dng Tr. i vi cc kt cui l in tr thun tu th Tr khng bao gi thp hn nhit mi trng xung quanh tr khi n c lm ngui. Cn lu rng i vi cc u cui l in khng (chng hn cc b khuch i thng s khng c lm ngui) hay cc thit b c nhiu nh khc th Tr c th thp hn 290K rt nhiu. Ta cng c th biu din nhiu u ra Nout ca mt b khuch i ph thuc vo nhit nhiu hiu dng ca n nh sau: Nout = A(Ni + Nai) = A(kTgf + kTrf) = Ak(Tg + Tr) f


Trang 96

trong : Tg l nhit ca ngun Nhit nhiu ng dn sng ng dn sng khc vi b khuch i ch n ch gy tn hao v nhiu. Ta xt mt ng dn sng ch c tn hao nh hnh:

Hnh 6.7

Gi thit ng ny c phi hp tr khng ti ngun v ti. L l tn hao cng sut c xc nh nh sau

Vy h s khuch i l A = 1/L (nh hn mt). Gi s nhit ca tt c cc phn t l Tg. Tng cng sut nhiu u ra l: Nout = kTgf v u ra ca mng ch l thun tr ti nhit Tg. Tng cng sut ngc v mng phi cng bng Nout m bo cn bng nhit. Nhc li rng cng sut nhiu c th kTgf ch ph thuc vo nhit , bng thng v phi hp tr khng. C th coi rng Nout gm hai thnh phn, Ngo v ANLi nh sau: Nout = kTgf = Ngout + ANLi trong : Nout = AkTgf l thnh phn cng sut nhiu u ra do ngun nhiu gy ra v ANLi l thnh phn cng sut nhiu do mng tn hao gy ra; NLi l nhiu ca mng quy i u vo. Kt hp hai phng trnh trn ta c: kTgf = AkTgf + ANLi Gii phng trnh trn tm NLi ta c:

ggL

LgLi

TLTA

AT

fkTfkTA

AN

)1(1

1

TL l nhit nhiu hiu dng ca ng dn sng. Chn nhit tham kho Tg = 290K, ta c th vit: TL = (L -1)290K Do , ta c th biu din h s nhiu hiu dng ca ng tn hao nh sau:

LTNF L 290

1


Trang 97

Nhit nhiu ca ni tng Ta c h s nhiu ca m tng ni tng nh sau:

12121

3

1

21 ...

1...

11

m

m

AAANF

AANF

ANFNFNF

Tng nhit nhiu trong trng hp ny nh sau:

12121

3

1

21 ...

1...

11

m

mtol AAA

TAA

TA

TTT

Hnh 6.8 cho thy mt t chc mch in hnh trong ng phi tn hao L c ni vi b khuch i c h s nhiu NF.

Hnh 6.8 Trong trng hp ny ta c: NFtol = L+L(NF-1) = L NF v h s nhiu ca phi l L v khuch i ca n l 1/L nn: Ttol = (L NF - 1)290K Ta cng c th vit nhit tng ca ng phi v b khuch i nh sau: Ttol = (L NF L + L - 1)290K = [(L-1)+L(NF-1)]290K = TL + LTr i vi cc h thng thng tin trn mt t NF thng c s dng. Cc h thng thng tin v tinh thng s dng khi nim nhit nhiu Nhit nhiu h thng Hnh 6.9 cho thy s ca mt h thng cha cc phn t gy nh hng nhiu nht my thu: anten, phi v b tin khuch i.

Hnh 6.9


Trang 98

Nhit nhiu h thng l tng nhit ca tt c cc phn t chnh ng gp vo nhiu my thu: TS = TA+Ttol trong : TA l nhit nhiu ca anten v Ttol l tng nhit nhiu ca phi v b tin khuch i. Vit li phng trnh trn ta c: TS = TA + TL + LTr = TA + (L - 1)290K + L(NF - 1)290K = TA + (LNF - 1)290K Nu L.NF c cho dB th TS c dng: TS = TA + (10L.NF/10 - 1)290K 6.2.3. T s tn hiu trn nhiu Ba thng s thng c s dng nh gi t s tn hiu trn nhiu l: sng mang trn nhiu (C/N hay Pr/N), sng mang trn mt nhiu (C/N0 hay Pr/N0) v nng lng bit trn mt ph nhiu (Eb/N0). Quan h gia cc thng s ny nh sau: Pr/N0 = (Pr/N) [dB] + 10lg(f) Eb/N0 = (Pr/N) [dB] - 10lg(Rb/f) trong : Pr l cng sut thu sng mang (C); Rb l tc bit; Eb l nng lng bit; f l rng bng tn. Eb = PrTb= Pr/Rb C/N0 v Eb/N0 khng ph thuc vo tn s thng c s dng so snh hiu sut ca cc h thng khc nhau. C/N ph thuc vo rng bng tn ca mt h thng cho trc (chng hn b lc my thu). Ta c: Pr/N0 = EIRP + Gr/T - LP - k [dB/Hz] Lu rng h s khuch i anten thu v nhit nhiu h thng c kt hp chung thnh mt thng s v i khi t s ny c gi l nhy my thu. 6.2.4. D tr Phading Vic phn tch qu ng truyn cho php cn i cc tn hao v li cng sut trong qu trnh truyn dn c th a ra mt lng d tr cng sut cn thit m bo truyn dn trong iu kin khng thun li m vn m bo cht lng truyn dn yu cu. Lng cng sut d tr ny c gi l d tr ng truyn hay d tr phaing v c xc nh nh sau:

req

b

r

b

NE

NEM

00

[dB]

trong : M l d tr ng truyn hay phainh, (Eb/N0)r, (Eb/N0)req l t s nng lng bit trn mt ph cng sut nhiu thu v yu cu. T s theo yu cu c xc nh theo BER yu cu. V tn hiu thu hu ch y thng l sng mang c iu ch nn ta thng ni n t s sng mang trn nhiu (C/N) hay (Pr/N) l t s SNR:

]./[][]/.[][]/[][][0

HzKdBWkdBLsbitdBRdBNEKdB

TGdBEIRPdBM pb

req

br


Trang 99

thay k = -228.6 dBW/K.Hz vo ta c:

]./[6.228][]/.[][]/[][][0

HzKdBWdBLsbitdBRdBNEKdB

TGdBEIRPdBM pb

req

br

Nu xt c tn hao cc phn t ni my pht n anten pht v t nhit tham chun TR = 290 K th ta c th vit li phng trnh (6.66) nh sau:

]/[204]/.[][

][][][][][][][][

0

11

HzdBWsbitdBRdBNE

dBNFdBLdBLdBLdBGdBGdBWPdBM

breq

b

prtt

trong : Pt, Gt, L1 l cng sut, h s khuch i v suy hao cc phn t ni anten pht. Gr, L2 l h s khuch i v suy hao cc phn t ni anten thu. -kTR = 204(dBW/Hz) NF l h s nhiu


Trang 100

PHAN 2: BAI TAP (3 tiet) 1. Mot song mang c ieu che bi mot tn hieu d lieu bang c s cc e tao ra mot tn hieu BPSK, s(t) = 10cos[ct + Dpm(t)], trong o m(t) = 1 tng ng vi d lieu nh phan {10010110}. Tb = 0.0025s va c = 1000. S dung Matlab ve dang song tn hieu BPSK va pho FFT cua no vi cac ch so ieu che so sau:

a. h = 0.2 b. h = 0.5 c. h = 1

2. Gia s d lieu ngau nhien 4800 bit/s c gi i tren mot kenh truyen thong dai bang tn hieu BPSK. Tm dai thong truyen dan BT sao cho ng bao pho giam xuong t nhat 30dB khi ben ngoai dai thong nay. 3. Calculate the baud rate for the given bit rate and typeof modulation:

a. 2000 bps, FSK b. 4000 bps, ASK c. 6000 bps, 2-PSK d. 6000 bps, 4-PSK e. 6000 bps, 8-PSK f. 4000 bps, 4-QAM g. 6000 bps, 16-QAM

4. Draw the constellation diagram for the following: a. ASK, amplitudes of 1 and 3 b. 2-PSK, amplitudes of 1 at 0 and 180 degrees

5. The data points of a constellation are at (4,0) and (6,0). Draw the constellation. Show the amplitude and phase for each point. Is the modulation ASK, PSK, or QAM? How many bits per baud can one send with this constellation? 6. ng xung v tinh ti tn s 12 GHz lm vic vi cng sut 6 W v h s khuch i anten 48,2 dB. Tnh EIRP dBW.

(a) 36 dBW; (b) 46 dBW; (c) 50 dBW; (c) 56 dBW 7. Tnh ton h s khuch i anten parabol ng knh 3 m lm vic ti tn s 12 GHz, coi rng hiu sut mt m bng 0,55.

(a) 47,9 dBi; (b) 48,9 dBi; (c) 50,9dBi; (d) 51dBi 8. Khong cch gia trm mt t v v tinh l 42.000 km. Tnh tn hao trong khng gian t do ti tn s 6 GHz.

(a) 190,4 dB; (b) 200,9 dB; (c) 210,9 dB; (d) 211,9dB 9. ng truyn v tinh lm vic ti tn s 14 GHz c tn hao phi bng 1,5 dB v tn hao khng gian t do bng 207 dB. Tn hao hp th kh quyn bng 0,5 dB, tn hao nh hng anten bng 0,5 dB, tn hao lch cc c th b qua. Tnh tng tn hao ng truyn khi tri quang.

(a) 199,5 dB; (b) 209,5 dB; (c) 210,5dB; (d) 211,5dB 10. Khi tnh ton qu ng truyn ti tn s 12 Ghz, tn hao trong khng gian t do l 206 dB, tn hao nh hng anten l 1 dB, tn hao hp th kh quyn l 2 dB. T s Gr/T ca my thu l 19,5 dB/K v tn hao phi l 1 dB. EIRP bng 48dBW. Hy tnh t s sng mang trn mt ph cng sut tp m.

(a) 86,10 dHHz-1; (b) 87,10 dHHz-1; (c) 88,10 dHHz-1; (d) 90 dHHz-1


Trang 101

11. A transmitter has an output of 2W at a carrier frequency of 2GHz. Assume that the transmitting and receiving antennas are parabolic dishes each 1.8 m in diameter. Assume that the efficiency of each antenna is 0.55 a. Evaluate the gain of each antenna b. Calculate the EIRP of the transmitted signal in units of dBW c. If the receiving antenna is located 25km from the transmitting antenna over a free-space path, find the available signal power out of the receiving antenna in units of dBW. 12. A receiving preamplifier has a noise figure of 13dB, a gain of 60dB, and a bandwidth of 2MHz. The antenna temperature is 490K, and the input signal power is 10-12W. a Find the effective temperature, in kelvin, of the preamplifier b Find the system temperature in kelvin 1 c. Find the output SNR in decibels.

CHƯƠNG 6: TRUYỀN DẪN VÔ TUYẾN TÍN HIỆU SỐ...

Documents

Transcript of CHƯƠNG 6: TRUYỀN DẪN VÔ TUYẾN TÍN HIỆU SỐ...