cau 1: mo hinh khao sat he thong anten thang:

xet N chan tu a1->aN co truc dat vuong goc voi mat phang hinh ve. cac chan tu nay co pha  nam tren truc Ot va duoc dat cach deu nhau mot khoang d

hinh ve

+ chenh lech duong di giua hai chan tu lien tiep la:

⌂R=d.cos teta voi teta la goc hop boi truc Ot va diem khao sat

*chenh lech duong di cua chan tu 2 so voi chan tu 1 la: ⌂R2=d.cos teta

*chenh lech duong di cua chan tu N so voi chan tu 1 la: ⌂Rn=(N-1).d.cos teta

- goi I1 la dong dien kich thich cho phan tu 1 va In la dong dien kich thich cho phan tu n

   In= I1.an   voi an= |an|.e^j txi n                  (n=1->n)

 an: trong so       txi n: bien doi pha

theo nguyen ly nhan do thi phuong huong ta co ham phuong huong cua he anten thang la:

     f (teta,φ)= f1(teta,φ),fk(teta,φ)

<=> f(teta)= f1(teta).fk(teta)        (vi dang huong voi φ)

   voi    f1(teta,φ): ham phuong huong cua mot chan tu

            fk(teta,φ): ham phuong huong cua mang (he so mang)

neu f1(teta)=1 (dang huong)

=>   f teta≡ fk teta

ham phuong huong duoc tinh theo cong thuc sau:

         f teta= (n=1->N)Σan.e^j txi n..e^(-j(n-1).k.d.cos teta)=  (n=1->N)Σan.e^j txi n..e^(-j(n-1).2π/lamda.d.cos teta)= (n=1->N)Σan.e^j txi n..e^(-j(n-1).π.D.cos teta)

voi D=d/(lamda/2)): khoang cach chuan hoa so voi nua buoc song

tuy thuoc vao cac gia tri cua an va txi n ta co cac he anten khac nhau

+ an=1 voi moi n; kich thich dang bien

txi n= 0 voi moi n: kich thich dong pha

->ta co he anten he anten dang ph, dang bien, tao bup song hep

+ an=1 voi moi n: dang bien

 txi n=(n-1).txi=(n-1)π.cos teta M   voi teta M: goc lai tia mong muon

-> ta co he anten ULA dieu khien pha va dieu khien bien do vua co kha nang quet bup song ,vua co the nen cuc dai phu

cau 2: dac tinh he anten thang dong pha, dang bien:

he anten thang goi la he anten dong ph, dang bien neu co:

  + bien do dong dien kich thich anten nhu nhau:  an=1 voi moi n

+ pha dong dien kich thich anten la nhu nhau : txi n=0 voi moi n

ta co ham phuong huong cua he anten thang la:

   f teta= (n=1->N)Σan.e^j txi n..e^(-j(n-1).π.D.cos teta)

voi D=d/(lamda/2): khoang cach chuan hoa giua cac chan tu so voi nua buoc song lamda/2

thuong su dung he anten co D=1<=> d=lamda/2

f teta= (n=1->N)Σan.e^j txi n..e^(-j(n-1).π.cos teta)

=> voi he anten thang dong bien, dang pha

 f teta= (n=1->N)Σe^(-j(n-1).π.cos teta)

ham phuong huong chuan hoa cua he anten thang dong pha,dang bien la:

f teta= |f teta| / max f teta     voi max [f teta] = N khi teta =90

=> f teta=| (n=1->N)Σe^(-j(n-1).π.cos teta)/N|

dat α= π.cos teta

<=> f teta = 1/N.|sin(N.α/2)/sin(α/2))

dang cua do thi phuong huong cua he anten thang dong pha, dang bien nhu sau ( trong he toa do decac)

HINH VE

tu gian do huong ta thay nhu sau

- gian do huong co N diem khong, giua 2 diem khong co 1 diem cuc dai.

co (N-1) diem cuc dai, cuc dai chinh (bup song chinh ) o teta=90

- cac cuc dai khac nhau la cuc dai phu, cuc dai phu thu nhat nam gan bup song chinh nhat co bien do SLL ≈ -13,26db ≈  0,217 ≈2/3π    (SLL: muc cuc dai)

bup song chinh (main lobe) tai teta =90 co do rong bup song o muc suy giam 3db la teta≈100/ND[do]

voi D=d/(lamda/2) khoang cach chuan hoa so voi lamda/2

thuong D=1 => teta ≈ 100/N [do]

N: so chan tu

cau 3: dac tinh cua he anten thang ULA (dang bien dieu khien pha)

he anten thang goi la dang bien, dieu khien pha (ULA)

neu co: + dong kich thich dang bien: an=1 voi moi n (cua so bien chu nhat)

              + goc pha kich thich bien doi: txi n =(n-1) txi = (n-1)π.cos teta M

   voi tetaM: goc lai tia mong muon

ham phuong huong cua he anten thang ULA la:

   f(teta) = (n=1->N)Σan.e^j txi n..e^(-j(n-1).π.cos teta)=(n=1->N)Σe^(j(n-1).π.cos tetaM)..e^(-j(n-1).π.cos teta)

ham phuong huong chuan hoa cua he anten thang ULA  la:

    F teta= |F teta|/ max(fFteta) = 1/N*|(sin (Nα/2)/sin(α/2)|

dang cua do thi phuong huong nhu sau

N=10, teta=60 do

hinh ve

Tu gian do ta thay 

So phan tu anten cua he thong dac trung cho kha nang tao bup song hep.

Teta(M) la tham so lai tia,lam lech bup song ra khoi huong phap tuyen 

Khi teta # 90 do thi gian do huong mat tinh doi xung

Bup sonbg chinh co do rong o muc suy giam 3dB la teta3 =100/(ND*sin(tetaM)

teta(M) goc lai tia mong muon, teta(M) chi bien doi trong 1 gioi han nhat dinh

Cuc dai phu thu nhat nam gan bup song chinhnhat co bien do SLL=-13,26dB

cau 4:dac tinh cua he anten Dolph- chebyshev

trong truong hop dieu khien ca bien do kich thich va pha kich thich an= f(n) va  txi n=(n-1)πcos tetaM

(voi tetaM: goc lai tia mong muon ) ta se co mot he anten thang vua co kha nang quet bup song vua co kha nang nen cuc dai phu (dieu khien an)

  de dieu khien bien do trọng số an nguoi ta co nhieu thuat toan nham phan bo bien do trọng số nhu phân bo bien do dang cua so chu nhat, dang Hamming, dang taylor-kaser, dang Dolph-chebyshev(thuat toan Dolph-chebyshev)

Ham phuong huong cua he anten Dolph-chebyshev la:

f teat =(n=1->N)Σa(n).e^(j¥n)*.e^(-j(n-1).π.cos teta)= (n=1->N)Σan..e^(-j(n-1).π.(cos teta- cos tetaM)

tetaM: goc lai tia mong muon

-> Ham phuong huong chuan hoa la:

 f teta= |f teta|/ max[f teta] = 1/N.a(n).|(sin(N.α/2))/sin(α/2)| voi α=π.(cos teta-cos tetaM)

dang cua do thi phuong huong cua he anten dolph- chebyshev nhu sau

HINH VE

o tren, ta ve do thi phuong huong ung voi N=10, tetaM=60 va muc nen cuc dai phu R=20dB

ta thay voi he anten Dolph-chebyshev ta co the nen cuc dai phu mot cach tuy y va ke ca quet bup song

- o he anten thang co bien do suy giam 13,26 dB, trong khi do voi he anten dolph- chebyshev ta co the dieu khien duoc muc suy giam nay.

cau 5: dac tinh cua he anten thang

he anten phang la he anten gom cac phan tu buc xa (o day su dung chan tu) co tam pha sap xep thanh hang (M hang) va (N cot) trong mat phang

xet mot he thong phang nhu hinh ve

hinh ve

  so phan tu xep doc Ox la M moi phan tu cach nhau dx

  so phan tu xep doc Oy la N moi phan tu cach nhau dy

Gia su phan tu duoc kich thich boi cac dong dien co bien do giong nhau,sai pha cua dong dien cua hai phan tu doc Ox la txi x docj Oy la txi y

tai diem khao sat P(R,φ,teta) voi R khoang cach so voi diem khao sat.

- ham phuong huong cua he anten thang la (chuan hoa)

     F(NM nho) (teta,φ)= F1 (teta,φ).F (M nho) (teta,φ).F (N nho) (teta,φ)

+ ham phuong huong cua moi hang F(M nho) (teta,φ)

   F(M nho) (teta,φ)=1/M. |(sin M.(αx/2))/(sin(αx/2))|

 voi αx= k.dx.cos (teta x)+ txi x

cos (teta x)= sin teta.cos txi

+Ham phuong huong cua moi cot: F(N nho) (teta,φ)

F(N nho) (teta,φ)=1/N. |(sin N.(αy/2))/(sin(αy/2))|

 voi αy= k.dy.cos (teta y)+ txi y

cos (teta y)= sin teta.cos txi

+ ham phuong huong cua tung thanh phan la F1(teta,φ)

   F1(teta,φ)=gom . .

. sin teta=> dipol

sin teta. [cos((k.l/2).cos teta)-cos(kl/2)]/(sin teta)^2   => chan tu doi xung

he so dinh huong cua he thong o huong buc xa cuc dai la Dmax (Dmax dat cuc dai theo truc Oz ung voi teta=0)

co: Dmax=D1maxM.N

D1max= max[(|E1|^2.R(o nho)^2)/60P(Σ1 nho)]

+P (Σ1 nho):cong suat buc xa trung binh cua mot phan tu

 +E1: cuong do truong

voi dipol D1 max= 1,5, Dmax=1,5.M.N

do rong bup song la:

+trong mat phang xOz: teta3≈100/Ax=100/M.Dx

+ trong mat phang yOz: teta3≈100/Ay=100/N.Dy