7 Bai Tap Tham Khao Thiet Ke Anten

8/12/2019 7 Bai Tap Tham Khao Thiet Ke Anten

1/25

Bi tp 1: Anten chn t i xngf0= NS2 + TS4 + 220 [MHz];

d = 2 mm1. Thit k anten chn t i xng cng hng ti tn s f0;2. Thay i d = 10 mm v so snh vi kt qu d = 2 mm;3. Xc nh Rvtrong trng hp d = 2 mm v d = 10 mm;

4. Xc nh Gmaxtrong trng hp d = 2 mm v d = 10 mm;5. Xc nh 20.5(E) trong trng hp d = 2 mm v d = 10mm;

6. Xc nh di thng ca anten trong trng hp d = 2 mm v

d = 10 mm vi VSWR 3;7. Lp bng so snh cc kt qu t c.


2/25

d = 2 mm d = 10 mmf0 278 MHz 278 MHz

Lthc 95.57% (/2 ) 92.96% (/2)Rv 70.98 [] 71.87 []

Gmax 2.134 [dBi] 2.132 [dBi]2 0.5(E) 78 78

f1f2(BW)

(228328)MHz

ff

= 14.39%ff

= 20.5%

Bng so snh kt qu


3/25

Nhnxt ktqumphngdavobng. ng knh ca anten tng th BW rng hn, y l

phngphp mrngdithng cho anten chntixng.

Chiudi thccaanten gimkhi ngknh tngln. Trvo xpxbngnhau. rngmccnh sng chnh theo mcnacng sut

khng i. Hstngch hunhkhng thay i.


4/25

Bi tp 2: Anten chn t khng i xng t trn mtphn x v hnf0= NS2 + TS4 + 220 [MHz];d = 2 mm

1. Thit k anten chn t khng i xng t trn mtphn x v hn cng hng ti tn s f

0;

3. Xc nh Rv;4. Xc nh Gmax;5. Lp bng so snh vi anten chn t i xng.


5/25

Anten chn t

khng i xngAnten chn t i

xng

f0 278 [MHz] 278 [MHz]

Lthc 95.07% (/4) 95.57% (/2)

Rv 35.94 [] 70.98 []

Gmax 5.147 d[Bi] 2.134 [dBi]

Bng so snh kt qu


6/25

Nhnxt ktqumphngdavobng. Chiu di thc ca anten chn t khng i xng v

anten chntixnggimihn5% v cutrc ca2anten gingnhau.

Trvo gimi1 naso vianten chntixngvchiudi gimi .

Hstngch tngln khi ttrn tmphnxrngvhnv nng lng tptrung theo mthng,pht x1nakhng gian pha trn.


7/25

Bi tp 3: Anten Khungf0= NS2 + TS4 + 220 [MHz];

d = 2 mm1. Thit k anten Khung trn c ng knh DD = 0/20;2. Thit k anten Khung trn c ng knh DD = 0/;3. Thit k anten Khung trn c ng knh DD = 20/;

4. Lp bng so snh cc kt qu Tr vo, Phn b dng, ctrng hng v Gmax ca 3 loi anten Khung trn.


8/25

DD = 0/20 DD = 0/ DD = 20/

Zv 0.15881 + j223.835 [] 119.144 - j94.7237 [] 160.723 -j121.471 []

Phn

bdng

ctrng

hng

Gmax 1.693 [dBi] 3.461 [dBi] 2.793 [dBi]


9/25

Nhnxt Chu vi nh rt nhiu so vibc sng th c trng

hng vung gc vi trc ca anten ging c trnghng ca anten xon c chu vi nh rt nhiu so vi

bcsng. Chu vi xpxbngbcsng ctrnghngtrng vi

trc ca anten hay mtphngcha anten v gingctrnghngcaanten xonc chu vi xpxbngbcsng.

Chu vi ln hn rt nhiu so vibc sng c trng

hngpht xchch1 gc no . Hstngch so snh tnganten nh. Trvo tngkhi chu vi tngln.


10/25


11/25

Cc kt qu t c

f0 278 [MHz]Rv 6.438 []

Gmax 4.833 [dBi]

TH 3D

Phn b dng


12/25

Tm ngknh cavng xonDD thamn iukinphnbdng gingvichntkhng ixngttrn tmphnxrngv hn.Tcl phnbdng max tiimcpngunv gimdn

vucuicaanten. Tta iuchnhDD cnghng:Cch 1. iuchnhDD.1. Khi DD tngthphnotng(Xv)2. Khi DD gimthphnogim(Xv)

Cch 2. iuchnhditnchatnsyu cu(tm ckhongdi tnch tnsyu cu th chuynsang kho st tns theoyu cubi).1. f tngth DD gim.

2. f gimDD tng.iuchnhDD thamn cc iukintrn.Trong bi ny vitnsf = 278 MHz th tm cDD = 8.435mm.


13/25

Bi tp 5: Anten xon t trong khng gian t dof0= NS2 + TS4 + 220 [MHz];d = 1 mm, H = 100 mm, N = 40;

1. Thit k anten xon anten cng hng ti tn s f0(tm ng knh DD ca vng xon);2. Xc nh R

v;

3. Xc nh Gmax;4. Xc nh th c trng hng ca anten;5. Xc nh phn b dng ca anten;

6. Lp bng cc kt qu t c v so snh vi antenxon t trn mt phn x v hn vi N=20 nh trong bitp 4.


14/25

Anten xon trong khnggian t do

Anten xon trn mtphn x v hn

f0 278 MHz 278 MHz

Rv 3.337 6.438 []

Gmax 1.776 [dBi] 4.833 [dBi]

TH 3D

Phn b dng


15/25

Tngtnhbi 4 ta cngitm DD thamn phnbdngging chn t i xng tc l gia anten xon l max vgimdnv 2 u.Trong bi ny ta tm DD = 8.285 mm.

Nhtxt. Trvo caanten xonttrong khng gian tdo tngln

khi ttrn tmphnxrngv hn. Hstngch tngln 3.057 dB khi ttrn tmphnx

v nnglngbcx1 nakhng gian. u im ca anten xon kch thc anten nh hn rt

nhiuso vianten chntnasng ticng 1 tns. Nhc im ca anten xon l tr vo rt nh cnphi

tngtrvo ln.


16/25

Bi tp 6: Anten Yagif0= NS2 + TS4 + 220 [MHz];

d = 2 [mm]; s chn t dn x 3;1. Thit k anten Yagi anten cng hng ti tn s f0(Xv(f0) = 0);

2. Tm Rv

(f0

);

3. Tm Gmaxti tn s f0;4. Tnh 20.5(E) v 20.5(H) ti tn s f0;5. Tnh di thng ca anten vi VSWR 3;

6. Lp bng so snh vi anten chn t na sng.


17/25

Anten Yagi Anten chn t nasng

f0 278 MHz 278 MHz

Lthc 93.96% (/2) 95.57% (/2)

Rv 70.91 [] 70.98 []

Gmax 7.318 [ dBi] 2.134 [dBi]

20.5(E) 64 78

20.5(H) 98 360

f1f2(BW)

(228328)MHz

ff

= 15.47%ff

= 14.39%


18/25

Cc bciuchnhanten yagi cnghngv trvotngngnhanten chntixng.1. Gimchiudi chntchngt(5-10)%.2. Thay ikhongcch cc chntchngvichnt

phnxnmtrong khong(0.150.25).3. Tng t nh th cng thay i khong cch chn t

chnghay (chntdnxutin) vichntdnx(0.10.35).

4. Thay ichiudi chntphnxt(0.480.52),dnx(0.360.46).


19/25

Nhnxt ktqumphng Anten Yagi c trkhng uvo tngngnhanten

Dipole. Anten Yagi c rng cnh sng chnh theo mc na

cng sut trong mtphng E rng hn anten Dipole vtnh nhhngth anten Yagi tthnanten Dipole.

Anten Dipole bc x v hng trong mtphng H cnanten Yagi chbcxtheo 1 hng.

Dithng ca2 anten tngngnhau. Hstngch caanten Yagi lnhn5,184 dB (3,3 ln)

so vianten Dipole . Phngphp citincaanten Yagi v hstngch. Nhcimanten Yagi so vianten Dipole l kch thc

cngknh.

2 (E) l h


20/25

tm gc 20.5 (E) m trng vng xa ln v chn mt ctXY,XZ,YZ xem.Nunhn thyngvng trn chaanten th lmtphngE (bi ny mtphngct l YZ). Gc theta bin thin v

phi =90 . V mPOSTFEKO thay itheta v phi nh.

Bi v dny lyanten xonminh ha. Lu ccbnphixem ttrcanten l trcOX hay OY hay OZ. Khi gc theta v phi skhcnh. Khng ckch OK xem xong th cancel.


21/25

tm gc 20.5(H) m trng vng xa ln v chn mt ctXY,XZ,YZ xem. Nu nhn thy ng vng trn vung gc vi

anten th l mt phng H (bi ny mt phng ct l XY). Gctheta =90 v phi bin thin. V m POSTFEKO thay i theta v

phi nh. Khng c kch OK xem xong th cancel.


22/25

Bi tp 7: Anten Loga-Chu kfmin= NS2 + TS4 + 220 - 50 [MHz];fmax= NS2 + TS4 + 220 + 50 [MHz];f0= NS2 + TS4 + 220 [MHz];d = 2 [mm]; s = 5 [mm];

1. Thit k anten Loga-Chu k VSWR 3 trong di tns [fmin, fmax];2. Tnh 20.5(E) v 20.5(H) ti tn s f0;3. Tm Gmaxti tn s f0;

4. Lp bng so snh vi anten Yagi.


23/25

Anten loga chu k Anten yagi

f fmin fmax(228 328) Mhz

f0=278 Mhz

20.5(E)

66

64

20.5(H) 120 98Gmax 5.728 [ dBi] 7.318 [ dBi]


24/25

Ccbciuchnhanten loga chu kthamn VSWR 3. Cho ditnkhost tnh ditnso vitnstrung tm

ra bao nhiu. Ly kt qu va tnh c chia cho 10%(anten chntnasng c ditnkhong10%). T tm cschntlnhnhocbngktquchia.

Thay i gc anpha t (1090)xem c thay i g so

vi2 chntuv cuicahsVSWR 3. Thay ichiudi Lmin (Lmax). Thay i(chu k ktcucaanten) t0.8 0.96. Lu ccbnc ththay ickhongchia tnsnh

xem c thay ig khng nVSWR. Tnghocgimschntnh. Ttm quy lutthay inh.


25/25

Nhn Xt

Hstngch caanten loga chu kgim1.59 dB so vianten yagi.

rng cnh sng chnh theo mc na cng sut trongmtphngE v mtphngH caanten loga chu krng

hnanten yagi v tnh nhhng th anten yagi tthnanten loga chu k.

u im ca anten loga chu k so vi anten yagi v dithng rng.

Nhc im ca 2 anten trn l cng knh so vi antenchntnasng.

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7 Bai Tap Tham Khao Thiet Ke Anten