bai bao cao an ten

8/3/2019 bai bao cao an ten

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LI NI U

Ngy nay cng vi s pht trin mnh m cc ng dng khoa hc k thut

trong cng nghip in t , c bit trong cng ngh in T Vin Thng thcc thit b trong lnh vc Vin Thng cng c ch to cng nhiu.V c bit

l cc ng dng ca n trong cc nghnh kinh t v i sng hng ngy angpht trin mnh m v rng ri.

Tuy nhin p ng c nhu cu ngy cng nhiu v phc tp cacng ngh in T th nghnh in T Vin Thng phi ngin cu tm ra gii

php ti u nht.c bit trong nhng nm gn y s ngi s dng mngVin Thng di ng ngy cng nhiu v ch trng chnh sch ca Nh Nc

ta, cc cng ty cn phi thay i,nng cao a thit b cng ngh cao, phngphp iu khin thit b trong nghin cu v trong ng dng. l nhim v ca

nghnh in T Vin Thng phi gii quyt. gii quyt c vn ny th Nh Nc ta cn phi c nhng i ng

thit k ng o v ti nng.Sinh vin nghnh in T Vin Thng tng laikhng xa ng trong i ng ny.Do cn trang b cho mnh mt trnh v

tm hiu bit su rng.Chnh v vy mn thc tp AnTen rt cn thit cho misinh vin Din T Vin Thng , l bi kim tra kho st tng hp ca sinhvin v cng la iu kin cho sinh vin nghnh in T Vin Thng t tm hiuv nghin cu kin thc v AnTen.Qua y em cng rt bit n Thy Minh v

C Hin tn tnh ch bo cho em em c th nm vng c nhng kinthc v mn thc tp AnTen ny v l nhng kin thc qu bu gip em t

tin hn khi ra trng i lm vic.Mc d c gng rt nhiu song do hiu bit kin thc cn hn ch kinh

nghin thc t cn t v trong qu trnh lm bo bo cn nhng thiu st nhtnh.V vy em mong c s ch bo v ng gp kin ca thy c v ccban sinh vin.

Hon thnh kha thc tp ny khng nhng gip em c thm c thm

kin thc hn v mn hc ny m cn gip em mt phng php lm vic mi

ch ng hn , linh hot hn v c bit l phng php lm vic theonhm.Qu trnh lm thc tp l mt thi gian thc s c ch cho bn thn em vnhiu mt.


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Bi M u : Lm quen Vi Cc Thit B Thu Pht Anten VCch Tin Hnh Thc Hnh

- H thng thu pht sng cao tn RF :+ My pht RF: My pht cung cp tn hiu RF truyn n anten pht, n

bao gm 2 mch dao ng, mt cng tc ti tn s 915MHz (1GHz), mch cnli cng tc ti tn s 10.5GHz (10GHz). Hai mch ny c th hot ng ch CW hoc ch iu ch AM.Khi nh v anten: Khi nh v anten cha m t lm quay anten thu, khichuyn ng ny lm gi cho anten thu.

+Khi giao din tip nhn d liu: Khi ny lm nhim v lin kt bnh v vi my tnh c cha phn mm iu khin. Khi giao din tip nhn dliu cha b vi x l trung tm ca h thng.Khi ngun: c nhim v cp ngun cho ton b h thng.Phn mm m phng s bc x ca Anten : Phn mm LVDAM-ANT(The

LabVolt Data Aquisition and management for Antennas) l phn mm iukhin v tr gc quay ca anten trn my thu ng thi ghi li nhng cc mc tnhiu nhn c trn anten thu.

+ Sinal level : Mc tn hiu nh nht l -30dB v ln nht l 0dB , ta cchnh suy hao cho tn hiu quan st tt

+ Antenna position : V tr tng i gia Anten thu v Anten pht- V Anten bc x tt nm trong vng trn qu o -10 dB n -5 dB- B quay ca Anten th trng vi trcCc anten l cc thit b chuyn tip gia ng dn sng hoc cc ng dytruyn dn v khng gian t do, c dung nhn cc sng t khng gian t

do hoc pht x cc sng c truyn dn.Ta lm quen cc php o v so snh s khc nhau ca mt s loi anten ngin nh chn t, chn t gp v chn t vng dt ng cc loi anten dngtrong thc hnh nh anten Dipole, anten Loa, anten Yagi v cc loi antenny s c kho st ti cc tn s cng tc khc nhau.Chn t l mt trongnhng ngun bc x c s dng kh ph bin trong k thut anten.- Da vo kt qu thc hnh thu c ta phn tch, tnh ton cc gi tr v sosnh vi gi tr l thuyt.


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Bi 1: MU BC X CAANTEN DIPOLE NABC SNG TI

TN S 1GHz

(2

P!l ; GHzf 1! )

I.Mc ch :

- Mn thc tp k thut AnTen nhm giup cho sinh vin hiu v nm vngnhng kin thc lc hc l thuyt khng nhng vy n cn gip sinh vin ccvn sau :

+ Lm quen vi cc thit b thu pht sng , lm quen vi phn mm LAB VOLT m phng kh nng bc x ca AnTen.

+ Kho st anten Dipole na bc sng tn s 1GHZ.T l thuyt v

thc t rt ra c cc thng s k thut ca anten ny .+ Qua cn rn luyn cho sinh vin k nng thc hnh v k nng tng

hp kin thc hc trong phn l thuyt.+ Bit c mu bc x ca anten Dipole na bc sng l nh th no.+ Rn luyn cho sinh vin c tnh t gic v c trch nhim vi nhng

dng c thc hnh cng nh trong vic thc hnh.

II. Phn l thuyt1.Khi nim v anten na bc sng :

- Anten Dipole na bc sng l anten n gin nht trong cc loi anten,ngm :

+ Nui bng dy song hnh

+ C 2 nhnh P o/42. Cu to :

Anten dipole na bc sng bao gm hai vt dn c hnh dng tu (dy dnhnh tr, hnh chp ...) c kch thc ging nhau v t thng hng trong khnggian, gia c ni vi ngun cao tn.

Cu to v phn b dng in ca chn t dipole3. Mu bc x :

a.Khi nim kiu bc x : Phn b tng i ca cng bc x nh l

hm ca hng bc x trong khng gian.


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b. Kiu bc x Dipole na bc sng :

Mu bc x trn mt phng H Mu bc x trn mt phng EIII. Phn thc hnh

Mc ch ca bi thc hnh ny l tm mu bc x ca dipole na bc sng

trong mt phng E v H.1. Th no lmt phng E vmt phng H ? Gii thch.a. nh ngha : - Mt phng E l mt phng trong in trng l cc i.

- Mt phng H l mt phng trong t trng l cc i.

b.Gii thch :Mu bc x ca 1 anten c tnh khng gian ba chiu tuy nhin tin cho

vic nghin cu ngi ta thng xt mu bc x lm 2 mt phng E v mtphng H.

V d :

2.Anten l g ?

- Theo IEEE nh ngha anten l phn ca h thng truyn hay nhn cthit k bc x hay nhn sng in t.Ni cch khc anten ly tn hiu RF


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c sinh ra bi radio v bc x n vo trong khng kh hay anten c th nhnsng in t cho radio.

3.nh ngha sphn cc ca anten :S phn cc ca anten l s nh hng ca cc vc t trng in t E ti mt

vi im trong khng gian. Nu Vct E gi nguyn s nh hng ca n timi im trong khng gian th l s phn cc tuyn tnh; Cn nu n quay

trong khng gian, th l s phn cc trn hoc elip. Trong hu ht cc trnghp, s phn cc ca sng c bc x l tuyn tnh, theo chiu ng hoc

ngang. mt khong cch ln thch hp so vi ng-ten ( Khong 10 ln bcsng), sng trng xa c th c coi l sng phng .

4. Tnh P :Ta c : c=3.108(m) v f=915(MHz)

Bc sng ca anten l :8

6

3.100.32( )

915.10

cm

f

P ! ! !

di ca anten Dipole na bng sng l :0.32

0.16( )2 2

l mP

! ! !

5. Da vo mu bc x o c. Tnh gc na cng sut. So snh vi gi trtnh c bng l thuyt v bng phn mm :

a)Khong cch gia 2 anten thu v anten pht l 1m :+Phn cc ngang (E) :

Gc na cng sut ca phn cc E l : 76.970


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Theo l thuyt th gc na cng sut ca phn cc E l :nh ngha gc na cng sut:

Gc na cng sut l gc gia hai hng m ti cng bc x gim mtna(gim 3dB) so vi v tr cc i.Tnh gc na cng sut ca phn cc E theo l thuyt nh sau: Nhn vo nt sau iu chnh 2 tia sao cho curs1 v curs2 v 0(dB) richnh 2 tia gim 3dB so vi v tr cc i nh hnh sau:

Curs1 l mc cng sut ca tia 1 gim i tnh t im cc i, curs2 l mccng sut ca tia 2 gim i tnh t im cc i.

Ta thy Diff=76 r y l gi tri ca gc na cng sut theo l thuyt,gi tr

ca gc na cng sut theo thc hnh l 76,97 r .Nh vy gi tr thc hnh gnbng gi tr l thuyt vi sai s nh.


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+Phn cc dc (H) :

Gc na cng sut ca phn cc H bng phn mm l 00

Gc na cng sut ca phn cc H bng l thuyt l :

Curs1 l mc cng sut ca tia 1 gim i tnh t im cc i, curs2 l mccng sut ca tia 2 gim i tnh t im cc i.


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Vy gc na cng sut tnh theo l thuyt l:360-11=349 r ,tnh theo thc hnh

ta c HPBW=334.37 r .Kt qu cho thy chnh kch gia thc hnh v l thuytl hi cao.iu ny do curs1 v curs2 chng ta ly trn chng trnh cha chnhxc.V n cng th hin rng gc na cng sut khi anten phn cc H l khngr rng cng nh tnh nh hng trong trng hp ny khng c cao.

Nhn xt : Khi phn cc E th ta thy tnh nh hng ca anten Dipole caohn khi phn cc H. Ta thy gc na cng sut khi phn cc E l

HPBWE=76.970 nn tnh nh hng ca n cao.

b) Khong cch gia 2 anten thu v anten pht l 1.25m+Phn cc ngang (E) :

Gc na cng sut ca phn cc E bng phn mm l : 81.570

Gc na cng sut ca phn cc E bng l thuyt l


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+Phn cc dc (H) :

Gc na cng sut ca phn cc H bng phn mm l 00

Gc na cng sut ca phn cc H bng l thuyt :


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Nhn xt :

- Ta thy gc na cng sut theo l thuyt v thc hnh tng ng nhau vc sai s nh khng ng k. Mt phn do nhiu v suy hao tn hiu nn dnn sai s

- Gc na cng sut ta tnh c bng l thuyt v dng phn mm LVDAM-ANT c s sai khc nhau nhng s sai khc ny khng ng k. l do tnhiu b nhiu trong qu trnh truyn trong phng thc hnh. Gc na cng sut

cho ta bit c tnh nh hng ca anten. Gc na cng sut cng ln th tnhnh hng yu, gc na cng sut cng nh th tnh nh hng ca anten cngcao. l l do m v sao cc chuyn gia khi lp t cc anten v tinh c gngsao cho gc na cng sut l b nht, cng gn bng 0 cng tt. Gc na cngsut s tng khi khong cch gia hai anten pht v thu tng. y cng l tiuch ng ch khi chng ta lp t mt h thng anten.

- Khong cch gia 2 anten thu v pht cng xa th tn hiu thu c cng nhv suy hao ng truyn cng ln, cng b nhiu v ngc li nu khongcch cng gn th tn hiu thu c cng tt, suy hao cng gim, nhiu cng

t.- Trong lc thc hnh ta nhn thy rng : tn hiu thu c l tt nht khi trc

ca anten dipole trong lc xoay trn vung gc vi trc ca anten YAGI

UDA. Bng chng l ti hai v tr ny ta thu c hai bp sng chnh ln. Cc vtr cn li tn hiu thu c gim dn v khi trc ca anten dipole trng vi trcca anten YAGI UDA th ta khng thu c tn hiu. y l v tr m chng tacn lu v trnh khi lp t mt h thng anten.

- Khi phn cc E, anten dipole c tnh nh hng cao hn so vi khi phn cc

H do gc na cng sut khi phn cc E nh hn gc na cng sut khi phn ccH (E=76.97,H=334.37) . Do vy khi ng dng anten dipole vo thc t, ta cnphi phn cc ngang hay dc cho ph hp. V d khi cn pht sng cho mt khuvc a l nh hay mt v tr c nh no ta phn cc ngang cn khi ph sngtrn din rng ta dng phn cc dc.


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Tm li, khi lp t mt h thng anten ta cn quan tm ti cc vn sau y :

+ Khu vc a l, khong cch a l khi lp t cc anten pht v anten thu.+ Cch phn cc cho anten sao cho ph hp.

+ V tr, hng t anten pht v anten thu.+ a hnh, chng ngi vt cn.....

Nhn xt chung :

- Khi tin hnh th anten pht c nh cn anten thu quay mt gc 360 .- Khi 2 anten thu v pht u phn cc E th ta thy tn hiu thu c anten

thu l 1 bp hnh s 8 vi khong cch gn nhau th tn hiu thu cng r t b sidng. y l anten c tnh nh hng do chng ta cn phi xc nh nh

hng ca anten cng cao th kh nng thu tn hiu cng ln. Khi 2 anten thu vpht phn cc H th ta thy tn hiu thu c l hnh trn ging nh anten ng

hng vi HPBWH=00

cho ta thy gc na cng sut l ln nht kh nng thutn hiu mi hng.

- Khong cch gia 2 anten thu v pht cng xa th tn hiu thu c cng nhv suy hao cng ln v ngc li nu khong cch cng gn th tn hiu thu

c cng tt v suy hao gim.IV.nh gi kt lun :

- Qua bui thc tp ny sinh vin c th tip cn vi cc loi anten v phnmm thu cc mu bc x qua nhng mu bc x sinh vin c th

phn tch c cc thng s ca anten .- Bui thc tp cn gip sinh vin cng c c nhng kin thc hc

phn l thuyt khng nhng vy m cn c th bit c s ging vnhau khi hc l thuyt v lc thc tp .

- Qu trnh thc tp rn luyn cho sinh vin cch lm vic theo nhm vcch phn tch cc thng s khi nhn c mu bc x

- Gip sinh vin bit cch lp t v cch s dng phn mm phn tch


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Bi2: BC X CAMING NG DN SNG TI TN S 10 GHz

I. Mc ch :

- Kho st mu bc x ca ng dn sng t rt ra thng s k thut.- S dng thnh tho vi phn mm m phng.

- Bit c cch lp t ca ng dn sng trong thc t thu c tin hiu lnnht.

- Hiu r hn tnh nng ca cc loi anten c th khi chng ta thay cc ngdn sng v tn s khc nhau chng ta vn nm c cch lp t n.

- Qua rn luyn cho sinh vin kh nng tng hp v thc hnh vi nhng lthuyt hc.

- Rn luyn tnh t gic v tinh thn trch nhim trong bui thc tp.- So snh gia l thuyt v l thuyt.

II. Phn l thuyt :- Sinh vin phi nm c nhng kin thc c bn v l thuyt bao gm nhng

phn sau :1. Cu to :

Bc x ca ming ng dn sng c cu to t ng dn sng ch nht, ditc dng ca sc in ng t vo ng, phn b in p dc khe c dng hnhsin. Nh vy phn b in trng dc theo khe tun theo quy lut ca sngng. Khi nng lng truyn n ming ng chng s bc x ra khng gian. Khi

m rng ming ng theo cc hng khc nhau s nhn c mt bc x khcnhau.

Cu to ca ng dn sng

2. Mu bc


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Thc tp AnTen SV:Nguyn Hng Vn

Mu bc x trong ta cc Mu bc x trong ta vung gc

3. Chun b:Sinh vin cn nm vng mt s thng s c tnh c bn ca antensau y:

Hiu sut bc x (%)

0

radP

PL ! (2.1)

Trong : P0 l cng sut truyn n anten (watts)Prad l cng sut c anten bc x (watts)

Cng bc x (watts trn steradian):Trong steradian l n v ca gc c v ton b khng gian c mt gc c l

4T . Chng ta c th nh ngha rng cng bc x trung bnhavg

J l:

4rad

avg

PJ

T! (w/ste) (2.2)

H s nh hng D: c nh ngha l t s gia cng bc x cc i

theo mt hng cho vi cng bc x trung bnh ng hng vi cng mtcng sut pht.

li ca anten: Vi anten c hiu sut l 1 th li ca anten cng chnh lh s nh hng, tuy nhin thng thng hiu sut ca anten l nh hn 1 v ta

nh ngha li ca anten theo biu thc sau:G DL! (2.3)

Gc c ca anten (ste): a;

L gc c cn thit bc x ton b cng sut P rad ti mc cng bc x cci.


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ax.rad a mP J! ; (2.4)

Vi nh ngha nh vy ta cung c th nh ngha li h s nh hng l:4

a

DT

!;

(2.5)

Din tch hiu dng: li ca anten cng c nh ngha l:

2

4e

G AT

P! (2.6)

Vi Ae l din tch hiu dng ca anten.Hng s F/B

L t s gia bp sng chnh ca anten v bp sng ngc. Thng thng bp sngngc l bp sng khng mong mun v gy nhiu ti bp sng chnh.

F/B=main lobe (dB) - back lobe (dB) (2.7)

II. Phn thc hnh1.Cu to ng dn sng hnh ch nht :

ng dn sng hnh ch nht c thit din l hnh ch nht.ng dn sng cuto t cc ng kim loi rng hoc t cc ng in mi rng hay c.Cc ng cpin nh cp ng trc coi nh l ng dn sng.

ng dn sng2.Nguyn l dn sng trong ng dn sng :

- Sng lan truyn trong ng dn sng , c th coi l do b phn x qua li gia ccthnh phn ng ( phn x trn b mt hay phn x ton phn trn b mt in mikhin cho nng lng sng in t c dn trong lng ng ).- Chiu bc x in trng ca ng dn sng c chiu song song vi chiu rng vvung gc vi chiu di ca ng dn sng.3. So snh mu bc x ca Anten 1 v2 :

- Thc hnh lp rp anten loa v ng dn sng t bc 1 n bc 9 .Anten loatruyn i tn hiu v ng dn sng nhn tn hiu.Khong cch truyn v thu tn hiul r = 1m ,t ngun dao ng RF 10GHZ v tr tn hiu cc i (MSP) l 0 .


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- Quan st mu bc x trn phn mm LVDAM- ANT ta thu c kt qu sau :

Mu bc x 1 Mu bc x 2Trong :

+ Mu bc x 1 l mu bc x ca ng dn sng khng lin tc phn cc ngang(mt phng E) vi MSL = -20,3 (dB) HPBW = 71 .

+ Mu bc x 2 l mu bc x ca ng dn sng lin tc phn cc ngang (mtphng E) vi MSL = -2,0 (dB) HPBW = 65 .

* Gii thch s ging v khc gia 2mu bc x trn :

+ Ging nhau :C 2 bc x trn u hn cc ngang v thu tn hiu l ng dn sng hnh ch nht ,khong cch gia 2 anten cng 1m v tr tn hiu cc i (MSP) l 0 .

+ Khc nhau :Kt ni ng dn sng lin tc ng dn sng thu c mu bc x ln cn kt nikhng lin tc th mu bc x nh.

+ Gii thch :

- Anten 1 c kh nng bc x nh hn s bc x ca Anten 2 l v : Anten 1s dng ng dn sng khng lin tc do vy khi thu tn th tn hiu trong ng

dn sng 1 phn b phn x ngc tr li do gp b mt phng chnh v vym tn hiu Anten thu c l rt nh iu ny ng ngha vi vic kh nng bc x ca chng l rt nh .Khng nhng vy khi ta ni ng dn songkhng lin tc lm cho vic phi hp tr khng gia ng ny vi ng kia

khng ging nhau hoc khng gn ging nhau.Phn x ca sng thu ctn hiu nh nn mu bc x nh .


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- Cn i vi Aten 2 khi ni ng dn sng lin tc cc ng ny c tr khnggn ging nhau nn vic phi hp tr khng d dng hn .H s phn x ln

xp x bng 1 do truyn sng tt v mu bc x ln hn so vi kt nikhng lin tc.Khng ch vy m ton b tn hiu Anten thu c khng b

phn x tr li nn mu bc x ln hn nhiu so vi mu bc x ca Anten14.Tnh P :

Ta c : c=3.108(m) v tn s f=10 (GHz)

Bc sng ca anten l :8

6

3.100.03( )

10.10

cm

fP ! ! !

5.Khong cch gia 2 anten thu vpht l 1m :a) Phn cc ngang (E)

Gc na cng sut ca phn cc E l : HPBWE = 64.640

b)phn cc dc (H) :


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Gc na cng sut phn cc H l : HPBWE = 63.330

H s nh hng D l :26000 26000

6.35W . W 64,64.63,33E H

DHPB HPB

! ! !

Din tch hiu dng Ae l : 2 240.03A 6.35 4,55.10

4 4eD

P

T T

! ! !

Hiu sut mingPA

L l :

Ta c din tch vt l ca anten loa l :Chiu rng ca n bng R= 2.5(cm)Chiu di ca n bng L=1.2(cm)Din tch vt l ca anten loa l :

Ap=RxL=1.2x2.5= 3(cm2)

Hiu sut ming l :4

4

4,55.101

3.10pe

A

p

A

AL

! !

Nhn xt :- Khi anten thu v pht u phn cc E hoc H th tn hiu thu c u c

mt bp sng chnh v nhng bp sng ph. Kh nng thu tn hiu ca anten cng

tt khi nh hng ca anten cng cao nhng nh hng ca anten phthuc vo din tch hiu dng ca anten do din tch hiu dng v bc sng

cng ln th inh hng cng cao. tng tnh nh hng ca anten ng dnsng ta cn m rng cc kch thc ming ca chng.

- Kt qu mu bc x phn cc E v phn cc H ca anten loa ny tng iging nhau bao gm mt bp chnh v mt s bp ph. Cc bp ph ny chnh thnh do nhiu ca mi trng truyn sng gy nn. Gc na cng suttrong phn cc E c ln hn gc na cng sut trong phn cc H nhng khngng k. i vi anten loa khi phn cc H th mc cng sut cc i nhn c

ln hn v gc na cng sut c c b hn so vi phn cc E => khi phn

cc H anten loa c mc nh hng cao hn so vi khi phn cc E.- Hiu sut ming ca anten trong trng hp ny l 1.5 cho thy din tch hiudng ln hn din tch vt l . Nguyn nhn ny l do sai s ca h s nhhng D.

- Ta c2

4eD A TP

! v


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Do khi bc sng cng ln th nh hng ca anten cng b dn n

gc na cng sut cng ln => tn hiu thu c km. V vy ta cn phithay i tn s hot ng ca anten thay i gc na cng sut sao cho

ph hp.- Anten kiu mng ng dn sng c nhc im l hng tnh km v c h s

phn x sng trong ng dn sng kh ln ( 3.02,0 z}V ). gim h s phn x

cn s dng cc thit b phi hp c bit. tng hng tnh ca anten kiuming sng, cn m rng cc kch thc ca ming sng (Theo sch l thuytv k thut Anten Phan Anh).- Theo sch l thuyt v k thut Anten Phan Anh ta c nng lng cao tn

c truyn theo ng dn sng ti c ca loa di dng sng phng. y, mtphn nh nng lng s phn x tr li, cn i b phn tip tc truyn theothn loa di dng sng phn k, ti ming loa. Ti ming loa, phn ln nnglng c bc x ra khng gian ngoi, mt phn s phn x tr li. S phn xsng t c loa s cng ln khi gc m ca loa cng ln, cn s phn x tming loa s cng nh khi kch thc ca ming loa cng ln. Do khi tnh

ton anten loa cn phi chn gc m v di R ca loa cho thch hp chocc sng bc cao nht pht sinh ti ming loa v c loa s suy gim nhanh vkhng lm mo dng phn b trng ming loa. Mt khi bit c quy lutphn b ca trng kch thch ming loa s tnh c trng bc x ca anten

theo l thuyt.III.nh gi thc tp :

- Bui thc tp ny gip sinh vin tip cn vi loi anten loa dng thp vcc thng s c th ca n.

- Khng nhng vy sinh vin cn bit c nhng thng s c th ca chngkhi a vo hot ng v s dng loi anten nay .T bit c nhng unhc im ca n.


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Bi3: LI CAANTEN LOA HNH THP

I.Mc ch :

- o mu bc x t tnh li ca anten ny .- Mc ch chnh ca bi l gip sinh vin lm quen vi anten loa dng thp v tin

hnh o c, tnh ton li ca loi anten ny.- Gip sinh vin rn luyn k nng cng nh kh nng vn dng kin thc hc

vo thc hnh.- Gip sinh vin gn gi hn vi cc loi anten thng dng nh anten Dipole ,anten

loa,anten YAGi GUDa.- Qua cng ren luyn cho sinh vin tnh t gic v c trch nhim trong cng

vic ca mnh.

II.Phn l thuyt :1. Cu to :

Khi m rng kch thc ming ca ng dn sng, ta s nhn c cc loi antenloa khc nhau. Trn thc t chng c cc loi anten loa hnh thp, hnh nn, loa H,

loa E. Trong bi ny chng ta s kho st loa hnh thp, n c xem l t hp caloi loa E v H. Do c hai mt iu m nn sng bc x c dng cu.

Anten loa dng thp


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2. Kiu bc x

Kiu bc x trn mtphngH Kiu bc x trn mtphngE3. Cc l thuyt c lin quan.

Suy hao ca khng gian t doKhi sng in t truyn trong khng gian t do s b suy hao nng lng, gi tr

suy hao ny ph thuc vo khong cch v bc sng.4

20logF

rL

T

P! (dB) (3.1)

Trong r, P ln lt l khong cch v bc sng c cng n v o.Vi cng 1 anten truyn sng n 2 im vi cc khong cch ln lt l r1 v r2th s chnh lch suy hao gia hai im c th tnh c l:

A1,2 = 20log 21

r

r

(dB) (3.2)

o li ca Anten

C nhiu cch o li ca anten. Trong cch n gin nht gi lphng php s dng anten chun, phng php so snh hoc phng php thay

th. Bng cch tin hnh so snh cng sut nhn c ca mt anten chun (Pref)vi anten cn kim tra (Ptest). li ca anten cn o (Gtest) tm c theo cng

thc:

testte st r ef

ref

pG G

p! (3.3)

Nu tnh theo n v (dB) th:Gtest=Ptest - Pref + Gref (dB) (3.4)

Ngoi ra, khi bit c chnh xc cng sut pht v thu ta cng c th tnh c li theo cng thc:

0

4 recPrGP

T

P! (3.5)


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Trong : G l lir l khong cch gia anten pht v anten thu

Prec v P0 ln lt l cng sut thu v phtP l bc sng tnh cng n v vi r

Mt s cng thc v Anten loa dng thp.

Anten loa dng thp trong mt phng H

Ta thy rng: lH2= R1

2 +2

2

A

(3.6)

V thc nghim chng minh rng anten s ti u khi tha mn:1

3A RP!

Tng t nh trong mt phng H, trong mt phng E ta c:

lE2= R2

2 +2

2

B

(3.7)

V anten ti u khi v ch khi: 22B RP!

Anten loa dng thp trong mt phng EThc nghim chng minh rng, dng sng khi ra khi ming anten loa khng phi

l dng phng m l nhng ng cong. Chnh iu ny gy nn s dch pha dnn li pha.

B

lE

R

R

a A

lH

R

R


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S li pha ny c m t bng h s li ng truyn chun ha s v t2

8E

E

Bs

lP P

(! !

2

8H

H

At

lP P

(! !

Cc gi tr ny c quan h vi lE v lH theo biu sau:

T nhng kt qu trnh by chng ta c th tnh li ca anten theo cngthc sau:

10 ( ) ( )( ) 10.08 10.log . E dB H dBA B

G dB L LP P

!

- (3.9)

4.Cc phng php xc nh li ca Anten :- C 4 phng php xc nh li ca Anten l :

+ Phng php dng 1 anten tham chiu+ Phng php o s liu kch thc ca anten+ Phng php xc nh cng sut bc x v cng sut ngun php

+ Phng php o rng ca gc na cng sut

(

(3.8)


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II. phn thc hnh1.Anten loa hnh thp dng ln khong cch gia 2 anten thu v pht l

80(cm) :+Phn cc ngang(E) :

Gc na cng sut ca phn cc E l 20.60

+Phn cc dc(H) l

Gc na cng sut ca phn cc H l 24.090


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2.Anten loa hnh thp dng ln khong cch gia 2 anten thu v pht l160(cm)

+Phn cc ngang (E) :


+Phn cc dc H :

Gc na cng sut ca phn cc H l :23.180


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3.Thay i anten hnh thp dng nh lm anten thu v khong cch gia 2

anten l 80(cm)+Phn cc E :

Gc na cng sut ca phn cc E l : 29.770+)Phn cc H

Gc na cng sut ca phn cc H l : 39.250


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4.Thay i anten loa dng nh lm anten pht v anten dng ln lm anten

thu v c khong cch l 80(cm)+Phn cc E :


+Phn cc H :

Gc na cng sut ca phn cc H l : 23.560


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Ta c li ca anten l :

Trng hp 1 : 2 anten loa thu pht to :

+phng php da vo kch thc vt lTheo l thuyt ta c : lh = 11cm, lE = 9.4cm, A=9.1cm. B=7.3cm

Ta c :2 20,073

0.248. . 8.0,03.0,094E

Bs

lP! ! !

Tra bng 1-31 ta c :LE(dB)= 1.1(dB)

Ta li c :2 20,091

0.328. . 8.0,03.0,11H

At

lP! ! !

Tra bng 1-31 ta c LH(dB)=0.75(dB)

Vy li ca anten trong trng hp ny l :

( ) 10.08 10log( . ) ( ) ( )

0.091 0.073( ) 10.08 10log( . ) 1.1 0.75 16.9( )

0.03 0.03

E H

A BG dB L dB L dB

G dB dB

P P!

! !

+phng php s dng mt anten mu lm anten tham chiu l :

0

4( ) rec

PrG dB

P

T

P!

Ta c cng sut pht l : P0=-2.4(dB)Cng sut nhn c l : Prec=-19.3(dB) li ca anten l

19.3

10

2.4

10

4 3.14 0.8 1047.8 ( ) 10lg 47.8 16.8( )

0.0310

G G dB dB

v v! ! ! !

+Phng php nh ngha li : li ca anten l

26000 2600052.6 17.1( )

. 20,99.23,56E HG D dB

HPBW HPBW! ! ! !


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Trng hp 2 :Dng phng php tham chiu ta c li ca anten loa nh l :

Ta coi anten loa ln l anten tham chiu cef

17rG dB! v ef 4.34rP dB! (kt qu ny

ly trong ln o khi anten pht v thu u l anten loa ln, khong cch gia 2 loala 0.8m).Loa nh l anten cn o vi

est tG cn o v

es7.04t tP dB! (kt qu ly trong ln o

khi anten pht l loa ln v anten thu l loa nh,khong cch gia 2 loa cng l0.8m.

Ta tnhest t

G theo cng thc sau: eses ef

ef

t tt t r

r

PG G

P! v hay

est tG (dB)= es ef ef ( ) ( ) ( )t t r r P dB P dB G dB =-7.04+3.34+17=13.3(dB).

III.nh gi nhn xt :- Ta thy li ca anten cng ln th kh nng thu tn hiu ca anten cng

tt . Khi ta dng ba phng php tnh li ca anten th cho kt qu tngng nhau. li ca anten cng ln cho ta thy c tn hiu thu tt hnv t b nhiu hn. V tn hiu nhiu l tn hiu mong mun do cn phi

tng li gim nhiu v gim suy hao tn hiu.- Vi c 4 phng php trn dng tnh li ca mt anten loa th cc kt

qu u cho thy rng chng c gi tr gn bng nhau .iu ny chng tvic s dng 1 hoc 4 phng php trn u c nhng tt nht l s dngphng php no l ti u nht v hiu qu nht .

- li ca anten loa ph thuc vo khong cch gia hai anten thu v pht.Khong cch cng ngn th gc na cng sut cng nh v li cng cao.Khong cch cng xa th gc na cng sut cng ln v li cng b.

- c bit c s khc nhau v li kh r rt khi ta so snh vic cho anten

loa nh thu v anten loa ln pht vi vic cho anten loa nh pht v antenloa ln thu. C th khi anten loa nh lm anten thu th gc na cng sut ln

hn so vi khi anten loa nh lm anten pht, do li cng nh hn. Vy

li ca mt h thng anten loa ph thuc vo kch thc ca anten loa v

cch sp xp cc loi anten loa.


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Bi 4: CCCHN T2

P,P V 3

2

P

I.Mc ch:- Gip sinh vin rn luyn tnh t gic v tinh thn trch nhim trong qu trnh

thc tp.

- Mc ch chnh ca bi ny l nghin cu mu bc x ca ca cc dipole2

P, P

v 32

P . t thy c s ph thuc ca tng tr u vo theo di

dipole.Chnh s thay i tng tr ny tc ng n hiu sut bc x ca anten.

- Rn luyn k nng cng nh kh nng vn dng kin thc hc vo qu trnhthc hnh.

I. Phn l thuyt

1. Cu to :Cc chn t

2

P, P v 3

2

P l mt trong nhng ngun bc x c s dng kh

ph bin trong k thut anten. Trong , chn t2

P chng ta thc hnh bi 1,

trong bi ny chng ta s kho st cc kt lun sau y: Tr khng u vo bin i theo P .

Kho st, mu bc x, h s tnh hng khi thay i P .

Hnh 1 Cu to vphn b dng in ca chn t2

P

Hnh 2 Cu to vphn b dng in ca chn tP


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Hnh 3 Cu to vphn b dng in ca chn t 32

P

2. Mu bc x

Mu bc x trn mtphngE Mu bc x trn mtphngH

Chn t2

P

Mu bc x trn mtphngH Mu bc x trn mtphngE

Chn tP


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Mu bc x trn mtphngH Mu bc x trn mtphngE

Chn t 32

P

3. Tr khng vo ca cc Dipole

Vi cc dipole c di l2

P, P v 3

2

P th in p, dng in bc x lun bo

m ng pha nn tr khng vo ca cc anten ny kh nh. Tng tr u vo ch

yu l thnh phn tr khng v thnh phn in khng gn bng 0. Vi dipole2

P

v

3

2

P

, Zin}

73 + 0j (;

)Bng sau cho ta cch tnh tr khng u vo ca cc dipole:

di L Tng tr vo Zin

04

LP

2

20L

TP

4 2L

P P

2.4

24.7L

TP

0.637.2

LP

P 4.17

11.14

LT P

Khi di dipole thay i dn n tng tr vo cng thay i lm mt tnh ngpha dn n tng tr vo tng v gim hiu sut bc x.II. Phn thc hnh1.khong cch gia hai anten thu vpht l 1(m)


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a.Anten Dipole2

P :

+Phn cc ngang (E) :

+Phn cc dc (H) :

:

b. Anten Dipole P :+Phn cc ngang (E) :


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+Phn cc dc (H) :

c.Anten Dipole 32

P


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+Phn cc ngang (E) :

+Phn cc dc (E) :

Bc sng ca anten diople l :

Ta c : c=3.108(m/s) v tn s f =915(MHz)Bc sng ca anten Dipole l :


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8

6

3.100.32( )

915.10c

mf

P ! ! !

Chiu di ca anten Dipole2

P l : 0.32 0.16( )2 2

L mP

! ! !

Tra bng 1-35 ta c :

Tr khng vo ca anten Dipole2

P l

Zin=73+0jChiu di ca anten Dipole P l L= P =0.32(m)

Tra bng 1-35 ta cTr khng vo ca anten Dipole P l

Zin=1550-1220j

Chiu di ca anten Dipole 3

2

P l : L= 3

2

P =0.48(m)

Tra bng 1-35 ta c :

Tr khng ca anten Dipole 32

P l : Zin=73+0j

Nhn xt : o chiu di cc chn t ca cc loi anten khc nhau nn ta tnh c Zin

ca cc loi anten khc nhau.

Sai s ca cc thng s do qu trnh do chiu di vt l ca cc chn t

khng tht chnh xc hay do nh hng ca mi trng.

Da vo bng s liu h s nh hng trn ta thy anten no c chiu di ln

hn th c nh hng cao hn.vy anten diole ph thuc vo in tr vo

Zin.

Khi di dipole thay i dn n tng tr vo cng thay i lm mt tnh

ng pha dn n tng tr vo tng v gim hiu sut bc x. Anten dipole c chiu di cng tng th gc na cng sut cng gim =>

li cng tng, tnh nh hng cng cao => hiu sut bc x cng cao.Tm li i vi anten dipole ty theo tng v tr a l khc nhau, khu vc

pht v nhn sng khc nhau m nn chn loi anten dipole cho ph hp. Chnghn khi cn pht sng cho mt khu vc nh th ch cn dng anten c chiu di nh

nhng i vi nhng khu vc ln hn th chiu di cn phi tng ln thu ctn hiu tt nht. Tuy nhin trong thc t vic la chn loi anten dipole pht v


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thu sng in t ngoi khng gian s b hn ch do s cn n mt anten c chiu

di ln, nh vy s rt cng knh, tn km v phc tp.

*H s nh hng cho anten dipole2

P l

Ta c HPBWE=73.33 ,HPBWH=180(khi tnh ton nu mu bc x qut trn mt

phng H c gc na cng sut bng 0 ,th khi tnh ton ta ly gc na cng sutl 180 o)

=> D= 26000/( HPBWE* HPBWH)=1.97*H s nh hng cho anten dipole P l

Ta c HPBWE=53.89 , HPBWH=180=> D= 26000/( HPBWE* HPBWH)=2.68

*H s nh hng cho anten dipole 3

2

P l

Ta c HPBWE=34.14, HPBWH=180 D= 26000/( HPBWE* HPBWH)=4.23NHN XT:Da vo kt qu tnh ton ch nh hng ta thy:antendipole c chiu diln hn s c inhhngln hn =>tnh nhhng

cao hn. Vy tnh nhhng ca antenph thuc vo di ca antenIII.nh gi kt lun :

- Qua bui thc tp sinh vin nm c nhng thuc tnh ca cc loi antenv qua c th bit c li ch ca chng v nhng vn thng gpkhi lp t.

- Trong qu trnh thc tp cn gip sinh vin bit c nhng thng s khi

hc l thuyt khc vi khi chng ta tp l bao nhiu v t c th rut rac nhng kt lun cng nh nh gi v loi anten.


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Bi 5: CHN T DIPOLE GP2

P

I.Mc ch :- To c hi cho sinh vin tip xc vi loi Anten mi qua sinh vin c th hiu

v nm r hn v cc loi Anten chn t.- Gip sinh vin c k nng thc tp cng nh xu chui kin thc hc vn

dng thc hnh mt cch nhanh chng v d dng.- Rn luyn cho sinh vin tnh t gic v c trch nhim hn trong qu trnh lm

thc tp.II. L THUYT LIN QUAN

1. Cu to :

Chn t vng dt bao gm hai chn t song song ni vi nhau to nn mt vngdt. Ta xt chn t vng dt nh mt chn t i xng na sng. V vy, c dngin ln gp i so vi dng in trong chn t nhnh. Khi in tr vo cachn t vng dt tng gp 4 ln so vi in tr ca chn t na sng :

vd

RIP .21 2

0! (5.1)

22

20 4.22

1PP RRRIP vd !

! (5.2)

2. th phng hng

th tnhhng trn mtphngH th tnhhng trn mtphngE


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III.PHN THC HNH

1.Anten Dipole gp2

P :

`+ Trng hp c balun :Phn cc E :

Phn cc H :


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+ Trng hp khng c balun :

Phn cc E :

+Phn cc H :


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2.i anten Dipole gp2

P thnh anten Dipole2

P

+Phn cc E :

3.i anten Dipole2

P thnh anten gp2

P c gn thmmt thanh nhm sao

cho

thanh ny cch dipole gp2

P vi balun mt khong4

P v tin hnh thu nhn

mu bc x.+Phn cc H


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li ca anten Dipole gp2

P khi c balun l

G(dBi)= -1.59-(-10.09)= 8.5(dBi)G(dBd)=1.9G(dBi)=1,9.(8.5)= 16.15(dBd)

li ca anten Dipole gp 2P

khi khng c balun l :G(dBi)= 2.81-1.59= 1.22(dBi)

G(dBd)=1,9.(1,22)= 2.381(dBd)Nhn xt :

- Anten Dipole gp2

P khi c balun s nhn tn hiu tt hn khi khng c balun.

Anten Dipole gp2

P khi c balun s c thm thnh phn dn sng do li s

tng tng ng vi vic tn hiu thu c s tt hn v t b nhiu hn do li

tng. Khi antan Dipole gp2

P khng c balun th li s gim v khng c thnh

phn balun nn kh nng thu tn hiu yu hn v c kh nng xy ra hin tng

nhiu v gy ra suy hao tn hiu.- i vi ipole gp /2 : khi khng c balun gc na cng sut nh hn (72.37 v

65) so vi khi c balun, do m li ca anten khi khng c balun s ln hn.- Khi chng ta thay i dipole gp /2 thnh dipole /2 th c s thay i ng k

v gc na cng sut. Khi phn cc E, gc na cng sut c nh hn so vi gcna cng sut ca anten dipole /2 khi c balun nhng li ln hn gc na cng

sut ca anten dipole /2 khi khng c balun. Vy khi phn cc E anten nndng anten dipole gp /2 khng c balun th li thu c s tt nht. Khi phn

cc H th i vi anten dipole /2 gc na cng sut rt ln (ln n 169.61), cctrng hp cn li gc na cng sut bng 0) => khi s dng anten dipole phn

cc H nn dng anten dipole gp th li cng sut thu c s tt hn so vivic s dng anten dipole khng gp.

IV.nh gi nhn xt :- Qua bui thc sinh vin bit c ng dng ca anten l nh th no trong

qu trnh thu v pht sng.- Khng nhng vy bui thc tp cn lm cho sinh vin bit c li ch ca

vic lm thc tp theo nhm l nh th no.


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Bi 6: ANTEN YAGI - UDA

I.Mc ch :

- Kim nghim s ph tuc ca li bp sng na cng sut ca Anten Yagi-Udatheo s phn t ca Anten.

- Gip sinh vin rn luyn c k nng nhy bn hn trong vic thc hnh cngnh k nng tng hp kin thc.

- Gip sinh vin nm chc v hiu c qu trnh hot ng ca phn mm Lab-Volt.

- C c hi tip xc c vi loi Anten thng dng hin nay khng nhng vycn gip sinh vin hiu r hn v tnh nng cng nh nguyn l hot ng ca loi

Anten s dng chn t.

- Gip sinh vin rn luyn tnh t gic v tinh thn trch nhim trong qu trnh lmthc tp phng thc tp.

II. PHN L THUYT1. Cu to

Cu to ca anten Yagi gm mt chn t ch ng (chn t c cp ngun),thng thng l chn t vng dt na sng, mt chn t phn x th ng v mt

chn t dn x th ng.

Cu to ca anten Yagi -Uda


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b. th phng hng

th phng hng trn mt phng H th phng hng trn mt phng E

III.PHN THC HNH

1. Anten ch s dng thnh phn iu khin( Dipole2

P )

2.Anten thu gm 1 thnh phn iu khin v 1 thnh phn phn x vikhong cch gia 2 thnh phn l 0.2P :


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3.Anten thu gm 1 thnh phn iu khin v 1 thnh phn dn sng. Khongcch gia chng l 0.2P

4.Anten thu l anten Yagi -Uda 3 thnh phn vi:A = 178 mm;B = 154 mm;C = 146 mm;1 = 87 mm;2 = 61 mm;


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5.Anten thu l anten Yagi - Uda bn thnh phn c cc thng s nh sau:

A = 178 mm;B = 154 mm;C = 146 mm;D = 144 mm;l = 88 mm;2 = 66 mm;3 = 85mm;

6.Anten thu l anten Yagi - Uda c 5 thnh phn vi cc thng s nh sau:


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A = 178 mm;B = 154 mm;C = 146 mm;D =144 mm;E = 142mm;1 = 94mm;2 = 75mm;3 = 100 mm;4 =108 mm

7.Anten thu l anten Yagi - Uda c 6 thnh phn vi cc thng s nh sau:

A = 178 mm;B = 154 mm;C = 146 mm;D =144 mm;E = 142mm;F = 140 mm;1 = 66 mm;2 = 70 mm;3 = 105mm;4 = 110 mm;5 = 120 mm

Hon thnh s liu cho bng sau:


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S thnh phn G (dBd) G (dBi) HPBWE (0) F/B

1 68.680 0

2 9.445 0.55 68.630 0

3 12.371 2.09 60.190 1.41

4 16.171 4.09 63.850 9.71

5 18.584 5.36 47.940 17.48

6 19.762 5.98 41.510 24.1

7 21.415 6.85 32.720 24..97

Nhn xt :- Nhn vo bng s liu ta thy khi ta tng nhiu chn t cho anten theo tng

trng hp th ta thy li ca anten cng tng do kh nng thu tn hiu caanten thu cng tt, r v s b t nhiu v gim suy hao tn hiu hn. Khi ta tng

thm chn t ca anten cng nhiu th li ca anten cng tng theo nhng ntng n mt mc no th li ca anten s tng bo ha v khng tng na.

Do ta cn mc thm chn t ca anten sao cho hp l.- Khi cng tng s lng anten dipole ghp thnh anten YAGI - UDA th gc

na cng sut HPBW cng gim, li ca anten cng tng ln, t s F/B cngtng. Do tn hiu ta thu c s c cht lng tt hn, suy hao ng truyn

gim dn v si dng ca tn hiu t i, tnh nh hng ca anten tng ln ngk.

- Mt anten yagi uda c s chn lng chn t cng nhiu th nng lng in tthu c s cng ln, tn hiu s thu c tt hn.

- Ta cng cn phi nhn nhn rng khi cng tng s lng thanh anten dipole lnth s hng nhn c tn hiu ca anten YAGI-UDA ny t dn nhng b lihng nhn c tn hiu th bp chnh c gi tr ln tc l hng bc x cc ica anten cng cao.- Khi lp t mt h thng anten YAGI-UDA nn ti phng hng sao cho

ti hng ta thu c nhiu tn hiu nht, hng ny cng s l duy nht vcng l hng c bc x cc i.* Tm li i vi anten YAGI-UDA khi mun lp t mt h thng anten cn philu ti cc vn sau y:+ Phng hng, v tr, cc chng ngi vt ngn cn s thu v nhn sng.+ S lng chn t ca anten.


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IV.nh gi v kt lun :

- Qua bui thc tp ny cng nh kha thc tp ny gip cho sinh vin bitc tm quan trong ca thit b anten trong i sng cng ngh v t

nhn thc c vai tr trch nhim ca mnh trong vic vn hnh cc hthng.

- Gip sinh vin v vn dng c nguyn l bc x sng in t truyn trongkhng gian t do v thu sng lm c s cho vic phn tch hiu c thng

s c bn ca anten nguyn tc cu to ca cc loi anten thng dng- Kha hc ny nhm trang b cho sinh vin hiu v vn dng c nhng

kin thc hc vo thc t.- Trong kha thc tp ny th Anten Yagi l loi anten nh hng rt ph

bin bi v chng kh d ch to. Cc anten nh hng nh yagi thng

cung cp vng bao ph nhng vng kh vi ti hay nhng ni cn vngbao ph ln hn vng bao ph ca anten omni-directional.Nn qua cnggip cho sinh vin bit dc cch vn dng v lp t anten no l thch hp

vi nhng a hnh c th.- Qua ton kha thc tp ny trang b sinh vin nhng kin thc c bn v

anten cho sinh vin t tin khi bc ra lm vic v t nhng kin thc sinh vin s tm hiu su hn v anten khng b ng khi gp cc loi

anten mi.

- Khng nhng vy kha thc tp cn cho sinh vin nhng kinh nghim thctp tht qu bu t s lm vic tht tt nhng kha thc tp tiptheo.


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