Bai Giang Mon Hoc Ky Thuat Sieu Cao Tan.diendandaihoc.vn

BI GING MN HC K THUT SIU CAO TNChng 1: GII THIU 1. Khi nim, quy c cc di tn s sng in t 2. M hnh thng s tp trung v thng s phn b. 3. Lch s v ng dng Chng 2: L THUYT NG DY TRUYN SNG. 2.1 M hnh mch cc phn t tp trung cho ng dy truyn sng 2.2 Phn tch trng trn ng dy 2.3 ng truyn khng tn hao c ti kt cui 2.4 Gin Smith 2.5 B bin i bc sng 2.6 Ngun v ti khng phi hp tr khng 2.7 ng truyn tn hao Bi tp chng Chng 3: MNG SIU CAO TN 3.1 Tr khng, in p v dng tng ng 3.2 Ma trn tr khng v ma trn dn np 3.3 Ma trn tn x 3.4 Ma trn truyn (ABCD) 3.5 th dng tn hiu Bi tp chng Chng 4: PHI HP TR KHNG V IU CHNH 4.1 Gii thiu 4.2 Phi hp tr khng dng cc phn t tp trung (mng L) 4.3 Phi hp tr khng dng dy chm 4.4 B ghp bc sng 4.5 L thuyt phn x nh 4.6 B phi hp tr khng a on dng nh thc 4.7 B ghp di rng v tiu chun Bode Fano Bi tp chng Chng 5: CHIA CNG SUT V GHP NH HNG 5.1 Gii thiu 5.2 Cc c trng c bn 5.3 B chia cng sut hnh T 5.4 B chia cng sut Wilkinson 5.5 Ghp nh hng ng dn sng 5.6 Cc b lai (ghp hn tp) Bi tp chng Chng 6: CC B LC SIU CAO TN 6.1 Gii thiu 6.2 Cc cu trc tun hon 6.3 Thit k b lc dng phng php thng s nh 1Su t m b i: www.daihoc.com.vn

6.4 Thit k b lc dng phng php tn hao chn 6.5 Thit k b lc SCT 6.6 Mt s loi b lc thng gp Bi tp chng Chng 1:

GII THIU

1. Khi nim: Khi nim siu cao tn c hiu ty theo trng phi hoc quc gia, c th t 30 MHz 300 GHz (1) hoc 300MHz 300 GHz (2),, hoc 1 GHz 300 GHz (3) Cc di tn s AM pht thanh 535 1605 kHz L band 1 2 GHz V tuyn sng ngn 3 30 MHz S band 2 4 GHz Pht thanh FM 88 108 MHz C - band 4 8 GHz VHF TV (2 4) 54 72 MHz X band 8 12 GHz VHF TV (5 6) 76 88 MHz Ku band 12 18 GHz UHF TV (7 - 13) 174 - 216 MHz K band 18 - 26 GHz UHF TV (14 - 83) 470 - 894 MHz Ka band 26 - 40 GHz L vi ba 2.45 GHz U band 40 60 GHz * V tn s cao di microwaves nn l thuyt mch c s khng cn hiu lc, do pha ca p dng thay i ng k trong cc phn t (cc phn t phn b). * Thng s tp trung: l cc i lng c tnh in xut hin hoc tn ti mt v tr xc nh no ca mch in. Thng s tp trung c biu din bi mt phn t in tng ng (phn t tp trung Lumped circuit element), c th xc nh hoc o c trc tip (chng hn R, C, L, ngun p, ngun dng). * Thng s phn b: (distributed element) ca mch in l cc i lng c tnh in khng tn ti duy nht mt v tr c nh trong mch in m c ri u trn chiu di ca mch. Thng s phn b thng c dng trong lnh vc SCT, trong cc h thng truyn sng (ng dy truyn sng, ng dn sng, khng gian t do) Thng s phn b khng xc nh bng cch o c trc tip. * Trong lnh vc SCT, khi so snh c vi kch thc ca mch th phi xt cu trc ca mch nh mt h phn b. ng thi khi xt h phn b, nu ch xt mt phn mch in c kch thc ( z , t ) R zi ( z , t ) L zi ( z , t ) ( z + z , t ) = 0 t

(2.1a)(2.1b)

i ( z , t ) G z ( z + z , t ) C z

( z + z , t ) i ( z + z , t ) = 0 t

Ly gii hn (2.1a) v (2.1b) khi z ( z , t ) i( z , t ) = Ri ( z , t ) L z t i ( z , t ) ( z , t ) = G ( z , t ) C z t

0 =>

(2.2a) (2.2b)

y l cc phng trnh dng time domain ca ng dy (trong min thi gian), cn c tn l cc phng trnh telegraph.

4Su t m b i: www.daihoc.com.vn

Nu v (z, t) v i (z, t) l cc dao ng iu ha dng phc th (1.2) V ( Z ) zI ( Z ) z

= ( R + j L ) I ( Z )

(2.3a)

= ( G + j C )V ( Z )

(2.3b)

Ch : (2.3) C dng tng t hai phng trnh u ca h phng trnh Maxwell

E = j H H = j E2) S truyn sng trn ng dy D thy c th a (2.3 a,b) v dngd 2V ( Z ) zd 2 I (Z ) z

2V ( Z ) = 0

(2.4a) (2.4b)

2 I (Z ) = 0

Trong l hng s truyn sng phc, l mt hm ca tn s. Li gii dng sng chy ca (2.4) c th tm di dng :V ( Z ) = V o+ e Z + V o e ZI ( Z ) = I o+ e Z + I o e Z

(2.5a) (2.5b)

T 2.5b c th vit di dng :

I(Z )

Vo+ Z Vo Z = e e Zo Zo

(2.6)

Chuyn v min thi gian th sng in p c th c biu din bi :

( z , t ) = Vo+ cos( t z + + ) e z + Vo cos( t + z + ) ezTrong : l gc pha ca in p phc Vo , Khi bc sng c tnh bi :Vn tc pha :p = = f =2

(2.7)

(2.8)

(2.9)


3) ng dy khng tn hao: (2.7) l nghim tng qut cho ng dy c tn hao vi hng s truyn v tr khng c trng c dng phc. Trong nhiu trng hp thc t tn hao ng dy rt b, c th b qua khi c th coi R = G = 0 v ta c = + j = ( R + j L )(G + j C ) = j LC=> = 0, = LC

(2.10)

Tr khng c trng:

Z0 =Khi :

L C

l mt s thc

(2.11) (2.12a) (2.12b)(2.13)

V ( Z ) = V o+ e j Z + V o e j Z+ I ( Z ) = I o e jZ + Io e jZ

=

2

=

2

p =

1 = LC

LC

(2.14)

2.2 TRNG TRN NG DY Trong tit ny chng ta s tm li cc thng s R, L, G, C t cc vector trng v p dng cho trng hp c th l ng truyn ng trc.1, Cc thng s ng truyn Xt on dy ng nht, di 1m vi cc vect E, vect H nh hnh v - S: Din tch mt ct ca dy - Gi thit V0e j z v I0e j z l p v dng gia cc vt dn. - Nng lng t trng trung bnh tch t trn 1m dy c dng

Wm =-

H .H 4s

*

ds => L =

I0

2

H .Hs

*

ds ( H / m )

(2.15)

Tng t in nng trung bnh tch t trn n v chiu di l:

Wl =

E * E .E ds => C = 4 V0 s

2

* E .E ds ( F / m ) s

(2.16)

- Cng sut tn hao trn mt n v chiu di do dn in hu hn ca vt dn kim loi l: 6Su t m b i: www.daihoc.com.vn

Pc =Vi Rs = 1

Rs 2=

C1 +C2

* H .H dl

(Gi thit H nm trn S)

S

2

l in tr b mt ca kim loi

Theo L thuyt mch =>

R=-

Rs I02

C1 +C2

H .H ''2

*

dl ( / m)

(2.17)

Cng sut tn hao in mi trung bnh trn n v chiu di l :

Pd =

E .ES

*

ds

' '' ' Vi '' l phn o ca hng s in mi phc = j = (1 jtg ) Theo LTM => li G l:

ds ( S / m ) (2.18) V0 S 2, V d: Cc thng s ng dy ca ng truyn ng trc trng ca sng TEM trong ng truyn ng trc c th biu din bi :2

G=

''

E .E

*

V z E= 0 e , b ln a

I H = 0 e z , 2

= ' j '' ,

= 0 .r

( v l cc vector n v theo phng v ) =>

L=

2 b 1 b dd = ln (H / m) 2 a (2 )2 0 a 2

2 ' (F / m) C = b ln a

R=

Rs 1 1 ( + )( / m ) 2 a b

2 " G= ( S / m) b ln a

* Cc thng s ng truyn ca mt s loi ng dy d D L cosh1 ( ) W 2aC 'Cosh 1 ( D / 2a)

'Wd


RG

Rs a

2 Rs W

'Cosh 1 ( D / 2a)

"Wd

3, Hng s truyn sng, tr khng c tnh v dng cng sut - Cc phng trnh telegraph (2.3 a,b) c th thu c t h phng trnh Maxwell - Xt ng truyn ng trc trn c sng TEM c c trng bi: Ez = Hz = 0 v = 0 (do tnh i xng trc) H phng trnh Maxwell x E = - j H (2.19a) (2.19b) xH=jE vi = j (c tn hao in mi, b qua tn hao in dn) (2.19) c th c trin khai thnh: E E 1 + +z ( E ) = j ( H + H ) z z H H 1 +z ( E ) = j ( E + E ) + z z

(2.20a) (2.20b)

V thnh phn z phi trit tiu nn :E = f( z )

g H = (z)

(2.21a)

(2.21b)

- iu kin bin EQ = 0 ti = a, b => EQ = 0 ti mi ni t (2.20a) => H = 0; khi c th vit li :E = jH z H = jE z

(2.22a) (2.22b)

T dng H (2.21b) v (2.22a) =>E = hz

(2.23) (2.24a) (2.24b)b a

- S dng (2.21b) v (2.23) =>h( z ) = jg ( z ) z g ( z ) = jh( z ) z

=> - in p gia hai vt dn c dng:V( z ) = b

=a

E ( , z )d = h( z ).ln

(2.25a)

- Dng in ton phn trn vt dn trong ti = a c dng: 8Su t m b i: www.daihoc.com.vn

I(z) =

2

=0

H (a, z )a.d = 2 .g ( z )

(2.25b) (2.26a) (2.26b) (2.27)

- Kt hp gia (2.24) v (2.25) =>

V ( z ) = jLI ( z ) z I ( z ) = (G + jC )V ( z ) z 2 E

* Hng s truyn sng :+ 2 E = 0 Z 2 2 = 2 => = + j

Vi mi trng khng tn hao => = j vi = = LC (2.28) * Tr khng sng : E = = = Z = (2.29) H Vi l tr khng ni ca mi trng * Tr khng c tnh ca ng truyn ng trcb b E ln ln V0 a = a = Z0 = = I0 2H 2

a 2

ln

b

(2.30)

* Dng cng sut (theo hng lan truyn Z) c th dc tnh qua vector Poynting:1 1 P = E H .dS = 2S 2 =0 = a2 b * V0 I 0

2 2 ln

b a

* .d .d = V0 I 0

1 2

(2.31)

(2.29) trng vi kt qu ca l thuyt mch. iu ny chng t cng sut c truyn i bi s lan truyn ca trng in t gia hai vt dn.

2.3 NG TRUYN KHNG TN HAO C TI KT CUI1, H s phn x in p:

- Xt ng truyn khng tn hao c ti u cui vi tr khng ZL. Khi s xut hin sng phn x trn ng truyn. y l c trng c s ca cc h phn bGi thit c mt sng ti c dng: V0+ e j z c pht bi mt ngun nh x min Z V 0 = L V0 + ZL + Z0 V0 V0

* nh ngha h s phn x bin in p :=

V0 Z L Z 0 = V0+ Z L + Z 0

(2.33)

Khi => V( Z ) = V0+ e jz + e jz

I(Z )

[ V [e = Z+ 0 0

]

(2.34a)

jz

+ e jz

]

(2.34b)

- Sng p v dng dng (2.32) l chng cht ca sng ti v sng phn x, gi l; sng ng. Ch khi = 0 mi khng c sng phn x. nhn c = 0 th ZL = Z0, khi ta ni ti cn bng tr khng (ph hp tr khng) vi ng dy (hay ti phi hp) 2, T s sng ng: (SWR: Standing ware ratio) - Dng cng sut trung bnh dc theo ng truyn ti im Z:+ 1 1 V0 2 * Pav = Re V( Z ) .I ( Z ) = Re 1 * e 2 jz + e 2 jz 2 2 Z0

[

]

2

{

}

=>

+ 1 V0 2 Pav = 1 2 Z0

2

(

)

(2.35)V0+2

- Nhn xt: Dng cng sut trung bnh bng const ti mi im trn ng truyn. Cng sut ton phn t trn ti Pav bng cng sut sng n cng sut phn xV0+ 2Z 02 2

2Z 0

tr i

nu = 0 cng sut tiu th trn ti cc i (gi thit my

pht c phi hp tr khng vi ng dy sao cho khng c sng phn x t min Z < 0.) - Khi ti khng phi hp vi tr khng (mismatched) s c tn hao quay ngc (return loss RL): RL = - 20 lg ( dB) (2.36) + Nhn xt: o Vi ti phi hp ( = 0 ) RL = dB o Vi ti phn x ton phn (= 1) RL = 0 dB - Khi ti phi hp ( = 0) th bin in p V(z)= V0+= const, ng dy c gi l phng (flat). - Khi ti khng phi hp tn ti sng phng x xut hin sng ng (bin p trn ng dy khng bng hng). 10Su t m b i: www.daihoc.com.vn

T (2.34a) V ( Z ) = V 0+ 1 + .e

j ( 2 l )

(2.37)

Trong : - l : khong cch tnh t ti z = 0 - : pha ca h s phn x = .e j => Nhn xt: + Bin in p dao ng theo ta + V( Z ) = Vmaxe j ( 2 l ) =1

= V0+ 1 +

(2.38)

+ Nu tng th t s Vmax/Vmin tng theo, do Vmax/Vmin c th dng o s mt phi hp tr khng (mismatch) ca ng dy, gi l t s sng ng (Standing ware ratio, SWR):SWR = Vmax 1 + = Vmin 1

(2.39)

hay Voltage_SWR, hay VSWR Nhn xt: + 1 SWR , SWR = 1 matched Load + Khong cch gia hai cc i lin tip l:l = 2 2

=

2

+ Khong cch gia 2 cc tr lin tip ll= 2

=

4

vi :bc sng =

2

+ nh ngha (2.31) v c th tng qut ha cho mi im l trn ng dy nh sau: vi = l T s thnh phn phn x trn thnh phn ti l:( l ) = V0e jl = ( 0 ) e jl + j l V0 e

(2.40)

Vi (0) l h s phn x ti Z = 0 cho bi (2.31) - V dng cng sut bng const, m bin in p thay i theo l tr khng vo ca on dy l + ti phi thay i. => nh ngha tr khng vo ca on dy l + ti nhn theo hng thun


Z in =

V( l ) I (l)

=

1 + e 2 jl Z0 1 e 2 j l

(2.41)

Dng (2.31) =>Z in = Z 0 Z L + jZ 0 tgl Z 0 L + jZ L tgl

(2.42)

3, Cc trng hp c bit: a) Ngn mch u cui: ZL = 0 t (2.31) => = 1 t (2.37) => SWR = + t (2.32) => V( Z ) = 2 jV0 sin z

(2.43a)

I (Z )

2V0+ = cos z Z0

(2.43b)

=> V= 0 ti u cui v I = max t (2.40) => r khng vo ca on dy l l:

Z in = jZ o tg l

(2.43c)

=> Zin thun phc, Zin = 0 khi l = 0, Z in = (h mch) khi l = 4 Zin bin thin tun hon theo l vi chu k 2 b) H mch u cui: Z L = , t (2.31) => = 1, SWR =

V( Z ) = 2V0+ cos zI (Z ) 2 jV0+ sin z = Z0

(2.44a)

(2.44b)

=> I = 0 ti Z = 0, V = Vmax , c) S thay i ca Zin(l) Z i n (l = /2) = ZL

Z in ( l ) = jZ o cot g l

(2.44c)

(2.45)

(t 2.40) on dy di nguyn ln na bc sng khng lm thay i tr khng ti bt k gi tr ca tr khng c trng. 12Su t m b i: www.daihoc.com.vn

Z02 Zi n (l = /4) = Z L

(2.46)

on bin i mt phn t bc sng v n bin i nghch o ZL d) Ghp hai ng dy : Dng ng dy c tr khng c trng Z0 nui ng dy c tr khng c trng khc Z1 Gi thit b qua sng phn x t ng dy Z1 ( tc n di hoc c kt cui bi ti c tr khng bng Z1) Khi : Z1 - Z0 =Z +Z1 0

(2.47)

Nhn xt: - Khng phi tt c cc sng ti u b phn x, mt s s truyn tip ln ng dy th hai vi bin xc nh bi h s truyn T - T (1.32a) vi z < 0 V( Z ) Z 0 V( Z ) Z >0 = V0+ e jz (2.48b) (B qua sng phn x trn ng dy 2) - Cn bng (2.46 a) v (2.46b) ti z = 0

[

]

Z1 - Z0 2Z1 T=1+=1+Z +Z = Z +Z (2.49) 1 0 1 0 - H s truyn gia hai im ca mt mch thng c biu din theo dB, gi l tn hao chn (IL: Insertion loss) IL = - 20 lg T (dB) Ph ch: - T s bin theo n v Nepers (Np) V1 lnV (Np) 2 - T s cng sut theo Np: P1 ln P (Np)2

(2.50)

1Np tng ng vi t s cng sut = e2 1Np = 10 lg e2 = 8,686 dB


2.4 GIN SMITH- Gin Smith, do P. Smith a ra nm 1939 ti Bell Telephone Laboratories, l phng php th c dng rng ri nht cho cc bi ton v tr khng v cc hin tng trn ng dy truyn sng. 1. th Smith: Thc cht l th cc ca h s phn x in p . - Gi s c th c biu din di dng cc (theo bin v pha) = e j . Khi mi gi tr c biu din bi 1 im trong h ta cc. Z - Trong ta Smith ngi ta dng tr khng chun ha Z = Z thay Z. 0 - Vi ng dy khng tn hao c kt ni vi ti ZL th h s phn x c th c vit qua tr khng chun ha nh sau:= ZL 1 = e j ZL +1

(2.51)

ZL Vi ZL = Z l tr khng ti chun ha. t quan h ny 0

ZL =

1 + e j 1 e j1 r2 i2

(2.52)

- Nu t = r + j i v zL = rL + j xL th t (2.50) rL =

(1 r )2 + i2

(2.53a) (2.53b) 2

xL =

(1 r )2 + i2 1 + i2 = 1+ r L 2 2

2i

- Vit li (2.51) di dng phng trnh ng trn : r r L 1 + rL 2

(2.54a) (2.54b)

(r 1) + i 1 = 1 x xL L

2

y l cc phng trnh ca 2 h ng trn trong mt phng r, i - (2.54a) biu din h cc ng trn in tr v (2.54b) biu din h cc ng trn in khng. * V d: Vi rL = 1 ng trn (2.54a) c tm ti r = 0,5, i = 0, bn knh bng 0,5. * Ch : - Tt c cc ng trn in tr (2.54a) u c tm nm trn trc honh (i = 0) v i qua im (1, 0) hay im = 1 bn mp phi ca gin . - Tm ca cc ng trn in khng (2.54b) nm trn trc ng i qua im (1, 0) hay ng r = 1 v cng i qua im (1, 0) hay im = 1. - Cc ng trn (2.54a) v (2.54b) lun vung gc nhau. * ng dng: Gin Smith c th dng gii bng th phng trnh (2.42) cho tr khng ng dy. 14Su t m b i: www.daihoc.com.vn

Z in =

Vi l h s phn x ti ti u cui l l chiu di on dy. - D thy (2.55) c dng tng t (2.52) ch khc s hng gc pha trong . Do Zin nu c th e j ti ti th tr khng vo chun ha Z nhn vo on dy l 0 cc th tm c bng cch quay im tha mn h (2.54) i theo chiu kim ng h 1 gc 2l quanh tm ca gin . (Bn knh gi nguyn v ln khng i dc theo chiu ng dy.) - d thc hin cc php quay ni trn, trn gin Smith c thang chia theo n v bc sng theo 2 hng. V l thang tng i nn ch c s khc nhau theo bc sng gia 2 im trn gin mi c ngha. V d 1: Cho ti c tr khng ZL = 130 + j 90 () kt cui ng dy 50 c chiu di 0,3 . Hy tm h s phn x ti ti v h s phn x ti u vo on ng dy, tr kghng vo, h s SWR v RL. ZL Gii: Tr ti chun ha zL = Z = 2,60 + j 1,8 0 Tm giao im ng trn rL = 2,60 v xL = 1,8 trn gin M dng compa o on OM ri i chiu vi thang c = 0,6 SWR = 3,98, RL = 4,4 dB ko di on OM c c gc pha ca h s phn x ti ti theo vng chia ngoi gin : 21,80 v vng trn bn knh OM Tm v tr ca tia OM v vng chia theo bc sng hng v ngun pht (WTG: Wavelengths toward generator) cho gi tr 0,22 . di chuyn im 0,22 i mt on 0,3 v pha ngun s cho gi tr 0,52 ,gi tr ny ng vi 0,02 .V tia t tm 0 qua im 0,02 ,tia ny ct vng trn bn knh OM ti im ng vi Zi n = 0,255 + j 0,117 sau Z i n = Z0 Zin = 12,7 + j 5,8 () Gc pha ca ti u on ng dy l 165,80. 2. Gin Smith vi tr khng v dn np kt hp: - Gin Smith c th s dng cho dn np chun ha theo cch tng t nh vi tr khng chun ha v c th dng chuyn i gia tr khng v dn np. - Tr khng vo ca on ng dy bc sng kt cui ti ZL l Zi n = 1/ZL, y l c s chuyn i mt tr khng chun ha vi mt dn np chun ha. - rng mt on bin i tng ng vi php quay 1800 quanh tm ca gin , do im i xng tm ca 1 im tr khng (hoc im dn np) s l mt im dn np (hay im tr khng) tng ng ca cng mt on dy c ti kt cui. Vy cng mt gin Smith c th dng tnh tr khng v dn np cho cng mt bi tan. - trnh nhm ln, c th dng gin Smith kp bao gm c gin tr khng v gin dn np, c dng tng t nhau ch l hnh nh i xng tm ca nhau. 15Su t m b i: www.daihoc.com.vn

1 + e 2 jl Z0 1 e 2 jl

(2.55)

V d 2: Cho ti ZL = 100 + j 50 kt cui ng dy c tr khng c trng 50 . Tm dn np ca ti v dn np vo ca on ng dy 0,15 . Gii: + Zl = 2 + j 1. c th tin hnh nh cc bc v d 1 ri quay gc /4 trong gin tr khng, sau quay gc 0,15 . + Cng c th v im zL ri c yL tng ng theo thang ca gin dn np: yl = 0,40 j 0,20 yL YL = yL . Y0 = Z = 0,008 j 0,004 (S) 0 Sau trn thang WTG tm im tham chiu tng ng 0,214 ,di chuyn on 0,15 cho n 0,,364 , v tia qua im ny ri c im ct vi vng trn SWR cho gi tr y = 0,61 + j 0,66 Y = 0,0122 + j 0,0132 (S)

2. 5 BIN I BC SNG1) Tr khng: Gi thit ti thun tr RL kt cui on /4 c tr khng c trng cn tm Z1 sao cho = 0 ti u vo ca n (on )Z in = Z 1 RL + jZ 1tgl Z1 + jRL tgl

(2.61) (2.62)

V

Z 12 2 => Z in = l = , = 4 4 RL

(2.63) = 0 cn c Z in = Z 0 => Z1 = Z 0 RL => Khng c sng ng trn feedline (SWR = 1). 2) p ng tn s: V d: Xt ti RL = 100 ghp vi ng truyn 50 qua b ghp hy v th bin ca h s phn x theo tn s chun ha f/f0 vi f0 l tn s m ti chiu di on ghp bng /4 Z 1 = 50.100 = 70,71 Gii:=

Z in Z 0 Z in + Z 0

vi Zin l hm ca tn s cho bi (2.46). p 4 f 0 f = 2f 0

l =

2 0 2f = 4 p


2. 6 MY PHT V TI KHNG PHI HP TR KHNG - Xt trng hp tng hp khi my pht v ti khng cn bng tr khng ving truyn Z0. Tm iu kin cng sut my pht truyn n ti t cc i.Z in = V( l ) I (l)=

Vi

l =

in p trn ng dy c th vit d dng V( Z ) = V0+ e jz + l e jz

Zl Z0 Zl + Z0

Z + jZ 0 tg l 1 + l e 2 jl Z0 = Z0 L 2 j l Z 0 + jZ L tg l 1 l e

(2.67) (2.68)

[

]

(2.69)

-

V0+ c th tm c nh iu kin bin ti z = l Z in V( l ) = Vg = V0+ e jl + l e jl Z in + Z g

[

](2.70)

=>-

V0+ = Vg

Z in 1 jl Z in + Z g e + l e jl

Dng (2.67)

Z0 e j l V = Vg Z 0 + Z g 1 g l e 2 jl+ 0

(2.71) (2.72)

Vi

g =

Z g Z0

Z g + Z0

H s sng ng trn ng dy.SWR = 1 + l 1 l2

(2.73)2

-

Cng sut t vo ti v ng truyn1 P = Vg 2 1 Z in Re Z in + Z g Z in

(2.74)

t Z in = Rin + jX in v Z g = R g + jX g =>P= 1 Vg 22

(R2

in

+ R g ) + (X in + X g )2

Rin

2

(2.75)

a) Ti phi hp vi ng truyn: Z in = Z 0 v P = Vg1 2

Zl = Z0 , l = 0, SWR = 1o 2 + Rg ) + X g 2

(Z

Zo

(2.76)

b) My pht phi hp vi ng truyn c ti kt cui: Z l , l, Z 0 c chn sao cho Z i n = Zg 17Su t m b i: www.daihoc.com.vn

Zi n - Zg =Z +Z =0in g

(2.77)

(Lu : c th tn ti sng ng trn ng truyn nu l 0)P= 1 Vg 22 2 2 4 Rg + X g

(

Rg

)

2

(2.78)

Nhn xt: Cng sut (2.78) c th nh hn cng sut (2.76). Cu hi: + Tr khng ti th no l ti u? + Tr khng vo ng truyn th no l ti u? * Phi hp lin kt: Gi thit Zg c nh, tm Zin P t cc di sau s suy ra Zl khi bit l. Cho o hm ca P theo phn thc v phn o ca Zin= 0 iu kin phi tm. T (2.75) P 2 2 2 = 0 => Rg Rin + (X in + X g ) = 0 Rin

(2.79a)(2.79b)

P = 0 => 2Xin (Xin + X g ) = 0 XinT (2.79a,b) =>

Rin = Rg , Xin = Xg*

Zin = Zg Hay (2.80) c gi l iu kin phi hp tr khng lin ktKhi cng sut ri trn ti l cc i. (t 2.75)P=2 1 1 Vg 2 4 Rg

(2.80)

(2.81)

Nhn xt: - Cng sut (2.81) ln hn (2.76) v (2.78) - l, g, c th khc khng. V mt vt l iu c ngha l trong hin tng a phn x c th xy ra hin tng ng pha dn ti cng sut ln hn khi ch c sng ti. - V phng din hiu qu th t hiu qu bcao c iu kin phi hp tr khng (Zl = Z0) hay iu kin phi hp lin kt (Z i n = Zg*) vn cha . chng hn khi Zg = Zl = Z0 ch c cng sut ca pht ri trn ti tc hiu sut l 50%. Hiu sut ny ch c th c ci thin nh gim Zg nh c th c. Bi tp chng 1. Cho ng truyn c L = 0,2 H/m, C = 300 p F/m, R = 5 /m v G = 0,01 S/m. Hy tnh hng s truyn sng v tr khng c trng ti tn s 500M Hz. Hy xt trng hp khng hao tn. 2. Cho mt hnh T CMR m hnh ny dn ti cng phng trnh Telegraph. 3. Mt ng truyn ng trc bng Cu vi bn knh vt dn trong l 1mm v ngoi l 3mm. Lp in mi c r = 2,8 vi gc tn hao 0,005. Tnh R, L, G, C ti tn s 3 GHz, tnh Z0 v vp. 18Su t m b i: www.daihoc.com.vn

4. Tnh v v th h s suy gim ca cp ng trc bi 3 theo dB trong khong tn s t 1 MHz ti 10 GHz. 5. Cho ng truyn khng tn hao c chiu di in l = 0,3 kt cui ti ZL = 40 + j 20 (). Tm L, SWR trn on l v Z i n (l + ti) 6. Cho ng truyn khng tn hao kt cui ti 100 . Tm Z0 nu bit SWR = 1,5 7. Mt my pht v tuyn c ni vi angten c tr khng 80 + j40 qua cp ng trc 50 c th cung cp 30W khi ni vi ti 50 th cng sut t vo angten l bao nhiu 8. Gin Smith c th tnh a, SWR trn ng truyn b, TL, c, YL d, Z i n (l + ti) e, Khong cch t ti n im c Vmax u tin . f, Vmin u tin v hnh 9. Dng gin Smith tm on ng truyn 75 ngn mch u cui ngn nht c: a, Z i n = 0 b, Z i n = c, Z i n = j 75 d, Z i n = - j 50 e, Z i n = j 10


Chng III:

L THUYT MNG SIU CAO TN

3.1 TR KHNG, IN P V DNG IN TNG NG1) in p v dng in tng ng tn s siu cao cc php o p v dng rt kh thc hin, tr khi mt cp u cui c xc nh r rng. iu ny ch thc hin c vi ng truyn sng TEM (cp ng trc, mch vi di) V hnh * Trn hnh v l dng ng sc in trng v t trng ca 1 ng truy sng TEM gm 2 vt dn Theo nh nghaV = E dl+

I=

C+

H .dl

* Vn s tr nn kh khn hn khi kho st ng dn sng. - Xt ng dn sng ch nht nh hnh v. Mode truyn sng ch yu l TE10: Cng thc (v hnh) j a x j z E y (x , y ,z ) = A sin e a (3.4.a ) = Ae y ( x , y , z )e j zH x( x, y, z ) = j a A sin

xa

e jz = Ah x ( x , y ) e jz

(3.4.b )(3.5)

S dng (3.1) cho (3.4.a)V= ja

A sin

xa

=>y

e jz dy

Nhn xt: Dng in p (3.5) ph thuc vo v tr x cng nh di ca ng ly tch phn theo hng trc y. Vy gi tr in p chnh xc l bao nhiu? Cu tr li l khng c gi tr in p chnh xc hiu theo ngha duy nht hoc thch hp cho mi ng dng. Vn trn pht sinh tng t cho dng in v tr khng khi sng khng phi l sng TEM. * C rt nhiu cch nh ngha in p, dng in tng ng v tr khng cho sng khng phi TEM v th khng duy nht. Tuy nhin c mt s nhn xt sau: + in p v dng ch c nh ngha cho mt mode dn sng c th v c nh ngha sao cho in p t l vi in trng ngang, cn dng in t l vi t trng ngang. + c c s dng tng t nh p v dng trong l thuyt mch, in p v dng cn c nh ngha sao cho tch ca chng cho ra dng cng sut ca mode truyn sng.20Su t m b i: www.daihoc.com.vn

+ T s p trn dng cho mch sng chy n l cn bng tr khng c trng ca ng truyn. Tr khng ny c th chn bt k, thng chn bng tr khng sng ca ng truyn. * Vi mt mode ng dn sng bt k cc thnh phn trng ngang c th c biu din:E t ( x , y , z ) = e ( x , y ) ( A + e j z + A e j z ) = H t ( x , y , z ) = h x , y A + e jz A e jz = e x , y + jz (V e + V e jz ) c1 h( x , y ) c2

(3.6a )(3.6b )

(

)

(I

+

e jz I e jz

)

Trong A+, A- l bin ca sng ti v sng ngc; e, h l cc thnh phn trng ngang ca mode c quan hh (x, y ) = a z e( x, y )

(3.7 )

vi : tr khng sng. T (3.6,a,b) c th nh ngha p v dng tng ng:V ( z ) = V + e j z + V e j z I ( z ) = I + e j z I e j z

(4.8a ) (3.8.b )

V+ V Vi + = = 0 I I

Nhn xt: - nh ngha (3.8) bao hm quan h t l gia p v dng tng ng vi in v t trng ngang. - Cc hng s t l c cho cc mi quan h ny l:C1 = V+ V I+ I = , C2 = + = A+ A A A

- Dng cng sut ca sng ti:P+ =1 + A 22

e h * .a z ds =s

V +I + * e h *.a z ds 2C1C 2 * s

(3.9)

cng sut

1 P+ = V *I + * th phi c 2 C1C 2 * = e h * .a z dss

(3.10 )

- Tr khng c trngZ0 = V + V C1 = = C2 I+ I

(3.11) (3.12 ) a21

Nu mun c 0 = :tr khng sng ( TE hoc TM ) ca mode truyn th :C1 = C2

( TE hoc TM )

Su t m b i: www.daihoc.com.vn

gii (3.10) v (3.12) => C1 , C 2 => in p tng ng v dng tng ng V d: Cho mode TE10 trong ng dn sng ch nht E y = A + e jz + A e jz sin a 1 + jz A e jz sin H = A e TE a

(

)

(

)

V ( z ) = V + e j z + V e j z I ( z ) = I + e j z I e j z

=

1 (V + e jz V e jz ) 0

1 P = V +I + * 2 1 P + = E y H x ddy 2 s= ab A+ 4 Z TE2 2 1 1 = V + I + * = A + C1C 2 * 2 2

Nu chn 0 = TE th

V + C1 = = TE I + C2

=>

C1 =

ab 2 ab 1 C2 = TE 2

2) Khi nim tr khng: C cc dng tr khng sau: - Tr khng ni ca mi trng = / ch ph thuc vo mi trng v bng tr khng sng ca sng phng.

- Tr khng sng ZVV =

Et 1 = c trng cho cc dng sng (TEM, TE, TM) Ht Yvv

v c th ph thuc vo loi ng truyn hoc ng dn sng, ph thuc vt liu v tn s hot ng.1 - Tr khng c trng Z0 = = y0L C l t s p trn dng cho cc sng chy.

V p v dng l xc nh duy nht cho sng TEM Z0 cng xc nh vi sng TEM. * Quan h gia cc c trng tr khng v nng lng trng EM tch t v cng sut tiu tn trong mng 1 ca.1 P = s E H * ds 2 = Pl + 2 j (Wm We )


Vi Pl: phn thc ca P Biu th phn cng sut trung bnh tiu tn trn mng, Wm, We. Biu th nng lng t trng v in trng tch t trong mng. - Nu nh ngha e v h l cc vect trng ngang chun ha trn mt kt cui ca mng, sao choEt ( x, y, z ) = V( z ) e ( x , y )e jz H t ( x , y , z ) = I ( z ) h ( x , y ) e jz

Vi se h *.ds = 1 th P = Khi in = R + jx =

Vy : -

V VI Pl + 2 J (Wm We ) = 2 = 1 2 I I I 2 P = 1 2 I 2 Phn thc ca in ,R lien quan n cng sut tn hao Pl

1 1 VI * e h * ds = 2 VI * 2s

Phn o X lien quan n nng lng tng cng tch t trong mng Nu mng khng tn hao th in thun o vX= 4 (Wm We ) I2

=

Dng cho ti cm khng (Wm > We ) m cho ti dung khng (Wm < We )

3.2 MA TRN TR KHNG V MA TRN DN NP1)Ma trn tr khng v ma trn dn np: V in p v dng c nh ngha ti cc im khc nhauca mng SCT,nn c th dng ma trn tr khng v ma trn dn np theo kiu LT mch rng buc nhng i lng ny vi nhau. iu ny s gip xy dng mch tng ng cho mng SCT bt k, phc v cho vic thit k cc phn th ng nh cc b ghp, cc b lc. v hnh - Xt mng SCT N cng ty , cc cng c th l dng ng dy truyn sng hoc ng truyn tng ng vi mt mode truyn dn sng n. Nu mt cng no v mt vt l c nhiu mode truyn th c th thay tng ng bng mt s cng n mode tng ng.

- Ti cng th n ty in p v dng tng c dngVn = Vn+ + Vn+ In = In In

(3.24a ) (3.24b )

(dng 3.8 vi ta Z = 0 ) Ma trn tr khng c nh ngha:


V V V

1 2

N

Hay vit gn hn

[V ] = [ ][I ]

=

11 21

12 22

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

1 N 2 N

N

1

N

2

NN

I I I

1 2

N

(3.25)(3.26)

Tng t cho ma trn dn np R rng T (3.25) => =

[I ] = [Y ][V ] [Y ] = []1I k = 0, k j

(3.27 )

Vi I

(3.28)

- (3.28) c ngha l Zi j c th tm c khi cp dng Ij cho cng th j, cc cng cn li h mch v o th m mch ti cng th i, cn li Z i j l tr khng truyn gia cng i v j. - Z i i l tr khng vo ti cng i khi tt c cc cng khc h mch. - Tng t:

Y =

Ii Vj

V k = 0 ,k j

(3.29)

2) Cc trng hp c bit: - Vy mt mng n cng ty s c th 2N2 i lng c lp, hay bc t do. (ng vi phn thc v o ca cc Zi j). - Nu mng l thun nghch, tc khng cha cc mi trng khng thun nghch (nh ferrile hay plasma) hoc cc linh kin tch cc, th Z i j = Z j i v Y i j = Yj i. - Nu mng l khng tn hao th Z i j v Y i j l cc i lng thun o.

3.3 MA TRN TN X1) Ma trn tn x: Xt mng N cng nh trong mc trc. nh ngha ma trn tn x tha mn quan h sau: V hnh:V 1 S 11 S12 ....S 1N V1+ 1 V 2 = V S ..........S V + NN N N N1 + V = [S ] V

Hay gn hn =>S = Vi

[ ]

[ ]

(3.40 )

V j+

Vk+ = 0 , k j


- Tc l Si j c th c tm khi t vo cng j mt sng ti c in p V+j v o bin in p sng phn x Vi- t cng i, khi tt c sng ti cc cng khc cho bng zero (hay kt cui vi ti phi hp trnh phn x). - Si i chnh l h s phn x nhn vo cng i khi tt c cc cng khc kt cui vi ti phi hp. - S i j cn gi l h s truyn t cng j ti cng i khi tt c cc cng khc kt cui vi ti phi hp. - C th chng manh rng ma trn [ S ] c th c xc nh t [ Z ] hoc [ Y] v ngc li. - Trc tin gi thit rng tr khng c trng ca tt c cc cng, Z o n, l ging nhau. (Trng hp tng qut s c cp sau). tin li cho Z o n = 1. T (3.24) Vn = V+n + V-n (3.42a) + + (3.42b) I n = In - I n = V n - V n T (3.25) v (3,42) [ Z ] [ I ] = [ Z ] [ V+ ] - [ Z ] [ V- ] = [ V ] = [ V+ ] + [ V- ] tc l c th vit ( [ Z ] + [ U ] ) [ V- ] = ( [ Z ] - [ U ] ) [ V+ ] (3.43) Vi [ U ] l ma trn n v So snh (3.43) vi (3.40) (3.44) [ S ] = ( [ Z ] + [ U ] )1 ( [ Z ] - [ U ] ) Z11 - 1 - Vi mng mt cng: S11 = Z + 1 , y chnh l h s phn x nhn vo ti 11 vi tr khng vo chun ha Z11. - biu din [ Z ] theo [ S ] c th vit li (3. 44): [Z] [S] + [U] [S]= [Z]- [U] (3.45) [ Z ] = ( [ U ] - [ S])- 1 ( [ U ] + [ S ] 2) Mng thun nghch v mng khng tn hao. a,Mng thun nghch: -T (3.42, a, b ) => Vn+ = (Vn + I n ) Hay1 2

[V ] = 1 ([ ] + [U ])[I ] 2+

(3.46)a

Vn =

Hay

[V ]

1 (Vn I n ) 2 1 = ([] [u ])[I ] 2

(3 .46 b )(3.47 )

chuyn v (3.47 ) => [S ]t = {([] + [U ]) } ([] [U ])t V [U ]t = [U ] v [Z ] i xng [Z ]t = [Z ] nn [S]t = ([Z ] + [U ]) 1 ([Z ] [U ]) [S ] = [S ]t t 3.44 1 t

-T (3.46 ) => [V ] = ([] [U ])([] + [U ])1 [V + ] => [S ] = ([] [U ])([] + [U ])1


Vy [S ] l ma trn i xng b,Mng khng tn hao: Cng sut trung bnh tiu th trn mng phi bng khng. Gi thit tr khng c trng bng n v cho tt c cc cngPav =t t 1 1 t Re [V ] [I *] = Re V + + V V + * V * 2 2 t t t t 1 = Re V + V + * V + V V + * V V * 2 t 1 + t + 1 = V V * V V *=0 2 2

{

}

{[ ] [ ] ([ ] [ ] )}

{[ ] [ ] [ ] [ ] [ ] [ ] [ ] }[ ][ ](3.49 )t

[ ][ ]

v [V + ] [V ]* +[V ] [V + ]* c dng A-A* nn l thun o do Re { } = 0t

Trong (3.49 ) s hng = [V + ] [V + ]* biu th cng sut n tng cng ,s hng 1 t V V * l cng thc phn x tng.V mng khng tn hao nn 2 cng sut trn 2

{

}

[ ][ ]

1 2

t

phi bng nhau ,Tc l

[V ] [V ]* = [V ] [V ]*+ t + t

(3.50 )

[V ] = [S ][V + ]+ t +

Hay [S]* = {[S ]t } vy [S] l ma trn unita - khai trin (3.51) =>

=>nu [V

+

] 0

[V ] [V ]* = [V ] [S ] [S ] * [V ]*+ t t +

=>

th [S ]t [S ]* = [U ]1

(3.51)

S=>

N

S Sk =1 k =1 N

k =1 N

ki

S* = S ki S* = 1 ki S* = 0 ki

, i, j

(3.52 )(3.53a )

ki

ki

vi i j

(3.53b )

- Tnh im ca mt ct bt k vi lin hip phc ca n bng n v. - Tnh im ca mt ct bt k vi lin hip phc ca cc ct khc bng zero (trc giao) - Kt lun tng t cho cc hng ca ma trn tn x 3) Php dch mt tham chiu V cc thng s ca [ S ] lin quan n bin v pha ca sng n v sng phn x t mng, do mt phng pha tham chiu, tc l mt phng xc nh (Vn+, In+) hoc (Vn-, In-) phi c xc nh trc. Khi dch chuyn cc mt tham chiu ny th cc thng s S b bin i. 26Su t m b i: www.daihoc.com.vn

Xt mng SCT N cng cc mt tham chiu ban u nh x ti Z0 = 0. Vi Zn l ta dc theo ng truyn th n cp in cho cng n. Gi [ S ] l ma trn tn x vi tp hp cc mt tham chiu ni trn. [ S ] l ma trn tn x tng ng vi v tr mi ca cc mt tham chiu. (3.54a) [ V- ] = [ S ] [ V+ ] + [ V ] = [ S ] [ V ] (3.54b) + + j n (3.55a) trong : V n = V n e -j n (3.55b) V n = V n e Vi n = n l n c gi l di in ca php dch ca cng n - Vit (3.55a,b) di dng ma trn ri thay vo (3.54a) e j1 e j1 0 e j2 V = [S ] e j N + V = [S ]V

[ ]

e

j2

[ ]

[ ]

0 + V e j N

[ ]

- Nhn c hai v vi ma trn nghch o ca ma trn u tin bn v tri

[ ]

e j1 V =

e

j2

0 e j1 [S ] j N e

e

j2

0 + V j N e

[ ]

So vi (3.54b) e j1 S = 0 e j1 [S ] j N e 0 e j N

[ ]

e

j2

e

j 2

(3.56)

- D thy Sn n = e 2 n Sn n, c ngha l pha ca Sn n di 2 ln di in trong php dch mt tham chiu n, bi v sng truyn 2 ln qua di ny theo hng ti v hng phn x. 4) Cc thng s tn x tng qut Xt mng SCT N cng vi Z0 l tr khng c trng (thc) ca cng n, Vn+, Vnl bin sng ti v sng phn x. nh ngha :an = bn = Vn+ 0n Vn_ 0n

(3.57a ) (3.57b )

L cc bin sng mi cho cng n. 27Su t m b i: www.daihoc.com.vn

-T (9.42 a,b) =>

Vn=Vn+ + Vn = (an + bn )

( 3.58a ) ( 3.58b )

In =

1 1 (an bn ) Vn+ + Vn = 0n 0n

(

)

Cng sut trung bnh ri trn cng n:Pn =

1 1 2 2 * * Re { n I n } = Re an bn + bn an bn an V 2 2

{2

(

)}

= an bn* * (v bn an bn an thun o)

1 2

2

1 2

( 3.59 )

C th ni cng sut trung bnh ri trn cng bng cng sut sng n tr cng sut sng phn x. - Ma trn tn x tng qut c nh ngha [b] = [S ][a ] ( 3.60 ) Trong

S ij =

bi a

a k = 0 ,k j

( 3.61)

- (3.61) c dng tng t (3.41) cho mng vi tr khng c trng ng nht ti tt c cc cng. Dng (3.57) v (3.61) =>

S ij =

v i 0 j V j+ 0 jV k+ = 0 , k j

( 3.62 )

Cng thc ny cho bit cch chuyn t cc thng s S cho mng vi tr khng c trng ng nht (V-i/V+j) thnh cc thng s S cho mng ni vi cc ng truyn c tr khng c trng khng ng nht.


3.4 MA TRN TRUYN (ABCD)Cc mng SCT thng gp trong thc t bao gm mt mng 2 cng hoc dy cascade ca cc mng 2 cng. Cc ma trn c trng (S, Z, Y) ca dy cc mng 2 cng bng tch cc ma trn 2 x 2 (ABCD) ca mng 2 cng. 1) Ma trn ABCD: c nh ngha cho mng 2 cng nh sau:V1 = AV 2+ BI 2 I1 = CV2 + DI 2

Hay

V1 A B V2 I = C D I 2 1

( 3.63 )

* Ch : Quy c du I2 ra khi cng 2 l tin li cho vic kho st mng cascade. - Khi c 2 mng kt ni cascadeV1 A1 I = C 1 1

B1 V2 D 1 I 2

( 3 .64 a ) ( 3.64 b )

V 2 A2 I = C 2 2V A

B 2 V 3 D2 I 3 B V

3 2 => 1 = 1 1 2 I I1 C1 D 1 C2 D2 3

B A

( 3 .65 )

Hay * Ch : A B A1 C D = C 1B1 A2 D 1 C2 B2 D2

- Th t nhn ma trn phi ging th t cascade. - C th xy dng mt th vin cc ma trn ABCD cho cc mng 2 cng c s v dng php phn tch cc mng phc tp thnh cascade ca cc mng c s. ` Bng 3.1 Cc thng s ABCD ca mt s mng c s quan trng.2) Quan h gia (ABCD) v [ Z ] T (3. 25), (3. 63) vi quy c du ca I2 nh trn=>V1 = I111 I 2 12 V2 = I1 21 I 2 22

( 3 .66 )

A=

v1 V2

I 2 =0

=

I 1 11 11 = I 1 21 21v2 = 0

( 3 .67 a )

B=

v1 I2

v2 = 0

=

I 1 11 I 2 12 I2

= 11

I1 I2

v2 = 0

12

= 11

12 21 I1 22 12 = 1 22 21 I1 21

( 3 .67 b ) 29


C=

v1 V2 I D= 1 I2

I 2 =0

=

v2 = 0

I1 1 = I 1 21 21 I = 2 22 21 = 22 I2 21

( 3.67 c )

(3.67d)

* Nu mng l thun nghch th Z12 = Z21 v AD BC = 1 3) Cc s tng ng cho mng 2 cng Xt chuyn tip gia mt ng truyn ng trc v mt ng vi di vi cc mt tham chiu nh hnh v t1, t2. - Do s gin on v mt vt l ca chuyn tip, nng lng in, t trng c th b tch t ti chuyn tip v gy ra cc hiu ng phn khng. Cc hiu ng ny c th o c hoc c phn tch l thuyt nh s hp en ca mng 2 cng nh hnh v. M hnh phn tch ny c th s dng cho cc trng hp ghp gia cc loi ng truyn khc nhau hoc cc ch gin on ca ng truyn nh s thay i nhy bc ca rng hoc cong - Thng ngi ta thay hp en bng s tng ng cha mt s cc phn t l tng. C rt nhiu cch, y s kho st mt cch ph bin v hu dng nht. - S dng quan h: [ V ] = [ Z ] [ I ] v [ I ] = [ Y ] [ V ] v nu mng l thun nghch th Z12 = Z21 v Y12 = Y21 v mng c th c biu din theo s hnh T hoc TT nh hnh v. V hnh - Nu mng l thun nghch th s c 6 bc t do (phn thc v o ca 3 thng s). - Mt mng khng thun nghch s khng th c biu din b s tng ng dng cc phn t thun nghch.

3.5 CC TH TRUYN TN HIU1) nh ngha: Cc phn t c bn ca gin l node v nhnh: - Node: Mi cng i ca mng SCT c 2 node ai v bi. Node ai l sng ti v bi l sng phn x t cng. - Nhnh: Mt nhnh l mt ng trc tip gia mt node a v mt node b, biu th dng tn hiu t node a n node b. Mi nhnh c mt thng s S kt hp hoc mt h s phn x.


Sng ti vi bin a 1 c tch thnh 2, phn qua S11 (v ra khi cng 1 nh mt sng phn x b1) v phn truyn qua S21 ti node b2. Ti node b2 sng ra khi cng 2. Nu c mt ti vi h s phn x zero c ni vo cng 2 th sng ny s ti phn x mt phn v i vo mng ti node a2. Mt phn s ti phn x ra khi cng 2 qua S22 v 1 phn c th c truyn ra khi cng 1 qua S12. Cc trng hp c bit: + Mng mt cng: + Ngun p:2) Phng php phn tch th dng tn hiu: + Lut 1: (Lut ni tip) Hai nhnh m node chung ca chng ch c 1 sng vo v mt sng ra (cc nhnh ni tip) c th kt hp thnh mt nhnh n vi h s bng tch cc h s ca cc nhnh ban u. (3. 69) V3 = S32V2 = S32 S21 V1 + Lut 2: (Lut song song) Hai nhnh gia hai node chung (2 nhnh song song) c th kt hp thnh 1 nhnh n c h s bng tng cc h s ca hai nhnh ban u. (3.70) V2 = SaV1 + SbV1 = (Sa + Sb).V1

+ Lut 3: (Lut vng n) Khi mt nhnh bt u v kt thc ti mt node c h s S, th c th trit tiu nhnh bi vic nhn cc h s ca cc nhnh nui node vi 1/(1 S)

S V2 =S 21V1 +S 22V2 V2 = 21 V1 1 S22 (3.71) V3 = S33V2 V3 = S32V2 S 32 S 21 V1 (3.72) V3 = 1 S 22+ Lut 4: (Lut tch) Mt nt c th tch thnh 2 nt c lp khi v ch khi bt k mt s kt hp no ca cc nhnh vo v ra (khng phi l cc nhnh vng n) u dn ti nt ban u.


Chng IV: PHI HP TR KHNG V TUNING 4.1 M U:Chng ny p dng cc l thuyt v k thut cc chng trc cho cc bi ton thc t trong KT SCT. Bi ton phi hp tr khng thng l mt phn quan trng ca qu trnh thit k h thng SCT. - Matching network thng l khng tn hao l tng v thng c thit k sao cho tr khng nhn vo matching network bng Z0 trit tiu phn x trn ng truyn, mc d c th c a phn x trn on Matching network v Load. * Mc tiu phi hp tr khng: - Ly c cng sut cc i trn ti, gim thiu cng sut tn hao trn ng truyn. - i vi cc phn t nhy thu, phi hp tr khng tng t s tn hiu / nhiu ca h thng (anten, LNA, ) - Phi hp tr khng trong mt mng phn phi cng sut (mng nui anten mng) s cho php gim bin v li pha. * Nu ZL cha phn thc khc 0 th mng phi hop Tn khng lun c th tm c. C nhiu phng n phi hp, tuy nhin cn theo cc tiu ch sau: + phc tp: n gin, r, d thc hin, t hao tn. + rng bng: cn phi hp tr khng tt trong mt di tn rng, tuy nhin s phc tp hn. + Lp t: Ty vo dng ng truyn hoc ng dn sng quyt nh phng n phi hp TK. + Kh nng iu chnh: trong 1 s trng hp c th yu cu MN hot ng tt khi ZL thay i.

4.2 PHI HP TR KHNG VI CC PHN T TP TRUNG (L NETWORKS)1) Gii thiu: - Dng n gin nht ca PHTK l dng khu L, s dng 2 phn t in khng phi hp 1 ti ty vi ng truyn c 2 cu hnh kh d.

- Nu tr khng ti chun ha zL= ZL/Z0 nm trong vng trn 1 + j x trn gin Smith th hnh v (4.2a) c dng, nu khng th dng (h4.2b). - Cc phn t in khng trong hnh 4.2 c th l C hoc L ty thuc vo ZL. Do c 8 kh nng xy ra. - Nu tn s nh v / hoc kch thc mnh nh th c th dng cc t v in cm thc (c th n 1 GHz). y l hn ch ca mch L.


2) Li gii gii tch: (dng cho computer aided design program, hoc khi cn c chnh xc cao hn so vi phng php dng Smith chart) ZL - Xt mch (h 4.2a), t ZL = RL + j XL, v zL = Z nm bn trong ng trn 0 1 + j x (r = 1), nn RL > Z0. - Tr khng nhn vo matching network c ti pha sau phi bng Z0, tc l: 1 Z0 = j X + j B + 1/(R + j X ) (4.1) L L - Tch phn thc v phn o ca (4.1) (4.2a) B (X RL XL Z0) = RL Z0 (4.2b) X (1 B XL) = B Z0 RL - XL

=>

B=

XL

RL

Z0

2 2 RL + XL Z0RL

2 2 RL + XL Z 1 X Z X= + L 0 0 B RL BRL

(4.3a)

(4.3b)

Nhn xt: T (4.3) c 2 li gii kh d cho B v X, c 2 li gii u kh d v mt vt l (B < 0 cun cm B > 0 t, X > 0 cun, X < 0 t). Tuy nhin c mt li gii c th gy ra gi tr nh hn ng k ca cc phn t in khng v c th l li gii thch hp hn cho rng di tt hn hoc h s SWR trn on gia b phi hp TK v ti nh hn. * Vi (h 4.2b) (RL < Z0): Dn np nhn vo matching networrk phi bng 1/Z0 hay 1 1 = R + j (X + X ) (4.4) Z0 L L B Z0 ( X + XL) = Z0 - RL (4.5a) (4.5b) X + XL = B Z0 RL

* phi hp ZL vi ng truyn Z0= th phn thc ca tr khng vo MN phi bng Z0, phn o = 0 MN c s bc t do t nht bng 2, l 2 gi tr ca cc phn t in khng.

4.3 PHI HP TR KHNG DNG ON DY CHM (Single Stub tuning)1) Khi nim: - u im: khng dng cc phn t tp trung d ch to; dng shunt stub c bit d ch to cho mch ghi gii (microstrip) hoc mch di (stripline) - Hai thng s iu chnh l khong cch d v Y hoc Z. - Chng hn vi h4.3a nu dn np nhn vo on dy cch ti 1 khong d c dng Y0 + j B th dn np ca dy chm s c chn l j B.


- Vi h4.3b nu tr khng ca on dy ni ti, cch ti on bng d, l Z0+jX th tr khng dy chm ni tip (series stub) c chn l jX. 2) Shunt Stub: V d: Cho ZL = 15 + j 10 (), thit k hai mng phi hp dng 1 dy chm mc song song ghp vi ng truyn 50 . Gi thit tn s phi hp l 2 GHz v ti gm c 1 in tr v 1 cun ni tip. gii: (phng php dng Smith chart) - Tm im zL = 0,3 + j 0,2. - V ng trn SWR tng ng v chuyn i thnh dn np yL (ly i xng tm ca im zL) - Khi dch trn ng dy th khng i nn tng ng vi php dch chuyn trn ng SWR. Y0 + j B ) - ng SWR ct vng 1 + j b ti 2 im y1, y2 (y0 = Y 0 - Khong cch d c cho bi 1 trong 2 gi tr tng ng trn thang WTG d1 = 0,328 0,284 = 0,044 d2 = (0,5 0,284) + 0,171 = 0,387 (0,284 tng ng vi yL) y1 = 1 j 1,33 y2 = 1 + j 1,33 dn np dy chm cho li gii y1 l j 1,33 v li gii y2 l j 1,33. - Nu dy chm h mch th chiu di ca n c tm bi vic dch chuyn t y = 0 theo mp ngoi ca gin (g = 0) v pha ngun pht n im j 1,33 l1 = 0,147 l2 = 0,353 - nghin cu s ph thuc tn s ca 2 li gii trn, cn tm gi tr ca R v L tn s cho trc (2 GHz): R = 15 , L = 0,796 nH. Sau v th theo f (GHz). 1 * Phng php gii tch: t ZL = Y = RL + j XL L - Tr khng on ng truyn d c ti ZL kt cui

Z = Z0

(RL + jXL ) + jZ0t , t = tgd Z0 + j(RL + jXL )t

(4.7)

2 RLt (Z0 XLt)(XL + Z0t) RL (1+ t 2 ) 1 +j Y = = G + jB = Z0 2 2 (4.8) RL + (XL + X0t)2 Z0 RL + (XL + X0t)2 Z

[

]

1 - Chn d (tc t) sao cho: G = Y0 = Z , t (4.8) 2 X L RL (Z 0 RL ) + X L Z 0 t= , vi R L Z 0 RL Z 0 2

[

0

]

(4.9)

- Nu RL = Z0 th t = XL/2Z0 ->34Su t m b i: www.daihoc.com.vn

1 1 2 tg t , t 0 d = 1 ( + tg 1t ), t > 0 2

(4.10)

- tm chiu di on dy chm l, dng t trong (4.8b)B v l suy ra t BS =- B. Vi dy chm h mch => l0 B 1 B 1 = tg 1 ( ) tg 1 ( S ) = (4.11a) Y0 2 Y0 2 Vi dy chem. h mch => lS Y 1 Y 1 = tg 1 ( 0 ) tg 1 ( 0 ) = (4.11b) B 2 BS 2 Nu cc chiu di trong (4.11a,b) c gi tr m th chiu di cn tm s c c nh cng thm on /2. 3) Dy chm ni tip: V d: Ghp ZL= 100 + j80() vo ng truyn 50 dng mt dy chm h mch mc ni tip.Tn s hot ng 2GHz, ti gm 1 in tr v 1 cun mc ni tip. Gii: Theo phng php dng gin Smith - Tm im tr khng chun ha ZL = 2 + j1,6 , v vng SWR. - Vi trng hp dy chm ni tip dng gin tr khng - ng trn SWR ct vng 1+jx ti 2 im Z1, Z2. - i chiu trn thang WTG d1 = 0,328 0,208= 0,120 d2 = (0,5 0,208) + 0,172 = 0,463 - Tr khng chun ha (1) z1 = 1 j 1,33 (1) z2= 1 + j 1,33 - (1) yu cu on chm c tr khng j 1,33. di ca 1 dy chm hmch c th tm c khi xut pht t z = . Dch chuyn dc theo mp ngoi ca gin (T= 0) v pha ngun ti im j 1,33 l1 = 0,397 l2 = 0,103 = 0,25 0,147 = 0,5 0,103 . * kho st s ph thuc vo tn s ca SWR cn tnh ra R = 100 v L = 6,37 nH ri v li s mch dng kt qu trn. V hnh

* Phng php gii tch: t YL=

1 = GL + BL Zl

- Dn np vo on d c ti kt cui : 35Su t m b i: www.daihoc.com.vn

Y = Y0

(GL + jBL ) + jY0t , t = tgd Y0 + j(GL + jBL )tZ = R + jX = 1 Y

(4.12)

=> tr khng vo :Vi

GL (1+ t 2 ) R= 2 GL + (BL + Y0t)22 GL t (Y0 tBL )(BL + Y0t ) X= 2 Y0 GL + (BL + Y0t )2

(4.13a)

[

]

(4.13b)

- Cn tm d sao cho R = Z0 = 1/Y0 t (4.13a) Y0 (GL Y0)t2 2BL Y0 t + (GLY0 GL2 BL2) = 0 2 BL GL (Y0 GL ) 2 + BL Y0 t= , vi GL Y0 GL Y0

[

]

(4.14)

t =- T t => d :

BL , vi GL = Y0 2Y0

1 1 2 tg t , t 0 d = 1 ( + tg 1t ), t > 0 2 - Dng t v (4.13b) => cm khng X, yu cu XS = -X => + Dy chm ngn mch : lS X 1 X 1 = tg 1 ( ) tg 1 ( S ) = Z0 2 Z0 2 + Dy chem. h mch : l0 Z 1 Z 1 = tg 1 ( 0 ) tg 1 ( 0 ) = X 2 XS 2

(4.15)

(4.16a)

(4.16b)

4.4 B GHP MT PHN T BC SNG- Cc b ghp nhiu on c th dng tng hp cc b phi hp tr khng hot ng nhiu di tn mong mun. - B ghp ch dng cho ta thun tr . - Mt ti phc c th c chuyn thnh ti thun tr bi vic s dng mt on ng truyn c chiu di thch hp gia ti v b phi hp, hoc dng on dy chm ni tip hoc song song ph hp. K thut ny thng dn ti thay i s ph thuc tn s ca ti tng ng v gy ra gim rng bng ca s phi hp tr khng.


Trong tit ny chng ta s kho st rng bng thng nh l mt hm ca s mt phi hp tr khng lm tin cho cc b ghp nhiu khu phn sau. (4.25) Z1 = Z 0 Z l Khi tn s f f0, th di in l 0/4, khi tr khng vo ca on ghp l :Z in = Z 1Z Z 0 Z 1 (Z l Z 0 ) + jt Z 1 Z 0 Z l - H s phn x = = in Z in + Z 0 Z 1 (Z l + Z 0 ) + jt Z 1 2 Z 0 Z l Z l Z 0 = Z l + Z 0 + 2 jt Z 0 Z l 1 = {1 + 4Z 0 Z L (Z L Z 0 ) 2 sec 2 }12 Zl Z0 = cos 2 Z0Zl2

Z L + jZ 1t Z 1 + jZ L t

( 4.26 )

( (

) )

( 4.27 ) (4.28) (4.29) ( 4.30)

[

]

- Gi m l gi tr bin cc i c th chp nhn c th rng bng ca b ghp c nh ngha l : = 2 m 2 2 Z0Zl m cos n = Zl Zo 1 2m

(4.31) (4.32)

rng bng t i

f f thng c biu din theo %:100 (%) fo fo

rng bng ca b ghp tng nu ZL Z0 Ni sng non TEM (ng dn sng) h s truyn khng cn l hm tuyn tnh ca tn s do tr khng sng s ph thuc tn s.iu ny lm phc tp hn cc c trng ca b ghp . Tuy nhin trong thc t rng bng ca b ghp thng nh sao cho khng nh hng n kt qu. nh hng ca cc in khng xut hin do s khng lin tc (s thay i kch thc ng truyn) c th c khc phc bi s iu chnh di ca on ghp.

4.5 B GHP DI RNG (Multisection matching Transformer)1) L thuyt phn x nh: Xt h s phn x ton phn gy bi s phn x ring phn t mt s gin on nh. a.B ghp 1 khu:


1 =

Z 2Z 1 Z 2+Z1

(4.34) ( 4.35) (4.36)

2 = 1 Z Z 3 = l 21 Z l+Z 2

C th tnh h s phn x tng = 1 + 12 21 3 e 2 h 2 3 e 2 jn

1 + 3 e 2 j = 1 + 1 3 e 2 j

(4.40)

* Nu s gin on gia cc tr khng Z1, Z2 v Z2, ZL l nh, th : (4.41) 1 .3 0 nu ZL > Z0; n < 0 nu ZL 33) l 4 m f 2( f 0 f m ) = =2 f0 f0 1 = 2 coqs 1 m 2 A 4 1

N

( 4.55)

4.6 TIU CHUN BODE FANO - Cc tiu chun Bode Fano cho cc dng tr khng ti khc nhau cho bitgii hn l thuyt ca gi tr h s phn x cc tiu c th c: - Gi s mun tng hp 1 mng phi hp vi p ng ca h s phn x nh hnh v (a). Khi nu dng mch ti RC (a) th

ln dw = ln 0

1

1m

dw

( 4.79)

m

= w ln

1 < m RC

- Vi ti RC c nh, w tng khi m tng - m ch = 0 khi w =0 - nu R tng v hoc C tng cht lng phi hp gim tc l mch High-Q kh phi hp hn Lowen_Q V ln1 t l vi tn hao ngc (return loss, dB) ti u vo ca mng phi

hp (MN), (4.79) c th xem nh l yu cu rng din tch gia ng cong tn hao ngc (RL) v ng = 1 (RL = o dB) phi nh hn hoc bng 1 hng s. Du = xy ra (trng hp ti u) khi ng RL c iu chnh sao cho = m trn ton bng thng v = 1 trong min cn li. iu ny ch c th c vi s phn t trong MN l v cng.


Chng V: CHIA CNG SUT V GHP NH HNG5.1 GII THIU- Cc b phn chia cng sut v ghp nh hng l cc cu phn SCT th ng dng chia hoc ghp cng sut. - Vi b chia cng sut, mt tn hiu vo c chia thnh 2 hay nhiu tn hiu c cng sut nh hn. Cc b chia c th l cc cu phn 3 hoc 4 cng, c hoc khng c tn hao. - Cc mng 3 cng thng c dng T v dng cho chia cng sut - Cc mng 4 cng thng dng cho ghp nh hng hoc hn tp. - B chia cng sut thng c dng chia cn bng (3dB) - Cc b ghp nh hng c th c thit k cho vic chia cng sut ty , cn cc b ghp hn tp thng dng cho chia cng sut cn bng. - Cc b ghp hn tp thng c gc lch pha gia cc cng ra l 900 (quadrature) hoc 800 (magic T). - C rt nhiu loi ghp ng dn sng v chia cng sut c khm ph v nghin cu ti MIT Radiation Labotory trong nhng nm 40 th. - n nhng nm 50 th, 60 th chng c pht trin dng cho cng ngh ng truyn di v vi di.

5.2 CC C TRNG C BNTrong phn ny s s dng l thuyt ma trn tn x rt ra nhng c trng c bn ca cc mng 3 v 4 cng, v nh ngha cc khi nim: cch ly, ghp v tnh nh hng l nhng i lng c bn c trng cho cc b ghp v chia hn tp. 1) Mng 3 cng (T Junctions) - L dng n gin nht ca cc b chia cng sut, gm 2 cng ra v 1 cng vo. - Ma trn tn x c 9 phn t c lp V hnh S11 S12 S13 [S ] = S21 S22 S23 (5.1) S31 S32 S33 - Nu cu phn l th ng v khng cha cc vt liu bt ng hng th phi l thun nghch v [S] phi i xng. 41Su t m b i: www.daihoc.com.vn

- Thng trnh tn hao cng sut, cn phi c kt cu khng tn hao v c phi hp tr khng tt c cc cng, tuy nhin iu ny l khng th thc hin c.


* Tht vy nu tt c cc cng u phi hp th Si i = 0, i =1,3. 0 S12 S13 [S ] = S21 0 S23 (5.2) S31 S32 0 - Nu mng l khng tn hao th t iu kin (3.53) ma trn tn x phi l unita S12 2 + S13 2 = 1 2 2 S12 + S 23 = 1 (5.3a,b,c) 2 2 S13 + S 23 = 1 * S 13 . S 23 = 0 * S 23 . S 12 = 0

(5.3d,e,f)

S . S 13 = 0* 12

Cc iu kin (5.3d-f) -> S12, S23, S13 = 0 -> mu thun - Vy mng 3 cng khng th ng thi thun nghch, khng tn hao v phi hp tr khng ti tt c cc cng (gi tt l phi hp). - Nu mng khng thun nghch th S i j S j i v iu kin phi hp tr khng ti cc cng v khng tn hao c th c tha mn, mng c gi l mch vng, cu to t cc vt liu bt ng hng (nh ferrite). - C th chng minh rng bt k mt mng 3 cng khng tn hao, phi hp, phi khng thun nghch (tc l 1 mch vng Circulator): + ma trn : 0 S12 S13 [S ] = S21 0 S11 (5.4) S31 S32 0 + iu kin khng tn hao =>

S .S 32 = 0 * S 21 .S 23 = 0 * S12 .S13 = 0* 31

(5.5a,b,c)

v

S12 2 + S13 2 = 1 2 2 S12 + S 23 = 1 2 2 S13 + S 23 = 1

(5.5d,e,f)

=> Hoc S12, S23, S13 = 0 , S21 = S32 = S13 = 1

(5.6a)

hoc S21, S32, S13 = 0 , S12 = S23 = S31 = 1

(5.6b)

=> Sij S ji , i,j = 1 3 , tc mng l khng thun nghch * Mt trng hp khc c th xy ra l mt mng khng tn hao, thun nghch th ch c 2 trong 3 cng l phi hp. - Gi s cng 1 v 2 l phi hp, khi :


0 S12 S13 [S ] = S12 0 S23 S13 S23 S33 khng tn hao cn c :* S13 .S 23 = 0 * * S12 .S13 + S 23 .S 33 = 0 * * S 23 .S12 + S 33 .S13 = 0

(5.7)

(5.8a,b,c)

S12 2 + S13 2 = 1 2 2 S12 + S23 = 1 2 2 2 S13 + S23 + S33 = 1

(5.8d,e,f)

Cc phng trnh d-e => S13 = S23 nn t (5.8a) => S13 = S23 = 0. Do

S12 = S33 = 1* Nhn xt: Mng bao gm 2 cu phn tch bit, mt phn c phi hp 2 cng, 1 phn khng phi hp, 1 cng * Trng hp mng 3 cng c tn hao th c th thun nghch v phi hp; y l trng hp ca b chia tr tnh. 2) Mng 4 cng (Cc b ghp nh hng)

0 S12 S13 S14 S 12 0 S23 S24 [S ] = S13 S23 0 S34 S14 S24 S34 0

(5.9)

Vi mng thun nghch, cc cng u phi hp - Nu mng khng tn hao, s c 10 phng trnh t iu kin ca ma trn unita. Chng hn xt tch ca hng 1 v hng 2, hng 3 v hng 4:* * S13 .S 23 + S14 .S 24 = 0 * * S14 .S13 + S 24 .S 23 = 0* * Nhn (5.10a) vi S 24 , (5.10b) vi S13 , tr ln nhau =>

(5.10a,b)

* S14 ( S13 S 24 ) = 0

2

2

(5.11)

Tng t cho hmg (1,3); (2,4) =>* * S13 .S 23 + S14 .S 34 = 0 * * S14 .S12 + S 34 .S 23 = 0

(5.12a,b)

Nhn (5.12a) vi S12, (5.12b) vi S34 v tr nhau =>

S 23 ( S12 S 34 ) = 0

2

2

(5.13) 44


a) Nu S14 = S23 =0, ta c b ghp nh hng * T tch ca cc hng vi chnh n =>

S12 S13 S12 S 24 S13 S 34 S 24 S 342 2 2

2

2 2 2 2

=1 =1 =1 =1(5.14a,b,c,d)

S13 = S24 v S12 = S34 => * Vic gin c tip theo c thc hin bi vic hcn goc pha tham chiu trn 3 trong 4 cng. gi s chn S12 = S34 = ; S13 = ej v S24 = ej vi v l cc s thc, v l cc hng s pha cn tm (1 trong 2 c chn trc ty ). - Tch chp hng 2 v 3 =>* * S12 .S13 + S24 .S34 = 0

(5.15) (5.16)

=> Quan h gia hng s pha :

+ = + 2 n

Trong thc t thng xy ra hai trng hp : 1,Ghp i xng: = = bng nhau ), Khi :

2

( pha ca cc s hng c bin cchn

0 . j 0 [S ] = 0 j 0 j 0

j 0 0

0 j 0

(5.17)

2,Ghp phn i xng: = 0, = ( pha ca cc s hng c bin c chn ngc pha), khi :

0 0 0 0 [S ] = (5.18) 0 0 0 0 Ch : - 2 dng b ghp ch khc nhau vic chn cc mt tham chiu. - Cc bin , tun theo chng trnh :

(5.19) => Ngoi gc pha tham chiu, mt b ghp nh hng l tng ch c 1 bc t do b) Nu S13 = S24 v S12 = S34

2 + 2 =1


- Nu chn pha tham chiu sao cho S13 = S24 = v S12 = S34 = j (tho 5.16) th t (5.10a) =>* * (S23 + S14) = 0 v t (5.12a) => (S14 S23) = 0

+ Nu S14 = S23 = 0 -> li gii tng t cho php nh hng.

+ Nu = = 0 , tc l S12 = S13 = S24 = S34 = 0 , y l trng hp ca mng 2 cng ring bit. * Kt lun: Bt k mng 4 cng thun nghch khng tn hao v phi hp u l 1 b ghp nh hng. * Hot ng ca b ghp nh hng: - Cng sut cung cp vo cng 1 c ghp ti cng 3 vi h s ghpS132 = 2, phn cn li ca cng sut cung cp c ly n cng 2 vi h sS122 = 2 = 1 - 2. Trong b ghp nh hng l tng, khng c cng sut no c ly ra cng 4 (Isolated port) + Cc i lng c trng cho b ghp nh hng: - ghp (Coupling) = C =10lg(P1/P3)=-20lg (dB) (5.20a) - nh hng (Directivity) : (5.20b) D = 10lg(P3/P4) = 20lg(/S14) (dB) - cch ly (Isolation) : (5.20c) I = 10lg(P1/P4) = -20lgS14 (dB) => I = C + D (dB) (5.21) * B ghp hn tp : l trng hp ring ca b ghp nh hng vi h s ghp l 3dB hay = = 1 . C 2 dng ghp hn tp tng ng gc lch cng 2 v 3 l 22

vi :

0 1 j 0 1 1 0 0 j [S ] = 2 j 0 0 1 0 j 1 0 V gc lch pha 1800 gia ng 2 v 3 v ghp bt i xng . 0 1 1 0 1 1 0 0 1 [S ] = 2 1 0 0 1 0 1 1 0

(5.22)

(5.23)


5.3 B CHIA CNG SUT T - JUNCTION1) Gii thiu: T Junction powerdivider l trng hp n gin ca mng 3 cng, c th s dng cho chia cng sut hoc cng cng sut v c th c thc hin cho hu ht cc dng mi trng ng truyn. 2) B chia khng tn hao: - C s tch t nng lng do s gin on ti junction, dn ti nng lng tch t c th quy cho dn np tp trung B.

- iu kin phi hp tr khng u vo (Z0) 1 1 1 Yin = jB + + = (5.24) Z1 Z2 Z0 - Nu cc ng truyn l khng tn hao th cc tr khng c trng l thc, tc B = 0 v 1 1 1 + = (5.25) Z1 Z2 Z0 - Trong thc t B thng b nh cc phn t in khng (trong di tn s hp). - Cc gi tr Z1, Z2 c th c chn thay i t s chia cng sut. C th dng cc on 1/4 thay i cc tr khng ng ra (Z1, Z2) - Nu cc ng ra c phi hp th ng vo s c phi hp, nhng s khng c s cch ly gia 2 cng ra v s c s mt phi hp khi nhn vo cc cng ra. V d: Tm Z1, Z2 mt b chia T khng tn hao c Z0 = 50 v cng sut c chia theo t l 2/1. Tnh h s phn x nhn vo cc cng ra. 3) B chia tn hao: (b chia tr tnh) Mt b chia T c tn hao c th phi hp ti tt c cc cng mc d cc cng ra c th khng c cch ly. Hnh bn minh ha mt b chia dng cc in tr tp trung, c chia u cho 2 cng ra (- 3 dB) . Quan nim rng tt c cc cng u c kt ni vi Z0 th tr khng Z nhn vo cc in tr Z0/3 theo sau bi cc ng ra l: Z 4Z Z = 0 + Z0 = 0 (5.26) 3 3 Vy tr khng vo ca b chia l : Z 2Z Z in = 0 + 0 = Z 0 (5.27) 3 3 Tc l li vo phi hp vi feed line. V mng l i xng cho tt c cc cng nn phi hp ti tt c cc cng, tc l S11 = S22 = S33 = 0 Ti tm ca mng : 2Z 0 2 3 V = V1 = V1 (5.28) 2Z Z0 3 + 0 3 3


V2 = V3 =

1 V1 2

(5.29)

=>

S21 = S31 = S23 = 1 2

- Cng sut pht ra mi cng thp hn cng sut vo 6 dB. - Ma trn tn x:0 1 1 1 [S ] = 1 0 1 2 1 1 0

(5.30)

C th chng minh [S] khng unita - Cng sut u vo :

1 V12 Pin = 2 Z0- Cng sut cc u ra :

(5.31)2 1

1 P2 = P3 = 2

(12 V ) = 1 PZ0 2

in

(5.32)

=> Mt na cng sut cung cp b tn hao trn cc in tr.

5.4 B CHIA CNG SUT WILKINSON1) Gii thiu: Dng cho mch di hoc vi di. V hnh C th phn tch mch wilkinson bng cch tch thnh 2 mch n gin hn bng k thut phn tch mode chn l. 2) Php phn tch mode chn l: n gin, c th chun ha tt c cc tr khng theo Z0 v v li (h.b) vi cc ngun th ti cc cng ra. Hai in tr ngun c gi tr chun ha bng hai mc song song cho 1 in tr gi tr 1, biu th tr khng ca ngun phi hp. on /4 c tr khng c trng, chun ha Z v tr shund c gi tr chun ha r (vi chia cn bng z = 2 v r = 2). nh ngha: Hai mode ring r ca s kch thch mch (h5.4.2): mode chn vi Vg 2 = Vg 3 = 2V v mode l vi Vg 2 = - Vg 3 = 2 V. Khi chng chp 2 mode s c kch thch vi Vg2 = 4, Vg3 = 0, t tm ra cc thng s S ca mng. a. Mode chn: Vg 2 = Vg 3 = 2 Ve2 = Ve3 v khng c dng qua cc in tr r/2 v qua ngn mch gia cc input ca 2 ng truyn ti cng 1 c th tch i mng (h5.4.2) vi vic h mch ti nhng im ni trn c s sau:


Khi nhn vo cng 2 thy tr khng

Z2 Z = 2e in

(5.33)

V ng truyn ging nh on /4 .Vy, nu Z = 2 th cng 2 s phi hp e vi mode chn, V2e = 1V v Z in = 1 .Tip theo s tm V1e t phng trnh ng truyn. Nu t x = 0 ti cng 1 v x = -/4 ti cng 2 th in p trn on ng truyn c th c vit: V( x ) = V + (e jx + e jx ) Vi V2e = V = jV + (1 ) = 1V 4 +1 V1e = V (0) = V + (1 + ) = jV 1= 2 2 2+ 2 => V1e = jV 20 0

(5.34)

H s phn x c nhn ti cng 1 v pha in tr chun ha 2 nn

(5.35)

b. Mode l: Vg = Vg3 = 2V => V2 = V3 v c mt in p khng dc theo on gia ca mch (h54.2) do c th tch bng cch ni t ti 2 im trn machj ct gia ca n c s sau: - Nhn vo cng 2 thy tr khng r/2 v on ng truyn song song l /4 v ngn mch ti cng 1 (nn c th xem nh h mch ti cng 2). Vy cng 2 s c phi hp nu chn r = 2. Khi V20 = 1V v V10 = 0 . Vi mode kch thch ny tan b cng sut ri trn r/2, khng c cng sut ti cng 1 3) Trng hp cc cng 2 v 3 kt cui vi ti phi hp: Tng t nh mode chn v V2 = V3 s tng ng V s (Khng c dng chy qua tr chun ho 2 nn c th b nh h.b) 2 1 Z in = 2 =1 (5.36) => 2 4) Cc b chi Wilkinson khng cn bng v N way - V s + cng thc

( )

- Gi s

P3 = K2 P2

Cc phng trnh thit k sau c th s dng:

1+ K 2 Z03 = Z0 K3

(5.37a)

Z02 = K 2 Z03 = Z0 K(1+ K 2 )1 R = Z0 K + K

(5.37b)

(5.37c)


Nu K = 1 b chi cn bng. Cc ng ra phi hp vi cc tr

R2 = Zo K, R3 =

Z0 . Cc b ghp phi hp c th c dng chuyn i cc tr K

khng ra ny. * Cc b chia Winkinson cng c th c thit k c N way divider hoc combiner nh hnh v. Mch ny c th phi hp ti tt c cc cng vi s cch ly gia tt c cc cng. Hn ch ca mch l yu cu c in tr ngang khi N 3, l hn ch khi ch to dng planar. Winkinson divider c th thc hin vi cc on bc thang tng rng bng

5. 5 GHP NH HNG NG DN SNG.1) Gii thiu:Cc b ghp nh hng l cc mng 4 cng c cc c trng c bn - Cng sut ti ti cng 1 s ghp ti cng 2 (through port) v ti cng 3 (coupled port) nhng khng ti cng 4 (isolated port). - Tng t, cng sut ti cng 2 s qua cng 1 v 4, khng qua 3 . - T s cng sut ghp t 1 n 3 l C: ghp (5.20a). - Cng sut r t 1 n 4 l I: cch ly (5.20c) - nh hng D = I C (dB) l t s cng sut ti cng ghp v cng cch ly. - B ghp l tng c nh ngha c I v D = , l b ghp khng tn hao v phi hp tt c cc cng. - B ghp nh hng c th c nhiu dng: ghp ng dn sng, ghp hn tp (3 dB, quadrature hoc magic T) . 2) B ghp l Bethe: c tnh nh hng ca tt c cc b ghp nh hng c c l nh s dng cc sng hoc cc thnh phn sng ring r, ng pha cng ghp v trit tiu nhau cng cch ly. Phng php n gin nht l dng 2 ng dn sng c chung 1 l nh trong vch ngn chung gia 2 ng, b ghp nh vy gi l Bethe hole coupler.

* Nguyn l hot ng: L ghp c th thay bng cc ngun bt x tng ng, gm cc moment in v t. Moment in v moment t dc bc x sng c tnh cht i xng chn v moment t ngang bc x sng i xng l. Bng cch iu chnh bin tng i ca cc ngun ny c th lm trit tiu bc x theo hng ca cng cch ly v tng cng bc x theo hng cng ghp. iu ny c th c thc hin nh iu chnh thng sS (h5.51a) v (h5.51b).


* Cu hnh song song (h5.51a). Gi thit c sng TE10 n cng 1,

Ey = AsinHx =

xa

e jz

(5.38a)(5.38a)(5.38a)

A x j z sin e Z 10 a x j z jA cos e Hz = aZ 10 a

Vi Z10 = k00 : tr khng sng ca mode TE10 - Bin ca sng ti v sng v ca ng dn sng bn di l : 0m 2 s 2 s jA + 2 s A10 = 2 sin + 2 2 cos2 0e sin P a Z10 a a a 10

(5.40a)

0m 2 s 2 s jA 2 s 0e sin + 2 sin 2 2 cos2 A = P a Z10 a a a 10 10

(5.40b)

+ Nhn xt: Bin sng ti cng 4 ( A10 ) ni chung khc vi bin sng ti cng . trit tiu cng sut ti cng 3 ( A10 ) cn iu kin:

+ A10 = 0 sin

sa

=

0 2 = 2 2 4 k0 a 2(2 a 2 ) 02

(5.41) (5.42a) (5.42a)

Khi h s ghp l : H s nh hng l :

C = 20 lg D = 20 lg

A (dB) A10A10 (dB) + A10

* Cc bc thit k Dng (5.41) tm S (v tr ca l) Dng (5.42) xc nh r0 (bn knh l) tha mn D, C cho trc. * Cu hnh xin: L t ti v tr S = a/2, iu chnh , trit tiu sng n cng 4. Trong trng hp ny in trng khng thay i theo nhng thnh phn t trng ngang thay i theo h s cos , do c th dng (5.40) vi vic thay m = mcos . Khi vi s =a 2

+ A10 =

jA 0e 0 2 m cos P Z10 10

(5.43a)

A10 =

0m jA 0e + 2 cos P Z10 10 k02 cos = 2 2C = 20 lg 4k 2 r 3 A = 20 lg 0 0 (dB) A10 3ab

(5.43b)

iu kin A = 0 ->

+ 10

(5.44)(5.45) 51

H s ghp :


V d: Thit k b ghp bethe song song cho di bng tn x - ng dn sng hot ng 9 GHz, h s ghp 20dB Gii: Cc hng s cho X band waveguide ti 9GHz, a = 0,02286m , b = 0.01016 , 0 = 0,0333m , k0 = 188,5m-1 , = 129m 1 , Z10 = 550 , P10 = 4,22.10-7 m2/ . a 1 T (5.41) => s = sin 0,972 = 9,69mm

(5.42) =>r03 => r0 = 4,15mm.

20 A 6 = 10 20 = 10 => t (5.40) => r0 theo iu kin : 0,1 =1,44.10 A10


Chng VI: B LC SIU CAO TN 6.1 GII THIUnh ngha: B lc siu cao tn l 1 mng 2 cng dng iu kin p ng tn s 1 v tr xc nh trong h thng SCT, bao gm cc loi tng t nh b lc tn s thp ng dng: bao gm tt c cc dng thng tin SCT, radar, cc h thng o dc v thy in. Lch s: T u th chin II, bi Mason, Sykes, Darlington, Fano, Lawson v Richards. - u nhng nm 503, cc nh nghin cu Stanford Research Institute ng dng phng php thng s nh nghin cu cc b lc SCT. - Hin nay hu ht cc b lc SCT c thit k s dng cc phn mm CAD trn c s phng php tn hao chn. - y vn l lnh vc ang c nghin cu mnh vi vic nghin cu tng hp b lc vi cc phn t phn b, ng dng siu dn nhit thp v cc linh kin tch cc. - Cc cu trc tun hon c cp trc tin do cc ng dng trong cc h thng sng chm, khuch i sng chy v do chng c p ng lc chn di, l c s cho phng php thng s nh. - Cc phng php thng s nh v tn hao chn u s dng m hnh cc phn t tp trung do vi cc b lc SCT, cc phng php ny cn phi c s iu chnh cho cc phn t phn b, chng hn dng cc tr khng bc thang v cc ng ghp hoc cc b copng hng ghp.

6.2 CC CU TRC TUN HON1) Gii thiu: - Mt ng truyn hoc mt ng dn sng v hn mc ti c chu k vi cc phn t in khng c gi l mt cu trc tun hon. - C th c nhiu dng, ty thuc vo mi trng ng truyn. - Thng cc phn t ti c to thnh t cc ch gin on trong ng truyn. chng c th c m hnh ha nh l cc in khng tp trung mc ngang ng truyn nh hnh v:


2) Phn tch cu trc tun han v hn: Xt cu trc m hnh nh (h6.2.2), mi cell n v chiu di d c dn np shunt qua im gia ca cell, b l dn np chun ha so vi Z0. Coi ng truyn l mt Cascade ca cc mng 2 cng ging nhau. in p v dng in ti 2 pha ca cell th n c quan h: Vn A CVn+1 (6.1) I = B DV n+1 n

Ch : A, B, C, D l cc thng s ma trn cho dy Cascade ca mt on ng truyn d/2, mt dn np shunt b v mt on ng truyn d/2, do t bng (3.1) A B cos 2 j sin 2 1 0 cos 2 j sin 2 C D = j sin cos jb 1 j sin cos 2 2 2 2 b b b j (sin + cos ) (cos 2 sin ) 2 2 = b b b j (sin + cos ) cos sin 2 2 2

(6.2)

Vi = kd * Vi sng truyn theo hng +Z phi c :

V( z ) = V( 0 ) e zI ( z ) = I ( 0 ) e zVi mt phng pha tham chiu ti z =0 - Ti cc nt :

(6.3a) (6.3b)

Vn+1 = Vn e d I n+1 = I ned=.> =>V n A I = C n A e d C

(6.4a) (6.4a)

B V n +1 V n +1 e d = D V n +1 I n +1 e d

B Vn +1 D e d Vn +1

(6.5) (6.6) (6.7)

Cho li gii khng tm thng th phi c : AD + e 2d ( A + D)ed BC = 0 AD BC =1 =>Coshd =

* Nu = + j =>

A+ D b = cos sin 2 2

b Coshd = CoshdCoshd + j sinhd.sin d = cos sin 2 => = 0 hoc = 0 (V v phi thun thc)

(6.8)


+ Trng hp 1: = 0, o : trng hp khng suy gim (sng) v c nh ngha l gii thng ca cu trc. Khi (6.8) b Coshd = cos sin (6.9a) 2 c th gii tm nu ln ca v phi 1, v khi s c v s gi tr tha mn (6.9a). + Trng hp 2: 0, = 0,: sng b suy gim theo chiu di ng truyn, y l gii chn (stop band) ca cu trc. V ng truyn l khng tn hao nn cng sut b phn x ngc tr li u vo ca ng truyn t (6.8)

b Coshd = cos sin 1 2

(6.9b)b 2

- (6.9b) ch c mt li gii > 0 cho sng chy dng. Nu cos sin 1 th (6.9.b) thu c t (6.8) bng cch cho = . Khi tt c cc ti tp trung trn ng truyn u l cc on 2 do tr khng vo ging nh trng hp = 0. * Vy ty thuc vo tn s v gi tr dn np chun ha m ng truyn ti tun hon c th l Pass band hoc Stopband v do c th xem nh l mt b lc. in p v dng ch c ngha ti cc u cui ca Unit cell. Sng p v dng lc ny c tn l cc sng bloch, tng ng nh cc sng n hi lan truyn qua mng tinh th tun hon. + nh ngha: Tr khng c trng ti cc u cui ca cell n v

ZB = Z0

V n +1 I n +1

(6.10)

-

(V cc Vn+1 l cc i lng chun ha) Cc ZB c tn l cc tr khng Bloch. A e d V n + 1 + BI T (6.5) =>

(

)

n +1

= 0

V t (6.10) =>

ZB = ZB =

(

BZ 0 A e d

) BZ0

t (6.6) =>

2A A D m

( A + D )2 4

(6.11)

Vi cc cell n v i xng , A = D

ZB =

BZ0 A2 1

(6.12)

Cc li gii tng ng tr khng c trng cho cc sng chy dng v m. Vi mng i xng, cc tr khang ny ng thi c chp nhn v khi chiu ca I n + 1 c nh ngha ngc li tr khng dng. T (6.2) B lun thun o - nu = 0, 0 => ZB thc - nu = 0, = 0 => ZB o 55Su t m b i: www.daihoc.com.vn

3) Cu trc tun hon c kt cui: ZL Gi s cu trc hot ng Passband

Vn = V0+ e jnd + V0e jnd

(6.13a)

In = I eVi

+ jnd 0

+I e

jnd 0

V0+ jnd V0 jnd = +e + e ZB ZB

(6.13b) (6.14a) (6.14b)

Vn+ = V0+e jnd : sng ti Vn+ = V0+e jnd : sng phn x

=>

Vn+ Vn Vn = V +V , I n = + + ZB ZB+ n n

(6.15)

- Ti ti (n = N) :+ VN VN + + VN = V +V = ZL I N = ZL B B + N N

(6.16)(6.17)

V = V

n + n

ZL = ZL

+ ZB ZB

1 1

+ Nu cell dn v l i xng (A = D) ZB = ZB = ZB =>

=

ZL ZB ZL + ZB

(6.18)

6.3 THIT K B LC BNG PHNG PHP THNG S NH1) Tr khng nh v hm truyn cho cc mng 2 cng: Xt mng 2 cng ty nh hnh v: nh ngha: + Zi1: Tr khng vo ti cng 1 khi cng 2 kt cui vi zi2. + Zi2: Tr khng vo ti cng 2 khi cng 1 kt cui vi zi1. Vy c 2 cng u phi hp khi cng kt cui vi cc tr khng nh ca chng. Chng ta s tm biu thc cho Zi1, Zi2 theo ABCD: V1 = AV 2 + BI 2 (6.22) I 1 = CV 2 + DI 2 Tr khng vo ti cng 1 khi cng 2 kt cui vi Zi2 :Z in1 = V1 AV2 + BI2 AZi 2 + B = = I1 CV2 + DI2 CZi 2 + D

(6.23)

(V V2 = Z i 2 I 2 ). AD BC =1 => 56Su t m b i: www.daihoc.com.vn

V2 = DV1 BI 1 I 2 = CV1 + AI 1

(6.24) (6.25) (6.26)

=>

Z in2 =

V2 DV1 BI 1 DV1 + BI 1 = = CV1 + AI 1 I2 CZ i1 + D

- Zin1 = Z1 , Zin2 = Z2 => Z i1 D B = Z i 2 ( A CZi1 ) =>Z i1 = AB , Zi2 = CD BD AC

(6.27)

DZ V Z in2 = in1 A

Nu mng i xng (A=D) th Zi1 = Zi2 * Hm truyn in p : xt mng nh (h.6.3.2)

B V2 = DV BI1 = D V1 1 Zi1 (V V1 = I1 Z i1 ) =>

(6.28)

V2 B D = D = AD BC Zi1 A V1

(

)

(6.29a)

I2 V A AD BC = C 1 + A = D I1 I1+ H s

(

)

(6.29b)

D nghch o nhau (6.29a) v (6.29b) v c gi l t s chuyn A

i ngc. + Phn cn li c nh ngha l h s lan truyn ca mng

e = AD BC=>

(6.30)

cosh = AD

(6.31)


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