Tai Lieu Ly Thuyet Mach

8/9/2019 Tai Lieu Ly Thuyet Mach

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HC VIN CNG NGH BU CHNH VIN THNG

L THUYT MCH

(Dng cho sinh vin ho to i hc txa)

Lu hnh ni b

H NI - 2006


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HC VIN CNG NGH BU CHNH VIN THNG

L THUYT MCH

Bin son : ThS. NGUYN QUC DINH


3/204

LI GII THIU

L thuyt mch l mt trong s cc mn csca k thut in t, vin thng, tng

ho, nhm cung cp cho sinh vin kh nng nghin cu cc mch tng t, ng thi n lc s l thuyt phn tch cc mch s. Vi ngha l mt mn hc nghin cu cc hthng to v bin i tn hiu, ni dung csl thuyt mch (basic circuits theory) ch yui su vo cc phng php biu din, phn tch, tnh ton v tng hp cc h thng in tov bin i tn hiu da trn m hnh cc cc thng s & cc phn t hp thnh in hnh.

Tp bi ging ny ch yu cp ti l thuyt cc phng php biu din v phn tchmch kinh in, da trn cc loi phn t mch tng t, tuyn tnh c thng s tp trung, cth l:

- Cc phn t & mng hai cc: Hai cc thng, c hoc khng c qun tnh nh phnt thun tr, thun dung, thun cm v cc mch cng hng; hai cc tch cc nh cc ngunin p & ngun dng in l tng.

-Cc phn t & mng bn cc: Bn cc tng h thng cha RLC hoc bin p ltng; bn cc tch cc nh cc ngun ph thuc (ngun c iu khin), transistor, mchkhuch i thut ton...

Cng c nghin cu l thuyt mch l nhng cng c ton hc nh phng trnh viphn, phng trnh ma trn, php bin i Laplace, bin i Fourier... Cc cng c, khinim & nh lut vt l.

Mi chng ca tp bi ging ny gm bn phn: Phn gii thiu nu cc vn chyu ca chng, phn ni dung cp mt cch chi tit cc vn cng vi cc th dminh ha, phn tng hp ni dung h thng ha nhng im ch yu, v phn cui cng ara cc cu hi v bi tp rn luyn k nng. Chng I cp n cc khi nim, cc thng scbn ca l thuyt mch, ng thi gip sinh vin c mt cch nhn tng quan nhng vn m mn hc ny quan tm. Chng II nghin cu mi quan h gia cc thng s trngthi ca mch in, cc nh lut v cc phng php cbn phn tch mch in. Chng

III i su vo nghin cu phng php phn tch cc qu trnh qu trong mch. ChngIV trnh by cc cch biu din hm mch v phng php vc tuyn tn s ca hm

mch. Chng V cp ti l thuyt mng bn cc v ng dng trong nghin cu mt s h

thng. Cui cng l mt s ph lc, cc thut ng vit tt v ti liu tham kho cho cng vicbin son.

Mc d c rt nhiu c gng nhng cng khng th trnh khi nhng sai st. Xin chn

thnh cm n cc kin ng gp ca bn c v ng nghip.

Ngi bin son


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THUT NGVIT TT

AC (Alternating Current) ch dng xoay chiu.

ADC (Analog Digital Converter) b chuyn i tng t -s.

DC (Direct Current) ch dng mt chiu.

FT (Fourier transform) bin i Fourier

KTT B khuch i thut ton.

LT (Laplace transform) bin i Laplace.

M4C Mng bn cc.

NIC (Negative Impedance Converter) b bin i trkhng m.


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Chng 1: Cc khi nim v nguyn l cbn ca l thuyt mch

5

CHNG 1

CC KHI NIM V NGUYN L CBN CA LTHUYT MCH

GII THIU

Chng ny cp n cc khi nim, cc thng s v cc nguyn l cbn nht ca lthuyt mch truyn thng. ng thi, a ra cch nhn tng quan nhng vn m mn hc nyquan tm cng vi cc phng php v cc loi cng c cn thit tip cn v gii quyt cc vn. C th l:

Tho lun quan im h thng v cc mch in x l tn hiu.

Tho lun cc loi thng s tc ng v thng ca mch di gc nng lng.

Cch chuyn m hnh mch in t min thi gian sang min tn s v ngc li. Cc thng s ca mch trong min tn s.

ng dng min tn s trong phn tch mch, so snh vi vic phn tch mch trong minthi gian.

NI DUNG

1.1 KHI NIM TN HIU V MCH IN

Tn hiu

Tn hiu l dng biu hin vt l ca thng tin. Th d, mt trong nhng biu hin vt l cacc tn hiu ting ni (speech), m nhc (music), hoc hnh nh (image) c th l in p v dngin trong cc mch in. V mt ton hc, tn hiu c biu din chnh xc hoc gn ng bihm ca cc bin c lp.

Xt di gc thi gian, mc d trong cc ti liu l khng ging nhau, nhng trong tiliu ny chng ta s thng nht v mt nh ngha cho mt s loi tn hiu ch yu lin quan nhai khi nim lin tc v ri rc.

Tn hiu lin tc

Khi nim tn hiu lin tc l cch gi thng thng ca loi tn hiu lin tc vmt thigian.N cn c gi l tn hiu tng t. Mt tn hiu x(t) c gi l lin tc v mt thi giankhi min xc nh ca bin thi gian t l lin tc.

Hnh 1.1 m t mt s dng tn hiu lin tc v mt thi gian, trong : Hnh 1.1a m tmt tn hiu bt k; tn hiu ting ni l mt th din hnh v dng tn hiu ny. Hnh 1.1b mt dng tn hiu iu ha. Hnh 1.1c m t mt dy xung ch nht tun hon. Hnh 1.1d m t tnhiu dng hm bc nhy n v, k hiu l u(t) hoc 1(t):

=

=

=

Rc th mch in khng thi xng c. Nu Ra = 0 th iu kin s l Rb =

Rd v mch tr thnh i xng v mt hnh hc. Cn nu Rc = Ra th Rd = v mch cng trthnh i xng v mt hnh hc.

- nh l Bartlett - Brune

Ni dung: Bn cc i xng v mt hnh hc bao gicng c th thay th bng s cu tngng ( cn gi l hnh X, hnh 5-19). Trkhng ZI bng trkhng vo ca na bn cc i xng

khi ngn mch cc dy dn ni hai na bn cc v cun dy th cp ca bin p 1:1, cn i vi

cc dy dn cho v bin p 1: -1 th phi hmch. Trkhng ZII bng trkhng vo ca na bncc i xng khi hmch cc dy dn ni hai na bn cc v cun dy th cp ca bin p 1:1,

cn i vi cc dy dn cho v bin p 1: -1 th phi ngn mch.

U21U

ZI

ZIIZII

ZBIB

Mng bncc i xng

I1 I2

1 2

124

U U


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Chng 5: Mng bn cc v ng dng

Ni dung nh l Bartlett-Brune c minh ho trn hnh 5-20:

1/2

bncc ixng

ZI

1/2

bncc ixng

ZII

Hnh 5-20: Minh ha cch tnh cc trkhng ca s cu

Trong nh l trn chng ta thy s c mt ca bin p, y l mt trong s cc phn t bn cc

cbn ca mch in. Bin p l tng theo nh ngha l mt bn cc c cch in mt chiu

gia ca vo v ca ra v c h phng trnh c trng:

U n U

In

I

2 1

2 1

1=

=

.

(5-48)

M hnh bin p l tng minh ho trn hnh 5-21a. B phn ch yu ca bin p thc gm hai

cun dy ghp h cm vi nhau, nu b qua in trca cc cun dy th bin p c v nhhnh 5-21b (n l t s vng dy gia cun th cp v scp)

1:n

U2

I2I1

U1

1:n

U2

I1I1

U1

Hnh 5-21bHnh 5-21a

i vi bin p l tng ta c:

Nu n=1 th : (5-49)U U

I I

2

2 1

=

=

1

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Nu n=-1 th: (5-50)U U

I I

2

2 1

=

=

1

Vy bin p 1:1 tng ng vi bn cc c hai dy dn song song hnh 5-22a, cn bin p 1:-1

tng ng vi bn cc c hai dy dn cho nhau nh hnh 5-22b.

U2

I1 I2I2I1

U1U2U1

Hnh 5-22bHnh 5-22a

U2

I2I1

U1

ZI

ZIIZII

ZIHnh 5-23

By gi ta s xt ti quan h gia cc thng s

trong s cu ca bn cc i xng. Nh ta

bit, i vi bn cc i xng ch cn xc nh

hai thng s, chng hn hai thng s l z11 v

z12. Trong s tng ng cu ca bn cci xng (hnh 5-23) ta c:

zU

IZ Z

I

I II11

1

1 02

1

2= = +

=

( ) (5-51)

zU

IZ Z

I

II I12

1

2 01

1

2= =

=

( ) (5-52)

Nh vy suy ra mi quan h ngc li:

ZI = z11 - z12 (5-53)

ZII = z11 + z12 (5-54)

Sau y ta xt mt th d vng dng ca nh l Bartlett-Brune.

Th d 5-4: Hy xc nh cc thng s zij ca mch in hnh 5-24a.

Gii: Theo kt qu tnh c t cc th d trc, ta bit mt s cch gii:

-Cch 1: Tch mch in trn thnh hai mng bn cc thnh phn mc ni tip-ni tip vi nhau.Xc nh cc thng s zij ca cc bn cc thnh phn, sau tng hp li thnh cc thng s zij

ca bn cc.

-Cch 2: Xc nh cc zij trc tip theo nh ngha trong h phng trnh trkhng c tnh cabn cc.

zR R R R R R

R R11

1 2

1

2

2= 1

+ + +

+

( ) ( )

z

R R R R

R R12

2

2 1

1

2

2=

+ +

+

( )

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-By gi ta s dng cch dng nh l Bartlett-Brune gii bi tp ny. Trc ht ta bi

ly mt na bn cc (hnh 5-24b), sau tnh ZI v ZII:

Z Z

RR

RR

R.R

R RI Vngm= =

+=

+

1

1

1

1

2

2

2

RI1 I2

R1

R

U2U1 R2

Hnh 5-24a

R

R1/2R1/2

R

ZV 2R2 2R2

Hnh 5-24b

Z Z R R II Vhm= = + 2 2

z Z ZR.R

R RR R

R R R R R R

R RI II11

1

1

2

1 2

1

1

2

1

2 22

2

2= + =

++ + = 1

+ + +

+( ) [ ]

( ) ( )

z Z Z R R R.R

R RII I12 2

1

1

1

2

1

22

2= = +

+=( ) [ ]

R R R R

R R

2

2 1

1

2

2

+ +

+

( )

Vy kt qu ny hon ton trng vi kt qucch trn.

5.1.6 Bn cc c tiTrong mc ny ta s cp ti cc thng

s ca bn cc khi ni bn cc vo gia

ngun v ti (hnh 5-25). Gi s Z1 l tr

khng ca ngun tn hiu ca 1, cn Z2

l tr khng ca ti ca 2 ca M4C,

trong :

Z2

Z1

E U2

Mng bncc c ti

Hnh 5.25

U1

I1 I2

Z1 =R1+jX1

Z2 =R2+jX2

a. Trkhng vo M4C:

Trkhng vo ca ca 1:

22221

12211

222

211

1

11

aZa

aZa

Zz

zZz

I

UZV +

+=

+

+== (5-55)

Trkhng vo ca ca 2:

11121

12122

111

122

2

22

aZa

aZa

Zz

zZz

I

UZV

+=

+

+== (5-56)

Trng hp ring khi ca 2 b ngn mch hoc hmch th trkhng vo ca 1:

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22

121

a

aZ nmV =

21

111

a

aZ hmV = (5-57)

Tng t nh vy, khi ca 1 b ngn mch hoc hmch th trkhng vo ca 2:

11

122

a

aZ nmV =

21

222

a

aZ hmV = (5-58)

b. Hm truyn tin p ca M4C:

2112222111

2122

.))((

.)(

zzZzZz

zZ

E

UpK

++== (5-59)

Trng hp ring: khi Z1=0, ta c:

222

21

12211

2

211222211

212

1

2

/1.)(

.)(

Zy

y

aZa

Z

zzZzz

zZ

U

UpKu +

=

=+

== (5-60)

Th d 5-5: Cho M4C nh hnh v 5.26a

R1

C U2U1

Hnh 5.26a

R2

+ Xc nh cc thng s aij ca M4C.

+ Vnh tnh c tuyn bin ca hm truyn t

in p)(

)()(

1

2

jU

jUjT = khi u ra M4C c Zt=R2.

+ Nhn xt tnh cht ca mch (i vi tn s).

Gii:

Theo nh ngha, d dng tnh c ma trn thng s truyn t:

+

++

=1

1][

2

1

2

2121

RpC

RR

CpRRRR

A

/T(j)/

R2/(2R1+R2)

0

Hnh 5.26b

Hm truyn t in p c tnh theo biu thc:

CRjRRR

R

aZa

ZjT

t

t

2121

2

1211 2

)(

++

=

=

c tuyn bin nh tnh nh hnh v 5.26b.

Nhn xt: y l mch lc thng thp, vng tn

s thp tn hiu vo v ra ng pha, vng tn s

cao tn hiu ra chm pha so vi tn hiu vo mt

gc /2.

c. Hstruyn t, lng truyn t ca bn cc

Nu t ngun l tng ta c th ly c cng sut ln bt k, th vi ngun khng l tng c

th d dng chng minh cng sut tc dng ln nht ti c th nhn c l:

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PE

R0

2

14=

.(5-61)

Cng sut tiu th trn ti u ra M4C c tnh theo cng thc:

2

2

2

2 R

UP = (5-62)

-H struyn tca bn cc theo nh ngha i vi mch thng:

2 0

2

=P

P>1 (5-63)

T c th rt ra: = =P

P

E

U

R

R

0

2 2

2

12

C th vit li biu thc trn theo hm ca tn s phc p:

1

2

22)(

R

R

U

Ep

= (5-64)

H s truyn t tnh theo cng thc trn ch dng cho cc mch thng, c trng cho mch

in tng qut ngi ta phi s dng thm biu thc ca hm truyn t in p nu mc

trc.

Ta c th vit li h s truyn t cho mch in tng qut:

( )( ). ( ) .

. . .p

z R z R z z

z R R=

+ + 11 1 22 2 12 21

21 1 22 (5-65)

Nh vy h s truyn t v hm truyn t in p t l nghch vi nhau. Trong cc mch

khuych i v tch cc th K(j) ln hn 1, cn trong cc mch thng th (j) ln hn 1. Hs truyn t l mt hm phc v c th biu din theo bt k loi thng s no ca bn cc da

theo bng quan h gia cc thng s.

Xt ring i vi trng hp bn cc i xng, trong trng hp R1 = R2:

( )( ).(

( ).

pZ R Z R

Z Z R

I II

II I

=)+ +

(5-66)

- Lng truyn tc vit di dng lgarit t nhin ca h s truyn t:

g j a( ) ln ln .arg( ) ( ) ( ) jb = = + = + (5-67)

trong a( ) ln = gi l suy gim, o bng Npe (Nu tnh theo xiben th

a( ) . log , = 20 dB); cn b() = arg() gi l dch pha, o bng rad.

d. Cc thng ssng (cc thng sc tnh) ca M4C

Trc ht ta xt ti khi nim phi hp trkhng trong l thuyt ng dy, khi c ngun tcng in p E vi ni tr trong l Zic mc vo ti c trkhng Zt (hnh 5-27a) . c s

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phi hp trkhng m bo khng c s phn x tn hiu th phi tho mn iu kin: Zt =Zi, khi

cng sut trn ti s l:

PE R

R X

i

i i

0

2

2 24=

+

.

( ) Zi

ZtE

Hnh 5-27a

(vi Zi =Ri + jXi ).v h s phn x khi PHTK s l:

rZ Z

Z Z

t i

t i

=

+= 0

Z20

Z10

Zv2=Z20Zv1=Z10E

Hnh 5-27b

M4C

By gi ta xt mng hai ca nh hnh 5-27b.

c s phi hp trn c hai ca (tc khng

c phn x) th cn phi c hai iu kin:

-Vi ti ca 2 l Z20 th trkhng vo ca

1 phi l Z10,

-Vi ti ca 1 l Z10 th trkhng vo ca

2 phi l Z20.

Ni mt cch khc, iu kin c s phi hp trkhng c hai ca l:

(5-68)

=

=

202

10

ZZ

ZZi

trong Z10 gi l trkhng sng ca ca 1 v tnh theo cng thc:

Za a

a a10

11 12

21 22

=.

.(5-69)

v Z20 gi l trkhng sng ca ca 2 v tnh theo cng thc:

Za a

a a20

22 12

21 11

=.

.(5-70)

Khi Bn cc c phi hp trkhng c hai ca th h s truyn t c gi l h s truynt sngv k hiu l 0:

011 10 22 20 12 21

21 10 202=

+ + ( ).( ) .

. . .

z Z z Z z z

z Z Z (5-71)

hay 012 10 22 20 11 21 10 20

10 20

12 21 11 222

=+ + +

= +a Z a Z a a Z Z

Z Za a a a

. . . .

. .. . (5-72)

Lng truyn t lc ny s l lng truyn t sng:

( )g a a a a j a0 0 12 21 11 22 0 0 0 0= = + = + = +ln ln ln .arg( ) jb (5-73)

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trong : a 0 = ln 0 gi l suy gim sng, o bng Npe.

b0 = arg(0) gi l dch pha sng, o bng rad.

e. Mi quan hgia cc loi thng sca bn cc:

Z Z ZV ngm V hm10 1 1= . ; Z Z ZV ngm V hm20 2 2= . (5-74)

thgZ

Z

Z

Z

V ngm

V hm

V ngm

V hm

0

1

1

2

2

= = (5-75)

Trong ZV1ngm: trkhng vo ca ca 1 khi ngn mch ca 2.

ZV1hm: trkhng vo ca ca 1 khi hmch ca 2.

ZV2ngm: trkhng vo ca ca 2 khi ngn mch ca 1.

ZV2hm: trkhng vo ca ca 2 khi hmch ca 1.

Cc thng s sng Z10, Z20, g0 hon ton xc nh bn cc tuyn tnh c thng s tp trung, th

ng v tng h. T cc thng s sng ta c:

zZ

thg11

10

0

= yZ thg

11

10 0

1= 0

20

1011 .chg

Z

Za =

0

2010

12

.

shg

ZZz =

02010

12..

1

shgZZy = 0201012 .. shgZZa = (5-76)

zZ

thg22 20

0= y Z thg22 20 0

1

= 02010

21 ..

1shg

ZZa =

0

10

2022 .chg

Z

Za =

f. Cc thng ssng ca M4Ci xng

Nu l bn cc i xng vi s tng ng l

mch cu (hnh 5-28), khi : ZI

ZIIZII

ZIHnh 5-28

Z Z Z ZV hm V hm I II1 2 12

= = +( )

Z ZZ Z

Z ZV ngm V ngm

I II

I I

1 2 2= = +.

.

I

T suy ra trkhng sngc tnh:

21

1202010 .

a

aZZZZZ III ==== (5-77)

Nu cc trkhng ca mch cu l cc phn ti ngu, ngha l:

Z Z R constI II = =02

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khi Z0 = R0, trkhng sng ca mch cu trong trng hp ny khng ph thuc vo tn s.

H struyn t sngca mch cu c tnh theo cng thc:

0DXI I II II I II

II I I II

Z Z Z Z Z Z

Z Z Z Z=

+ +

( . ).( .

( ). .

)(5-78)

t qZ

Z

I

II

= (5-79)

Khi : 01

1DX

q

q=

+

(5-80)

Mt khc, trong M4C i xng c phi hp trkhng, Z10 = ZV1, do :

2

1

210

20

2

02

.2 U

U

U

E

Z

Z

U

EDX

=== (5-81)

ng thi lng truyn t sng c xc nh theo biu thc:

DXDXDXDX bjaU

Uj

U

U

U

Ug 00

2

1

2

1

2

100 .)arg(.lnlnln +=+===

(5-82)

Th d 5-6: Xc nh cc thng s sng ca mch in hnh 5-29.

Gii: Ta xc nh cc trkhng vo ca 1:XL2=2XL1=1

XC=3U2U1

Hnh 5-29

ZV1ngm=jXL1 nt [jXL2 // (-jXC)] = 7j

ZV1hm=jXL1 nt (-jXC) = -2j

Vy trkhng sng ca 1 l:

Z Z ZV ngm V hm10 1 1= . = 14 .

Tng ti vi ca 2:

ZV2ngm=jXL2 nt [jXL1 // (-jXC)] =7

2

j

ZV2hm

=jXL2

nt (-jXC) = -j

Vy trkhng sng ca 2 l:

Z Z ZV ngm V hm20 2 2= . =7

2

Lng truyn t sng ca mch c tnh theo cng thc:

thgZ

Z

Z

Z

V ngm

V hm

V ngm

V hm

0

1

1

2

2

= = = =7

23 5j , .

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Th d 5-7: Cho mt bn cc i xng c trkhng sng Z0 = 1000, lng truyn t sng

g0 12

= +

j , trkhng ti Zt = 1000. Bn cc mc vo ngun c Em =100V, in trtrong ca

ngun l 1000. Hy tnh in p v dng in ca 2.

Gii: Theo bi, Zt = Zi= Z0 , nh vy bn cc i xng ny c phi hp trkhng c haica. Theo l thuyt phn tch ta c:

gU

Ujb j0

1

2

0 12

= + = +ln

Vy ln lnU

U

E

U

1

2 221= = suy ra U

E

eV2

2

50

2 718 5= = =

. ,,

v b U U0 1 2 2= =

suy ra

U E2 2

=

Vy ta c U ej

2218 5=

, . )

(V

Th d 5-8: Cho M4C nh hnh 5-30, cho bit R = 1n v chun, C = 1 n v chun.

a. Xc nh cc thng s sng ca M4C.

b. Tnh h s truyn t (p) khi mc M4C trn vo ngun v ti vi cc gi tr Ri = Rt = R0 = 1n v chun.

Gii:

a. y l bn cc i xng, nn c th p dng nh l Bartlett-Brune a v bn cc hnh X

vi cc thng s:C C

R R

R/22CU2U1

Hnh 5-30

ZI =[ C // R ]=1

1p +

ZII =[(C nt R) // (C nt R)] =p

p

+ 12

Vy trkhng sng ca bn cc l:

Z Z Z Z ZI II10 20 0= = = . =1

2p

H s truyn t sng c tnh theo cng thc:

01

1DX

q

q=

+

=p p

p p

+ +

+

1 2

1 2(trong q

Z

Z

I

II

= )

b. Trong trng hp ny khng cn s phi hp trkhng nn h s truyn t ca mch c

tnh theo cng thc:

( )( ).( )

( ).

( ).(p

Z R Z R

Z Z R

p p

p

I II

II I

=)+ +

=

+ +

+0 0

02

2 3 1

1

133


136/204


By gita bin i (p) v dng cha cc thnh phn chun:

( )( ).( )

.

( )(/

)

pp p

p

p p

p=

+ +

+=

+ +

+

2 3 1

12

12

11 3

12 2

c tuyn (j) trong trng hp ny gm c mt thnh phn tng ng vi h s k, hai thnhphn ng vi im khng nm trn trc -, v mt thnh phn tng ng vi im cc l cpnghim phc lin hp nm trn trc o.

5.2 MNG BN CC TUYN TNH KHNG TNG H

Tr li h phng trnh c trng ca bn cc tuyn tnh, khng cha ngun tc ng c lp

gm c hai phng trnh tuyn tnh, thun nht:

a11U1 + a12U2 + b11I1 + b12I2 = 0

a21U1 + a22U2 + b21I1 + b22I2 = 0

T hai phng trnh trn ta c th lp nn 6 h phng trnh c tnh. Mi mt h phng trnh

c tnh ca bn cc tng ng vi mt tp thng sc tnh. Trong phn trc ta nghin cu

cc h phng trnh c tnh ca bn cc vi gi thit v s tng h ca mch in. By gita

s xt gc tng qut hn, tc l trong mch c th tn ti cc phn t khng tng h. Lc

ny cc iu kin tng h:

z z g y

a h

12 21 12 21

211

= = =

= =

g y

h b = -1

21 12

12

s khng c tho mn, nh vy mch tng ng ca bn cc khng tng h cn phi xcnh bi bn phn t (tng ng vi bn thng s). a s cc mch khng tng h l tch cc,

do trong phn ny cng s xt mt s phn t tch cc.

5.2.1 Cc ngun c iu khin

Bn cc khng tng h cn c bn phn t biu din, trong c t nht mt phn t khngtng h. C mt loi phn t khng tng h, tch cc c nhc ti trong chng I, l

ngun iu khin. c trng ca ngun iu khin l cc thng s ca n chu siu khin bi

mch ngoi V bn thn n cng l mt bn cc khng tng h. C th n c chia thnh:-Ngun p c iu khin bng p (A-A), hnh 5-31a. Sc in ng ca ngun Eng lin h vi

in p iu khin U1 theo cng thc:

Eng =kU1 (5-83)

-Ngun p c iu khin bng dng (A-D), hnh 5-31b. Trong sc in ng ca ngun Eng

lin h vi dng in iu khin I1 theo cng thc:

Eng =rI1 (5-84)

-Ngun dng c iu khin bng p (D-A), hnh 5-31c. Trong dng in ngun Ing lin h

vi in p iu khin U1 theo cng thc:

Ing =gU1 (5-85)

134


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-Ngun dng c iu khin bng dng (D-D), hnh 5-31d. Dng in ngun Ing lin h vi

dng iu khin I1 theo cng thc:

Ing =I1 (5-86)

5.2.2 Cc s tng ng ca mng bn cc khng tng h, tch cc

Tt c cc loi M4C khng tng h, tch cc u c th biu din tng ng c cha nguniu khin. Ta s biu din s tng ng ca bn cc vi s c mt ca ngun iu khin.

a. S tngng gm hai trkhng v hai ngun iu khin

I2I1

U2y11 y22

I12U2 I21U1U1

Hnh 5-32b

I2I1

U2rI1U1

A-D

I2I1

U2kU1U1

A-A

I2I1

U2gU1U1

D-A

I2I1

U2I1U1

D-D

Hnh 5-31 M hnh ha cc ngun c iu khin

Nu xut pht t h phng trnh trkhng:I2I1

U2

Z11 Z22

Z12I2 Z21I1

U1

Hnh 5-32a

U z I z I

U z I z I

1 11 1 12 2

2 21 1 22

= +

= + 2

2

ta s biu din c s tng ng ca

bn cc nh hnh 5-32a.

Nu xut pht t h phng trnh dn np:

I y U y U

I y U y U

1 11 1 12 2

2 21 1 22

= +

= +

th s tng ng ca bn cc s biu din

c nh hnh 5-32b.

Tng t nh vy cng c th biu din

mng bn cc khng tng h theo h

phng trnh hn hp H nh hnh 5-32c.

I2I1

U2

h11

h22h12U2

h B21BUB1B

U1

135Hnh 5.32c


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b. S tngng gm ba trkhng v mt ngun iu khin

Cc s c thc thnh lp t cc s chun hnh T v hnh bng cch gn ni tipngun in p iu khin vo mt trong ba nhnh ca s hnh T, hoc mc song song ngun

dng iu khin vo mt trong ba nhnh ca s hnh . Nh vy s c rt nhiu cc trng hpc th, nhng trong thc t thng gp l cc s hnh 5-33, tng ng vi cc h phng trnh

trkhng v dn np:

U z I z I z I

U z I z I z I z I

1 11 1 12 2 12 1

2 21 1 22 2 12 1 12

= +

= + 2 2

2

I y U y U y U

I y U y U y U y U

1 11 1 12 2 12 1

2 21 1 22 2 12 1 12

= +

= +

Hnh 5-33

Z11-Z12I1 I2

U2

Theo cc s trn, nu z12 = z21 hoc y12 = y21 th cc s ny li trv dng bn cc tng h bit. Sau y ta xt mt s phn t phn tng h, tch cc.

5.2.3 Mt s bn cc khng tng h, tch cc thng gp:

a. B bin i trkhng m (NIC)

K hiu ca b bin i trkhng m nh hnh 5-34. H phng trnh c trng ca NIC l h

phng trnh hn hp:

(5-87)U kU

I kI

1

2 1

=

=

-Nu k = 1, ta s c: U UI I

1 2

2 1

==

U1Z12

Z22-Z12

(Z21-Z12)I1 (y21-y12)U1

y22+y12

I1 I2

y11+y12U2U1

-y12

INIC

k=1

I1 I2

U1 U2

Hnh 5-34

UNIC

k= -1

I1 I2

U1 U2

136


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theo quy c v du ca bn cc, in p hai ca s cng chiu cn dng in hai ca s

ngc chiu, phn t NIC trong trng hp ny c k hiu l INIC.

-Nu k = -1, ta c:U U

I I

1 2

2 1

=

=

trng hp ny in p hai ca s ngc chiu cn dng in hai ca s cng chiu, phn tNIC vi k=-1 c k hiu l UNIC.

T ta rt ra: [ ] ,Hk

kNIC =

0

0[ ]

/

/G

k

kNIC =

0 1

1 0

i vi NIC cc h phng trnh trkhng v dn np khng c ngha.

Trkhng vo ca 1 khi mc ti ca 2:

ZU

I

kU

I

k ZV11

1

2 2

2

2= = = . t (5-88)

Nh vy NIC ng vai tr l mch bin i trkhng m. Chng hn nu ti l dung khng th

u vo tng ng l dung khng m.

b. Transistor -ICIE

B

CE

Hnh 5-35

IB

Transistorc coi l mt bn cc tch cc. Hnh 5-35 lk hiu chiu dng in trong transistor PNP. Dng

Emitterc phn phi gia Base v Collector, tho mn

h thc:

>

==

=

==

11

)1(

998,098,0

B

C

EB

E

C

I

I

II

I

I

(5-89)

Dng Emitter ch yu c xc nh bi in p UBE , ngoi ra cn ph thuc vo in pCollector, t dng IC cng ph thuc mt t vo in p UCE.

-T cc tnh cht , c th c nhiu cch biu din s tng ng ca transistor, ty thucvo tng iu kin lm vic c th (tuyn tnh/ phi tuyn, tn s cng tc, hay cch mc mch) vyu cu tnh ton m ngi ta s dng s tng ng thch hp. min tn hiu nh, tn s

thp, ngi ta hay dng s tng ng hn hp H vi hai ngun iu khin ( ni trn),

hoc dng s tng ng vt l vi mt ngun iu khin nh hnh v 5-36a.

rECE

I1=IE I2=-IC

U2U1 rB

B

rC

IE

rECE

I1=IE I2=-IC

U2U1 rB

B

rC

rmIE

137Hnh 5-36a Hnh 5-36b


140/204


Trong s ny c ngun dng ph thuc IE . Cc in trtrn s l cc in trvi phn ca

cc thnh phn dng xoay chiu c bin nhm bo on lm vic tuyn tnh v c xcnh bi h cc c tuyn ca transistor. in trrE c gi tr vi m n vi chc m, rB khong

vi trm m, trong khi r

B

C c gi tr cao (t hng trm kn vi M). Ngun dng cng c

thc thay th bi ngun p nh hnh 5-36b, vi eng= rC.IE = rm.IE, trong rm = .rC.

Tu theo cch chn u vo v u ra, c th c ba loi mch khuch i transistor:

-S bazchung (hnh 5-37a). Di y l

ma trn trkhng ca transistor tng ng vitrng hp ny:

rECE

B

I1=IE I2=-IC

U2U1

rB

rC

rmIE

Hnh 5-37a

[Z]r r r

r r r r BCE B B

B M B C

= ++ +

rBCB

E

I1 I2

U2U1 rE

rC

rmIE

Hnh 5-37b

-S Emitter chung (hnh 5-37b). Di y

l ma trn trkhng ca transistor tng ng

vi trng hp ny:

[ ]Zr r r

r r r r r EC

E B E

E M E C M

=+

+

-S collector chung (hnh 5-37c). Di y l

ma trn trkhng ca transistor tng ng:rB

EB

C

I1 I2

U2U1

rE

rC

rmIE

Hnh 5-37c

[ ]Z r r r r r r r r

CC

C B C M

C E C M

= + +

Trong thc t, ty vo ch phn cc

bng cc ngun mt chiu, transistor c th

c ng dng lm cc mch kha,

mch khuch i, mch bin i tn s...

Trong hnh 5-38 l mt th d mch khuchi tn hiu s dng transistor mc Emitter

+E

C3

0

R4R2

C2

1n

C1

R1

Q1

UrRt

R3

Uv

138Hnh 5-38: mch khuch i mc E chung


141/204


chung ghp RC. Vic la chn cc gi tr linh kin bn ngoi m bo sao cho transistor lm vic

trong min khuch i.

Cc ng dng c th ca transistor sc nghin cu chi tit trong cc hc phn k tip.

c. Mch khuch i thut ton:Mch khuch i thut ton l mt trong nhng bn cc khng tng h, tch cc in hnh. Tn

gi ca mch l dng ch nhng mch khuch i lin tc a nng c ni trc tip vi nhau,

c h s khuch i ln, trkhng vo ln v trkhng ra nh, v vi cc mch phn hi khc

nhau th mch khuch i thut ton s thc hin nhng chc nng khc nhau. K hiu v c

tuyn vng hl tng ca mch c v trn hnh 5-39.

Ura

AU0

U-U0U0

Hnh 5-39: K hiu v c tuyn truyn t ca KTT

I2

UU2UraI1

U1

+

A

_

P

N

ch tuyn tnh, mch khuch i vi h s khuch i A>0 s cho in p u ra:U A U A U Ura = = . ( ) 2 1 (5-90)

Nu U1 = 0 th Ura = A.U2 ngha l in p ra ng pha vi in p vo, do u vo (+) c

gi l u vo khng o pha (P).

Nu U2 = 0 th Ura = -A.U1 ngha l in p ra ngc pha vi in p vo, do u vo (-)

c gi l u vo o pha (N).

Mch khuch i thut ton s l l tng khi h s khuch i A bng , dng in cc u

vo bng khng, trkhng vo l , trkhng ra bng khng. Trong thc t h s khuch i camch l mt s hu hn, ng thi ph thuc vo tn s. M hnh ca mch thc t m t trong

hnh 5-40, trong c tuyn tn s ca A c th coi nh c dng gn ng:

139


142/204


A pA

p( ) =

+

0

0

1

(5-91)

A.(UP UN)

P

N

Ura

Zra

Zvao

20lgA,dB

Am

C0

Hnh 5-40: M hnh tng ng KTT v c tuyn tn s hm truyn t ca n

Mch khuch i thut ton c rt nhiu cc ng dng trong thc t cch tuyn tnh v phi

tuyn nh cc b so snh, khuch i cc thut ton x l, lc tch cc, dao ng...

gi cho mch lm vic min tuyn tnh th ngi ta phi tm cch gim mc in p vo (U)sao cho in p ra khng vt qua ngng bo ha dng VH hoc bo ha m VL. iu ny c

th thc hin c nhcc vng hi tip m trong mch.

Th d 5-9: Hy xt chc nng ca mch inhnh 5-41a. Z2

Hnh 5-41a

Z1I1N

U=0UVUra

_

+

Gii:

Nu coi KTT l l tng v lm vic trong

min tuyn tnh th ta c:

U = 0

v khi im N c gi l im t o.

Dng in vo: IU

Z

U

Z

V r

1

1 2

= = a

T ta rt ra: UZ

Z Ura V=

2

1; K p

Z

Z( ) = 2

1

-Nu Z1, Z2 l thun trth chc nng ca mch l khuch i o pha.

-Nu thay Z1 l thun tr, Z2 l thun dung khi hm truyn t ca mch:

K ppCR

( ) = 1

Ngha l mch trn thc hin chc nng ca mch tch phn. Trong min tn s mch ng vai trl b lc thng thp tch cc bc 1.

-Nu thay Z1 l thun dung, Z2 l thun trth:

K p pR ( ) = C

140


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Ngha l mch trn thc hin chc nng ca mch vi phn. Trong min tn s mch ng vai tr

l b lc thng cao tch cc bc 1.

Th d 5-10: Hy xc nh hm truyn t in p ca mch in hnh 5-41b, coi KTT l ltng v lm vic trong min tuyn tnh.

Hnh 5-41b

Z2Z1

N

U=0

UVUra

+

_

I

Gii:

Dng in chy trong nhnh hi tip:

211 ZZ

U

Z

UI raV

+==

Hm truyn t ca mch l:

1

21)(Z

Z

U

UpK

V

ra +==

Nh vy, bng vic thay i tnh cht ca nhnh hi tip m mch c th thc hin c cc thutton ng dng khc nhau. l mt vi th d v tnh a nng ca loi linh kin ny

nghin cu su hn v cc ng dng ca mch khuch i thut ton v transistor, c bit l

cc thng s ca mch tng ng trong mi cch mc, hc sinh cn c thm trong cc gio

trnh v cc ti liu tham kho ca cc hc phn k tip.

5.3 MNG BN CC C PHN HI

Mng bn cc c phn hi l mt dng kt cu ph bin ca cc h thng mch. Trong mt

phn tn hiu ra sc a quay v khng chu vo. M hnh tng qut ca mng bn cc c

phn hi nh hnh v 5-42:

Gi

thit: M4C ban u c h s truyn t h:

Mng bn ccK

Khu h.tip

YX

Xht

Xv

Hnh 5-42: M hnh tng qut M4C c phn hi

VX

YK

= (5-92)

Khu phn hi c h s hi tip:

Y

Xht

= (5-93)

141


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Nh vy, h thng kn (c phn hi) s c h s truyn t mi:

.1 K

K

X

YKht

== (5-94)

Trong trng hp hi tip m (tn hiu hi tip lm suy yu tn hiu vo), khi 11 >

K , tr

s hm truyn t ca h kn s nh hn so vi h h.

Trong trng hp hi tip dng (tn hiu hi ti p lm tng cng tn hiu vo), khi

11 >

K , khi tr s hm truyn t ca h kn s ch ph thuc vo khu hi tip. thng l trng hp hi tip m su.

Nu xt ti kt cu v cc thng s tham gia, ngi ta chia hi tip thnh cc loi sau:

+Hi tip ni tip in p: tn hiu hi tip ni tip vi tn hiu vo v t l vi in p u ra. M

hnh ca n c minh ha nh hnh 5.43a.

+Hi tip ni tip dng in: tn hiu hi tip ni tip vi tn hiu vo v t l vi dng in u

ra. M hnh ca n c minh ha nh hnh 5.43b.

I1

U1

K

I2

I1

U2U1

I2

Iht

Uht U2

I2

U2

Hnh 5.43a

K

I2

I1

U2U1

I2

Iht

Uht U2

I2

U2

I1

U1

Hnh 5.43b

+Hi tip song song in p: tn hiu hi tip song song vi tn hiu vo v t l vi in p ura. M hnh ca n c minh ha nh hnh 5.43c.

K

I2

I1

U2U1

I2

Iht

Uht U2

I1

U1

I2

U2

Hnh 5.43d

K

I2

I1

U2

U1

I2

Iht

Uht U2

I2

U2

Hnh 5.43c

I1

U1

142


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+Hi tip song song dng in: tn hiu hi tip song song vi tn hiu vo v t l vi dng inu ra. M hnh ca n c minh ha nh hnh 5.43d.

5.4 MT SNG DNG L THUYT MNG BN CC

Ni dung chnh phn ny l nhng ng dng da trn l thuyt ca mng bn cc, c bit i suvo cc ng dng ca mng bn cc thng v tng h.

5.4.1 Mng bn cc suy gim

Mng bn cc suy gim c thnh ngha mt cch tng qut l cc mch chia in p chnh xc

m khng lm thay i ni trtrong Ri ca ngun. Mch suy gim phi tho mn cc yu cu sau:

-Mch suy gim phi l bn cc i xng vi trkhng c tnh bng in trtrong ca ngun.

-Kt cu n gin v tnh ton d dng, ng thi khng yu cu dch pha gia tc ng vo v

p ra, ngha l truyn t c tnh:

g = a >0 (5-95)

p ng c yu cu ny th cc phn t ca b suy gim phi l cc thun tr. Cc phn t

ca b suy gim c tnh ton theo cc s chun ca bn cc nh sau:

a. S hnh T(hnh 5-44a):

RR

sha

i

3 = (5-96)R2I1 I2R1

U2U1 R3

Hnh 5-44a

R R Rtha

Rsha

i1 2= = i (5-97)

b. S hnh (hnh 5-44b):

GR shai

3

1= (5-98)

G2

I1 I2

G1 U2U1

G3

Hnh 5-44b

G GR tha R shai i

1 21 1= = (5-99)

Th d 5-11: Hy tnh mch suy gim lm vic vi ngun c in trtrong l Ri=600, suy gimc tnh l 2,75 Npe.

Gii: Theo cc iu kin ca bi ton:

Ri = 600 ; a = 2,75Npe.Vy cc phn t ca mch suy gim theo s hnh T l:

143


146/204


RR

sha sh

i3

600

2 7577= = =

, ;

R RR

tha

R

sha th sh

i i1 2

600

2 75

600

2 75522= = = =

, ,

Tng t bn c th tnh cc phn t ca mch suy gim theo s hnh .

5.4.2 Mng bn cc phi hp trkhng

Khc vi bn cc suy gim, nhim v ca bn cc phi hp trkhng l kt hp vi ngun lm thay i ni tr trong (Ri1) ca ngun thnh gi tr mi (Ri2), hoc ngc li, bin i tr

khng ti thnh trkhng ngun. Do c im ch yu ca bn cc phi hp trkhng l tnh

khng i xng. Ngoi ra, yu cu khi kt hp vi ngun th truyn t c tnh ca n l thun

o:

g = jb (5-100)

Vi cc yu cu ny, cc phn t ca b phi hp trkhng c tnh ton theo cc s chunca bn cc nh sau:

a. S hnh T(hnh 5-45a):

Z jR R

b

i1 i

3

2= .

sin(5-101)

Z2I1 I2Z1

U2U1 Z3

Hnh 5-45a

Z jR R

b

R

tgb

i i i

1

1 2 1= (sin

) (5-102)

Z jR R

b

R

tgb

i i i

2

1 2 2= (sin

) (5-103)

b. S hnh (hnh 5-45b):

Y jR R bi i

3

1 2

1=

sin(5-104)

Y2

I1 I2

Y1 U2U1

Y3

Hnh 5-45b

Y jR R b R tgbi i i

1

1 2 1

1 1= (

sin) (5-105)

Y jR R b R tgbi i i

2

1 2 2

1= (

sin)

1(5-106)

Th d 5-12: Hy tnh mch phi hp trkhng gia ngun c in tr trong l 5000 v ti

75. Gi sin p in p ra chm pha hn in p vo 450.

Gii: Theo cc iu kin ca bi ton ta c:

Ri1 =5000; Ri2 =75; b = /4 [rad/s].

Vy cc phn t ca mch phi hp trkhng theo s hnh T l:

144


147/204


Z jR R

bj j

i i

3

1 2 5000 75

2 2866= = =

.

sin

.

/

Z jR R

b

R

tgbj j

i i i

1

1 2 1 5000 75

2 2

5000

14134= = = (

sin) (

.

/)

Z jR R

b

R

tgbj j

i i i

2

1 2 2 5000 75

2 2

75

1791= = =(

sin) (

.

/)

Tng t bn c th tnh cc phn t ca mch phi hp theo s hnh .

5.4.3 Mch lc thng LC loi k

a. Khi nim chung

Mi mch c cha cc phn tin khng sao cho trkhng ca n ph thuc vo tn su c

th coi nh c tnh cht chn lc i vi tn s. Mt cch nh tnh c thnh ngha mch lc tn

s l nhng mch cho nhng dao ng c tn s nm trong mt hay mt s khong nht nh (gil di thng) i qua v chn cc dao ng c tn s nm trong nhng khong cn li (gi l di

chn). V mt kt cu, mch lc tn s l tng l mt bn cc c suy gim c tnh tho mn:

=

chandaitrong

thongdaigtron0)(a (5-107)

Hay ni mt cch khc, h s truyn t in p ca mch lc tn s tho mn:

==

chandaitrong0

thongdaigtron1)(

1

2

U

UK (5-108)

c tnh tn s K( ) ca mch lc l tng biu th trong

hnh 5-46. Vi mch lc th ng, tnh cht chn lc l

tng ch c thc hin khi cc phn t xy dng nn

mch l thun khng, ng thi ti phi hp trong di thng

l thun tr. Chng ta s xt cc mch lc m s ca n

c dng hnh ci thang nh hnh 5-47a, kt cu ny gip cho

mch lc lm vic n nh do n c s dng rt rng

ri trong thc t.

/K()/

1

2

1 di thng

Hnh 5-46

Za Za Za

Zb ZbZbZb

T

Hnh 5-47a

phn tch mt mch lc phc tp, thng dng phng php ct thnh nhng on nhngin theo cc s hnh T hoc hnh , hnh thun hoc hnh ngc (hnh 5-47b) kt ni vinhau theo kiu dy chuyn.

145


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Za/2Za/2

Zb 2Zb 2Zb

Za Za/2

2Zb

Za/2

2Zb

Hnh 5-47b

Cc s hnh T v hnh thng c s dng nghin cu v mt l thuyt mch lc. Ccthng sc tnh ca hai loi s ny c tnh theo cc cng thc:

a

bad

Z

ZZTZ

41

2)( += (5-109)

a

bbd

Z

ZZZ

41

12)(

+

= (5-110)

a

b

a

b

T

Z

Z

Z

Z

gth2

1

41

,+

+

= (5-111)

b. iu kin di thng ca mch lc

Vi kt cu cc phn t to thnh Za, Zb cho, cn xc nh iu kin v di thng (hay dichn) ca mch lc. Trong di thng ta phi c:

a

g jb

=

=

0

Rt ra hai iu kin trong di thng:

Th nht: Cc phn t Za, Zb l thun khng.

Th hai: v)(TZd )(dZ phi thun tr.

v iu kin ny s tng ng vi:

a

b

Z

Z41 hay 0

41

b

a

Z

Z(5-112)

y l iu kin di thng ca mch lc c kt cu hnh ci thang.

Ti tn sc ca mch lc, ta s c:

1)(

)(4 =

ca

cb

Z

Z

(5-113)

c. Mch lc loi k

Mch lc loi k l loi mch lc thun khng ni trn c cc phn t tho mn iu kin:

146


149/204


Za.Zb = k2 (5-114)

(trong k l mt hng s thc)

tho mn iu kin trn, n gin nht l chn cc nhnh Za, Zb l cc phn t thun khng m

trkhng c tnh cht ngc nhau. Sau y ta xt c th loi mch lc ny.

d. Cu trc ca mch lc loi k- Mch lc thng thp:

Za = jLa ;b

bCj

Z

1= (5-115)

La/2 La/2

Cb

La

Cb/2Cb/2

Hnh 5-48

Hnh 5-48 m t mt mt lc hnh T v hnh ca mch lc thng thp.

Tn s ct ca mch lc c xc nh theo cng thc:

14

)(

)(4

2==

baca

cb

CLZ

Z

Rt raba

c CL

2

= (5-116)

- Mch lc thng cao:

a

aCj

Z

1= ; Zb = jLb (5-117)

Hnh 5-49 m t mt lc hnh T v hnh ca mch lc.

2Ca Ca2Ca

2Lb2LbLb

Hnh 5-49


14)(

)(4 2 == ab

ca

cb CLZ

Z

Rt ra

ab

cCL2

1= (5-118)

147


150/204


- Mch lc thng di:

)1

(a

aaC

LjZ

= ;)

1(

1

b

b

b

LCj

Z

= (5-119)

tho mn iu kin ca mch lc loi k, cn c:

0

11==

bbaa CLCL(5-120)

Hnh 5-50 m t s mch lc.

Ca


2

0

0

4

)(

)(4

=

ab

baa

b

CL

CLZ

Z

tb

a

a

b

C

C

L

Lp == (5-121)

Rt ra ppc 1(02,1 += (5-122)

Di thng ca mch lc thng di: c1c2

V ta c quan h sau:

==

=

pCL ba

cc

cc

012

2

021

22

(5-123)

- Mch lc chn di:

)1

(

1

a

a

a

LCj

Z

= ; )1

(b

bbC

LjZ

= (5-124)

Tng t tho mn iu kin ca mch lc loi k, cn c thm iu kin:

011 ==

bbaa CLCL(5-125)

CbCb Cb

La

Hnh 5-50

Lb

LaCa

LbLb

148


151/204


Hnh 5-51 m t s mch lc chn di.


2

0

0

4)(

)(4

=

ba

ab

a

b

CL

CL

Z

Z

Cng tb

a

a

b

C

C

L

Lp == ; v

pp

16

1' = (5-126)

Rt ra''

02,1 1( ppc += (5-127)

Di thng ca mch lc thng di c hai khong: c1 ; c2

V ta cng c quan h:

==

=

'

012

2

021

22

1p

CL abcc

cc

(5-128)

e. Tnh cht ca mch lc loi k

Ta s xt trkhng c tnh v truyn t c tnh ca tng loi mch lc.

- i vi mch lc thng thp

* Xt trkhng c tnh ca mt lc hnh T (hnh 5-52a):

2

2

12

412

)( ca

a

bad

LjZZZTZ =+=

-Trong di chn ( > c): Z(T) mang tnh in cm.

Cb

Hnh 5-51

Lb

Cb Cb

Lb

LaLa

CaCa

Lb

La/2 La/2

Cb

Hnh 5-52a

-Trong di thng ( < c): Z(T) mang tnh in tr vc tnh theo cng thc: Z (T)

L

C

a

b

R()

c0

Hnh 5-52b

)(1.)(2

2

R

C

LTZ

cb

ad ==

S ph thuc ca Z(T) theo tn sc biu th tronghnh 5-52 b.

149


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* Xt trkhng c tnh ca mt lc hnh (hnh 5-53a):

2

2

1

12

41

12)(

cb

a

b

bdCj

Z

ZZZ

=

+

= La

Cb/2Cb/2

Hnh 5-53a-Trong di chn ( > c): Z() mang tnh in dung.

-Trong di thng ( < c): Z() mang tnh in tr vc tnh theo cng thc:

Z ()

L

C

a

b

R()

c0

Hnh 5-53b

)(

1

1.)(

2

2

R

C

LZ

c

b

ad =

=

S ph thuc ca Z() theo tn sc biu th trong

hnh 5-53 b.

* By gita xt sang truyn t c tnh:

-Trong di thng ( < c): suy gim c tnh a =0, khi:

2

2

2

2

,

21

1

.

c

c

T tgbjgth

== hay

2

2

2

2

21

1

c

c

tgb

=

-Trong di chn ( > c): in p trn ca ra gim nh mt cch ng k sao cho lc khngcn ti s dch pha gia n vi in p vo. Ngi ta quy c l b gi nguyn gi tr ca n

ti c, sao cho sang di chn tgb =0 v thg=tha. Khi :

(b)

(a)

aTT, bTT

c0

Hnh 5-54

2

2

2

2

21

1

c

c

artha

=

Hnh 5-54 biu din s ph thuc ca a v btheo tn s trong cc di khc nhau.

- i vi mch lc thng cao

* Xt trkhng c tnh ca mt lc hnh T (hnh 5-55a):

2

2

12

141

2)(

caa

bad

CjZ

ZZTZ

=+= 2Ca2Ca

Lb

Hnh 5-55a

-Trong di chn ( < c): Z(T) mang tnh in dung.

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-Trong di thng ( > c): Z(T) mang tnh in trv c tnh theo cng thc:

)(1.)(2

2

R

C

LTZ c

a

bd ==

Z (T)

LCa

b

R()

c0

Hnh 5-55b

S ph thuc ca Zd(T) theo tn s c biu thtrong hnh 5-55 b.

* Xt trkhng c tnh ca mt lc hnh (hnh 5-56a):

2

2

1

12

41

12)(

c

b

a

b

bd Lj

Z

ZZZ

=

+

=

-Trong di chn ( < c): Z() mang tnh in cm. Ca

2Lb2Lb

Hnh 5-56aZ ()

L

C

b

a

R()

c0

Hnh 5-56b

-Trong di thng ( > c): Z() mang tnh in trv c tnh theo cng thc:

)(

1

1.)(

2

2

R

C

LZ

ca

bd =

=

S ph thuc ca Zd() theo tn s c biu thtrong hnh 5-56 b.

* By gita xt sang truyn t c tnh:

-Trong di thng ( > c): suy gim c tnh a =0, khi :

2

2

2

2

,

21

1

.

c

c

T tgbjgth

== hay

2

2

2

2

21

1

c

ctgb

=

-Trong di chn ( < c): ngi ta cng quy c b gi nguyn gi tr ca n ti c, sao cho sangdi chn tgb =0 v thg= tha. Khi :

(b)

(a)

-

aTC, bTC

c0

Hnh 5-57

2

2

2

2

21

1

c

cartha

=

Hnh 5-57 biu din s ph thuc ca a v b theo tn s trong cc di khc nhau.

- i vi mch lc thng di

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Xt mt lc hnh T v hnh ca mch lc thng di (hnh 5-58):

Hnh 5-58

CbLb

La/2La/2 2Ca2Ca

Cb /2 Cb/22Lb

LaCa

2Lb

Do vic tnh ton kh phc tp, nn y khng thc hin tnh ton trc tip m ch da vo tnhcht tng ng ca n i vi cc mch lc thng thp v thng cao trn cc on tn s khc

nhau. C th l:

-Trn on > 0 : nhnh Za mang tnh in cm, cn Zb mang tnh cht in dung, do mchlc thng di s tng ng nh mt mch lc thng thp.

Hnh 5-59a

L

C

a

b

Z (T)

R()

c20c10

thng thpthng cao

L

C

a

b

Z ()

R()

c20c10

thng thpthng cao

-Trn on < 0 : nhnh Za mang tnh in dung, cn Zb mang tnh cht in cm, do mchlc thng di s tng ng nh mt mch lc thng cao. Hnh v 5-59 biu din s ph thucca cc thng sc tnh ca mch lc thng di theo cc di tn s khc nhau.

Hnh 5-59b

(b)(a)

-

aTD, bTD

0 c2c10

Thng cao Thng thp

- i vi mch lc chn di

Xt mt lc hnh T v hnh ca mch lc chn di (hnh 5-60):

152


155/204


Cb

La/2La/2

2Ca2Ca

Lb

Hnh 5-60

Cb/2 Cb/2

2Lb

La

Ca

2Lb

Tng t nh mch lc thng di, da vo tnh cht tng ng ca mch lc chn di i vi

cc mch lc thng thp v thng cao trn cc on tn s khc nhau. C th l:

-Trn on > 0 : nhnh Za mang tnh in dung, cn Zb mang tnh cht in cm, do mchlc chn di s tng ng nh mt mch lc thng cao.

-Trn on < 0 : nhnh Za mang tnh in cm, cn Zb mang tnh cht in dung, do mchlc chn di s tng ng nh mt mch lc thng thp.

Hnh v 5-61 biu din s ph thuc ca cc thng sc tnh ca mch lc chn di theo cc ditn s khc nhau.

Hnh 5-61a

Z (T)

L

C

a

b

R()R()

0 c2c10

Thng caoThng thp

Z ()

L

C

a

b

R()

0 c2c10

Thng caoThng thp

R()

(a)

(a)

(b)

aCD, bCD

c1

c200

Hnh 5-61b

(b)-

Thng thp Thng cao

153


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Trn y ta xt cc tnh cht ca b lc loi k, trong cc thng sc tnh c nh ngha

da vo iu kin phi hp trkhng c hai ca. Nhng iu kin ny li rt kh thc hin, bi

v thng thng trkhng ti v ni khng ca ngun c gi tr l thun trcnh, hay nu c

ph thuc tn s th cng theo quy lut ring ca n. Trong khi trkhng c tnh ca mch

lc loi K cho d c tnh cht thun trtrong di thng nhng vn b ph thuc kh nhiu vo tn

s. V vy nhc im ca loi b lc ny l trkhng c tnh v s truyn t tn hiu bnhhng nhiu bi tn s.

Th d 5-13: Tnh cc phn t ca mch lcthng th p loi k c di thng t 0 n 1000Hz,

tr khng c tnh u di thng l 600. V

khu T v ca mch lc.

La=19mH

Cb/226,5nF

Cb/226,5nF

La/2

9,5mH9,5mH

La/2

Cb=53nF

Hnh 5-62

Gii: Theo cc gi thit ta c:

f

L C

ZL

C

c

a b

a

b

= =

= =

11000

0 600

d ( )

Rt ra

C n

L m

b

a

= =

= =

1

6 1053

6 1019

6

2

. .

.

F

H

Cc s mt lc thng thp c vhnh 5-625.4.4 Mch lc thng LC loi m

khc phc nhc im ca b lc loi k, ngi ta ci tin mt bc v mt kt cu t

c cht lng cao hn. Cc mch lc c gi l mch lc m.

a. Cc phng php xy dng b lc loi m

xy dng b lc m, ngi ta dng cc phng php chuyn t b lc loi k.

- Chuyn ni tip: Bao gm cc bc nh sau:

+Chn khu cbn hnh T v tnh ton da vo trkhng ca nhnh.+Gi li mt phn trn nhnh ni tip, sao cho trkhng ca n trthnh:

aa ZmZ .' = (vi m


157/204


a

ba

Z

ZZ 41

2+

a

ba

mZ

ZmZ '41

2+= (5-131)

Rt ram

ZZ

m

mZ bab +

=

4

1 2'(5-132)

Za/2Za/2

Zb

Za/2 = mZa/2Za

/2 = mZa/2

Zb

Hnh 5-63

Z Zd T d TK M( ) ( )=

Khu lc m c xy dng bng cch ny gi l khu lc m ni tip. N cng c kt cu hnh T.

Hnh 5-63 m t qu trnh chuyn ni tip va trnh by trn.

- Chuyn song song: Bao gm cc bc nh sau:

+Chn khu cbn hnh v tnh ton da vo dn np ca nhnh.

+Gi li mt phn trn nhnh song song, sao cho dn np ca n trthnh:

bb YmY .' = (vi m


158/204


Khu lc M c xy dng bng cch ny gi l khu lc M song song. N cng c kt cu hnh

. Hnh 5-64 m t qu trnh chuyn song song va trnh by trn.

b. Cc tnh cht ca mch lc loi m

Trong phn trn ta xt cch xy dng mch lc loi M t mch lc loi K, trong cn ch

rng iu kin cn bng trkhng c tnh ca cc khu loi K v loi M s lm cho hai loimch lc s c cng di thng. Tuy nhin iu cha th hin nhng ci thin ca mch lc loi

M so vi mch lc loi K mt cch thuyt phc. By gita hy xt ti cc thng sc tnh ca

mch lc M theo mt cnh nhn khc, trc ht l trkhng c tnh ca mt lc hnh trongcch chuyn ni tip (hnh 5-65).

Hnh 5-65

Za Za Za

Zb ZbZbZb

ZbZb

Zb

Zb

Za=mZa

'

'

2

'

'

''

41

1).

4

1(2

41

12

a

b

ba

a

b

bd

Z

Zm

ZZ

m

m

Z

ZZZ

M

+

+

=

+

=)(

trong nu ch n iu kin cn bng trkhng c tnh ta s c:

+

+

=

+

+

=b

a

a

b

b

a

b

bad

Z

Zm

Z

Z

Z

Z

Z

m

m

ZZ

m

mZ

M

4

11

41

2

41

).4

1(2

22'

)(

hay

+=

b

add

Z

ZmZZ

KM

4

11

2

)(

'

)( (5-137)

Kt qu trn ni ln rng, trkhng c tnh ca b lc loi M trong cch chuyn ni tip cn ph

thuc h s m. iu ny ch ra kh nng, nu chn m thch hp c th lm cho Zd(M) t ph thuc

vo tn s nht.

i vi trkhng c tnh ca mt lc hnh T trong cch chuyn song song (hnh 5-66) ta cng

c:

Yb Yb

Ya

Yb Yb

Ya YaYa

YbYb

Ya

Yb

T T

156Hnh 5-66


159/204


'

'

2'

'

''

''

)

' 4

1.)

4

1(2

14

12

14

12 b

a

ab

b

a

aa

ba

d Y

Y

m

YY

m

mY

Y

YZ

ZZ

Z ++=+=+=M(T

hay

a

ba

b

a

b

a

ab

d

Y

YmY

Y

Y

m

Y

Y

m

YY

m

mZ

.4

11

1.

2

41

41

.

)4

1(2

122

)'

+

+

=

+

+

=M(T

tc l

a

b

Tdd

Y

YmZZ

K

.4

11

1.

2)()

'

+

=M(T (5-138)

Kt qu trn cng ni ln rng, trkhng c tnh ca b lc loi M trong cch chuyn song song

ph thuc h s m.

C th ta xt b lc thng thp, c cc trkhng xut pht t loi K:

Za = jLa;b

bCj

Z

1=

-Theo cch chuyn ni tip s c b lc loi M,

tng ng:

Z ()

L

C

a

b

m=1(loi K)

m=0,6

m=0,4

R()

c0

Hnh 5-67a

+

=

=

b

ab

aa

mCjL

m

mjZ

mLjZ

1

4

1 2'

'

]).1(1.[

1

1.

2

2

2

2

2)

'

c

c

b

a mC

LZd

=M(

Hnh 5-67a l th biu din s ph thuc ca trkhng c tnh mt lc hnh mch lc thngthp ni tip theo gi tr ca m.

-Theo cch chuyn song song s c b lc loi M,

tng ng: Z (T)

L

C

a

b

m=0,4

m=0,6

R() m=1

c0

Hnh 5-67b

=

+

=

bb

a

ba

mCjY

mLjC

m

mjY

'

2' 1

4

1

157


160/204


]).1(1[

1.1.

2

22

2

2

)'

c

cb

ad

mC

LZ

=M(T

Hnh 5-67b l th biu din s ph thuc ca trkhng c tnh mt lc hnh T mch lc thng

thp song song theo m.Nh vy, nu chn m=0,6 th s cc khng c tnh ca cc mt lc nu trn s t ph thc vo tn

s nht. i vi mch lc thng cao cng c kt qu tng t.

By gita xt ti truyn t c tnh (g) ca mch lc loi M, trong ch yu xt n suy gim

c tnh (a). Khu lc M phc tp hn khu lc K, do trn cc nhnh ni tip v song song ca

mch lc c th xy ra cng hng lm hmch Ya hoc ngn mch Zb. Khi suy gim c

tnh s ln v cng, v vy cc tn s cng hng ny c gi . Chng l nghim ca ccphng trnh

04

12

' =+=mYY

mmY aba

hoc 04

1 2' =+

=m

ZZ

m

mZ bab

Hay l 114 2


161/204


Nhn xt:

Trong khong tn s gia c v , suy gim c tnh tng t 0 n . Do dc ca c

tuyn ph thuc vo b rng ca khong (c, ), m b rng ny li ph thuc vo m, t ta cth chn dc ca c tuyn mt cch tu theo m. y l mt u im ln ca mch lc M so

vi mch lc K. Tuy nhin khi i su vo di chn th suy gim c tnh li gim kh nh. y l

nhc im ca b lc M so vi b lc loi K.

5.4.5 B lc thng LC y

a. Nguyn tc thit kchung

Nguyn tc tnh ton mt b lc l phi m bo cc yu cu k thut, sao cho cht lng ca n

cng t ti l tng cng tt. Ni mt cch c th:

Khu M

m

Za/Zb/

Khu K

Za/Zb/

c

1/2 khu

M,

PHTK

m=0,6

Z

a / Z

b

1/2 khu

M,

PHTK

m=0,6

Z

a / Z

b

Zd(T,) Zd(T,) Zd(T,)

Ri

Rt

Zd(,T)

E

Hnh 5-69: B lc Lc y

Zd(,T)

-Suy gim c tnh (a) phi hon ton trit tiu trong di thng v rt ln trong ton b di chn.

-B lc phi c phi hp trkhng tt vi ngun v ti.

Trong thc t, p ng y cc yu cu k thut, thng phi xy dng cc b lc phc tp

gm nhiu khu khc nhau v c cc tnh cht b xung cho nhau. Nhn chung mt b lc nh vy

phi c hai khu khng i xng hai u lm nhim v phi hp trkhng vi ngun v ti, v

mt s khu lc i xng loi M hoc K (hnh T hoc hnh ) ni vi nhau theo kiu dy chuyn(hnh 5-69). Sau y ta i su vo cc khu trong b lc:

Khu lc M (i xng) c a vo m bo ra khi di thng suy gim c tnh tng rt

nhanh. Do c tnh cng i su vo di chn th suy gim c tnh ca n cng tng, do Khu

lc K (i xng) c a vo trc khu lc M khc phc nhc im v s gim ca suy

gim c tnh khi i su vo di chn ca khu lc M. Nh vy m bo cc khu ny c cng

di thng v s phi hp trkhng th khu M sc thc hin bng cch chuyn t khu Ktheo cch chuyn tng ng. H s m do tn s suy gim v cng quyt nh.

159


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Hai khu 1/2 M (khng i xng): c t hai u b lc phi hp trkhng gia b lc

vi ngun v ti. Do bn thn nhim v phi hp trkhng dn n n phi c tnh khng i

xng. Mt khc va m bo phi hp vi ngun v ti, ng thi va m bo phi hp u

ni n vi cc khu K v khu M pha trong b lc mt cch bnh thng, ngi ta to ra cc

khu ny bng cch: to ra khu M t khu lc K theo cch chuyn tng ng, vi h s m=0,6,

sau bi khu M va to trn ch gi li mt na. Vi h s m=0,6 th trkhng c tnhca vo v ca ra ca b lc sm bo thun trv n nh, m bo s phi hp trkhng

vi ngun v ti.

Vic ghp ni cc khu trong b lc sao cho nhn t ngoi vo c trkhng c tnh Z()=Ri=Rt

trong trng hp chuyn ni tip (hnh 5-70a) v Z(T)=Ri=Rt trong trng hp chuyn song song

(hnh 5-70b).

}Z b' {2Z b" }2Z b"

Z a'

2

Z a'

2

Z a"

2

Z a

2

Z a

2

Z a"

2

Ri

Rt

E

Hnh 5-70a

Zb

Z()Z

() Z(T) Z(T) Z(T)

Ri

Rt

E

Hnh 5-70b

2Ya 2YaY

a

Ya

Yb /2 b/2Yb/2Yb/2Yb /2 Yb/2

Z(T)Z

(T) Z() Z() Z()

b. Cch tnh ton b lc y

Thng thng cc s liu sau y sc cho trc: Di thng (tn s ct), trkhng c tnh

trong di thng, in trtrong ca ngun v in trti, tn s suy gim v cng, cc yu cu v

suy gim c tnh v phi h p tr khng ... u tin vic tnh ton khu K sc thc hin

trc, sau mi chuyn sang tnh ton cc khu M. Sau y l cc cng vic tnh ton cn thittrn cc loi b lc:

1. B lc thng thp:

- Khu lc K:

=

=

=

===

c

b

c

ba

c

ti

b

a

RC

R

CL

RRRC

L

2

2L

2

a

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-Cc khu lc M:

=2

c2

2-1=m

1

m

c

(Vi khu 1/2M th m = 0,6)

Hnh 5-71 l cu trc ca cc khu (K, M v 1/2M) ca b lc thng thp y trong cc trng

hp chuyn ni tip v chuyn song song.

Nu chuyn ni tip:

=

=

=

bb

ab

aa

mCC

Lm

mL

mLL

'

2'

'

4

1

La/2 La/2

Cb

Hnh 5- 71a

La/2La

/2

LbCb

La/2

2Lb

Cb/2

Nu chuyn song song:

=

=

=

ba

aa

bb

C

m

mC

mLL

mCC

4

1 2'

'

'

La

Cb/2Cb/2

Hnh 5-71b

La

Ca

Cb/2Cb

/2

La/2

Cb/2

2Ca

2. B lc thng cao:

- Khu lc K:

=

=

=

===

c

a

c

ab

c

ti

a

b

RC

R

CL

RRRC

L

2

1

2L

2

1

b

-Cc khu lc M:c

c m 2

22 -1=m1

=

(Vi khu 1/2M th m = 0,6)

Hnh 5-72 l cu trc ca cc khu (K, M v 1/2M) ca b lc thng cao y trong trng hp

chuyn ni tip v chuyn song song.

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Nu chuyn ni tip:

=

=

=

m

CC

Cm

mC

m

LL

aa

ab

bb

'

2

'

'

1

4

2Ca 2Ca

Lb

2C 2Ca

C

L

2Ca

Cb/2

2Lb

Hnh 5-72a

Nu chuyn song song:

=

=

=

m

CC

LmmL

m

LL

aa

ba

bb

'

2

'

'

14

Ca

2Lb2Lb 2Lb

2Lb

La

Ca

2Lb

La/2

2Ca

Hnh 5-72b

3. B lc thng di:

- Khu lc K:bbaa

ccCLCL

11. 21

2

0 ===

ba

ccCL

212 =

RRRCL

CL

ti

a

b

b

a ====

Rt ra21

12a

12 2C

2

cc

cc

cc

aR

RL

=

=

)(

2C

2

)(

12

b

21

12

cccc

ccb

R

RL

=

=

-Cc khu lc M:2

12

212

11

12

mmCL

cc

ba

=

=

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rt ra2

12

12 )(1

=

ccm

(Vi khu 1/2M th m = 0,6)

Nu chuyn ni tip:

m

CmLL aaa ==

'

a

' C

bb

b mCm

LL == 'b1

'

1 C

aab Cm

mL

m

mL

2

'

b2

2'

21

4C

4

1

=

=

Trong hnh 5-73a minh ho cch chuyn ni tip khu lc thng di .

Hnh 5-73a

CbLb

La/2La/2 2Ca2Ca

Cb1

Cb2

La/2La

/2 2Ca2Ca

Lb1

Lb2

Nu chuyn song song:

m

CmLL aaa ==

'

a1

'

1 C

bb

b mCm

LL == 'b

' C

bba Lm

m

Cm

m

C 2'

a2

2'

2 1

4

L4

1

=

=

Trong hnh 5-73b minh ho cch chuyn song song khu lc thng di.

163

Cb/2 Cb/22Lb

LaCa

2Lb

Cb/2 Cb

/22Lb

Ca2

Ca1

2Lb

La2

La1

Hnh 5-73b


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4. B lc chn di:

- Khu lc K:bbaa

ccCLCL

11. 21

2

0 ===

ab

ccCL2

112 =

RRRC

L

C

Lti

a

b

b

a ====

Rt ra21

12b

12

)(2C

)(2 cc

cc

cc

bR

RL

=

=

)(2

1

C

)(2

12a

21

12

cccc

cc

a R

R

L

=

=

-Cc khu lc M: 2122

12 1)(12

1mm

CLcc

ab

==

rt ra2

12

12 )(1cc

m

=

(Vi khu 1/2M th m = 0,6)

Nu chuyn ni tip:

aa

a mLm

CC == 'a

' L

m

LmCC bbb ==

'

b1

'

1 L

aab Lm

mC

m

mC

4

1L

1

4 2'b22

'

2

=

=

Trong hnh 5-74a minh ho cch chuyn ni tip khu lc chn di.

Cb

La/2La/2

2Ca2Ca

Lb

Hnh 5-74a

Cb1

Cb2

La/2La

/2

2Ca2Ca

Lb1

Lb2

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Nu chuyn song song:

aa

a mLm

CC == 'a1

'

1 L

m

LmCC bbb =='

b

' L

bba Cm

mL

m

mL

4

1C

1

4 2'a22

'

2

=

=

Trong hnh 5-74b minh ho cch chuyn song song khu lc chn di.

Cb/2 Cb/2

2Lb

La

Ca

2Lb L

Cb/2 Cb

/2

2Lb

La1

Ca1

Ca2

2Lb

a

Hnh 5-74b

La2

5.4.6 Mch lc tch cc

vng tn s thp, loi mch lc thng LC thng khng thch hp cho cc ng dng thc t

v s cng knh ca cc phn t trong mch v phm cht ca mch b suy gim kh nhiu, thay

vo l cc loi mch lc tch cc RC dng KTT.

a. Khi nim chung:

Hm truyn t tng qut ca mch lc tch cc RC c dng:

n

m

ppaa

ppbbpK

+++

+++=

1-n

1-n10

m

1-m

1-m10

pa+...

b+pb...)( , (n m) (5-140)

Bc ca mch lc l bc ln nht ca mu s (n). Thng thng n c quyt nh bi s lngin dung C trong cc vng hi tip ca mch. i vi mch lc tch cc RC, thng khi hm

mch c bc cng cao th nhy ca cc i lng c trng ca mch i vi phn t tch cc

cng tng mnh, sc ca c tuyn tn s cng tin dn n l tng.

Trong l thuyt tng hp mch, phng php thng dng xy dng mch lc tch cc RC l

phng php phn tch a thc v mc dy chuyn cc khu bc mt v bc 2. Gi s t hm

mch K(p) l phn thc hu t, khi c th phn tch ra thnh tch:

)().()().(

)().(

..)(

)(

)( 12

2

0 pKpFcpbpp

cpbpp

pkpD

pN

pKj j

jjj

i i

iii

r

=+++

+++

==

(5-141)

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-u tin tch ra hm F(p) c th thc hin bng mch thng RC. Trong cc im cc ca

F(p) phi l thc:

F pP p

Q p

P p

p jj

( )( )

( )

( )

( )= =

+

Trong Q(p) cha cc nghim thc l im cc thc ca K(p). Cn P(p) cha mt phn ccnghim ca N(p), v bc ca P(p) nh hn hoc bng bc ca Q(p). Khi F(p) c thc thc

hin bng cc phng php tng hp mch thng. Nu P(p) ch cha cc im khng thc th

c th thc hin bng mch hnh ci thang.

-Cn li K1(p) l t hp cc hm truyn bc hai v sc thc hin bng cc khu bc hai (cha

cc phn t tch cc) vi u im c in trra rt nh.

b. Khu lc tch cc RC bc 2:

Khu lc bc hai c mt ngha c bit quan trng v l khu cbn tng hp cc hm

bc cao bt k. Tng qut, khu lc bc hai tng ng vi hm truyn in p:

2

10

2

210 b)(ppaa

ppbbpKu ++

++= (5-142)

Hm mch ny hon ton c th thc hin c bng mch KTT vi cc vng phn hi v mchRC. Mch phn hi ca KTT c th l mt vng hoc nhiu vng.

M4C(a)

Mch phnhi(b)

C

I1b

I2a _

U1 U2+

Hnh 5-75: Khu lc c mt

-Khu dng phn hi mt vng: Hnh 5-75 m t mt khu tch cc RC c mt vng phn hi

m dng KTT; (a) l mch thng RC; (b) l mch phn hi.

vn hn hi

Vit li hm truyn di dng:

)(

)()(

pD

pNkpKu = (5-143)

Trong h s ca s hng bc cao nht N(p) v D(p) bng 1; D(p) l a thc Hurwitz c cc

nghim na mt phng tri; N(p) khng c nghim trn trc dng c th thc hin mchin c dy t chung. d dng thc hin hm mch bng khu mch bc hai, ngi ta thng

chn mt a thc ph P(p) c cc nghim thc, khng dng v bc i (tng qut, i = max bc N,

bc D) -1; C th chn bc i cao hn, nhng khi s linh kin s tng ln), sao cho:

)(

)()(

)(

)()()(

2

1

pP

pDk

pP

pNk

pDpNkpKu == (1)

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Theo h phng trnh dn np ca mch a ta c:

I2a= y21aU1 + y22aUC = y21aU1 (do C l im t o, UC =0 )

Theo h phng trnh dn np ca mch b ta c:

I1b = y11bUC + y12bU2 = y12bU2

Ch rng I1b = -I2a; v i vi mch thng tuyn tnh y12b=y21b, nn:

K pU

U

y

yu

a

b

( ) = = 2

1

21

21

(2)

T (1) v (2) ta rt ra:

y kN p

P pa21 1= .

( )

( ); y k

D p

P pb21 2=

( )

( );

k

kk1

2

= (5-144)

Nh vy mch a l s thc hin y21a. Mch b l s thc hin y21b. Cn k1 v k2 l cc hng s

sc tm ra khi thc hin mch RC. Cn y21a v y21b phi l cc hm cho php ca mch thng RC. R rng tu thuc vo vic la chn a thc P(p) ta c th c rt nhiu mch RC thc

hin hm truyn t trn. Vic chn mch no l ti u c da theo mt quan im thit k no.

-Khu c phn hi nhiu vng: S hnh 5-76 l mt th d khu bc hai c thc hin vinhiu vng phn hi.

Tu theo vic la chn cc phn t Y1, Y2,...,Y5 ta c th thc hin c hm mch K(p) c cc

chc nng mch khc nhau nh lc thng thp, thng cao, thng di, chn di ... Tuy nhin cu

trc ny khng thc hin c hm phn thc hu t bt k.

Y5Y4

Y1Y2

Y3

_

U1 U2+

Hnh 5-76: Khu lc c phnhi nhiu vng

Th d 5-14:

Xc nh chc nng ca mch in hnh 5-77a.

Gi thit vi mch l l tng v lm vic chtuyn tnh.

-E

+E-

+RU1 UB2

C

C

Hnh 5-77a

R

R

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Gii:

Tnh hm truyn t: Lp phng trnh trng thi ti cc nt theo nh lut Kirchhoff I, t rtra:

+ Trong min p:

22)()()(

222

1

2

++==

RCppCRRCp

pUpUpT

)( jT

1/2

0 0

Hnh 5-77b

+ Trong min :

jRCCR

RCjjT

22)(

222 +=

Gi tr bin :

2222222 4)2()(

CRCR

RCjT

+= , ti

RC

20 = th

2

1max)( =jT .

thnh tnh c dng nh hnh 5-77b. Nh vy y l khu lc tch cc thng di bc 2.

TNG HP NI DUNG CHNG V

c trng cho M4C c th dng cc loi thng s Z, Y, A, B, G, H. Mi loi gm c 4thng s. Vi mng bn cc tng h ta ch cn xc nh 3 thng s.

Cc thng sc tnh ( cc thng s sng) cng hon ton c trng cho M4C chPHTK ti cc ca ca M4C.

Da vo cc thng sc trng ca M4C cng vi ch ca ngun v ti, ta hon ton cth xc nh c cc tnh cht truyn t tn hiu t ngun ti ti thng qua M4C.

Khi phn tch , ngi ta thng trin khai cc M4C thnh cc s tng ng. Mng

tng h thng thng dng s tng ng hnh T, hnh (hoc hnh cu vi M4Ci xng). Mng khng tng h tch cc th vic trin khai thnh cc s tng ng kh

a dng, ty thuc vo iu kin lm vic v di tn cng tc cng vi cc khuyn co ca

nh sn xut.

Cc h thng phc tp chnh l s ghp ni ca nhiu khu li m thnh. Trong tn hiu u ra c thc t chc quay trvu vo nhm thay i cc tnh cht truyn t tn hiu

ca mch hoc to ra cc hiu ng c bit cho mch hoc xy dng nn cc mch to daong.

Tt c cc h thng to v bin i tn hiu u c th phn tch v tng hp da trn l thuytmng bn cc.

CU HI V BI TP CHNG V

5.1 Mng bn cc c cha diode l loi M4C:

a. Thng. c. Khng tng h

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b. Tng h. d. Khng tng h, tch cc.

5.2 Mng bn cc c cha transistor l loi M4C:



5.3 Transistor l loi M4C:



5.4 Mt mng bn cc tuyn tnh, bt bin, tng h tha mn:

a. y12= y21 c. z12= z21

b. a=-1 d. C 3 phng n trn u ng

5.5 Cng thc no di y ng vi M4C c ghp t n M4C n gin theo cch ghp nitip- song song?

a. c.1

n

k

k

YY= =

= 1

n

k

k

ZZ= =

=

b.1

n

k

k

HH= =

= d.1

n

k

k

AA= =

=

5.6 Cng thc no di y ng vi M4C c ghp t n M4C n gin theo cch ghp nitip-ni tip

a. c.1

n

kk

YY= == 1n

k

kZZ= ==

b.1

n

k

k

HH= =

= d.1

n

k

k

AA= =

=

5.7 Mng bn cc tuyn tnh, tng h, thng c th khai trin thnh s tng ng:

a. Hnh T c. Hnh cu

b. Hnh d. C ba phng n u sai

5.8 Mng bn cc tuyn tnh, tng h, thng v i xng c th khai trin thnh s tng

ng:

a. Hnh T c. Hnh cu

b. Hnh d. C ba phng n u ng

5.9 Mng bn cc i xng v s tng ng hnh cu c mi quan h:

a.11 12

11 12

I

II

Z Z Z

Z Z Z

=

= +b.

( )

( )

11

12

1

2

1

2

I II

II I

Z Z Z

Z Z Z

= +

=

c. C hai phng n trn u ng

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5.10 Cc trkhng sng ca M4C c thc tnh theo cng thc:

a.10 2 2

20 1 1

v ngm v hm

v ngm v hm

Z Z Z

Z Z Z

=

= c.

10 1 2

20 2 1

v ngm v hm

v ngm v hm

Z Z Z

Z Z Z

=

=

b.10 1 1

20 2 2

v ngm v hm

v ngm v hm

Z Z Z

Z Z Z = =

d.10 1 2

20 1 2

v ngm v ngm

v hm v hm

Z Z Z

Z Z Z = =

5.11 Trkhng sng ca mng bn cc i xng c thc tnh theo mch tng ng cu:

a.0 10 20

/I II Z Z Z Z Z = = = c. 0 10 20 I II Z Z Z Z Z = = =

b.0 10 20 I II Z Z Z Z Z = = = + d. 0 10 20 I II Z Z Z Z Z = = =

5.12 Xt mt ngun pht c ni trthun Zng=R0 v mt ti thun trZt= R0 . Khi :

a. cng sut trn ti t cc i.

b. khng c s phn x tn hiu t ti v ngun.

c. khng cn thm khu phi hp trkhng gia ngun v ti.

d. tt c cc iu trn u ng.

5.13 Khi tn s tn hiu vo mch lc thng thp tng, in p li ra s:

a. Gim c. Gi nguyn.

b. Tng d. Gn bng in p li vo.

5.14 lc ly di tn Audio (t 0 kHz n 20 kHz) v loi b cc tn s khc, phi s dng loimch lc no ?

a. Thng th p. c. Thng di.

b. Thng cao. d. Chn di.

5.15 V mt kt cu, mch in c hi tip ni tip dng in ph hp vi kiu ghp no?

a. ghp ni ti p-song song c. ghp song song-song song

b. ghp ni tip-ni ti p d. ghp song song-ni tip

5.16 Cho mng bn cc nh hnh v 5-78. Hy xc nh cc thng s hn hp Hij ca mng bncc

R1

R2U1 U2

I1 I2

Hnh 5-78

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5.17 Hy xc nh s tng ng hnh T ca mng bn cc nh hnh v 5-79.

R1M

U1 U2L2L1

**

Hnh 5-79

R2

5.18 Cho mng bn cc nh hnh v 5-80. Xc nh iu kin ca Zng v Zt c s phi hp trkhng trn c hai ca ca M4C.

Z1

U2U1

Hnh 5-80

Z2

R

C U2U1

Hnh 5-81

2R

5.19 Cho bn cc nh hnh 5-81:

a. Xc nh cc thng s yij ca M4C.

b. V nh tnh c tuyn bin v c tuyn pha ca hm truyn t in p

)(

)()(

1

2

jU

jUjT = khi u ra M4C c Zt=2R.

c. Nhn xt tnh cht ca mch (i vi tn s).

5.20 Cho mng bn cc nh hnh 5-82:

L

U2U1

Hnh 5-82

2R

Ra. Xc nh cc thng s aij ca M4C.

b. Vnh tnh c tuyn bin v c tuyn pha ca

hm truyn t in p)(

)()(

1

2

jU

jUjT = khi u ra

M4C c Zt=2R.


R

5.21 Cho bn cc nh hnh 5-83:

L 2R UB2BUB1B

Hnh 5-83

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a. Xc nh cc thng s yij ca M4C.

b. V nh tnh c tuyn bin v c tuyn pha ca hm truyn t in p

)(

)()(

1

2

jU

jUjT = khi u ra M4C c Zt=2R.


5.22 Thit k mch lc thng di loi k bit trkhng c tnh ti tn s trung tm bng 10k,di thng ca mch nm trong khong (10 - 12)kHz.

5.23 Tnh cc phn t ca mch lc thng di M vi cc s liu: Z(0)=600, fc1=10kHz,

fc2=12kHz, f1=9,5kHz, f2=12,8kHz.

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Hng dn tr li

HNG DN TR LI

CHNG 1: CC KHI NIM V NGUYN L CBN

1.1 M hnh ton hc ca mch in trong min thi gian c thc trng bi:

b. Mt h phng trnh vi phn hoc sai phn.

1.2 Hiu qu khi chuyn mt mch in analog t min thi gian sang min tn s l:

d. s thay th h phng trnh vi phn bng mt h phng trnh i s.

1.3 Trkhng ca phn t thun dung l :

b)1

C CZ jX j C

= =

1.4 Trkhng ca phn t thun cm l :

c) L LZ j L jX = = 1.5 Dn np ca phn t thun dung l :

b) C CY j C jB= =

1.6 Dn np ca phn t thun cm l :

d)1

L LY jj L

= = B

1.7 Trkhng tng ng ca on mch hnh 1.45.

a. Z=1-j5

1.8 Trkhng tng ng ca on mch hnh 1.46.

d. Y=5-j5 (S)

1.9 S tng ng ca on mch c trkhng Z= 2+j2 l hnh 1.47.b.

1.10 S tng ng ca on mch c trkhng Z =3-j2 l hnh 1.48.a.

1.11 S tng ng ca on mch c dn np Y=2+j5 (S) l hnh 1.49.b.

1.12 S tng ng ca on mch c dn np Y=3-j5 (S) l hnh 1.50.a.

1.13iu kin phi hp cng sut tc dng trn ti t cc i l:

d. Trkhng ti bng lin hp ca trkhng ngun (Zt =Rng-jXng ).

1.14 Trong mch in RLC ni tip, nu UL ln hn UC th:

a. Mch c tnh cm khng.

1.15 Ti im cng hng ca mch cng hng RLC ni tip:

c. Mch c tnh thun tr, dng vi p l ng pha.

1.16 H s phm cht Q ca mch cng hng RLC ni tip c th tng bng cch:b. Gim R.

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Hng dn tr li

1.17 Trkhng ca mch RLC song song ti tn s cng hng l

b. Cc i v thun tr.

1.18 Mch in hnh 1.51 c (nhiu nht):

d. 3 nt, 5 nhnh

1.19 Trkhng tng ng ca mch: jZtd 33 =

-Bin phc dng in trong mch:oj

m eI15.

2

1=

-in p trn Z1:oj

m eU30

1

=

-in p trn Z2:oj

m eU30

2 2=

1.5 2

Hnh 6-1

1s 1s 1s 1s

1.20 a.

)30sin(26)(1o

ttu =

)15sin(3)(1o

tti +=

)60sin(22

3)(2

otti +=

)30sin(22

3)(3

otti =

b. S tng ng chi tit theo cc tham s c dng nh hnh 6-1.

c. Cng sut tc dng:

P = U.I cos = 9W.

1.21 a.

)60sin(26)(1ottu +=

i1(t) =2.sin(t + 15o)

i2(t) =4/3.sin(t + 15o)

i3(t) =2/3.sin(t + 15o)

b. S tng ng chi tit theo cc tham s c dng nh hnh 6-2, n v l .

r=1

XC=3

XL=5

Hnh 6-2

XC=6

r=3 r=6

0,5

Hnh 6-3

0,5s

0,5s

0.5s

0,5s

1,5

c. Cng sut tc dng:

P = U.I cos = 6W.

1.22 a.

)30sin(22)( ottu =

)15sin(34)( otti +=

173


177/204

Hng dn tr li

)30sin(23

2)(1

otti =

)60sin(23

2)(2

otti +=

b. S tng ng chi tit theo cc tham s c dng nh hnh 6-3.c. Cng sut tc dng:

P = U.I cos = 4/3W.

CHNG II: CC PHNG PHP CBN PHN TCH MCH IN

2.1 Trong mt mch vng khp kn, tng i s cc st p trn cc nhnh:

d. bng khng.

2.2 Nu in p ngun cung cp v st p ca hai phn t bit, st p ca phn t th ba:

c. c th xc nh c bng cch p dng nh lut Kirchhoff vin p.

2.3 Nu tnh ton ca bn cho thy tng i s cc st p trong mt mch vng l khc khng th:

d. tnh ton ca bn cha ng

2.4 Cschnh ca phng php dng in vng da vo :

c. nh lut Kirchhoff vin p

2.5 Nu khi gii mch in thu c dng trong mt nhnh c gi tr m th:

a. Chiu ban u ca n l khng ng.

2.6 Khi phn tch mt mch in c Nn nt v Nnh nhnh bng phng php in p nt, th sphng trnh to ra l:

b. Nn-1 phng trnh c lp

2.7 Khi phn tch mch in tuyn tnh p dng nguyn l xp chng, th:

b. Ln lt ch gi li mt ngun, cc ngun cn li cn c loi b.

2.8 Csphn tch mch bng phng php ngun tng ng da vo :

a. nh l Thevenine- Norton

2.9 Trong mch hnh 2.25, in p ri trn R2.

d. 10 Vdc

2.10 Phng trnh khng ng i vi mch in hnh 2.26:

d. IR2 = I1+I2

2.11 Cc biu thc dng in vng cho mch:

( )

( )1 1 3 2 3 1

1 3 2 2 3 2

V V

V V

I R R I R E

I R I R R E

+ =

+ + =

2.12

174


178/204

Hng dn tr li

a. H phng trnh:

-Xt vng 1: IV1(R1+ZL1+ZC) - IV2.ZC = E1.

-Xt vng 2: -IV1ZC + IV2(R2+ZL2+ZC) =-E2.

b. Dng trong cc nhnh:

IL1 = IV1.

IL2 = IV2.

IC = IV1 IV2

2.13 Dng in trn cc nhnh theo phng php dng in vng:

1

2

3

1

2

1 2

4

3

1

R V

R V

R V V

I I A

I I A

I I I

= =

= =

= = A

2.14 Cc biu thc dng in vng:

( ) ( )

( ) (

1 1 1 2 1

1 2 2 2

V L C V C M

V C M V L C ) 2

I R jX jX I jX jX E

I jX jX I R jX jX E

+ + =

+ + =

2.15 Cc phng trnh vng:

1 1 1 2 1

1 2 2 2

1 1

1 1

V V

V V 2

I R j L I E j C j C

I I R j L E j C j C

+ + + =

+ + + =

2.16 Dng in trong cc nhnh bng phng php in p nt

-PTrnh in p nt:

UA(G1+G2+G3) = Ing1-Ing2

-Thay s tnh c:

UA = -1V.

-Vy ta c:

I1=0,2A.

I2=0,25A.

I3=0,05A.

2.17 Cc phng trnh nt:

1

1 1

2

2 3

1 1 1 1

1 1 1 1

A B

L C C

A B

C C

EU U

2

R jX jX jX R

EU U

jX R R jX

+ + =

+ + + = R

175


179/204

Hng dn tr li

2.18: Cc phng trnh nt:

1

1 2 2 2 2

4

2 2 4 2 2

1 1 1

1 1 1

A B n

ng

A B

U U IR R j L R j L

EU U j C

4

g

R j L R R j L R

+ = + +

+ + +

+ +

=

2.19 Et v Rt ca ngun tng ng:

Et= 5V; Rt = 5 Ohm.

2.20: Trkhng tng ng ca mch Thevenine:

Rt= R1// R2 // R3

2.21 Tnh dng in IR4 theo phng php ngun tng ng:

Nu tnh theo thevenine khi :

=

=

7

12

7

36

tdAB

hmAB

Z

VU

Vy theo s tng ng Thevenine ta c:

ARZ

UI

tdAB

hmABR

13

18

)27

12(7

36

4

4 =+

=+

=

Kt qu tng t nu ta trin khai theo Norton.

2.22 Trkhng tng ng Rt ca mch Thevenine.

Rt= 1250

2.23 Tnh dng in iR2 bng nguyn l xp chng:

Khi Ing1 tc ng (Eng4 b ngn mch): dng in trn R2 tnh c l 2A (chiu t A sang B).

Khi Eng4 tc ng (Ing1 b hmch): dng in trn R2 l 0,5A (chiu t B sang A).

Tng hp: dng in trn R2 l 1,5A (chiu t A sang B).

CHNG III: HIN TNG QU TRONG CC MCH RLC

3.1. Khi mi im cc ca hm mch F(p) nm bn na tri mt phng phc (khng bao hm trco), p ng f(t) s:

a. hi t v 0 khi t.

3.2. Khi mi im cc ca hm mch F(p) nm bn na tri mt phng phc, cng lm nm trntrc o, p ng f(t) s:

d. khng tin n v hn khi t.

3.3. Khi tn ti im cc ca hm mch F(p) nm bn na phi mt phng phc, p ng f(t) s:

176


180/204

Hng dn tr li

d. tin n v hn khi t.

3.4. Lut ng ngt ca cc phn t qun tnh c pht biu :

b. Trong cun dy khng c t bin dng in, trong tin khng c t bin in p,k c ti thi im ng ngt mch.

3.5. Hm gc UC(t) nu bit nh ca n:

b.31 1( )

6 2

t

CU t e= +

3.6. Hm gc iL(t) nu bit nh ca n l:

b.2 3( ) 2 3 2t tLi t e te e

= + + 3t

3.7 Xc nh uC(t):

-iu kin u:

UC(0) =90V.

-Ngt kho K, vit phng trnh cho mch:

)0(.)(

)().1

(11

CC UCR

pEpUpC

R+=+

Kt qu tm c:

44

6

10.5

10100

)10.5(

10.590)(

+=

+

+=

pppp

ppUC

-Chuyn v min thi gian:

t

C etU410.510100)( =

3.8 Xc nh uC(t) ?

-iu kin u:

UC(0) =60V.

-ng kho K. S dng cc phng hc tm nh ca p ng. Kt qu tm c:

)6

10(

10.560)(4

4

+

+=

pp

ppUC


t

C etU6

104

3030)(

+=

3.9 Xc nh uC(t):

-iu kin u:

UC(0) =100V.

-ng kho K. S dng cc phng hc tm nh ca p ng. Kt qu tm c:

177


181/204

Hng dn tr li

18

10

1090)(

6

+

+=

pp

pUC


tC etU

18

106

1090)( +=

3.10 Xc nh uC(t):

-iu kin u:

UC(0) =3V.

-Ngt kho K. S dng cc phng hc tm nh ca p ng. Kt qu tm c:

15

10

36)(

4

+

=

p

ppUC


t

C etU15

104

36)(

=

3.11 Xc nh uC(t):

-iu kin u:

UC(0) =5V.


5

10

510)(

7

+

=

pp

pUC


t

C etU5

107

510)(

=

3.12 Xc nh iL(t):

-iu kin u:

IL(0) =0,5A.


410

6/13/2)(

+=

pppIL


tL etI

410

61

32)( =

3.13 Xc nh iL(t):

178


182/204

Hng dn tr li

-iu kin u:

IL(0) =0,5A.

-Ngt kho K. Lp phng trnh cho mch. Kt qu tm c:

4

10

5,0)(

+

=p

pIL


t

L etI410

2

1)( =

3.14 Xc nh iL(t):

-iu kin u:

IL(0) =1A.


410

4/14/3)(

++=

pppIL


t

L etI410

4

1

4

3)( +=

3.15 Xc nh iL(t):

-iu kin u:

IL(0) =3A.

-Ngt kho K. S dng cc phng php hc tm nh ca p ng. Kt qu tm c:

410

21)(

++=

pppIL


t

L etI41021)( +=

3.16 Xc nh uC(t):

-iu kin u:

UC(0) =20V.

-ng kho K. S dng phng php in p nt hc tm nh ca p ng. Kt qu tmc:

55

5

10

515

)10(

10.1520)(

++=

++

=pppp

ppUC


179


183/204

Hng dn tr li

t

C etU510515)( +=

3.17. Gi thit h khng c nng lng ban u, tc uC(0-)=0:

2

0

2

0.1

/1)().()(

++==

p

CRp

CpXpHpU

Bin i Laplace ngc ta c p ng ra l:

+

+=

t

RCte

CRC

tut

RC000

1

0

22

2

0

sin1

cos

)1

(

1)(

3.18

a. Xc nh dng in i(t) sinh ra trong mch v in p UC(t).

- Trong khong )8.0(;0 x mst x =


184/204

Hng dn tr li

i(t)[mA]

t(ms)

20

-20

Hnh 6-4b

b. Khi phm cht ca mch tng ln 5 ln, lc qu trnh qu ca mch s b ko di hn sovi trng hp xt trn. iu ny lm cho trong cc khong tn ti v trng ca chu k xung,hin tng xy ra trong mch cha t n xc lp, do p ng ca chu k trc s ko dichng ln p ng ca chu k sau, lm mo dng tn hiu mt cch ng k.

3.19

a. Xc nh dng in i(t) sinh ra trong mch v in p UC(t).

- Trong khong )2(0 x mst x =


185/204

Hng dn tr li

b. Mch b lch cng hng:

+Trong giai on 0 xt


186/204

Hng dn tr li

2

2

21

1)(

ii

pppH

++

= , vi v 0


187/204

Hng dn tr li

Nhn xt: Mch lc thng cao. Vng tn s cao tn hiu vo v ra ng pha, vng tn s thp

tn hiu ra nhanh pha so vi tn hiu vo mt gc /2.

CHNG V: MNG BN CC VNG DNG

5.1 Mng bn cc c cha diode l loi M4C:

c. Khng tng h

5.2 Mng bn cc c cha transistor l loi M4C:

c. Khng tng h

5.3 Transistor l loi M4C:

d. Khng tng h, tch cc.

5.4 Mt mng bn cc tuyn tnh, bt bin, tng h tha mn:

d. C 3 phng n.

5.5 Vi M4C c ghp t n M4C n gin theo cch ghp ni tip- song song:

b.1

n

k

k

HH= =

=

5.6 Vi M4C c ghp t n M4C n gin theo cch ghp ni tip-ni tip:

c.1

n

k

k

ZZ= =

=

5.7 Mng bn cc tuyn tnh, tng h, thng c th khai trin thnh s tng ng:

a. Hnh T

5.8 Mng bn cc tuyn tnh, tng h, thng v i xng c th khai trin theo:

d. C ba phng n.

5.9 Vi mng bn cc i xng v s tng ng hnh cu:

c. C hai phng n trn u ng

5.10 Cc trkhng sng ca M4C c thc tnh theo cng thc:

b.10 1 1

20 2 2

v ngm v hm

v ngm v hm

Z Z Z

Z Z Z

=

=

5.11 Trkhng sng ca mng bn cc i xng c thc tnh theo mch tng ng cu:

c.0 10

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Tai Lieu Ly Thuyet Mach