Bi Thuyt Trnh Chuyn : Hp Cng Hng Hnh Ch Nht Thnh vin:Bi c H Hi
inh Tun Hi Nguyn Nht Anh Khi Gio Vin: Tn Tht Nghim I. Gii thiu
chung tn s thp thng dng mch cng hng LC, tuy nhin tn s siu cao khng
dng c v: Tn hao nhit rt ln Tn hao do bc x ng k Dn n: phm cht mch
gimkhng cn kh nng tch ly nng lng mt i tnh cng hng khng cn kh nng la
chn tn s I. Gii thiu chung Tm trng in t tn ti bn trong hp cng hng
bng cch: + Tm nghim phng trnh Maxwell vi cc iu kin b cho. + Tm cc i
lng c bn:- tn s cng hng ring,- phm cht ca hp cng hng ng vi cc dng
dao ng khc nhau trong hp Hp cng hng trong lnh vc siu cao l 1 th tch
in mi c bao kn bi b mt kim loi.Thnh kim loiin miSVI. Gii thiu chung
I. Gii thiu chung Tn hao do bc x xem nh khng c v dn in ca thnh kim
loi ln, th tch in mi gn nh c cch ly vi mi trng xung quanh. Tn hao
nhit nh v vt dn c gi tr ln ==> H s phm cht rt cao (tm ) HP CNG
HNG HNH CH NHT Dao ng ring (dao ng khng tn hao): HP CNG HNG HNH CH
NHT Trc ht ta xt hp cng hng ch nht l tng, tc l kim loi lm thnh hp c
dn in
Mi trng bn trong hp l in mi l tng Mode TE Mode TE iu kin Ex( z=o
) =0 C1 + C2 = 0 C1 = - C2 Thay vo cng thc trn: t: Vi p = 1,2,3.
Mode TE Cui cng, ta c kt qu: Tng t: Mode TE Tn s cng hng Ta c: = 2f
v Suy ra, tn s cng hng: Tn s cng hng Bc sng trong hp cng hng: C th
tn ti v s kiu dao ng ring loi in ngang, mi kiu c xc nh bng cc s m,
n, p v k hiu l TEmnp. Mode TM Lm tng t nh Mode TE, ta c dao ng ring
kiu t ngang TMmnp trong hp cng hng ch nht khng tn hao : Mode TM Tn
s dao ng ring, bc sng ring ca dao ng ring TMmnp ging nh TEmnp. Vi
Amnp, Bmnp l cc hng s ty , c xc nh t kt qu ca vic kch thch trng.
Nhn xt -Hp cng hng c v s tn s cng hngmnp ng vi v s kiu sng dao ng
ring.Khc vi mch cng hng LC ch c mt tn scng hng - mnp ph thuc vo kch
thca, b, c ca hp cng hng v ch cc ch sm, n, p xc nh kiu dao ng. Nhn
xt i vi t ngang, dao ng TM110 c tn s ring nh nht: i vi in ngang,
dao ng TE101 v TE011 c tn s ring nh nht:
Nhn xt -Phn b trng ca cc dao ng ring trong hp cng hng c dng sng
ng theo c 3 phng x, y, z -Cc thnh phn ngang ca in trng v t trng lch
pha nhau /2.V th c sau 1/4chu k khi nng lng trng in t cc i th nng
lng trng t bng 0 v ngc li. Nng lng cc i Vi iu kin cng hng Trong ng
dn sng hnh ch nht, ta bit h s pha biu th qua bc sng trong ng dn
sngtc dng: Vi p = 1,2,3. Dng dao ng no tha mn iu kin cng hng s c
bin rt ln trong hp, cn cc dng dao ng khc s b tiu hao rt nhanh Cng
sut tn hao trong thnh ng dn sng Trong thc t, mi trng dn in ca thnh
ng khng l tng; mi trng in mi khng cch in, khng l tng . Nng lng
trong hp cng hng suy hao dn.Nu khng kch thch, sau mt thi gian, nng
lng bin mt. a) iu kin b Leontovitch: Ta c : iu kin bin: Suy ra Cng
sut tn hao trong vt dn ngha vecto Poynting l mt cng sut trn mt n v
din tch. Vy tnh cng sut tn hao trong vt dn ta phi xt Vecto phc:
Vecto trung bnh: Mt cng sut trung bnh: Cng sut tn hao trong vt dn
Cng sut trung bnh qua mt tit din S: Ta c: v Suy ra: Cng sut tn hao
trong vt dn Cng thc tnh tn hao ca sng trong cc thnh phn ca hp cng
hng cho bi: Thay: Cng sut tiu hao trn thnh ng: Phm cht ca hp cng
hng H s phm cht Q c nh ngha nh sau: W: nng lng ti a tn tr hp cng
hng Q cng ln cng tt do W cng ln v Pd cng nh Phm cht ca hp cng hng
Vi: v Ta c: - l h s t thm - Hm l bin cc i ca t trng- V l th tch ca
hp cng hng - S l din tch thnh ng - 2,2 l cc thng s ca cc mi trng
thnh ng kim loi bao quanh hp cng hng. ng dng Trong ch dao ng t do:
dng lm hp ting vng kim tra cc trm pht xung Trong ch dao ng cng bc:
ng vai tr chn lc cho cc thit b thu pht o lng Trong cc dng c in t v
bn dn siu cao:to ra khng gian tng tc v trao i nng lng gia trng in t
v cc in t hoc l trng to hoc khuch i cc dao ng siu cao tn C th s dng
lm cc b lc x l tn hin v to xung cho vi iu khin. Bi tp p dng Bi 1:Hp
cng hng c kch thc a x b x l bn trong l khng kh. Cho bit dao ng in t
trong hp cng hng kiu TE101 c bin cng trng in cc i bng Em. - Tnh tn
s dao ng ring , cc thnh phn ca vector cng trng in v vector cng trng
t- Tnh nng lng trng in, nng lng trng t trong hp cng hng Ta c: y ta
xt trng TE101 nn ta c : Nh vy: 2 2 2m n pa b ltec| | | | | |= + +
|||\ . \ . \ .m=1n=0p=12 2 2 2 22 22 22 2 1 0 1 1 1 a +l = + + = +
= = a +la b l a l a l al| | | | | | | | | | |||||\ . \ . \ . \ . \
. Cc thnh phn ca vector cng trng in: t(*) ta c: 101200.sin cos sinx
zycE Ej x y zE Ak a a b le t t t t= =| |= |\ .1012mcjE Ak ae t| |=
|\ .sin sin cosy mx zE E ta lt te =Cc thnh phn ca vector cng trng
t: T (*) ta c: 010120.sin cos cosyxcHA x y zHk l a a b lt t t t t=|
|| |= | |\ .\ .sin cos sinmxE x zH tl a lt t tee=Tng t ta c: Ta c M
: Tng t ta c: cos sin sinmzE x zH tl a lt t tee= ( )21.2EVW t EdV c
=}sin sin cosy mx zE E ta lt te =( ) ( )2 2 2 2 2 2 21 1 1. sin sin
cos cos2 2 8E y m mV Vx zW t E dV E tdV E abl ta lt tc c e c e = =
=} }( ) ( )2 21sin8M mW t E abl t c e =Bi tp p dng Bi 2:Hp cng hng
c kch thc a x b x l bn trong l khng kh. Cho bit dao ng in t trong
hp cng hng kiu TE101 c bin cng trng in cc i bng Em Xc nh s phn b in
tch mt trn cc thnh hp cng hng .Tnh in tch tng cng trn mi thnh hp
cng hng Xc nh s phn b mt dng in mt trn cc thnh hp cng hng. Tnh cng
dng in mt tng cng trn 4 b mt bn ca hp cng hng ( x=0, x=a, z=0, z=l)
Ta c: i vi trng TE101: Nh vy: v ( ) . . ( ) E n S o c =sin sin cosy
mx zE E ta lt te =( )0 ( ) 0 x x a o o = = = =.0 .sin sin 0 0aEa at
t= = = Tng t: V: ( )0 ( ) 0 z z a o o = = = =( ) ( )0 sin sin cosmx
zy y b E ta lt to o c e = = = =( ) ( ) ( )20 040 0 cosa lmx zEalq y
q y b y dxdz tco et= == = = = = =}} Ta c: (1) (2) T (1) v (2) ta c:
()sJ n nS = ( ) ( )0 sin sin .ms s yE zJ x J x a t ia lt tee= = =
=( ) ( )0 sin sin .ms s yE xJ z J z l t il at tee= = = =( ) ( )20
02 0 2 0 4 .sina lms s y s yEacI J x i dx J z i dz tce et= = + = =}
}( ) ( )1 10 sin cos sin sin cosms s x zE x z x zJ y J y b t i ia a
a l c a lt t t t tee| |= = = = + |\ .Bi tp p dng Bi 3:Hp cng hng lp
phng cnh 6cmbn trong l khng kh. Cho bit dao ng in t trong hp cng
hng kiu TE101 c bin cng trng in cc i bng Em=2.104V/m .Tnh tn s dao
ng ring , bc sng ring , nng lng in t tng cng trong hp cng hng
