1 BOCOTH NGHIM SIU CAO TN V ANTEN GVHD:TRN TH HNG SVTH:PHM QUC THOI NGUYN VN THIN V VN QUYN NHM 41B LP 11DTLT Lap 2: Basic Tranmission Line in the Frequency Domain

2.1 Basic Transmission Line Model:

T l mt ng truyn khng tn hao vi cc thng s sau : Z0 l tr khng c trng TD l thi gian tr (time delay) ,chnh l chiu di ca ng truyn trn mt n v thi gian Gi : L l chiu di ng truyn, upl vn tc pha ca sng trn ng truynth: L= up. TD (2.1) ViLl t cm trn mt n v chiu di, v Cdung khng trn mt m v chiu di Th ta c: ' '1C Lup =''0CLZ =(2.2) (2.3) 2 2.1.1 A standard coaxial cable i vi cp ng trc thng thng RG-58, tr khng c tnh l Z0 = 50 v vn tc pha Up =2 / 3 c. (Lu : c = tc ca nh sng 3e8 = m / s) Cu hi 1: i vi ng dy truyn ti nh vy, cc in cm v in dung trn mt bng bao nhiu? Tr li: T cng thc (2.2) v( 2.3) ta suy ra V : i vi loi cp ng trc khng tn hao, cc cng thc sau y lin quan n in cm L v in dung C vi bn knh ca dy dn bn trong a v dy dn bn ngoi b: 3 Cu hi 2: i vi mt cp ng trc khc nhau, =0 v =30.Tnh b/anuZ0 = 50 ? Tr li:t cng thc tnh L,C ta suy ra

V vy:Cu hi 3: Nu b = 3 mm trong cu hi 2, th a bng bao nhiu? Tr li: Nu b = 3 mm trong cu hi 2, th 4 2.2A SPICE model of a transmission line problem. S dng li s mch trn,ch thay i TD nh l mt bin qut s dng PARAMATERS 0 0Load0ZG50InputPARAMETERS:delay= 5nsZL1000VG1Vac0VdcT1TD = {delay }Z0 = 50Cu hi 4: tn s 200 MHz, v vi Up = 2 / 3 c,bc sng trong ng truyn bng bao nhiu? Tr li:bc sng trong ng truyn c tnh bng cng thc: Cu hi5: Thi gian tr kt hp vi /16 bng bao nhiu?(Vi) Tr li:: Thi gian tr vi /16 l:

fLuLTpD= =S dng SPICE m phng cc trng thi AC n nh ca ng dy ny cho chiu di 0, /16, 2/16, ..., 15/16, .. 5 S dngExcel,to mt bng bin in p v bin dng in cc node Input v Load cho mi chiu di. 6 Cu hi 6:S dng PSPICE, Excel, hoc Matlab v s ln ca in p ti u vo nh l mt hm ca chiu di. T cc gi tr in p trn biu v mi quan h , xc nh VSWR, v t VSWR tnh |I|Tr li: minmaxVVVSWR =- kim tra cc thay i ca in p u vo l hm ca di, chng ta c th kim tra cc thay i ca in p u vo nh l mt hm ca thi gian tr. Chng ta c th m phng in p u vo nh l mt chc nng ca thi gian tr v thi gian tr v di ca ng dy lin quan vi nhau t cc cng thc:. Vi Up l mt hng s, L tng n ln cng nh tng Td n ln. V vy, kim tra s thay i ca in p u vo nh l mt chc nng ca Td l lm iu ny vi L(chiu di ca ng truyn). -Vi L = , ta c:

V vy, chng ta c th thit lp mt tham s trong PSpice vi gi tr 0 u v 5ns gi tr cui cng. Ngoi ra, chng ta kim tra L ti cc im c/16 cch u nhau. V vy, trong qu trnh qut tham s ta tng gi tr l 0,3125 ns nh trn (cu hi 5). S dng PSPICE v ln ca in p ti u vo nh l hm theo chiu di: 7 2333 . 333667 . 666minmax= = =mVmVVVVSWR3333 . 0311 21 211| | ~ =+=+= IVSWRVSWRT cc gi tr in p trn th dng sng v mi quan h:

Vmax = 666.667mV Vmin = 333.333mV

Cu hi7: S dng PSPICE, Excel, hoc Matlab v s ln ca u vo dng in nh l hmca chiu di.T cc gi tr dng in trn biu , xc nh VSWR, v t VSWR tnh|I|. Tnh lng in p v dng in ca VSWR v |I|? Tr li:-S dng PSPICE v ln ca dng in ti u vo nh l mt hm ca chiu di: -T cc gi tr dng in trn th dng sng, xc nh VSWR, v tnh ton VSWR t |I|. Imax = 13.3333mA Imin = 6.6667mA 26667 . 63333 . 13minmax= = =mAmAIIVSWR3333 . 0311 21 211| | ~ =+=+= IVSWRVSWR8 Cu hi 8: V ln ca tr khng theo chiu di dng PSPICE.. Tr li : ln ca tr khng u vo nh l mt hm ca chiu di V phn thc ca tr khng bng cch s dng PSPICE 9 V phn o ca tr khng bng cch s dng PSPICE Cu hi 9:Tnh ton trc tipI v VSWRs dng phng trnh (2.6) v (2.7) di y. So snh li kt qu o cc cu6, 7 v 8?Tr li: 3333 . 03150 10050 10000~ =O + OO O=+= IZ ZZ ZLL231131111=+=I I += VSWRcc kt qu trn ph hp vi kt qu o cu 6,7,8 10 Cu hi 10 :V ln in p ti ti nh mt hm ca chiu di. in p thay i nh th no vi chiu di?T , bn ngh ngun c giao t ti s thay i nh th no theo chiu di? Tr li:

- ln in p ti ti nh l mt hm ca chiu di: ln ca inp ti khn thay i theochiu di,t ngun c giao t ti cng s khng thay i theo chiu di. Ngunti tinhmt hm ca chiu di: 11 2.3 A shortcut, and more load impedances di in ca ng truyn ll, lufl lptt|2 2= =V vy, s thay i di ca ng truyn tl n 10l s nh hng n tn st 10f n f.

Cu hi 11: Nu ta c 1 mt cp ng trc nh trong cu hi 4, th tn s ca n /2, 2.5 l bao nhiu (ch rng ta khng c thay i chiu di vt l ca ng truyn) Tr li : Cp ng trc trong cu hi 4 c tn s f = 200MHz V vy, n c chiu di 21 tiMHz MHz f 100 200 .2121= =V n c chiu di 5 . 2 tiMHz MHz f 500 200 5 . 2 5 . 2 = =Cu hi 12:V dng sng bin ca in p vo ng vi cc chiu di (Lengths) khc nhau(c th iu chnh thay i tn s)?Biu ca bn c ph hp vi cu hi 6 khng? VSWR bng bao nhiu?Tr li: Dng sng ln ca in p ti u vo i vi "chiu di" khc nhau theo trc ngang. c v trong PSPICE nh sau: Biu ny ph hp vi biu trong cu hi 6. Tnh VSWR Vmax = 666.667mVVmin= 333.333mV 2333 . 333667 . 666minmax= = =mVmVVVVSWR12 Thay th ti 25 cho ti 100 . 0 00LoadT1Z0 = 50TD = 5nsInputVG1Vac0VdcRG50ZL250 Cu hi 13:V dng sng bin ca in p vo, v so snh vi trng hp trc ti 100 . T hnh v tnhVSWR?Tr li :Dng sngca bin in p ti u vo: Tnh VSWR: Vmax = 666.667mVVmin= 333.333mV

Ta thy trong hai trng hp 100 v 25 , th VSWR l bng nhau. Thay ti bng mt "mch ngn", c thti 0.001 0Load InputZL0.001RG500VG1Vac0VdcT1Z0 = 50TD = 5ns0013 Cu hi 14: V dng sng bin ca in p vo. T biu dng sng, Tnh VSWR.T cc phng trnh (2.6) v (2.7) Tnh VSWR. So snh hai kt qu? Tr li : - th dng sng bin ca in p ti u vo: Tnh VSWR t hai cch: + T th: Vmax = 1V Vmin= 0V + T cng thc (2.6) and (2.7):

= = =VVVVVSWR01minmax150 01 . 050 01 . 000 ~O + OO O=+= IZ ZZ ZLL =+=I I +=| 1 | 1| 1 | 111VSWRTa thy hai kt qu tnh c l bng nhau By gi ta thay bng mt ti h mch ,c th lZL= 1 M.0LoadT1Z0 = 50TD = 5ns0VG1Vac0VdcRG50ZL1Meg00Input14 Cu hi 15:V dng sng bin ca in p vo. T biu dng sng, Tnh VSWR.T cc phng trnh (2.6) v (2.7) Tnh VSWR. So snh hai kt qu? Tr li : - th dng sng bin ca in p ti u vo: Tnh VSWR t hai cch: T th: Vmax = 1VVmin= 0V T cng thc (2.6) and (2.7):

Vy hai kt qu tnh c l nh nhau = = =VVVVVSWR01minmax11000050999995050 1050 106600~OO=O + OO O=+= IZ ZZ ZLL =+=I I +=| 1 | 1| 1 | 111VSWR15 Lap 3: Transients on Transmission Lines 1. Gii Thiu: Trong bi Lab ny ta tp trung vo cc k thut c th hiu c bn cht ca ng truyn di s kch thch in p hnh SIN u vo. T c c ci nhn su sc hn v cho php thit k cc mch n gin.2. u cui ch c in tr: 2.1 Hm bc nhy v s phi hp tr khng ti To mt ng truyn c Z0=50,TD=25ns,vi u vo l tn hiu bc nhy 10u(t) ,Rg = 50 . Tr khng ti RL = 50 , nh hnh v Vg0T1Z0 = 50TD = 25ns0RG50ZL50Vsource0Vload Vg0Cu hi 1: V dang sng ca in p ngun(Source) v ti (Load) ca ng truyn trong khong thi giant = 050 ns.Dng hiu bit ca bn so snh vi kt qu,bn ng c hai khng?nu khng,ti sao? Cc h s phn x ti ngun l: 050 5050 5000=+=+= IZ RZ RgggCc h s phn x ti ti l: 050 5050 5000=+=+= IZ RZ RLLgVy Khng c sng phn x ca in p ti ngun v ti. ) ( 550 5050 . 10VZ RZ VVL gL gL=+=+==LVT dng sng trn v qua s liu tnh ton ta thyl thuyt v th l hon ton ging nhau 5(V) sau 25ns (thi gian tr ca ng truyn.) Tr li : 16 Thay RL=202.2 Hm bc nhy v khng phi hp tr khng ti 00Vload0Vg0RL20VgRg50VsourceT1Z0 = 50TD = 25nsO

Cu hi 2: V dang sng ca in p ngun(Source) v ti (Load) ca ng truyn trong khong thi gian t=0100ns ?Dng hiu bit ca bn so snh vi kt qu,bn ng c hai khng?nu khng,ti sao khng?Cu tr li gii quyt cho cu cui cng trong bao lu? in p ngun v cui ti trn ng truyn t=0100ns . Cc h s phn x ti ngun l 050 5050 5000=+=+= IZ RZ RgggCc h s phn x ti ti l: 7350 2050 2000 =+=+= IZ RZ RLLL) ( 550 5050 . 1000VZ RZ VVggl=+=+=+Tr li : 17 . Tit = T = 25ns, sng t n ti trng (nhn cui) z = l,v bi v khng ph hp vi sng phn nh bin : 073= = Ig) (715. V V Vl L l = I =+ V vy, in p trn ng dy (trong trng hp ny, in p ti) l tng ca hai sng:) ( 86 . 27207155 V V V Vl l L~ = = + =+ Ti t = 2T = 50ns, sng gi n cui cng z = 0, v bi v, khng c sng phn nh trong hnh thc ca mt ln sng vi bin in p. Nh vy, in p trn ng dy (trong trng hp ny, in p ngun)l : 0 = Ig02 =+V) ( 86 . 27207155 V V V Vl l S~ = = + =+ Ti t = 3T = 75ns, sng t n ti trng (nhn cui) z = l, v bi v ,khng ph hp to ra mt ln sng phn nh vi bin :073= = Ig) ( 0 .2 2V V VL= I =+ ) ( 86 . 2 V VL = Tng t nh vy, ta cng c:) ( 86 . 2 V VS =Nh vy,sau 50ns in p ti hai im cui ca ng truyn c cng cu tr li cui cng. . 18 2.3 Hm bc nhy,khng phi hp tr khng ti v ngun 00Vg VloadT1Z0 = 50TD = 25nsRg200Vg0Vsource0RL20Thay RL=20 ,Rg=200 . O O-in p cui ngun v ti ca ng truyn vi t = 0 ... 300 ns(cu 3) 19 Cu hi 3:V dang sng ca in p ngun(Source) v ti (Load) ca ng truyn trong khong thi gian t = 0300 ns ,tng t cu hi 2. Tr li :-in p cui ngun v ti ca ng truyn vi t = 0 ... 300 ns 6 . 050 20050 20000=+=+= IZ RZ Rggg7350 2050 2000 =+=+= IZ RZ RLLLTit = T = 25ns, sng t n ti trng (nhn cui) z = l, v bi vkhng ph hp vi sng phn nh bin : 073= = IL) ( 86 . 0762 ).73( . V V Vl L l~ = = I =+ V vy, in p trn ng dy (trong trng hp ny, in p ti) l tng ca hai sng:) ( 14 . 178762 V V V Vl l L~ = = + =+ Ti t = 2T = 50ns,sng gi n cui cng z = 0,v bi v, c sng phn nh trong hnh thc ca mt ln sng vi bin in p Nh vy, in p trn ng dy ( in p ngun)l : 6 . 0 = Ig) ( 63 . 02V V V V Vl l S~ + + =+ + Ti t = 3T = 75ns, sng t n ti trng (nhn cui) z = l, v bi v ,khng ph hp to ra mt ln sng phn nh vi bin :073= = Ig) ( .2 2V V VL+ I =) ( 85 . 0 V VL = 20 Bng gi tr tnh ton c: Nh vy, hnh v v cu tr li chnh xc,ph hp. in p cui: . vi T=225ns gii quyt cu tr li cui 21 2.4 Xung ngn Sdngngtruyncchiuditngcng25ns(Nndnghaingtruynkhcnhauccngtr khngctrnglZ0=50vTD=12.5ns).Rg=200vRL=20.amtxungtrong10nsnng truyn,vi vg(t) = 10(u(t)- u(t-10ns)) 0Vmiddle00VgT1Z0 = 50TD = 12.5nsRL20VgRg2000 0 0VsourceT1Z0 = 50TD = 12.5nsCu hi 4: v dng sng in p ngun,im gia,cui ti ca ng truyn t = 0100 ns,pht ho biu ,bn hiu hnh v in p khng? di ca cc xung m bn thy ti im gia ca ng truyn(the ghost pulse) t cui ti? ln ca cc xung bng bao nhiu? Dng sng: 6 . 050 20050 20000=+=+= IZ RZ Rggg7350 2050 2000 =+=+= IZ RZ RLLL) ( 250 20050 . 1000VZ RZ VVggl=+=+=+in p ti ngun, im trung tm, v cui ti ca ng truyn vi t = 0 ... 100ns Tr li : 22 Bng tnh ton, ta c bng sau: Vi t=12.5ns cho xung ghost cui ti Xung ghost c rng 12.5ns 23 2.5 Xung di hn Trong phn trn ta dng xung ngn(10 ns) so snh vi TD(time delay)=25ns ca ng truyn Cu hi 5:s dng tr khng ng truyn v tr khng ngun, ta nh ngha li mt ngun vg = +10 V vit = 0 20 ns, and V for t = 20 40ns. V dng sng in p ngun ,im trung tm(center),v cui ti ca ng truynvi t = 0100 ns?chuyn tip t cao n thp c hon ton r rng ti cui ti khng? Dng sng: 0Vmiddle00VgT2Z0 = 50TD = 12.5nsRL20VgRg2000 0 0VsourceT1Z0 = 50TD = 12.5ns - T th, chng ta c th thy rng s chuyn i t cao n thp l hon ton r rng cui ti. Tr li : 24 3.Reactive Termination 3.1 A step into a capacitive load S dng ngun vg(t) = 10u(t), Rg=25 ,ng truyn c Z0=50v TD= 25ns. VsourceCL1nRg250T1Z0 = 50TD = 25nsVgVg Vload000Cu hi 6: u cui ti l mt t in 1 nF . v in p ngun v cui ti ca ng truyn vi t = 0600ns. Do c s np v x ca t in,ta c th dng cng thc sau: Tr li : V Vinitial0 =V VFinal133 . 12 =V t V 4292 . 8 ) ( =ns t 50 25 75 = =89) (10 214 . 4)133 . 12 0133 . 12 4292 . 8ln(10 . 50) ln( = ==final initialfinal tV VV Vtt) ( 14 . 421010 . 214 . 498O = = =CR RCtt25

Cu hi 7:Lp li cu hi trn,nhngthay cun dy ti 0.25 H VsourceL0.25uH12Rg250T1Z0 = 50TD = 25nsVgVg Vload000Tr li : V Vinitial3 . 13 =V VFinal0 =V t V 51 . 1 ) ( =ns t 11 25 36 = =99) (10 056 . 5)0 3 . 130 51 . 1ln(10 . 11) ln( ===final initialfinal tV VV Vtt) ( 44 . 4910 . 056 . 510 . 25 . 096O = = = =ttLRLR26

C100pFVsourceL1uH12Rg250T1Z0 = 50TD = 25nsVgVg Vload0RL1k00Cu hi8:L pli cuhi trn ,s dng ti l mt mch song song gia R, L ,C vi RL=1000 , L = 1 H vC = 100 pF. Tr li : 27

4. Impedance Matching4.1. Pre-lab Assignment Cu hi9: Thit k mt b chuyn i sng vung ni ti 150 in tr ngun 75 , di trng thi ca ng truyn trong gii hn ca bc sng. Tr li : Tr khngvo: vi Chng ta c th chia t s v mu s cho v cho Ta c : Tr khng c trng: 28

4.2. Lab Assignment Cu hi10: Tm chiu di thc t ca mng kt ni sng vung thit k cu 9nu up = 2x108 (m/s) v tn s = 1 GHz M phng p ng tn s ca mch in bng cch qut tn s t 1Mhz n 3Ghz dng ngun sng sin 5Vpp vi in tr ngun 75 ,v in p vo v ti trn tn s.V ln ca h s phn x ca mng kt ni v tm rng bng ti || nh hn 0.2.Tr khng vo ca mng kt ni ti 1Ghz l bao nhiu? Tr li : in p vo v ti ng vi tn s:29 ln ca h s phn x ca mng kt ni:

|| nh hn 0.2 t608.1 Mhz n 1.388 Ghz v tn s t 2.607Ghz n 3Ghz.Tr khng vo ca mng kt ni ti 1Ghz l 75. 30