May Thu Bpsk

n 2

Thit k my thu BPSK

TRNG I HC TN C THNG KHOA IN T VIN THNG

BO CO MN HC N 2

THIT K MY THU BPSK

SVTH

L Duy Bnh ng Ngc Anh Hong Trng An Th.s Hong Mnh H

MSSV: 811089D MSSV: 811085D MSSV: 811079D

GVHD

GVHD : Th.s Hong Mnh H

n 2

Thit k my thu BPSK

Li cm n

Trc ht chng em xin cm n thy Hong Mnh H tn tnh hng dn cng nh gip chng em hon thnh n ny . Gip chng em cng c li nhng kin thc hc v nng cao hn kin thc ca mnh . Qua n ny chng em hiu r hn th no l iu ch s c bit l iu ch BPSK v nguyn l hot ng ca tng khi mach trong my thu BPSK .. Chng em cm n thy to iu kin thun li v gip chng em hon thnh tt ti ny ,c c nhng kin thc quan trng trong thc tin gip chng chng em cm thy vng vng hn cho mc tiu theo ui ca mnh .


n 2

Thit k my thu BPSK

ti : Thit k my thu BPSKCc thng s : Tn s : RF 1 MHZ , IF 200 KHz Cng sut thu t Anten : -50dbm Tr khng Anten : 50 Ohm Tc d liu : 10 kbp/s in p ngun : 12V Nhim v thit k : Nghin cu l thuyt v phng php iu ch BPSK a ra s khi. Tnh ton thit k cho tng khi M phng tng khi thit k Kt hp cc khi thnh mt h c th Phn tch , gii thch kt qu,a ra nhn xt Bo co


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Thit k my thu BPSK

Mc lc


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Thit k my thu BPSK

PHN 1 : S LC MY THUI. nh ngha may thu :May thu la thiet b au cuoi trong he thong thong tin vo tuyen ien. May thu co nhiem vu tiep nhan va lap lai tin tc co cha trong tn hieu chuyen i t may phat di dang song ien t. May thu phai loai bo cac loai nhieu khong mong muon, khuech ai tn hieu mong muon va sau o giai ieu che e nhan c thong tin ban au. May thu co rat nhieu tham so, nhng chung ta chu yeu ch xet cac ch tieu ky thuat c ban cua may thu nh sau : 1. o nhay :bieu th kha nang thu tn hieu yeu cua may thu. o nhay c xac nh bang sc ien ong cam ng toi thieu cua tn hieu tren Anten e am bao may thu lam viec bnh thng. No thng c o bang V. Muon nang cao o nhay cua may thu th he so khuyech ai (Av, Ai) cua no phai ln va mc tap am noi bo cua no phai thap (giam tap am tang au). sieu cao tan (f > 30MHz) o nhay cua may thu thng c xac nh bang cong suat ch khong phai bang sc ien ong cam ng tren Anten. 2. o chon loc :la kha nang chen ep cac dang nhieu khong phai la tn hieu can thu. Noi cach khac, o chon loc la kha nang la chon tn hieu ra khoi cac loai nhieu ton tai au vao may thu. 3. Chat lng lap lai tin tc : c anh gia bang o meo cua tn hieu (meo phi tuyen, meo tan so, meo pha). ay chu yeu ch xet o meo tang khuyech ai cong suat am tan e sao cho tn hieu ra loa t b bien dang so vi tn hieu a ti bo ieu che cua may phat. 4. Day tan cua may thu : la khoang tan so ma may thu co the ieu chnh e thu c tn hieu vi cac ch tieu k thuat theo yeu cau. Ngoai ra ta con phai xet en cac yeu to ch tieu ky thuat khac cua may thu nh Pra, tnh on nh bien o va tan so,

S o khoi tong quat cua may thu :


n 2

Thit k my thu BPSK


n 2chnh sau:

Thit k my thu BPSK

II. Phan loai may thu : may thu c phan lam 2 loai1. tiep :Mach K RF

May thu khuech ai trcGiai ieu che out K Am tan

Mach vao

-

Tang K cao tan co nhiem vu khuech ai trc tiep cac tn hieu a c chon t mach vao e a tn hien co mc ien ap tng oi ln cung cap cho tang giai ieu che, nh o giam bt o meo cua tn hieu am tan. May thu dang nay n gian, de dang lap rap, nhng o nhay va o chon loc van con kem, khong on nh . Viec nang cao o nhay va o chon loc cua may thu nay b han che sau : So tang khuech ai khong the tang len mot cach tuy y v : So tan cang tang th tnh on nh cua mach khuyech ai cao tan RF cang giam. - So tan cang tang th so mach cong hng cung tang, he thong ieu chnh cong hng phc tap, cong kenh, at tien. Tan so cao kho at c he so khuech ai ln. Tan so cang cao th dai thong cang rong, lam giam o chon loc cua may thu. Muon co dai thong hep phai dung mach cong hng co he so pham chat cao nhng co the gay meo tn hieu. Do khong dung c cac he thong cong hng phc tap nen khong co kha nang at ac tuyen tan so co dang ly tng. 2. May thu oi tan :

Mach vao

K cao tan

oi tan

K trung tan

Tach song

KCS Am tan

out

Dang may thu nay co them khoi oi tan va khoi K trung tan. Tn hieu cao tan a ieu che (AM, GVHD : Th.s Hong Mnh H

n 2

Thit k my thu BPSK FM, PM) nhan c t Antena qua mach vao (bo loc bang thong), qua bo khuech ai cao tan RF, c a vao bo oi tan e bien oi thanh mot tan so tn hieu khac goi la tan so trung gian, nhng qui luat ieu che khong oi ( con goi la trung tan). Tn hieu trung tan thng c gi co nh khi tan so tn hieu thay oi (thu cac ai khac nhau) va thng c chon thap hn tan so tn hieu au vao e tang o on nh tan so. Sau o c a vao bo khuech ai trung tan roi a en mach tach song. Thc chat cua bo oi tan la thc hien phep nhan tan so. No bao gom mot bo dao ong noi tao ra tan so trung gian va mot bo tron tan. Bo tron tan nay co the la mot phan t phi tuyen hay tuyen tnh co cac tham so bien thien tuan hoan.

So vi dang may thu khuech ai trc tiep, may thu oi tan co u iem sau : Co the dung nhieu tang khuech ai trung gian (8 10 tang) e co the at ti he so khuech ai toan may cao ma van am bao tnh on nh cua may thu (do tan so tn hieu c ha xuong trung tan).. Mach cong hng co ket cau n gian. Co the at o chon loc cao Acn1. Ben canh o th may thu oi tan van ton tai mot so nhc iem : Ket cau phc tap. Mc tap am noi bo cao do co khoi oi tan. Xuat hien mot so nhieu khong mong muon : nhieu anh, nhieu lot thang, nhieu tan so lan can,

III. Cac khoi trong may thu :1. Anten : Antena la mot thiet b khong the thieu cua may thu, khong co Antena th khong the thu c tn hieu. Nhieu vu cua Antena la tiep nhan nang lng song ien t do may phat phat ra roi a ti mach vao cua may thu. Trong thc te muon giam bt tap am ben ngoai (nhieu khong mong muon) th ta phai la chon Antena co o nh hng cao.


n 2 2.

Thit k my thu BPSK

Mach vao : Mach vao la mach ien noi lien Anten vi tang vao au tien cua may thu . No co nhiem vu chuyen tn hieu cao tan nhan c t Anten thu en tang au tien va am nhiem mot phan o chon loc cua may thu. Mach vao gom 3 phan: He thong cong hng (n hoac kep) co the ieu chnh en tan so can thu. Mach ghep vi nguon tn hieu cua mach vao (Anten). Mach ghep vi tai cua mach vao (tang K cao tan au tien). Cac ch tieu ky thuat cua mach vao : He so truyen at : la ty so gia ien ap ra cua mach vao ieu chnh cong hng mot tan so nao o va sc ien ong cam ng tren Anten (EA) :VO EA

AMV =

Vi AMV cang ln th he so khuyech ai chung toan may cang ln.

o chon loc : S e =

AO Af

Dai thong D Tan oan lam viec :pham vi ieu khien mach mach vao t fomin fomax (Amin Amax).

Cac mach ghep Anten vi mach cong hng vao : Mach cong hng ghep ien dung trong vi Anten : Mach nay co he so truyen at Av = const , nhng Av va Ap thap , he so mac mach phu thuoc vao tan so, do o thng th Cgh >> C e C quyet nh tan so cua mach cong hng. Dang mach nay thng dung trong cac bo khuech ai tan so trung gian.

C Q C g h 2

0


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Thit k my thu BPSK


n 2

Thit k my thu BPSK Mach cong hng ghep bien ap t ngau vi Anten :

Dang mach nay co o ghep rat chat ( A p 1 ), cho nen .Voi sieu cao tan th he so tap am nho. V vay cach ghep nay thng dung tan so sieu cao, ngoai ra con dung vi bo khuech ai dung Transistor e ton hao cong suat mach vao t. Mach cong hng ghep bien ap vi Anten : Dang mach nay dung nhieu trong may thu. o ghep gia Antena va mach cong hng quyet nh bi ho cam M ch khong phai do Lgh. Ghep ho cam th he so truyen at phu thuoc vao tan so, con o ghep khong phu thuoc vao tan so, nen thng dung trong mach khuech ai cao tan RF.

3. Mach khuech ai cao tan : Mach KCT co nhiem vu K tn hieu a ieu che cao tan en mot gia tr nhat nh e a vao mach oi tan.

Tai la ien tr :

He so K cua tang se tng oi ong eu oi vi moi tan so. GVHD : Th.s Hong Mnh H

n 2

Thit k my thu BPSK Tan so cang cao th he so K cang giam nhng khong nhieu. Neu tien ti tan so cat th he so K se giam nhanh.

Tai la cuon cam :

Tan so cang cao th he so K cua tang cang cao nhng tan so o cung khong the vt qua tan so cat, ben canh o con b han che bi ien dung cua cuon cam. Tai la mach cong hng :

He so K se ln nhat tan so cong hng cua mach vao fo, con cac tan so lan can th he so K giam dan. Tan so cong hng co the ieu chnh c nh trong may thu K trc tiep.

Tom lai he so K cua mach co tai la mach cong hng ln hn nhieu so vi mach co tai la ien tr, cuon cam. 4. Mach oi tan : oi tan la qua trnh dch chuyen tn hieu a ieu che dang cao tan thanh tn hieu co tan so thap hn (tan so trung tan khong oi) nhng van gi


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Thit k my thu BPSK nguyen cau truc pho cua no (dang tn hieu ban au). oi tan con goi la tron tan, co ky hieu dau nhan. Cac thong so c ban bo oi tan :I IF VS

o ho dan oi tan : g C = o li oi tan : AV =V IF VS I IF V IF

ien dan oi tan : g =

o li cong suat oi tan : PG = Mach K trung tan :

PIF PS

5.

La mach K cong hng co nhiem vu K tn hieu sau bo oi tan u ln e a vao bo tach song ( giai ieu che ) do mc tn hieu sau oi tan khoang 2 * f 0 * ZC 4 = 2 Chn C4 = 680 pF Tng t ZC3 > 2 * f 0 * ZC 3 = 2 Chn C3 = 68 pF1

Thit k my thu BPSK

16

*10 * 2700

= 58.94 pF

Vi f0 = 1 MHZ xem anten siu cao tn hat ng trong khong tn s hp nn chn cc thng s anten nh sau : CA = 50 pF rA = 50 LA = 10 uH in dn tng ca khung cng hng : g01 =1 1 = wo * Lo * Qo = Rt 1 1

1 2 *10 *10 *10 * 806 6

= 1,989.10-4 (1/)

Rt1 = g 1 = 5026.5 o in dn ca Anten : gA =4 * 50 * 50 * 10 12 1 4 * rA * CA = = = 10-3 (1/ ) 50 * 10 6 RA LA

Suy ra RA = 1K Tnh tan cc gi tr t trong mch : Cb1 = Cbe1 + CM1 M CM1 = Cbc1*(1 + gm1*Rt1) gm1 = 40*ICQ1 = 40*0.5*10-3 = 0.02 Cbc1 = 1 (pF) CM1= 10-12(1 + 0.02*5026,5) = 101,53 pF Suy ra Cbe1 = 2 * fT * rb ' e1 rbe1 = gm1 = = 5000 0.02 Cbe1 =hfe 1

hfe 1

100

100 2 * 300 *10 * 50006

= 10,61 pF

Cb1 = 101,53 + 10,61 = 112,14 pF Theo gi thuyt :

1C = 1

12

4

2

*

fo

* Lo

=

4

2

* (10 6) * 10 *10

2

6

= 2533 pF


n 2 M C1 = Ct1 + m2CA + Chn N1 = 2 Suy ra Ct1 = C1 - m2CA -

Thit k my thu BPSK

Cb '1

N

2 1

Cb '1

N

2 1

1 ,14 12 Ct1 = 2533 0.32*50 = 2500 pF 4

T h phng trnh :3 C2 7 C1 * C 2 Ct1 = = 2500 C1 + C 2

C1 =

(1) (2)

Th (1) vo (2) ta c : C2 = 42175.56 Suy ra

3 1 0 C2 2500. =0 7 7

C2 = 8333,3 pF C1 = 19444,4 pF

Chn

C1 = 22 nF C2 = 8,2 nF

li dng v p : li dng : Ai1 = -a1*gm1*R = 1 2 1

gm1 * R1 N1+

'

M R = rbe1*(N1) //RB1*(N1)2//Rt1//RA*m2

1

R1

'

=

1 rb ' e1 * N 12

1 RB1 * N 12

+

1

1 1 + RA * m Rt 1

2

R

'

1

1 1 1 = + + 5026 ,5 + 5000 * 4 27000 * 4

1 1000 * 0.32

Suy ra R1 = 88


n 2 Ai1 = 0.02 * 88 = - 0.88 2

Thit k my thu BPSK

Dng vo : ii = EA*m2*gA = 707*10-6*0.32*0.02 = 1,2726.10-6 A Dng ra : i01 = i1*A1 = 1,2726.10-6 *0.88 = 1,2.10-6 A p ra : Vo1 = io1*Rt1 = 1,2.10-6 *5026.5 = 6,03 mV


n 2 3) Tng khch i th 2:

Thit k my thu BPSK

8

N

3

4

1 2 V

d c0

C R 5

8

6 5 1

1

N

2

5

Q B P

2 F 3 6 3

C

54 8

C

6

R

4

R

6

C

7

0

0

PC max = 0.5 W VCEm ax

Chn TST Q2 : BFP420

fTI C maxh fe

= 25 V = 300 MHz = 25 mA = 100

Khi :f = f T 300 = = 3 MHz h fe 100

Suy ra : 0.3 f < f0 < 3 f Do mch lm vic tn s trung bnh Ch DC: I C 2 = 10 mA Q h fe 2 = 100 V = 0,7v VR6 = 0.1VCC= 0.1*12 = 1.2 V Suy ra R6 =VR 6 = IC 2 Q

1.2 10 *103

= 120

Chn R6= 120


n 2ICQ 2 VBB2= hfe 2 *RB2 + VBE2 + VR6

Thit k my thu BPSK

1 1 R6*hfe2 = *120*100 = 1200 1 0 1 0 10 * 10 3 Suy ra VBB2 = *1200 + 0.7 + 1,2 = 2,02 V 100 RB 2 1200 VBB 2 = 2,02 = 1442,8 R4 = 1 1 VCC 12

M RB2 =

Chn R4 = 1.5 K R5 =12 VC C *RB2 = 2,02 *1200 = 7128,7 VBB 2

Chn R5 = 7,5 K Ch AC : S tng ng :

1

Vbe1gm1

L2

g02

C5

Rb2N22

rbe2N22 g03 Vbe2gm2

1

L2

C8

Chn ZC7 2 * fo * ZC 6 = 2 Chn C6 = 1.5 nF GVHD : Th.s Hong Mnh H

1 *10 *1206

= 1,326.10-9

n 2 in dn tng ca khung cng hng : g02 =1 1 = wo * Lo * Qo = Rt 2 1

Thit k my thu BPSK

1 2 *10 * 10 *10 * 806 6

= 1,9894.10-4 (1/)

Rt2 = go 2 = 5026,5 Tnh tan cc gi tr t trong mch : Cb2 = Cbe2 + CM2 M CM2 = Cbc2*(1 + gm2*Rt2) gm2 = 40*ICQ2 = 0.4 Cbc2 = 1 (pF) CM2= 10-12(1 + 0.4*5026,5) = 2011,6 pF M Cbe2 = 2 * fT * rb ' e 2 rbe2 = gm 2 = 250 Cbe2 =h e2 f

hfe 2

100 2 * 300 *10 * 2506

= 212,2 pF

Cb2 = 2011,6 + 212,2 = 2223,8 pF Theo gi thuyt :

1C = 2

12

4

2

foN

Lo

=

4

2

(10 )

6

2

* 10 *10

6

= 2533 pF

M C2 = C5 +

Cb ' 22 2

Suy ra C5 = C2 -

Cb ' 2

N

2 2

= 2533 -

2223,8 4

Chn N2 = 2 Suy ra C5 = 2477,05 pF Chn C5 = 2,2 nF li dng v p : li dng : Ai2 = -a2*gm2*R = 2 2 2

gm 2 * R 2 N2

'

M R = rbe2*(N2) //RB2*(N2)2//Rt2 GVHD : Th.s Hong Mnh H

n 2

Thit k my thu BPSK =

1

1 rb ' e 2 * N 22

R1

' 2

+

1 RB 2 * N 22

+

1 Rt 2

R

' 2

=

1 1 1 + + 5026 ,5 250 * 4 1200 * 4

Suy ra R2 = 710,6 Ai2 = 0,4 * 710,6 = - 142 2

Dng ra : i02 = i01*A2 = 1,2.10-6 *142 = 1,704.10-4 p ra : Vo2 = io2*Rt2=1,704.10-4*5026,5 = 0,85 V

4) Mch dao ng : Ch DC01 2 C 2 1 V n d 1 2 c R g 11 2 n 1 R 8 2 C 9 Q 3 L 3 2 2 C 1 F 3 R 1 1 k P 0 4 22 0 C 1 1 1 2 L 2

1

2

B

0/ S1 2

1 I E2 C 1 o

1

0

1

Khi transistor hoat ong che o bao hoa th suit ap tren no la 0,2vVC = VC -0,2V=11,8V C C

=> VR10 =0,3. VCC =3,3.11,8= 3.45V Chon: I CQ 3 = 10mAVR10 3,54 Vay: R10 = I = = 354 CQ 3 10 .10 3

chon R10 = 390 Vi: RBQ 3 =1 1 ( h fe +1) R10 = 101 x 390 = 3939 1 0 10


n 2I CQ 3 = VBB 3

Thit k my thu BPSK 0,7 RBQ 3 hfe R10 +

VBB3 =

I

CQ 3

hfe

*RBQ3 + 0.7 + VR10 = 4,6339

R9 = RBQ 3

VCC 11,8 =3939 4,6339 = 10,03 K VBB 3

Chn R9 = 10 K R8 = RBB 3 VCC =6,48 K VCC VBB 3

Chon R8 = 6,8 K

CHE O AC:

hie = 1,4hfe

25 mv = 1,4.100 25 = 350 I CQ 4 10

Chon: Lch = 1 uHZin = RBQ 3 // hie =

3939 * 350 =321,44 3939 + 3501 2 * pi * f 0 * Z in

mat khac: Xng C ng >>

C ng >>

1 2 * 3,14 * 1,2 *10 * 321,446

= 4,126.10-10

Chon C ng = 4,7 nF Ta c w0 =

1 L 2.C t1 4 *2

Suy ra Ct =

fo

2

*L

= 17,6 nF


n 2 Chn C0=40 nF e mach hoat ong on nh ta co the chon:C b 'e = 1nF C12 = 10 .Cb ' e = 10 nF

Thit k my thu BPSK

Ta c :1 1 1 1 = + + Ctd C11 C12 Co

==>

1

C

=

1

11

C

-

1

t

C

-

1

12

C

o

Suy ra C11 = 45,8 nF Chn C11 = 47 nF li hi tip :

B=

C C

11 12

= 0,458

Ta c : A=

hfe * RC hie

Vi Rc = B2*(RB//hie) = 0,4582*321 = 67,33 Suy ra A =100 * 67 ,33 = 19,2 350

Suy ra A.B=0,458*19,2 = 8,8 > 1 Suy ra tha iu kin l mch dao ng ZC13 2.5v la +5v va

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