BÁO CÁO THÍ NGHIỆM SIÊU CAO TẦN VÀ ANTEN.pptx

8/10/2019 BO CO TH NGHIM SIU CAO TN V ANTEN.pptx

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Thnh vin thc hin: PHAN VN NG NGUYN DUY HI

PHAN NGUYN KHOA

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LAB 2: Basic Transmission Lines

in the Frequency Domain MC TIU: - S dng spice kho st sng sin trn ng truyn

khng tn hao- Lm quen vi sng phn x c bn t ti v so snh ccm phng vi gi tr tnh ton bng gin Smith

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M HNH NG TRUYN S DNG

NG TRC Xt mt cp ng trc thng dng RG-58 c:

Question 1: tnh in cm v in dung trn mt n v chiu di

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M HNH NG TRUYN S D

CP NG TRC Vi cp ng trc khng tn hao ta c cngthc linh gia L,C v bn knh trong a,bn knh ngoi bca cp

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M HNH NG TRUYN S D

CP NG TRCQuestion 2 : Cho mt cp ng trc khc vi

, . Tnh t s

= 4,235

Question 3 :Vi b= 3mm

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M HNH NG DY TRUYN SNG

DNG TRN SPICE

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ZG

50

ZL

100

T1

TD = {delay }

Z0 = 50

0 0

00

LOADINPUT

VG1Vac0Vdc

PARAMETERS:

delay = 5ns

Question 4: cho f = 200MHz vTm bc sng ?


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M HNH NG DY TRUYN S

XY DNG TRN SPICE Question 5: Cho chiu di ng truyn

Tm thi gian tr

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Question 6: V s cng ca in p ti u vo nh l mt hm ca chiu di. T cc gi tr in p trn biu v mi quan h, xc nh VSWR, v t VSWR tnh | |.

10Vi l chy t 0 n


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Question 7: V hm cng dng in u vo nh l mt hm ca chiu di L. t tnh v ||.

Nhn xt: VSWR v || c gi tr nh nhau theo 2 cch tnh

trn 11


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Question 8: V cng ca tr khng vo

ln ca tr khng u vo nh l mt hm ca chiu di

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Phn thc ca tr khng

Phn o ca tr khng

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Question 9: Tnh ton trc tip v VSWR s dng phng trnh (2.6) v (2.7) di y. So snh li kt qu o cc cu 6, 7 v 8?

Kt qu ny ging kt qu bi 6&8

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Question 10: V cng in p ti

Cng in p ti ti

Nhn xt: S thay i bin ca ln in p l rt nh mV. Do c th coi in p khng thay i theo chiu di

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Nng lng

Nhn xt: bin ca nng lng cng thay i rt nh mW. Do vy nng lng cng khng thay i theo chiu di ng truyn

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TR KHNG TI

Question 11: Nu ta c 1 mt cp ng trc nh trong cu hi 4, th tn s ca n ti chiu di /2, 2.5l bao nhiu? (ch rng ta khng c thay i chiu di vt l ca ng truyn)

ng trc trong cu hi 4 c tn s f = 200MHzV vy:

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Question 12: V bin ln ca in p u vo i vi chiu di khc nhau ( ch thay i tn s ). So snh vi kt qu question 6.Tnh VSVR trong trng hp ny

Nhn xt: Kt qu tng t question 6 T th ta c: Vmax=666.667mV

Vmin=333.394mV

18Vi l chy t /2 n 5 /2


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Question 13: V bin in p vo


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By gi cho ngn mch. Ngha l ZL = 0.001 (rt nh )

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Question14: V bin in p vo

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By gi cho h mch. Ngha l ZL = 1M (rt ln)

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LAB 3: Transients on Transmission

Lines Mc ch: Hiu c bn cht ca ng truyn di s kch

thch in p hnh SIN u vo Tm hiu s lan truyn ngn ca dy dn vi s tr gip

ca cc thng s trong SPICE.

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2. Kt cui vi in tr thun 2.1 Ti c phi hp

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Rg

50

RL50

T1

TD = 25ns

Z0 = 50Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 50nV3 = 10T4 = 50.001nV4 = 0

00

0 0

Vg Vsource Vload

Question 1: V dng sng ca in p ngun v ti ca ng truyn trong khong thi gian t = 050 ns. Gii thch dng tn hiu v c v so snh vi kt qu tnhton


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-H s phn x ti ngun l :

-H s phn x ti ti l :

-V khng c sng phn x,in p ti ngun v ti

Kt qu tnh ton trng vi kt qu hin th trn hnh v


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2.2 Ti khng phi hp Thay i tr khng ti thnh

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Question 2: V dang sng ca in p ngun v ti ca ng truyn trong khong thi gian t=0100ns ? Sosnh kt qu tmuc vi kt qu mong i c cu tr li cui cng.

Rg

50

RL20

T1

TD = 25nsZ0 = 50

Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 100nV3 = 10T4 = 100.001nV4 = 0

0

0

0

0

VsourceVg Vload


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-Ti t = T = 25ns, sng i n ti u cui , v do ti khng phi hp nn sng

phn x c bin

V vy in pti ti l tng ca haisng ti v sng phn x

Ti t = 2T = 50ns, sng i n my pht , v bi v khng c sng phn x

Nh vy, in p ngun l:

Ti t = 3T = 75ns, sng i n ti , v bi v ti khng phi hp tr khng nn

02

V


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Tng t nh vy, ta cng c :

32

)(86.2 V V S


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2.3 Ti v ngun khng phi hp tr khng

Thay i tr khng ti thnh RL = 20 , v thay i trkhng ngun Rg = 200 .

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Rg

200

RL20

T1

TD = 25nsZ0 = 50

Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 300nV3 = 10

T4 = 300.001nV4 = 0

0 0

0 0

Vg Vsource Vload

Question 3: V dang sng ca in p ngun v ti ca ng truyn trong khong thi gian t = 0300 ns.So snh kt qu tm uc vi kt qu mong i c cu tr li cui cng.


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-H s phn x ti ngun :

- H s phn x ti ti :


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Ti t = T = 25ns, sng i n ti u cui , v do ti khng phi hp nn sng

phn x c bin

V vy in pti ti l tng ca haisng ti v sng phn x

Ti t = 2T = 50ns, sng i n my pht, v do my pht khng phi hp nn sng phn x c bin

Nh vy, in p ngun l:

Ti t = T = 70ns, sng i n ti u cui , v do ti khng phi hp nn sng

phn x c bin

in pti ti : .Tnh ton tng t ta c bng thng k


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2.4 Ngun l xung vung ngn

a mt xung vung ngn10 nsvg(t) = 10( u(t)- u(t-10ns))n ng truyn

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Rg

200

RL20

T1

TD = 12.5ns

Z0 = 50Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 10nV3 = 10T4 = 10.001nV4 = 0T5 = 100V5 = 0

00

0

Vg

0

VsourceT2

TD = 12.5ns

Z0 = 50

0 0

Vmiddle Vload

Question 4: v dng sng in p ngun,im gia,cui ti ca ng truyn t = 0100 ns.Phc tho biu Bounce


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Bngcc gi tr tnh ton


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Bng cc gi tr tnh tonBiu bounce

Cn 12.5s xungghost n ti cui.

rng xung ghost l 12.5s


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2.5 Xung di

Question 5: s dng tr khng ng truyn v tr khngngun, ta nh ngha li mt ngun v g = +10 V vi t = 0 20ns, and V for t = 20 40ns. V dng sng in p ngun ,im gia, v cui ti ca ng truyn vi t = 0100 ns. ?C phi s chuyn i t cao ti thp hon ton cui ti?

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Rg

200

RL20

T1

TD = 12.5ns

Z0 = 50Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 20nV3 = 10T4 = 20.001nV4 = -10T5 = 40V5 = -10T6 = 40.001

V6 = 0

00

0

Vg

0

VsourceT2

TD = 12.5ns

Z0 = 5000

Vmiddle Vload


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T th, chng ta c th thy rng s chuyn i t cao n

thp l hon ton r rng vo cui ti.


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3. Kt cui tr khng ti phn ng 3.1 Ti dung khng

S dng ngun vg(t) = 10 u(t), R g =25 , ng truyn c Z0=50 v TD= 25ns.

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Question 6: u cui ti l mt t in 1 nF . v in p ngun v cui ti ca ng truyn vi t= 0 600ns.Do c s np v x ca t in,ta c th dng cng thc sau:

Rg

25

T1

TD = 25ns

Z0 = 50VgT1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 600nV3 = 10 00

0

Vg

0

Vsource

C1

1n

Vload


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3.2 Ti cm khng

Question 7:

Thay t in bng 1 cun cm c gi tr 0.25 H lp li vn trn

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Rg

25

T1

TD = 25ns

Z0 = 50Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 600nV3 = 100 0

Vg

0

Vsource

0

L10.25uH

1

2Vload


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Question 8: Thay ti bng 1 in tr RL=1000, L=1H,C=100pF mc song song lp li cu hi nh trn.

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Rg

25

T1

TD = 25ns

Z0 = 50Vg

T1 = 0V1 = 0T2 = 0.001nV2 = 10T3 = 600nV3 = 100 0

Vg

0

Vsource

0

L11uH

1

2Vload

R11000

C1100p


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4. Phi hp tr khngQuestion 9: Thit k b ghp c phi hp tr khng ti 150 ohm, v ngun 75 ohm.

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Question 10: Tm chiu di ng truyn trong cu 9. M phng p ng tn s ca mch t 1Mhz 3Ghz, s dng ngun sng Sin 5Vpp, tr khng ngun 75 ohm. V Vinput, Vload, bin h s phn x, tm bng thng khi | | 0.2.Tnh tr khng vo ca mng phi hp ti 1 Ghz .

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th Vin v Vload

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Bin ca h s phn x

51|| b hn 0.2 t 608.3 MHz ti 1.387 GHz v tn s t 2.607 GHz ti 3 GHz


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Tr khng vo ca mng phi hp ti tn s 1Ghz

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BÁO CÁO THÍ NGHIỆM SIÊU CAO TẦN VÀ ANTEN.pptx