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chan tu ĐX

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cau 1: dac tinh phan bo dong dien va dien tich tren chan tu doi xung

chan tu doi xung la cau truc gom hai doan vat dan co hinh dang tuy y, kich thuoc giong nhau dat thang hang rong khong gian va o giua duoc noi voi nguon giao dong cao tan

de bieu dien su phan bo dong va dien tich cua chan tu doi xung co hai phuong phap chinh xac la gan dung song thuong dung phuong phap gan dung (phuong phap duong day)

gia su: chan tu co chieu dai lban kinh a rat nho (dang hinh tru rat mong) sao cho thoa man 2a/lamda≤0.01

hinh ve

khi do phan bo dong dien tren anten la:

Iz(z)=Ib.sin[k.((l/2)- |z|)]

voi Ib: bien do dong dien o diem bung song dung

k=2π/lamda: hang so song

=> Iz(z)=Ib.sin[2π/lamda.((l/2)- |z|)]= Ib.sin(l/lamda.π - |z|/lamda.2π)

theo phuong trinh bao toan dien tich :

  dIz/dz + iw.Qz=0

voi Iz= 2πa.Jz  : bienj do dong dien tai toa do z cua chan tu

+ Jz: mat do dong dien mat

+ Qz: dien tich mat tren 1 don vi vi chieu dai chan tu

=> phan bo dien tich tren chan tu doi xung la:

  Qz1= kl(b nho)/iw.cos[k.(l/2-z)]    z>0

Qz2= -k.l(b nho)/iw.cos[k.(l/2+z)]    z<0

tu phuong trinh bieu dien su phan bo dong dien va dien tich ta co do thi

HINH VE

cau 2: trinh bay phuong phap tinh cuong do truong buc xa cua chan tu doi xung

xet chan tu doi xung dat doc truc z nhu hinh ve

HINH VE

coi chan tu la tap hop cua nhieu dipol dien ma tren do co dong dien Iz=const

xet 1 cap dipol dien rat nho dz voi vi tri +z va -z

khi do cuong do truong gay boi 2 cap dipol nay tai diem khao sat M(R,0teta,φ) la:

  dE = (dE+z) + (dE-z)= i/2.w.1/R.e^(ikR).Iz.dz/lamda.sin teta=  i/2.w.1/R.Ib.sin[(k.(l/2-|z|))]. dz/lamda.sin teta.[e^(ik.Rz)+e^(-ik.R-z)]

voi Rz=R-z.cos teta: khoang cach tu +z den M

       R-z=R + z.cos teta khoang cach tu -z den M

=> dE= i/2.1/R.Ib.1/lamda.sin teta.sin[k.(l/2-|z|)].e^(-ikR).[e^(ik.Rz)+e^(-ik.R-z)].w = i.A.W.e^(-ikR).sin[k.(l/2-|z|)].cos(k.z.cos teta).dz

voi A: he so tinh toan A=1/2.1/R.Ib.1/lamda.sin teta

khi do cuong do truong tong gay ra tai diem khao sat

E=⌠dE= (l/2->0)⌠i.A.w.sin[k(l/2-z)].cos(kz.cos teta).dz

tinh tich phan tren ta duoc ket qua nhu sau E= i.A.W.[cos(kl/2.cos teta)- cos(kl/)2

H≡Hφ=E/W=iB.Ib.e^(-ikR).1/R. cos(kl/2)

W=√µ/ε   : tro khang song moi truong  µ: do tu tham cua moi truong   ε: hang so dien moi

Ib: bien do dong dien o diem bung song dung

R: khoang cach tu tam chan tu toi diem khao sat

k=2π/lamda: hang so so,    lamda:buoc song

+l: do dai chan tu

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