Giải Bài Tập Siêu Cao Tần

5/20/2018 Gi i B i T p Si u Cao T n

1/18

GII BI TP SIU CAO TNCHNG 2:Bi 2.1: Cho ng truyn c 0.2 /L H m , 300 /C pF m , 5 /R m ,

0.01 /G S m . Tnh hng s truyn sng, tr khng c tnh ti 500MHz. Tnh likhi khng c tn hao (R=G=0).

Hng s truyn sng: R j L G j C

5 628.32 0.01 0.94 590 178, 93j j

24.30 89.47 0.22 24.30( / )j rad m

Tr khng c tnh: 0R j L

ZG j C

0

5 628.32668.41 0.15

0.01 0.94

jZ

j

0 25.85 0.08 25.85 0.04( )Z j

Khi khng c tn hao: (R=G=0)0 ; 24.33( / )LC rad m

0 25.82( )L

ZC

Bi 2.2: Chng minh phng trnh Telegrapher

p dng KVL:( , ) ( , )

( , ) ( , ) ( , ) ( , ) 02 2 2 2

R z L z i z t R z L z i z tu z t i z t i z t v z z t

t t

Chia 2 v cho z , ly lim 2 v khi 0z :( , ) ( , )

( , )v z t i z t

Ri z t Lz t

Hay:( ) ( , )

( )v z i z t

Ri z Lz t

p dng KCL:( , )

( , ) ( , ) ( , ) ( , )2 2

R z L z i z ti z t i z z t G z v z t i z t

t


2/18

( , )

( , ) ( , ) 02 2

R z L z i z tC z v z t i z t

t t

Chia 2 v cho z , ly lim 2 v khi 0z :( , ) ( , )

( , )i z t v z t

Gv z t C

z t

Hay:

( ) ( , )( )

i z v z t Gv z C

z t

Bi 2.5: Cp ng trc bng ng, ngknh trong 1mm, ngoi 3mm, 2.8r , gc

tn hao 0.005tg . Tnh R, L, G v C ti 3GHz, tr khng c tnh, vn tc pha. i vi cp ng trc:

774 10 1.5

ln ln 2.2 10 /2 2 0.5

bL H m

a

9

100

12 10 2.8

22 ' 36 1.42 10 /1.5

ln ln ln0.5

rC F mb b

a a

1 1 1 125.94( / )

2 2

sRR ma b a b

022 '' 2 ' 0.013 /

ln ln ln

rtgtgG S mb b b

a a a

0

5.94 4146.9

1547.33 0.20.013 2.68

R j L j

Z G j C j

0 39.34 0.1 39.34 0.069( )Z j

11113.83 179.64R j L G j C

105.42 89.82 0.33 105.42( / )j rad m

0.33; 105.42

81.8 10 ( / )pv m s

Bi 2.7: Cho ng truyn khng tn hao, chiu di in 0.3l , kt cui vi tiphc. Tm h s phn x ti ti, SWR, tr khng vo.Bit tr khng c tnh

0 75Z , tr khng ti 40 20LZ j .

H s phn x:

0

0

35 200.27 0.22 0.35 140.39

115 20

L

L

Z Z jj

Z Z j

1 1 0.352.08

1 1 0.35SWR


3/18

2

0.3 0.6l

00

0

40.74 21.95( )LinL

Z jZ tg lZ Z j

Z jZ tg l

Bi 2.8: ng truyn khng tn hao ktcui vi ti 100L

Z . Nu SWR=1.5.

Tm tr khng c tnh c th.1

1.5 0.21

SWR

0 0 0

0 0 0

100 1000.2

100 100

L

L

Z Z Z Z

Z Z Z Z

Vi0

Z thc.

0

0 0

00

0

1000.2

100 66.67

150100 0.2100

Z

Z Z

ZZ

Z

Bi 2.9: Mt my pht v tuyn ni vi Anten c tr khng 80+j40 vi cp ngtrc 50 . Nu my pht 50 c th cung cp 30W khi kt ni vi ti 50 , cungcp cho Anten l bao nhiu?

0

0

30 400.3 0.22 0.37 36.03

130 40

L

L

Z Z jj

Z Z j

2 2ef 1 30 1 0.37 25.893load inc r incP P P P WBi 2.10: Cp ng trc 75 , ng truyn c chiu di 2.0cm kt cui vi ti

37.5+j75 . Nu 2.56r , tn s 3.0GHz. Tm tr khng vo, h s phn x ti tiv ti u vo, SWR.

H s phn x ti ti:

00

0

37.5 750.08 0.62 0.62 82.87

112.5 75

L

L

Z Z jj

Z Z j

1 1 0.624.26

1 1 0.62SWR

883 10

1.875 10 ( / )2.56

p

r

cv f m s

p

20.0625 6.25 2.01m cm l l

20 0.62 82.87 1 4.02 0.62 147.36j ll e

0.52 0.33l j

2

0 2

1 0.52 0.33175 19.24 20.46 28.09 46.76

1 1 0.52 0.33

j l

in j l

jeZ Z j

e j


4/18

Bi 2.11: Tnh SWR, , RL cn thiu trong bng sau.

20lgRL

1 1

1 1

SWRSWR

SWR

/2010 RL

SWR RL(dB)

1.00 0.00 1.01 0.005 46.02

1.02 0.01 40

1.05 0.024 32.40

1.07 0.032 30.0

1.10 0.048 26.38

1.20 0.091 20.82

1.22 0.10 20

1.50 0.2 13.98

1.92 0.316 10.0

2.00 0.333 9.55

2.50 0.429 7.35

Bi 2.12: Cho ng truyn c 15gV Vrms , 75gZ , 0 75Z , 60 40LZ j

v 0.7l . Tnh cng sut cung cp cho ti theo 3 cch -Tm v tnh LP :

00

0

15 400.02 0.30 0.3 94.05

135 40

L

L

Z Z jj

Z Z j

2

22 2

0

1 15 11 1 0.3 0.6825

2 2 75

g

L

VP W

Z

-Tm inZ v tnh LP :

21.4l l

00

0

48.19 27.33LinL

Z jZ tg lZ Z j

Z jZ tg l

2 2

15Re 48.19 0.6809

123.19 27.33

g

L in

g in

VP Z

Z Z j

W

-Tm LV tnh LP :

0 z zV z V e e

00 1LV V z V

0 l lV z l V e e v 00

l lVI z l e e

Z


5/18

V dng in l lin tc nn:

0

2

g g

l

g

V V z l V I z l V

Z e

12

g

L l

VV

e

V ng truyn khng tn hao nn 0 nn 12

g

L l

VV

e

22 21 7.6867

Re Re 60 0.68182 72.11

gLL L L

L L

VVP Z Z

Z Z

W

Bi 2.14: Cho ng truyn nh sau 10gV Vrms , 50gZ , 0 50Z , 75LZ

v 0.5l . Tnh cng sut ti incP , cng sut phn x efrP , cng sut truyn qua

anstrP .

2

2l

00

0

Lin L

L

Z jZ tg lZ Z Z

Z jZ tg l

Mch tng ng l ngun ni vi gZ v inZ .

Cng sut ngun:2 2

our

1 1 100.4

2 2 50 75

g

s ce

g in

VP

Z Z

W

Cng sut tn hao trn gZ :2

2

os

1 1 1050 0.16

2 2 50 75l s gP Z I

W

Cng sut a vo ng truyn:2

2

ans

1 1 1075 0.24

2 2 50 75tr inP Z I

W

Cng sut ti:2

2

0

1 1 1050 0.25

2 2 50 50incP Z I

W

Cng sut phn x:2

2 0ef

0

0.01Lr inc incL

Z ZP P P

Z Z

W

Nhn xt:

ans ef tr inc r P P P

our ans oss ce tr l sP P P


6/18

Bi 2.15: Mt my pht kt ni vi ti vi 10gV Vrms , 100gZ , 0 100Z ,

80 40LZ j v 1.5l . Tm in p l hm ca z vi 0l z .

0

0

20 400.06 0.24 0.24 104.04

180 40

L

L

Z Z jj

Z Z j

0 j z j zV z V e e 00 1LV V z V

2 33 0

2l tg l

in LZ Z

10

80 40 4.71 1.18 4.85 14.04 ( )180 40

g

L in

g in

VV Z j j V

Z Z j

0 0

0

10100 5( )

200

g

g

VV Z V

Z Z

25 5 1j z j z j z j zV z e e e e

25 1 j zj j ze V z e e Khi :

ax 5 1 5 1.24 6.2( )mV V 2

1 2 2 0.355j l

e l z

( Ta phi chn sao cho z


7/18

Tm v tr tia qua Lz hng v pha my pht c c gi trtng ng. Di chuyn i 1

on 0.4. V tia t im ny qua tm gin c inz , 0.5 0.4inz j .

Suy ra: 0 50 0.5 0.4 25 20( )in inZ Z z j j

minV khi 0.326l .

maxV khi 0.076l .

Bi 2.18: Tng t bi 2.17 vi 40 30LZ j

Tm c SWR=2; 0.33 90 ; (0.8 0.6) / 50 0.016 0.012( )LY j j S ;

(1.86 0.42) 50 93 21( )inZ j j

minV khi 0.1245l .

maxV khi 0.3745l

Bi 2.19: Tng t bi 2.17 vi 1.8l Tm c SWR=2.45; 0.42 54.2 ; 0.01 0.008( )LY j S ;

(0.42 0.14) 50 21 7( )inZ j j

minV khi 0.326l .

maxV khi 0.076l

CHNG 4:Bi 4.7: Tm ma trn Z v Y ca mng 2 cng.

Mng hnh :Ma trn Z :

2

1 111 22

1 0 1

2

A A B

A A B A BI

A A B

Z Z ZV VZ Z

Z Z ZI Z ZV Z Z Z

2

1 2

221 12

1 01

2

A

A B A

A A B A BI

A A B

ZV

V Z Z Z Z Z

Z Z ZI Z ZV

Z Z Z

Ma trn Y :

2

1 111 22

1 01

A B

A B A BV

A B

I I Z ZY Y

Z ZV Z ZI

Z Z

2

1

221 12

1 10

1B

BV

V

I ZY Y

V V Z

Mng hnh T:Ma trn Z :


8/18

2

1

111 22

1 10

1 1

A B A B

A BI

IY YV Y Y

Z ZI I Y Y

2

1

221 12

1 10

1B

BI

I

V YZ ZI I Y

Ma trn Y :

2

1 111 22

1 11 0 2

A A B

A BV

A A B

Y Y YI IY Y

I IV Y Y

Y Y Y

( Ch : / /A BY Y th : td A BY Y Y )

2

1 2

221 12

1 01

21 1

A

A B A

A BV

A A B

YI

I Y Y YY Y

V Y YI

Y Y Y

Bi 4.9: Mng 2 cng c cc tham s sau:

1 10 0V 1 0.1 30I

2 12 90V 2 0.15 120I

Tm in p ti v in p phn x ti 2 cng nu tr khng c tnh 0 50Z .

1 1 1V V V

1 1 10

1I V V

Z

1 0 11

10 0 50 0.1 307.27 9.9 ( )

2 2

V Z IV V

1 0 11

10 0 50 0.1 303.10 23.8 ( )

2 2

V Z IV V

Tng t:

2 0 22

12 90 50 0.15 1209.44 101.5 ( )

2 2

V Z IV V

2 0 22

12 90 50 0.15 1203.33 55.7 ( )

2 2

V Z IV V

Bi 4.10: Tm ma trn tn x ca ng truyn khng tn hao. Chng minh cc matrn l unitary.

2

111

1 0

0

V

VS

V

( V sng ti2

V chnh l sng phn x1

V )

1

1 112

2 10

j l

j l

V

V VS e

V V e


9/18

2

2 121

1 10

j lj l

V

V V eS e

V V

1

222

2 0

0

V

VS

V

( V sng phn x2

V chnh l sng ti1

V )

Bi 4.11: Hai mng 2 cng c ma trn tn x l AS vBS . Chng minh tham

s21

S khi ni tng 2 mng trn l: 21 2121

22 111

A B

A B

S SS

S S

Ta c:11 A

yx

VVS

VV

v

2 2

y xBV V

SV V

2

221

1 0V

VS

V

. Khi 2 0V th: 2 21

B

xV S V v 11

B

y xV S V

21 1 22 21 1 22 11

A A A A B

x y xV S V S V S V S S V

22 11 21 1 22 11 2 21 21 11 1A B A A B B AxS S V S V S S V S S V

2

2 21 2121

1 22 110 1

A B

A B

V

V S SS

V S S

Bi 4.16: Cho mng 4 cng c ma trn tn x:

0.1 90 0.6 45 0.6 45 0

0.6 45 0 0 0.6 45

0.6 45 0 0 0.6 45

0 0.6 45 0.6 45 0

S

-Mng c tn hao khng?-Mng c thun nghch khng?-RL ti cng 1 khi tt c cc cng khc phi hp? -IL v pha gia cng 2 v 4 khi tt c cc cng cn li phi hp? -Tm h s phn x nhn ti cng 1 khi ngn mch cng 3 v cc cng khc phi

hp?Gii:

-i vi hng 1:2 2 2 2 2 2 2

11 12 13 14 0.1 0.6 0.6 0 0.73 1S S S S

Do mng c tn hao.-Mng khng thun nghch v ma trn S khng i xng.

-Khi cc cng 2, 3, 4 phi hp th h s phn x 11S v phn x ti cc cng khc

bng khng ( phi hp th khng c phn x ).20 log 20 log 0.1 20RL dB

-Khi cng 1 v 3 phi hp th h s truyn qua gia cng 2 v 4 l:

24 42 0.6 45T S S

20 log 20 log 0.6 4.44IL T dB


10/18

Gc pha: 45 -Khi ngn mch cng 3 v phi hp tr khng ti cng 2 v 4:

2 4 0V V ( V cng 2 v 4 phi hp)

3 3V V ( V cng 3 ngn mch )

1 11 1 12 2 13 3 14 4 11 1 13 3V S V S V S V S V S V S V

3 31 1 32 2 33 3 34 4 31 1V S V S V S V S V S V ( V cng 3 b ngn mch nn

33 0S )

11 11 1 13 31 1 1 11 13 31

1

VV S V S S V S S S

V

1 0.1 90 0.6 45 0.6 45 0.36 0.1 0.37 164.5j

Bi 4.19: Mng 2 cng c cc tham s ma trn tn x sau:

11 0.3 0.7S j ; 12 21 0.6S S j ; 22 0.3 0.7S j

Tm cc tham s tr khng tng ng nu tr khng c tnh0

50Z .

11 22 12 21

11 011 22 12 21

1 1 1.3 0.7 0.7 0.7 0.3650

1 1 0.7 0.7 0.7 0.7 0.36

S S S S j jZ Z

S S S S j j

11 2.24 52.24( ) 52.29 87.5 ( )Z j

12

12 21 0

11 22 12 21

2 0.7250

1 1 0.7 0.7 0.7 0.7 0.36

S jZ Z Z

S S S S j j

12 21 44.78( )Z Z j

11 22 12 21

22 0

11 22 12 21

1 1 0.7 0.7 1.3 0.7 0.3650

1 1 0.7 0.7 0.7 0.7 0.36

S S S S j jZ Z

S S S S j j

22 2.24 52.24( ) 52.29 87.5 ( )Z j

Bi 4.24: ng truyn gm 10 0gV V

, 50gZ

, 1 40 30Z j

, bin p 3:1,on dy

4

c 0 75Z , ti 60LZ . Dng ma trn ABCD tm in p LV trn

ti.3 0 0 75 750 2250 225

1 90 301 1 1

0 1 0 0 03 75 225

j jA B j

C D j j

1 2 2 2 L

L L

B BV AV BI A V A V

Z Z

31 10 0 1.34 4.01 10 ( )225750 2250

60

L

L

VV j VB j

A jZ

4.23 1.25 ( )LV mV

Bi 4.25: Tm ma trn [ABCD] theo 2 cch trc tip v ni tng. Tnh trc tip:


11/18

2

1

1

12 0

1

1

I

I ZV Y

A YZIV

Y

2

1 1

12 0V

V VB ZVI

Z

2

1 1

12 0I

I IC Y

IV

Y

2

1

2 0

1

V

ID

I

Ghp ni tng:1 1 0 1

0 1 1 1

A B Z YZ Z

C D Y Y

Bi 4.26: Chng minh ma trn dn np ca 2 mng 2 cng mc song song hnh cth tm c bng cch cng 2 ma trn.

Trng hp 1:Tra bng ta c:

1 1 B

A

YA

Y ; 1

1

A

BY

2

1 2 B

B

A

YC Y

Y ;

1 1 B

A

YD

Y

111

1

A BDY Y YB

; 1 1 1 112 211

1A

B C A DY Y YB B ; 122

1

A BAY Y YB

Vy: 1A B A

A A B

Y Y YY

Y Y Y

Tng t: 2C D C

C C D

Y Y YY

Y Y Y

1 2A B C D A C

A C A B C D

Y Y Y Y Y Y Y Y Y

Y Y Y Y Y Y

Trng hp 2:Tra bng ta c:

11

2

1 Z

AZ

; 12

1C

Z

2

11 1

2

2 Z

B ZZ

; 11

2

1 Z

DZ


12/18

1 1 211 22 21 1 2 12

D Z ZY Y

B Z Z Z

; 1 1 1 1 212 21 2

1 1 2 1

1

2

B C A D ZY Y

B B Z Z Z

1 2 2

2 2

1 2 1 1 2 1

1

2 1 2

2 2

1 2 1 1 2 1

2 2

2 2

Z Z Z

Z Z Z Z Z ZY

Z Z ZZ Z Z Z Z Z

3 33 2

3 3

1 1

1

0 1 1 1

Z ZA B ZY

C D

Z Z

1 2 2

2 2

3 1 2 1 3 1 2 1

1 2

2 1 2

2 2

3 1 2 1 3 1 2 1

1 1

2 2

1 1

2 2

Z Z Z

Z Z Z Z Z Z Z ZY Y Y

Z Z Z

Z Z Z Z Z Z Z Z

Bi 4.28: Tm cc tham s matrn tn x S cho ti ni tip v song song. i vitrng hp ni tip 12 111S S , v 12 111S S i vi trng hp song song. Gi s

tr khng c tnh l0

Z .

Trng hp ni tip:1

0 1

A B Z

C D

0 011

0 0 0

/

/ 2

A B Z CZ D ZS

A B Z CZ D Z Z

0 022

0 0 0

/

/ 2

A B Z CZ D ZS

A B Z CZ D Z Z

012 11

0 0

21 1

2 2

ZZS S

Z Z Z Z

Trng hp song song:1 0

11

A B

C DZ

0 0 0

110 0 0

/

/ 2

A B Z CZ D ZS

A B Z CZ D Z Z

0 0 022

0 0 0

/

/ 2

A B Z CZ D ZS

A B Z CZ D Z Z

012 11

0 0

21 1

2 2

Z ZS S

Z Z Z Z


13/18

Bi 4.30: Dng th tn hiu tm t s cng sut2 1

/P Pv3 1

/P P vi mng 3

cng c ma trn tn x nh sau:

12

12 23

23

0 0

0

0 0

S

S S S

S

Theo ma trn tn x th khng c tn hiu truyn t cng 1 sang cng 3 v ngc li ( v

13 31 0S S ), ti cc cng khng c phn x ( v 11 22 33 0S S S ). Ch c tn hiu

gia cng 2 v 3, gia cng 1 v 2. Do th tn hiu:

Suy ra:2

2 121 12

2 3 231

Sb a

S

2 122 12

2 3 231

Sa a

S

122 12

2 3 231

Sb a

S

3 12 233 12

2 3 231S Sb a

S

12 233 12

2 3 231

S Sa a

S

Ta c:

2 42 2 22 12

1 1 1 122

2 3 23

1 11

2 2 1

SP a b a

S


14/18

2 22 2 22 122 2 2 22 12 12

2 2 2 1 12 2 22 2 2

2 3 23 2 3 23 2 3 23

11 1 1

2 2 21 1 1

SS SP a b a a

S S S

2 2 2 2 22 2 212 23 3 12 23

3 3 3 12 22 2

2 3 23 2 3 23

1 1

2 2 1 1

S S S S

P a b aS S

2 22 22 122 2 2

2 22 22 2

1 1 12 3 23 2 12

1

1

SP a b

P a b S S

2 2 22 23 12 233 3 3

2 22 22 2

1 1 12 3 23 2 12

1

1

S SP a b

P a b S S

CHNG 5:Bi 5.3: Tr khng ti 200 160

LZ j phi hp vi ng truyn 100 dng

on dy chm song song h mch.Tr khng ti chun ha: 1 1.6Lz j

Bi 5.7: Cho ti 200 100LZ j phi hp vi ng truyn 40 dng on dy

chiu di l c tr khng c tnh1

Z . Tm l v1

Z .

11 0

1

40LinL

Z jZ tg lZ Z Z

Z jZ tg l

1 1 1 1200 100 40 4000 8000Z jZ Z tg l Z tg l j tg l Cn bng phn thc v phn o:

1 1

1 1

200 40 4000100 8000

Z Z tg lZ Z tg l tg l

1

1 1

25

100 8000

Z tg l

Z Z tg l tg l

1 1

2

25 25

100 25 320 4.1

Z tg l Z tg l

tg l tg l

Chn: 14.1 102.5tg l Z

Suy ra: 76.3 103.7 0.288l l

Bi 5.13: Thit k b bin i / 4 phihp ti 350 vi ng truyn 100 .

Tnh0

f

f

bit SWR 2 . Tn s 0 4f GHz .

Tr khng c tnh: 1 0 187.08LZ Z Z

1 2 1 1

1 2 1 3m

SWR

SWR


15/18

01

20 0

242 cos 0.71

1

Lm

Lm

Z Zf

f Z Z

hay 71%.

Bi 5.16: Thit k b ghp 4 khu phi hp ti 10 vi ng truyn 50 . Tnh

0

f

f

bit 0.05

m

.

04; 10 ; 50LN Z Z

0

1

0 0

12 ln 0.05

2

N L L

N

L

Z Z ZA

Z Z Z

1/

1

0

4 12 cos 0.67

2

N

mf

f A

hay 67%.

4 4 4 4

0 1 2 31; 4; 6; 4C C C C

n=0: 41 0 0 1

0

ln ln 2 ln 3.81 45.15N LZZ Z C Z

Z

n=1: 42 1 1 20


Z

n=2: 43 2 2 30


Z

n=3: 44 3 3 4

0

ln ln 2 ln 2.41 11.13N L

ZZ Z C Z

Z

Bi 5.18: Tnh0

f

f

cho b ghp N=1, 2 v 4 khu khi

0

1.5 6LZ

Z v 0.2m .

0

1

0 0

12 ln2

N L L

N

L

Z Z ZAZ Z Z

1/

1

0

4 12 cos

2

N

mf

f A

0/LZ Z N=1 N=2 N=4

A

0

f

f

A

0

f

f

A

0

f

f

1.5 0.101 1.821 0.051 1.821 0.013 1.822

2.0 0.173 0.785 0.087 1.095 0.022 1.339

3.0 0.275 0.474 0.137 0.826 0.034 1.136

4.0 0.347 0.372 0.173 0.723 0.043 1.050

6.0 0.448 0.287 0.224 0.627 0.056 0.965

CHNG 7:Bi 7.2: Tm nh hng, ghp, cch ly, RL ti cng vo khi tt c cc cngcn li phi hp ca mng 4 cng (Directional Couplers) c ma trn tn x sau:


16/18

0.05 30 0.96 0 0.1 90 0.05 90

0.96 0 0.05 30 0.05 90 0.1 90

0.1 90 0.05 90 0.05 30 0.96 0

0.05 90 0.1 90 0.96 0 0.05 30

S

2 2 23 13 1 4 14 1 2 12 1; ;P S P P S P P S P

nh hng:2

3 13 13

4 14 14

10lg 10lg 20lg 6.02P S S

D dBP S S

ghp: 1132

3 13

110lg 10lg 20lg 20

PC S dB

P S

cch ly: 1 1424 14

110lg 10lg 20lg 26.02

PI S dB

P S

Tn hao quay ngc: 1120lg 20lg 26.02RL S dB

Bi 7.4: Ngun pht 4W vo mng 4 cng c C=20dB; D=35dB, tn hao chnIL=0.5dB. Tm cng sut ra ( bng dBm ) ti cc cng.

i1

Pra dB: 110lg 6.021

PP dB dB

W

12 1

2

10lg 6.02 0.5 5.52 35.52P

IL P P IL dB dBmP

2 3.56P W

13 1

3

10lg 6.02 20 13.98 16.02P

C P P IL dB dBmP

3 39.99P mW 1

4 1

4

10lg 6.02 35 28.98 1.02P

D P P IL dB dBmP

4 1.26P mW

Bi 7.6: Mch suy gim tr tnh T v . Nu u vo v u ra phi hp vi0

Z , v

t s in p ra v in p vo l , tm cc phng trnh thit kcho1

R v2

R .

Nu 0 50Z , tnh 1R v 2R cho suy gim 3dB, 10dB, 20dB.

+i vi mch hnh T:Da vo bng ta nhn c ma trn truyn [ABCD]:

2

1 11

2 2

1

2 2

1 2

11

R RRA B R R

C D R

R R

i sang ma trn tn x S :


17/18

2 2

1 2 0 1 2 1 0 1 2 00 011 2 2

0 0 1 2 0 1 2 1 0 1 2 0

2/

/ 2

R R Z R R R Z R R ZA B Z CZ DS

A B Z CZ D R R Z R R R Z R R Z

V u vo phi hp vi0

Z nn khng c phn x ti u vo hay 11 0S .2 2

1 2 1 02 0R R R Z

Hay: 2 2

0 1 0 12 2 0 10 1 2 1 2

1 1

22 2

Z R Z RZ RZ R R R R

R R

H s truyn gia u vo v u ra l12

S .

2 0

22 2 2

0 0 1 2 0 1 2 1 0 1 2 0

2 2

/ 2

AD BC R ZS

A B Z CZ D R R Z R R R Z R R Z

2 0

2 2

1 2 0 1 2 1 0

2

2 2

R Z

R R Z R R R Z

2 0 2

2

1 2 0 0 1 2 0

R Z R

R R Z Z R R Z

0 1 0 10 1 2 0 1

1

1 11 1

2

Z R Z RZ R R Z R

R

1 0 1 1 01

2 11

R Z R R Z

2 02

2

1R Z

Cho 0 50Z :

( )dB 1

R 2

R

3 0.71 8.48 143.17

10 0.32 25.76 35.65

20 0.1 40.91 10.10

+i vi mch hnh :Da vo bng ta nhn c ma trn truyn [ABCD]:

22

1

2 2

2

1 1 1

1

21

RR

RA B

C D R R

R R R

i sang ma trn tn x S :

2 2

1 2 1 0 1 2 1 2 0 1 2 1 00 011 2 2

0 0 1 2 1 0 1 2 1 2 0 1 2 1 0

2/

/ 2

R R R Z R R R R Z R R R ZA B Z CZ DS

A B Z CZ D R R R Z R R R R Z R R R Z

V u vo phi hp vi0

Z nn khng c phn x ti u vo hay 11 0S .

2 21 2 1 2 02 0R R R R Z

Hay:

2 2

1 0 1 02 2 2

1 0 1 0 1 0

2 2R Z R ZR

R Z R Z R Z


18/18

H s truyn gia u vo v u ra l12

S .

2

1 022 2 2

0 0 1 2 1 0 1 2 1 2 0 1 2 1 0

2 2

/ 2

AD BC R ZS

A B Z CZ D R R R Z R R R R Z R R R Z

2

1 0

2 2

1 2 1 0 1 2 1 2 0

22 2

R ZR R R Z R R R R Z

1 2 02 2 22

1 1 1 0

2 1 11 1

R R ZR R R

R R R Z

1 0 02

1 0 1 0

21 1R Z ZR

R Z R Z

0 1 0 1 01

2 11

Z R Z R Z

2

2 0

1

2R Z

Cho 0 50Z :

( )dB 1

R 2

R

3 0.71 294.83 17.46

10 0.32 97.06 70.13

20 0.1 61.11 247.5

CHNG 8:Bi 8.4: Tnh tr khng nh v h s truyn ca mng.

Tr khng tng ng on mch LC:2

21 1

1td td

LC j CZ j L Yj C j C LC

Ma trn truyn [ABCD]:2

2

2

2

1 21 0

1 1

0 1 111

1

LCj L

A B j L LCj C

C D j CLC

LC

2

22

1

2

1 2

1 1 2

11

i

LCj L

AB LLCZ LCj C

CD CLC

2

2 2 2

2 2

11

1 2 1 2

1 1

i

BD j L LZ LC

LC j CAC C LC

LC LC

2

2

1 2cosh

1

LCAD

LC

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