Cso Do Luong Dtu

7/30/2019 Cso Do Luong Dtu

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HC VIN CNG NGH BU CHNH VIN THNG

BI GING MN

CSO LNG IN T

Ging vin: KS. Mnh H

in thoi/E-mail: 0913826568; [email protected]

B mn: K thut in t- Khoa KTDT1

Hc k/Nm bin son: Hc k 1/2009-2010


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www.ptit.edu.vnGING VIN: KS. Mnh HB MN: KTT - KHOA KTT1

Trang 2

Sch tham kho

1. C s k thut o lng in t, V Qu im, nh xut bnKHKT, 2001

2. o lng in-v tuyn in, V Nh Giao v Bi Vn Sng,

Hc vin k thut qun s, 19963. Electronic Test Instruments, Bob Witte, 2002

4. Radio Electronic Measurements, G.Mirsky, Mir Publishers,Moscow, 1978


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Trang 3

NI DUNG

CHNG 1. Gii thiu chung vo lng in t

CHNG 2.nh gi sai so lng

CHNG 3. Cc c cu chth trong my o

CHNG 4. My hin sng

CHNG 5.o tn s, khong thi gian v di pha

CHNG 6.o dng in v in p CHNG 7.o cng sut

CHNG 8. Phn tch ph

CHNG 9.o cc tham s ca mch in


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Trang 4

nh ngha Cc phng php o

Phng tin o v cc c tnh c bn ca phng tin o

Phn loi cc my o

Chng 1. Gii thiu chung v o lng in t


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Trang 5


1.1.nh ngha

o lng: l khoa hc v cc php o, cc phng php v cc cngc m bo cc phng php o t c chnh xc mong

mun

o lng in t: l o lng m trong i lng cn o cchuyn i sang dng tn hiu in mang thng tin o v tn hiu

in c x l v o lng bng cc dng c v mch in t.


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Trang 6


1.2. Cc phng php o1. Phng php o trc tip: dng my o hay cc mu o (cc

chun) nh gi s lng ca i lng cn o. Kt qu o chnhl tr s ca i lng cn o.

VD: o in p bng vn-mt, o tn s bng tn s-mt, o cngsut bng ot-mt,...

c im: n gin, nhanh chng, loi b c cc sai s do tnh

ton2. o gin tip: kt qu o khng phi l tr s ca i lng cn o,

m l cc s liu c s tnh ra tr s ca i lng ny.

V d: o cng sut bng vn-mt v ampe-mt, o h s sng chybng dy o,...

c im: nhiu php o v thng khng nhn bit ngay c kt

qu o

aX =

( )n

aaaFX ,...,,21

=


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Trang 7

3. Phng php o thng k: thc hin o nhiu ln mt i lng ovi cng thit b o v trong cng in kin o, kt qu o c tnh lgi tr trung bnh thngkca cacc ln o .

- c im: cho php loi tr cc sai s ngu nhin v thng dng khi

kim chun thit b o.

4. Phng php o tng quan: dng o cc qu trnh phc tp,khi khng th thit lp mt quan h hm s no gia cc i lng l

cc thng s ca mt qu trnh nghin cu. VD: tn hiu u vo vu ra ca mt h thng Thc hin bng cch xc nh khong thi gian v kt qu ca mt s

thut ton c kh nng nh c tr s ca i lng thch hp.

c im: cn t nht hai php o m cc thng s t kt quo cachng khng ph thuc ln nhau. chnh xc c xc nh bng di khong thi gian ca qu trnh xt.



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5. Cc phng php o khc: Phng php o thay th: Php o c tin hnh hai ln, mt ln

vi i lng cn o v mt ln vi i lng o mu. iu chnh hai trng hp o c kt qu chth nh nhau.

Phng php hiu s: Php o c tin hnh bng cch nh gihiu s tr s ca i lng cn o v i lng mu. (phng phpvi sai, phng php ch th khng, phng php b)

Phng php o thng: kt qu o c nh lng tr c tip trnthanh ca thit b ch th. Tt nhin s khc ca cc thang ny c ly chun trc vi i lng mu cng loi vi i

lng o. Phng php ch th s: i lng cn c o c bin i thnh

tin tc l cc xung r i rc. Tr s ca i lng cn o c tnhbng s xung tng ng ny



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Trang 9

1.3. Phng tin o v cc c tnh c bn1. Phng tin o l phng tin kthut thc hin php o, chng c nhng

c tnh o lng c qui nh.- Phng tin o n gin: mu, thit b so snh, chuyn i o lng- Phng tin o phc tp: my o (dng c o), thit b o tng hp v h thngthng tin o lng.

+ Mu: phng tin o dng sao li i lng vt l c gi tr cho trc vi chnh xc cao. Chun l mu c cp chnh xc cao nht. Chun l phng tin o

m bo vic sao v gi n v tiu chun.+ Thit bso snh: phng tin o dng so snh 2 i lng cng loi xemchng = , > , < + Chuyn io lng: phng tin o dng bin i tn hiu thng tin o

lng v dng thun tin cho vic truyn tip, bin i tip, x l tipv g i linhng ngi quan st khng th nhn bit trc tip c (VD: b Ko lng;bin dng, bin p o lng; quang in tr, nhit in tr,...)



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+ Dng co: phng tin o dng bin i tn hiu thng tin o lng

v dng m ngi quan st c th nhn bit trc tip c (VD: vnmt,ampe mt,...)

+ Thit b o tng hp l cc phng tin o phc tp, a nng dng kim tra, kim chun v o lng cc tham s phc tp.

+ H thng thng tin o lng: H thng mng kt ni ca nhiu thit b o,

cho php o lng v iu khin t xa, o lng phn tn...

Dng c o

Mc tng ha

Dng c okhng t

ng

Dng c ot ng

Dng ca tnhiu

Dng c otng t

Dng co s

Phng phpbin i

Dng co bin

i thng

Dng co bini cnbng

Cc i lngu vo

Dng co dng

in

Dng co tn s

...


Hnh 1.1 S phn loi tng quan thit b o


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Trang 11

2. Ccc tnh cbn ca phng tin o

Cc c tnh tnh: c xc nh thng qua qu trnh chun ho thit b.+ Hm bin i: l tng quan hm s gia cc i lng u ra Y v cc

i lng u vo X ca phng tin o, Y=f(X)+ nhy: l t s gia bin thin ca tn hiu u ra Y ca phng

tin o vi bin thin ca i lng o u vo X tng ng.K hiu:

+ Phm vio: l phm vi thang o bao gm nhng gi tr m sai s chophp ca phng tin o i vi cc gi tr o c qui nh+ Phm vi chth: l phm vi thang o c gii hn bi gi tr u v gi tr

cui ca thang o.

+ Cp chnh xc: c xc nh bi gi tr ln nht ca cc sai s trongthit b o. Thng c tnh ton bng gii hn ca sai s tng iquy i.

+ phn gii: Chnh l chia ca thang o hay gi tr nh nht c thphn bit c trn thang o (m c th phn bit c s bin i trnthang o)

dYS

dX=



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Trang 12

Cc c tnh ng: Phn ln cc thit b o khng p ng tc thingay khi i lng o thay i (do qun tnh, nhit dung hoc indung) s hot ng trng thi ng hoc trng thi giao thi ca

thit b o cng quan trng nh trng thi tnh.+ i vii lngo c 3 d ng thayi nhsau: Thay i c dng hm bc theo thi gian Thay i c dng hm tuyn tnh theo thi gian

Thay i c dng hm iu ha theo thi gian+ c tuyn ng ca thit b o: Tc p ng trung thc

Tnh tr Sai s ng



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1.3. Phn loi cc my oa) My o cc thng s v c tnh ca tn hiu:

VD: Vn mt in t, tn s mt, MHS, my phn tch ph, ...

x(t): tn hiu cn o a ti u vo my

Mch vo Thit bbin i

Thit bchth

Nguncung cp

Tn hiumang thngtin o x(t)

Hnh 1.2 Cu trc my o tham s v c tnh ca tn hiu



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Trang 14

Mch vo: truyn dn tn hiu t u vo ti Thit b bin i. Thchin tin x l tn hiu vo nh tin khuch i, suy gim, gii hn bngtn, lc nhiu, phi hp tr khng, nhng khng lm mt thng tin o.Mch vo thng l b K ti catt (Zvo cao), thc hin phi hp trkhng.

Thit b bin i: thc hin so snh v phn tch.C th to ra tn hiu cn thit so snh tn hiu cn o vi tn hiu mu.C th phn tch tn hiu o v bin , tn s, hay chn lc theo thi gian.Thng l cc mch K, tch sng, bin i dng in p tn hiu, chuyni dng nng lng,... Thit b chth: biu th kt qu o di dng thch hp vi gic quan giaotip ca sinh l con ngi hay vi tin tc a vo b phn iu chnh, tnhton,...

VD: Cc c cu ch th , ng tia in t, c cu ch th s dng LED 7on hay LCD 7 on Ngun cung c p: cung cp nng lng cho my v lm ngun to tnhiu chun



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b) My o c tnh v thng s ca mch in:

Mch in cn o thng s: mng 4 cc, mng 2 cc, cc phn

t ca mch in.S khi chung: cu to gm c ngun tn hiu v thit b ch

th, (hv)

VD: my o c tnh tn s, my o c tnh qu , my o hs phm cht, o RLC, my th n in t, bn dn v IC;my phn tch logic; my phn tch mng 4 cc ...



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Hnh 1.3 Cu trc my o tham s v c tnh ca mch in

Ngun tnhiu th

Thit b bini, xl tnhiu

Thit bch thNguncung cp

Mch cn otham s, c tnh

(a) Cu trc my o tham s v c tnh ca mch c ngun tn hiu th

(b) Cu trc my o tham s v c tnh ca mch c ngun tn hiu th c lp

Mch vo Thit b bin i,xl tn hiu

Thit bch thNguncung cp

Mch cno tham s,

c tnh

Ngun tnhiu th


Ch 1 Gii thi h l i t


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c) My to tn hi uo lng:

- B to sng ch: xc nh cc c tnh ch yu ca tn hiu nh dng vtn s dao ng, thn g l b to sng hnh sin hay xung cc loi

- B bin i: nng cao mc nng lng ca tn hiu hay tng thm xc

lp ca dng tn hiu, thn g l b Kin p, K cng sut, b iu ch,thit b to dng xung,...Cc my pht tn hiu siu cao tn thng khng c B bin i t gia

B to sng ch v u ra, m dng B iu ch trc tip khng ch

dao ng ch

Mchra

Thit bo

Nguncung cp

B iuch

B tosng ch

B bini

x(t)

Hnh 1.4 Cu trc my to tn hiu o lng


Ch 1 Gii thi h l i t


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-Mch ra: iu chnh cc mc tn hiu ra, bin i Zra ca my. N

thng l mch phn p, bin p phi hp tr khng, hay b ph tiCatt.

-Thit b o: kim tra thng s ca tn hiu u ra. N thng l vn mt

in t, thit b o cng sut, o h s iu ch, o tn s,...- Ngun: cung cp ngun cho cc b phn, thng lm nhim v bin i

in p xoay chiu ca mng li in thnh in p 1 chiu c nnh cao.

d) Cc linh kin o lng:cc linh kin l, ph thm vi my o to nn cc mch o cn thit.Gm cc in tr, in cm, in dung mu; hay cc linh kin ghp

gia cc b phn ca mch o (VD: b suy gim, b dch pha, b phnmch nh hng,...)


Ch 2 h i i l


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Trang 19

Chng 2.nh gi sai so lng

1. Khi nim v nguyn nhn gy sai s

2. Phn loi sai s: theo cch biu din sai s, theo s ph

thuc ca sai s vo i lng o, theo v tr sinh ra sais, theo qui lut xut hin sai s.

3. ng dng phng php phn b chun nh gi sai

s

4. Cch xc nh kt quo

5. Sai s ca php o gin tip

Ch 2 h i i l


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Trang 20

2.1. Khi nim & nguyn nhn sai s:* Khi nim sai s: l chnh lch gia kt qu o v gi tr thc ca

i lng o. N ph thuc vo nhiu yu t nh: thit b o, phngthc o, ngi o

* Nguyn nhn gy sai s:- Nguyn nhn khch quan: do dng c o khng hon ho, i lngo c b can nhiu nn khng hon ton c n nh,...- Nguyn nhn ch quan: do thiu thnh tho trong thao tc, phngphp tin hnh o khng hp l,...

2.2. Phn loi sai s* Theo cch biu din sai s:

- Sai s tuyti: l hiu gia kt qu o c vi gi tr thc ca i

lng o- Sai s tngi chn thc: l gi tr tuyt i ca t s gia sai stuyt i v gi tr thc ca i lng o

.100%ctthuc

X

X

=

thucdo XXX =


Chng 2 nh gi sai s o lng


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Trang 21

- Sai s tngi danh nh:

- Sai s tngi qui i: l gi tr tuyt i ca t s gia sai s tuyti v gi tr nh mc ca thang o.

cp chnh xc ca i lng o

Xdm= Xmax -Xmin : gi tr nh mc ca thang oNu gi tr thang o: 0Xmax Xdm=Xmax

* Theo sphthuc ca sai s vo i lngo:- Sai s im 0 (sai s cng) l sai s khng ph thuc vo gi tr i

lng o.- Sai s nhy (sai s nhn) l sai s ph thuc vo gi tr i lng

o

.100%dddo

X

X

=

%100.dm

qdX

X=




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* Theo vtr sinh ra sai s ta c sai sphng php v sai sphngtin o:-Sa i s phng php l sai s do phng php o khng hon ho

-Sa i s phng tin o lsa i s do phng tin o khng hon ho.Gm: sai s h thng, sai s ngu nhin, sai s im 0, sai s nhy,sai s c bn, sai s ph, sai s ng, sai s tnh.

Sai s c bn ca phng tin o l sai s ca phng tin o khi s

dng trong iu kin tiu chun

Sai sph ca phng tin o l sai s sinh ra khi s dng phng tino iu kin khng tiu chun

Sai s tnh l sai s ca phng tin o khi i lng o khng bin itheo thi gian

Sai s ngl sai s ca phng tin o khi i lng o bin i theo

thi gian




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Theo qui lut xut hin sai s: Sai s h thng

Sai s ngu nhin

2.2.1. Sai s h thng Do cc yu t thng xuyn hay cc yu t c qui lut tc

ng.

Kt qu o c sai s ca ln o no cng u ln hn hay bhn gi tr thc ca i lng cn o

VD:

+ Do dng c, my mco ch to khng hon ho

+ Do chn phng php o khng hp l, hoc li trong qutrnh xl kt quo,...

Do kh hu (nhit , m,...) khi o khng ging vi iu kin

kh hu tiu chun theo qui nh



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2.2.2. Sai s ngu nhin

Do cc yu t bt thng, khng c qui lut tc ng. VD:

+ Do in p cung cp ca mch o khngn nh

+ Do bin thin kh hu ca mi trng xung quanh trong qutrnh o

Tr s o sai: l kt qu cc ln o c cc gi tr sai khc qung, thng do s thiu chu o ca ngi o hay do cc tc

ng t ngt ca bn ngoi. Xl sai s sau khio:

+ i vi sai s h thng: x l bng cch cngi s gi trca

sai s h thng vo kt quo, hoc hiu chnh li my mc,thit bo vi my mu

+ i vi sai s ngu nhin: khng x l c, ch c th nhlngc gi trsai s ngu nhin bng l thuyt xc sut &

thng k



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2.3. ng dng phng php phn b chun nh gi sai s

Yu cu: - tt c cc ln o u phi thc hin vi chnh xc nh nhau- phi o nhiu ln2.3.1. Hm mtphn b sai s

- Tin hnh o n ln mt i lng no , ta thu c cc kt qu o c

cc sai s tng n g l x1, x2, ...,xn- Sp xp cc sai s theo gi tr ln ca n thnh tng nhm ring bit,vd: n1 sai s c tr s t 00,01; n2 sai s c tr s t 0,010,02; ...

- , ,... l tn sut ( hay tn s xut hin) cc ln o c cc

sai s ngu nhin nm trong khong c gi tr gii hn - Lp biu phn b tn sut:

n

n11

=n

n2

2

=

limn(x)=p(x)




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p(x) l hm s phn b tiu chun cc sai s (hm s chnh tc).

(hm Gauss) (1)22

)( xheh

xp

=

h : thng s o chnh xch ln ng cong hp v nhn(xc sut cc sai s c tr s b thln hn) thit bo c chnh

xc cao

Qui tc phn b sai s:

a. Xc sut xut hin ca cc sai s c tr s b th nhiu hn xc sutxut hin ca cc sai s c tr s ln.b. Xc sut xut hin sai s khng ph thuc du, ngha l cc sai s c tr

s bng nhau v gi tr tuyt i nhng khc du nhau th c xc sut

xut hin nh nhau.




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2.3.2. Sdng ccc sphn b nh gi kt qu o v sai s o

1. Sai s trung bnh bnh phng:+o n ln mt i lng X, cc kt qu nhn c l n tr s sai s c gi

tr nm trong khong gii hn x1 xn+ h khc nhau xc sut ca chng khc nhau+ h = const vi mt loi tr so xc sut sai s xut hin ti x1 v ln cn

ca x1 l:

tng t ta c:

11

21

2

dxeh

dp xh

=

22

22

2

dxeh

dpxh

=

n

xh

n dxeh

dp n22

=

.......................

x




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Xc sut ca n ln o coi nh xc sut ca mt s kin phc hp, do :

Pph= dp1. dp2... dpn

Tm cc tr ca h:

( ) ( )[ ] 02

22222

1

=

+

=

ii xhinn

xh

n

nph

exhh

eh

ndh

dP

02 22 = ixhn

Sai s TBBP ():

( )n

xxxh

n

dxdxdxeh

n ...21... 222

21

2 +++

= (2)

n

x

h

i=2

2

1(3)

n

xn

i

i== 1

2

(4)




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2. Trs trung bnh cng:

o X, thu c n cc kt qu o: a1, a2, ..., an Cc sai s cacc ln o ring bit: x1= a1-X, x2= a2-X, ..., xn= an-X Cc xi cha bit X cn o cha bit

Thc t chxc nh c tr s gn ng nht vi X (tr s c xc sutln nht):

(9)

3. Sai s d: Sai s mi ln o: xi =ai x cha bit v x cha bit. Sai s d l sai s tuyt i ca gi tr cc ln o ai vi :

Thc t:

n

a

n

aaaa

n

i

i

n

==

+++= 121

...

aaii =

Xa

a

0.1111

=== ====

n

i

i

n

i

i

n

i

i

n

i

i aaana (10)


Chng 2. nh gi sai s o lng


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Trang 30

(11)

4. Sai s TBBP ca : (12)

5. tin cy v khong tin cy:Xc sut ca cc sai s c tr s khng vt qu 1 gi tr cho trc

no , bng:

1

1

2

1

2

==

==

nn

xn

i

i

n

i

i

na

=a

( ) ( )

=


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Nu bit P, davobng hm s trong s tay tra cu v ton hay

(16) l khong tin cy, khong ny c xc sut cha ng tr s thc cai lng cn o X l . P l tin cy ca php nh gi.

Kt qu o:

(17)

m bo tin cy P =0,997 th ly t=3 ta c:

Quan h gia tin cy P, t, vi n >10 (bng 1)

( )t at

=

at=

atXa


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102 n

astaX =

7. Sai s cci v sai s th:Sai s cc i (n >10)

Sai s th: sai s |i| ca ln quan st no ln hn sai s cc i ( )th l sai s th.

8. Phn b student:Khong tin cy:

Gi tr ca ts c cho trong bng 2

= stM

tM =

( )102 n

102 ns a s aa t X a t < < +

M

g g g



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2. Tnh sai s d:

Kim tra: hay khng?

3. Tnh sai s TBBP:

4. Kim tra xem c sai s th?nu c sai s th th loi b kt qu o tng ng v thc hin libc 1-4 vi b kt quo mi, s ln o n mi.

5. Tnh sai s TBBP ca trs TB cng:

01

==

n

i

i

aaii =

1

1

2

= =

n

n

ii

n

a

=

2.4. Cch xc nh kt quo: Thc hin o n ln thu c cc kt

qu o: a1, a2, ..., an

1. Tnh trs trung bnh cng:

g g g

n

a

a

n

i

i== 1



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6. Xc nh kt qu o: vi

nu :

* Cch vit hng chs ca KQ o:

- Ly chcn ly vi 2 s sau du phy.- Ly phi ch ly ch s sao cho bc ca s cui ca n bcca hai con s ca .

VD: kt qu o l X = 275,24 1,08 th phi vit li l: X = 275,2 1,1

ataX = 10>n

102 n astaX =

ata

at

g g g

Bng 1 Gi tr t theo gi tr xc sut cho trc


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Bng 1. Gi tr t theo gi tr xc sut cho trc


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Bng 2.



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2.5. Sai s ca php o gin tipGi s X l i lng cn o bng php o gin tip; Y,V,Z l cc ilng o c bng php o trc tip

X = F(Y,V,Z)Y, V, Z l cc sai s h thng tng ng khi o Y, V, Z ; X l sai sh thng khi xc nh X

X + X = F(Y+ Y,V+ V,Z+ Z )

Cc sai s c gi tr nh nn:

( )F F F

+ X=F Y,V,Z + + +Y

X Y V ZV Z

F F FX= + +

YY V Z

V Z



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TH1: X = aY + bV + cZ

X = a Y + bV + cZ

TH2:

Thc t dng sai s tng i:

Xc nh sai s TBBP ca php o gin tip thng qua sai s TBBP cacc php o trc tip thnh phn

X

X Y V Z

= + +Y

= + +

Y V Z

X V Z

=

=KY V ZX

1 1 1=K Y V Z +K Y V Z +K Y V ZY V Z

2 2 2

X = + +Y V ZF F F

Y V Z

Chng 3. Cc b chth trong my o


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Nguyn tc hot ng chung ca c cu o

C cu chth kim: tin, in t C cu chth s: LED, LCD



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3.1. Nguyn tc hot ng chung ca c cu oBao gm 2 thnh phn c bn : Tnh v ng.

Hot ng theo nguyn tc bin i lin tc in nng thnh cnng lm quay phn ng ca n. Trong qu trnh quay lc csinh cng c hc mt phn thng lc ma st, mt phn lm bini th nng phn ng.

Qu trnh bin i nng lng trong CC c th hin theochiu bin i: dng in Ix (hoc Ux ) nng lng in t Wt,Wt s tng tc vi phn ng v phn tnh to ra F (lc) to

mmen quay (Mq) gc quay ; t l vi f(Ix) hoc = f(Ux)



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Gi s c cu o c n phn tnh in (mang in tch) v n cun dy.

Thng thng in p c a vo cun dy. Nng lng in tsinh ra c xc nh nh sau:

i : cun dyj : phn t mang in tch

: in dung v in p g ia 2 phn t tch in i v j.: dng in trong cc cun dy i v j.: in cm ca cundy i: h cm gia hai cun dy i v j

Nng lng in t sinh ra ph thuc vo in p, in dung, dngin, cun cm v h cm.

:ijM

,i jI

,ij ij

C U

iL

1 1

2 2

1 1 11 1

1 1 1

2 2 2

i n i nj n j nn

dt ij ij i i ij i ji i ij i j i

W C U LI M I I

= = = =

= = == =

= + +



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Tng tc gia phn tnh v phn ng to ra 1 momen quay bng sbin thin ca nng lng t trn s bin thin gc quay.

: s bin thin ca nng lng t: s bin thin ca gc quay

to r a s ph thuc gia gc quay v gi tr o; trong khi o ngita s dng thm l xo phn khng to ra momen phn khng chng li

s chuyn ng ca phn ng. D: l h s phn khng ca lxo

Kim ch th s dng li v tr cn bng khi

pk D=

qMpkM =1dt dt dW dW D

d D d

= =

dtW : ph thuc vo in p, dng in t vo cun dy.

d

dtdWqM =

dtdW

d



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3.2. C cu ch th o lng Cc c cu chth kim ng tia in t CRT C cu chth s (dng LED 7 on hay LCD 7 on)

Mn hnh ma tr n (LED, LCD, Flasma, OLED)

3.2.1 C cu ch th kim:

- Dng c o t in kiu nam chm vnh cu (TNCVC).- Dng c o kiu in t.- Dng c o in ng.

3.2.1.1. B ch thkiu t in: hot ng theo nguyn tc bin iin nng thnh c nng nh s tng tc gia t trng camt nam chm vnh cu v t trng ca dng in qua mtkhung dy ng



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1. Cu to:

- Phn tnh: gm 1 nam chm vnh cu (1),hai m cc t (2), 1 li st t (3). Gia (2)v (3) to thnh 1 khe hp hnh vnh

khuyn cho php 1 khung dy quay xungquanh v c t trng u hng tm (B)

- Phn ng: gm 1 khung dy nh (4) c

th quay xung quanh trc ca 1 li st t,1 kim ch th (5) c gn vo tr c cakhung dy, 1 l xo phn khng (6) vi 1u c gn vo tr c ca khung dy, ucn li c gn vi v my.

Hnh 3.1

nh v kim ng im `0` khi cha o th mt u ca l xophn khng trc c lin h vi mt vt chnh `0` chnhgia mt trc ca c cu o.



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2. Hotng:

Dng in trong cun dy ca c cu TNCVC phi chy theo mtchiu nht nh cho kim dch chuyn (theo chiu dng) t v tr`0` qua sut thang o.

o chiu dng in cun dy quay theo chiu ngc li v kimb lch v pha tri im `0`. Do cc u ni ca dng cTNCVC c nh du `+` v `-` cho bit chnh xc cc cnni. C cu TNCVC c coi l c phn cc.

Phng trnh m men quay v thang o: Khi c dng in I chy qua khung dy s to ra 1 t trng tng

tc vi t trng B ca NCVC to ra 1 mmen quay:

: bin thin ca t thng qua khung dy

B: t trng NCVC

== ddI

ddWM eq

= dSNBd ...



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N: s vng dy

S: din tch khung dyd : bin thin gc quay ca khung dy

Mq= I.B.N.SM men quay M

qlm quay khung dy, khi mmen phn khng do l

xo phn khng tc ng vo khung dy tngMpk= D. (3.5)

D - h s phn khng ca l xo

- gc quay ca kimKhi mmen quay Mq cn bng vi mmen phn khng Mp ca l xo thkim s dng li trn mt s ng vi mt gc no .

Mq = Mpk (3.6)

ISID

SNBDSI

...

....

0===

D

SNBS

..0 = l nhy ca ccu o



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3. cim ca ccu o t in:

u im: Thang o tuyn tnh c th khc thang o ca dng in I theo gc quay

ca kim chth nhy c cu o ln

Dng ton thang (Itt) rt nh (c A) chnh xc cao, c th to ra cc thang o c cp chnh xc ti 0,5% t chu nh hng ca in t trng bn ngoi.

Nhc im: Cu to phc tp, d b h hng khi c va p mnh Chuqu ti km do dy qun khung c ng knh nh Chlm vic vi dng 1 chiu, mun lm vic vi dng

xoay chiu phi c thm it nn in

ng dng: dng r t nhiu l mc cu chth cho cc dng c o in nh:

Vnmt, Ampemt, dng c o in vn nng,

cc php o cuc nbng

3 2 1 2 C i t h t th l l i t



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3.2.1.2. C cuin t: hot ng theo nguyn l: nng lng in tc bin i lin tc thnh c nng nh s tng tc gia ttrng ca cun dy tnh khi c dng in i qua vi phn ng cac cu l cc l st t

1. Cu to: c 2 loi

- Loi cun dy hnh trn.- Loi cun dy hnh dt

Hnh 3.3.1

+ Loi cun dy hnh trn:

- Phn tnh: l mt cun dy hnh tr trn,pha trong thnh ng c gn l st t mmun quanh

- Phn ng: gm mt l st t cng cun cong v gn vo tr c quay nm idin. Trn trc quay gn kim chth v l xophn khng



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+ Loi cun dy dt:

- Phn tnh: gm 1 cun dy dt, gia c 1 khe hp.- Phn ng: gm 1 a st t c gn lch tm, chmt phn nm trong khe

hp v c th quay quanh trc. Trn trc ca a st t c gn kim chth v lxo phn khng

Hnh 3.3.2 Ccu in t loi cun dy dt

2 Nguyn l hot ng chung:



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2. Nguyn l hotng chung:Khi c dng I in chy qua cun dy tnh s to ra mt nng lng t

trng

vi L l in cm cun dy, c gi tr tu thuc vo v tr tng i ca lst t ngv tnhS bin thin nng lng t trng theo gc quay to ra mmen quay trc quay kim chth quay

Khi kim chth quay mmen phn khng tng: Mpk=D.Ti v tr cn bng: Mpk= Mq

Gc quay ca kim chth t l vi bnh phng ca I qua cun dy

21

2

ttdW dL

D I

d d

= =

21 1

2

ttdW dL

ID d D d

= =

2

0IS=

=d

dL

DS

2

1

0,

21

2ttW LI=

ttq

dWM

d=



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3. cim ca CCin t:

+ u im:9 CC t in c th lm vic vi dng xoay

chiu.9

C cu to vng chc, kh nng chu ti tt.+ Nhc im:9 nhy km do t trng phn tnh yu9 Thang o phi tuyn9 chnh xc thp do d nh hng ca t

trngbn ngoi do tnhaost t ln

9 Tiu th nng lng nhiu hn c cu o tin.

+ ng dng: vn c dng nhiu trong ccng h o in p ln

Hnh 3.4 ng h oin p cao sdng CC

in t



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3.2.2. C cu ch th s

Vt cno

Tr s oc

Khong tcha cc xung

c tn s f

m xungtrong t

Hin thdi dng

ch s

kt qubin i

bin io

Hnh 3.5S khi ccu ch th s n gin

1. Nguyn l hotng chung: cc c cu o hin th s thng dngphng php bin i tr s ca i lng o ra khong thi gian c

lu t ph thuc tr s o cha y cc xung lin tip vi tn s nhtnh.Thit b chthm s xung trong khong thi gian t v th hin kt quphp m di dng ch s hin th.



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2. cim:

(a) Cc u im: chnh xc o lng cao. Chth kt quo di dng ch s nn dc.

C kh nng t chn thang o v phn cc Tr khng vo ln. C th lu li cc kt quo a vo my tnh. Dng thun tin cho o t xa.

(b) Cc nhc im: S phc tp Gi thnh cao

bn vng nhHin nay thit b o s thng s dng cc loi c cu chth s nh sau: C cu chth s dng it pht quang LED

C cu chth s dng LCD



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Hnh 3.6Thanh LED

3.2.2.1. B chths dngit pht quang (LED

LED l mt tip xc P-N, vt liu ch to u l cc lin kt canguyn t nhm 3 v nhm 5 ca bng h thng tun honMendeleev nh GaAs (LED c mu ), GaP (LED c mu hocmu lc), GaAsP (LED c mu hoc vng).

Khi LED c phn cc thun cc ht dn a s khuch tn tqua tip xc P-N (in t t do t n sang p, l trng t p sang n)chng gp nhau s ti hp v pht sinh ra photon nh sng. Cng pht sng ca LED tl vi dng in qua it .



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* LED 7 on:

Cc dng co hin th s thng dng b ch th 7 on sng ghpli vi nhau theo hnh s 8. Cc on sng l cc it pht quang. Khicho dng in chy qua nhng on thch hp c th hin hnh bt ks no t 0-9.

C 2 loi: LED 7 on sng Ant chungLED 7 on sng Catt chung

LED 7 on sng Catt chung: catt ca tt c cc it u c nichung vi im c in th bng 0 (hay cc m ca ngun). Tc ngvo u vo (ant) ca it mc logic 1 it sng.

LED 7on sng Ant chung: cc ant c ni chung vi cc dngca ngun (mc logic 1). Tc ng vo u vo (Catt) ca it mc

logic 0 it sng.

st p khi phn cc thun it l 1,2V v dng thun khi c chihp l l 20mA.

Nh i d t i l



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Nhc im: cn dng tng i ln.

u im: ngun in p mt chiu thp, kh nng chuyn mch nhanh,bn, kch tc b.

Hnh 3.7 Led 7 on

3 2 2 2 B ch th s dng tinh th lng (LCD)



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3.2.2.2. B chths dng tinh th lng (LCD)

Tinh th lng l tn trng thi ca mt vi hp cht hu c c bit.Cc cht ny nng chy 2 trng thi: lc u trng thi nng chylin tc, sau nu nhit tip tc tng th chuyn sang cht lngng hng bnh thng.

Pha trung gian gia hai trng thi ny l trng thi tinh th lng B ch th dng tinh th lng (LCD) thng c b tr cng theo dng

s 7 on nh b chth LED.

Hnh 3.8Cu to mi thanh LCD

Trn 2 tm thu tinh ngi ta ph mt lpkim loi dn in to nn 2 in cctrong sut, gia 2 lp kim loi l lp chtlng tinh th.

Khi ch th ch s, ngoi in p t vo 2in cc ca phn t cn cn ngun sngt pha tr c hay pha sau ca b ch thv phng.



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(b) ngun sngt sau: khi c tn hiu thtinh th lng c nh sng i qua to nnhnh s trn mn hnh. Mn hnh l tmphng en.

Ngun in cung cp l ngun 1 chiuhoc l ngun in p xung.

(a) ngun sngt trc: khi ctn hiu th tinh th lng c nhsng phn x t gng.

Hnh 3.9

Hnh 3.10



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VD: mt LCD 7 on (hnh 3.11)

u chung ca cc phn t ch th LCDni vi +E qua R.

Cc in cc ring ni vi cc u raiu khin.

Khi transistor T6 tt, U6a = 0 phn t 6khng chth. Khi T6 thng, U6a = +E kch thch

phn t 6 tr nn trong sut, cho nhsng i qua.

u im ca LCD: Ngun cung cp n gin, tiu th cng sut nh, c mW Kch tc b, ph hp vi cc thit b o dng mch t hp, kthut vi in t.

Hnh ch s kh r rng, ch to n gin.Nhcim ca LCD: Di nhit lm vic hp (100C-550C) Tui th cha tht caoTuy vy cc u im l c bn nn loi ny ngy cng c dng nhiu

Hnh 3.11

3 2 2 3 Mn hnh ma trn



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3.2.2.3. Mn hnh ma trn

a. Mn hnh ma tr n LED

b. Mn hnh ma tr n LCD Hnh 3.12Ma trn LED

Hnh 3.13- ng dng ca mn hnh ma trn LCD trong my o v mn hnh my tnh

Chng 4. My hin sng (xil)


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Nguyn l quan st tn hiu trn MHS: nguynl qut tt lin tc, nguyn l qut i, nguyn l ng

b

S cu to mt MHS in hnh

Mt s ch lm vic

MHS nhiu tia : MHS 2 knh dng ng tia in t

1 tia v CM in t

xil in t s

S



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4.1. Nguyn l quan st tn hiu trn MHS:

1. Phng php v dao ng ca tn hi u Mt tn hiu thng c biu din di 2 dng:

+ Hm theo thi gian: u = f(t)+ Hm s theo tn s: u =(f)

quan st dng sng, o cc c tnh v cc tham s ca tn hiu dng

mt my o a nng l MHS (xil). MHS l mt loi my v di ng theo 2 chiu X v Y hin th dng tn hiua vo cn quan st theo tn hiu khc hay theo thi gian. `Kim bt v ` camy l mt chm sng, di chuyn trn mn hnh ca ng tia in t theo quilut ca in p a vo cn quan st.

Ff

f-F f+F f f

UAM

0

Hnh 4.1- Tn hiu iu bin v phca n



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2. Cc loi xil:

xil tn thp, xil tn cao, xil siu cao tn xil xung (/T b) xil 2 tia; xil nhiu knh

xil c nh (loi tng t v loi s) xil s; xil c ci t VXL

3. Cng dng, tnh nng ca xil:

xil l mt my o vn nng, n c cc tnh nng: Quan st ton cnh tn hiuo cc thng s cng ca tn hiu:

+ o in p, o dng in, o cng sut

+ o tn s, chu k, khong thi gian ca tn hiu+ o di pha ca tn hiu+ v tng v o c c tnh ph ca tn hiu+ vc tuyn Vn-Ampe ca linh kin+ v tng, o c tuyn bin -tn s ca mng 4 cc

4 C h ti k th t h il



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4. Cc chtiu kthut ch yu ca xil:

Phm vi tn s cng tc: c xc nh bng phm vi tn s qut. nhy (h s li tia theo chiu dc): mV/cm

L mc in p cn thit a n u vo knh lch dc bng bao nhiu mV tia in t dch chuyn c di 1 cm theo chiu dc ca mn sng.

nhy cng c th c tnh bng mm/V. ng knh mn sng: xil cng ln, cht lng cng cao th ng knh

mn sng cng ln (thng thng khong 70mm-150 mm). Ngoi ra cn c h s li tia theo chiu ngang, tr khng vo,...

5. Ch qut tuyn tnh lin tc

a) Nguyn l qut ng thng trong MHS a in p ca tn hiu cn nghin cu ln cp phin lch Y, v in p

qut rng ca ln cp phin lch X. Do tc dng ng thi ca c hai in trng ln 2 cp phin m tia in tdch chuyn c theo phng trc X v Y.

Qu o ca tia in t dch chuyn trn mn s vch nn hnh dng ca inp nghin cu bin thin theo thi gian.



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Ch : in p qut l hm lin tc theo thi gian qut lin tcin p qut l hm gin on theo thi gian quti

b) Nguyn l qut tuyn tnh lin tc

in p qut tuyn tnh lin tc c tcdng li tia in t dch chuyn lp i lpli 1 cch lin tc theo phng ngang tl bc nht vi thi gian.

qut tuyn tnh lin tc cn phi dngin p bin i tuyn tnh lin tc (tngtuyn tnh hay gim tuyn tnh)

Hnh 4.2 - Qut t2 lin tc vi Tq = Tth

Chu k qut: Tq = tth + tng U



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C u qu q th ng

Thng thng: tng 15% tth tc l tng rt nh hn tth nn c th coi Tq tth, l tng: tng = 0 (Tq = tth)

Nu tn s qut cao, mn hunh quang c d huy mc cnthit th khi mi ch c Uq t vo cp phin X c mt ng sngtheo phng ngang. Khi c c Uth t vo cp phin Y v nu Tq = nTth

th trn mn xut hin dao ng ca mt hay vi chu k cain p nghin cu (Uth).

Nu Tq nTth th dao ng khng ng yn m lun di ng ri lon kh quan st. Hin tng ny gi l khng ng b (khng ng phagia U

qv U

th).

tng

tth

Uq

t

( )Nn

Hnh 4.3 - in p qut r ng ca thc t



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Thc t, tng 0. V tng


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H s khng ng thng ():

c nh quan st vi cht lng cao cn:

tng


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(a) Tq = Tth : tn hiu ch xut hin

trong mt t/g rt b ( mt cht

+ Kh thc hin ng b dao ng khng n nhHnh 4.5

7. Nguyn lng b:



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g y g

Khi quan st dng tn hiu trn MHS, i khi nh b tri, nhy,... l domt ng b.

*

Minh ha :

nh I, II, III l cc dao ng tng ng ti cc chu k qut tng ng.

N phn b ln lt t tri qua phi, do tnh cht lu nh ca mn hnhcc nh s m dn theo th t tng ng cm gic dao ng chuyn ng t tri qua phi.

*

tng t, cm gic d/ng chuyn ng t phi qua tri

( )Nn thqth nTTTn


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(minh ha ):

Dao ng ng yn nhng khng phn nh ng dng tn hiucn quan st m ch gm nhng on tn hiu khc nhau cn quanst m thi.

* Tq = nTth (minh ha Tq = Tth ),

Dao ng n nh v phn nh ng dng tn hiu cn quan st.

iu kin ng b: Tq = nTthQu trnh thit lp v duy tr iu kin ny l qu trnh ng b caMHS

Cc ch ng b:+ ng b trong: tn hiu ng b ly t knh Y ca MHS+ ng b ngoi (EXT)+ ng b li (LINE)

bTq3 33=

qT

Nn

U h



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0

0

Uq1

Uth

t

t

Tq1

Tth 2Tth 3Tth

Tth

0

Uq4t

Tq4

0

Uq2t

Tq2

0

Uq3t

Tq3

4.2. S cu to mt MHS in



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hnh

4.2.1. Cu to MHS: ng tia in t

Knh lch ng Y Knh lch ngang X v ng b Knh Z (khng ch sng)

*ng tia in t:+ l b phn trung tm ca MHS, sdng loi ng 1 tia khng ch bngin trng

+ C nhim v hin th dng sngtrnmn hnh v l i tng iu khinchnh (Uy, Ux, UG).

Mchvo v

phn

p Y

Tinkhuch

i

Dy

tr

Khuchi Y

ixng

Toxung

chun

K/i

ng bv todng

Toxungng b

Toin p

qut

i

lin

tc

Mchvo v

K X

K/i Xi

xng

Chncc tnh

K/iZ

Knh lch ng Y

Knh lch ngang X v ng b

Knh Z

Ti G ca CRTUZ

Ux Uqut

S3

S2

S1AC

DC

GND

Uth

Vpp

CH

EXTLINEAC

50Hz

Ub Ux

Uxb

1

23

CRTX1 X2

Y1

Y2

Hnh 4.7 - S khi MHS 1 knh dng ng tia in t

* Knh lch ng Y: c nhim v nhn tn hiu v o cn quan st bin i



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Knh lch ng Y: c nhim v nhn tn hiu v o cn quan st, bin iv to ra in p ph hp cung cp cho cp li ng Y1, Y2. Gm cckhi chc nng:

+ Chuyn mch kt niu v o S1: cho php chn ch hin th tn hiu.S1 ti AC: chhin th thnh phn xoay chiu ca Uth.S1 ti DC: hin th c thnh phn mt chiu v xoay chiu ca Uth.S1 ti GND: chquan st tn hiu ni t (0V).

+ Mch vo phn p Y: c nhim v phi hp tr khng v phn p tn hiu vo tng kh nng o in p cao. Thng dng cc khu phn p R-C mc lintip nhau, h s phn p khng ph thuc vo tn s. Chuyn mch phn pc a ra ngoi mt my v k hiu l Volts/Div.

+ Tin khuch i: c nhim v khuch i tn hiu, lm tng nhy chung caknh Y. Thng dng cc mch K c tr khng vo ln v c h s K ln.+ Dy tr: c nhim v gi chm tn hiu trc khi a ti K Y i xng,thng dng trong cc ch qut i trnh mt mt phn sn trc catn hiu khi quan st. Thng dng cc khu L-C mc lin tip.

+ K Yi xng: c nhim v K tn hiu, lm tng nhy chung ca knh Y,



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ng thi to ra in p i xng cung cp cho cp li ng Y1Y2.

+ To in p chun: to ra in p chun c dng bin , tn s bit trc,dng kim chun li cc h s lch tia ca MHS

* Knh lch ngang X v ng b: c nhim v to ra in p qut phhp v dng v ng b v pha so vi UY1, Y2 cung cp cho cp lingang X1X2

+ Chuyn mch ng b S2: cho php chn cc tn hiu ng b khc nhau.S2 ti CH: t ng b (Ub = Uth)S2 ti EXT: ng b ngoi (Ub=UEXT), tn hiu ng b c a qua u voEXT.

S2 ti LINE: ng b vi li in AC 50Hz (Ub=UAC50Hz) ly t ngun nui.+ Kng b v to dng: k/i tn hiu Ub ph hp v to r a dng xung nhnn cc tnh c chu k: Tx=Tb

+ To xung ng b: chia tn U v to ra xung ng b c chu k:



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+ To xung ng b: chia tn Ux v to ra xung ng b c chu k:Txb=nTx=nTb. Xung ny s iu khin b to in p qut to ra Uq rngca tuyn tnh theo ch qut i hoc qut lin tc v c chu k Tq=Txb.

+ Ki xng: Kin p qut v to ra in p i xng a ti cp lingang X

1X

2.

+ Mch vo v KX: nhn tn hiu UX v k/i, phn p ph hp.

+ Chuyn mch S3: chuyn mch la chn ch qut (qut lin tc, qut i)

+ B to in p qut: to in p qut lin tc (hoc qut i) a n cpphin X

* Knh iu khin ch sng Z: c nhim v nhn tn hiu iu ch

sng UZ vo, thc hin chn cc tnh v k/i ph hp ri a ti liiu ch G ca CRT.

4.2.2. Cu to ca ng tia in t:



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ng tia in t CTR (Cathode Ray Tube) l 1 ng thu tinh hnh tr c chn khng cao, u ng c cha cc in cc, pha cui loe ra hnh nnct, mt y c ph 1 lp hunh quang to thnh mn hnh. Cu to

gm 3 phn:

Hnh 4.8 - S cu to ca ng tia in t

A3 (Ahu)

Lp than ch

X1

X2

Y1

Y2

Mn chn

Si t F

Katt K

Li iu ch G

Ant hi t A1

Ant gia tc A2

EK

-2,05kVR2R1Rbright Rfocus

Sng in t H thng li tia Mn hnh

Mn

hunhquang

a) Mn hnh: - Lp hunh quang thng l hp cht ca Phtpho. Khi c



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Trang 78

in t bn ti mn hnh, ti v tr va p, in t s truyn ng nng chocc in t lp ngoi cng ca nguyn t Phtpho, cc in t ny s nhyt mc nng lng thp ln mc nng lng cao v tn ti trong 1 thigian r t ngn ri t nhy v mc nng lng thp ban u v pht ra

photon nh sng.- Mu sc nh sng pht ra, thi gian tn ti ca im sng ( d huy camn hnh) s ph thucvohp cht ca Phtpho (t vi s n vi s).

b) Sngin t: gm si t F, catt K, li iu ch M, cc ant A1,A2.Nhim v: tog ia tc v hi t chm tia in t- Cc in cc c dng hnh tr , lm bng Niken, ring Katt c ph mtlp xit kim loi y tng kh nng bc x in t.

- Cc in cc pha sau thng c vnh r ng hn in cc pha tr c vc nhiu vch ngn cc chm in t khng i qu xa tr c ng vichi t s d dng hn. Vi cu to c bit ca cc in cc nh vy s tora 1 t trng khng u c bit c th hi t v gia tc chm tia.

Ngun cp: UK = -2kV



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UKG = 0V-50VUA2 = 0VUA1 = 50V-300V

+ Li iu ch G c cung cp in p m hn so vi K v c ghpst K d dng cho vic iu chnh cng ca chm in t bn timn hnh.+ Chit p trn G (iu chnh in p) thng c a ra ngoi mt

my v k hiu l Bright hoc Intensity dng iu chnh sng tica dao ng trn mn hnh.+ Ant A2 (Ant gia tc) thng c ni t trnh mo dao ng khi in p cung cp cho cc in cc khng phi l in p i xng.

+ Ant A1 (Ant hi t) cng c chit p iu chnh a ra ngoi mtmy, k hiu l Focus, dng iu chnh hi t ca chm tia in ttrn mn hnh.



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Hnh 4.9Cu to CRT khng chbng in trng



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Trang 81

Mn hnh

C

Cc mt ng th

Lc tc dng ln in t

Chm in t

G A1A2

K

F

Hnh 4.11 Chm in tchuyn ng qua in trng gia ccin cc

c) H thng li tia: c nhim v lm lch chm tia in t bn ti mn hnh

y a=



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Trang 82

theo chiu ng hoc chiu ngang ca mn hnh.

Cu to gm 2 cp phin lm lch c t trc, sau v bao quanh trcca ng:

Cp li ng Y1Y2. Cp li ngang X1X2 .

*Xt lch ca tia in ttheo chiung:



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Trang 83

Khi Uy= 0, tia in t bn ti chnh gia mn hnh ti im C.

Khi Uy 0, in trng gia cc phin lm lch s lm lch qu o ca tiain t theo chiu ngvbn ti mn hnh ti v tr M, lch 1 khon g l y

Ly : khong cch t cp li ng n mn hnhly : chiu d i cacccp phin lm lchdy : khong cch gia 2 phin lm lchUA : in p g i a tc ca ng tia (ph thucvoUA2 v K)

2

y y yoy y

y A

U l Ly S U

d U= =

2

y yoy

y y A

l LySU d U

= = nhy ca ngtia in t

dy

A2

Uy Y1

Y2

ly

Ly

y

M

C

chm e-+

-

Mn hnh

Hnh 4.12

Tng t, lch ca tia in t theo chiu ngang:



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Trang 84

* Nguyn l to nh trn mn my hin sng:H thng li tia iu khin ng thi tia in t theo 2 trc: trc thngng v tr c nm ngang, ngha l ng thi a vo n ng tia int 2 in p iu khin UY v UX .

Gi s a vo knh Y v a ti cp li ng Y1Y2;in p qut a ti cp li ngang X1X2 in p trn cccp li tia nh sau:

2

x x xox x

x A

U l Lx S U

d U= =

sin .th mU U t=taqU .=

ySthUyyUyU == 21 oySyKyS =

4.2.3. Mt s ch lm vic:



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Trang 85

a. Qut lin t cng b trong (ngoi)

Dng quan st nh ca tn hiu lin tc theo thi gian v o cctham s ca chng.

S2 v tr CH (hoc EXT nu l ng b ngoi), S3 v tr 2

Tn hiu t li vo knh Y, qua Mch vo v b phn p Y c

khuch i ti mt mc nht nh, sau c gi chm li ri aqua B K Yi xng to 2 tn hiu c bin ln, o phanhau a ti 2 phin ng

b. Quting b trong



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Trang 86

Dng quan st v o tham s ca dy xung khng tun hon hocdy xung tun hon c hng ln.

S2

v tr CH, S3

v tr 1

Qu trnh hot ng: ging ch 1

c. Ch khuch i

Dng o tn s, gc lch pha, su iu ch, vc tnh Vn-Ampe ca it hoc dng lm thit b so snh. Hnh nhn c trnmn MHS gi l hnh Lixazu

S3 v tr 3

B to qut trong c ngt ra khi qu trnh hot ng. MHS lmvic theo 2 knh c lp X,Y v u vo X cng l u vo tn hiu

4.3. MHS nhiu tia



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Trang 87

Dng quan st ng thi nhiu qu trnh (tn hiu)MHS 2 tia c li vo cp phin lch ng tch bit(knh A, knh B): Mi knh c mch K lm lch ring Mt b to gc thi gian chung cho c 2 knh

MHS 2 knh dngng tia in t1 tia v CMin t Hai b K tn hiu vo ring cho knh A, knh B Mt b K lch ng cho c 2 knh. Tn hiu vo b K ny c

chuyn mch lun phin gia 2 knh B to gc thi gian (b to sng qut ngang) iu khin tn schuyn mch

Hnh 4.13

a) Phng php dng chuyn mch in t kiu



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Trang 88

lun phin (ALT mode):

Tn hiuv ob K lch ng c chuynmch lun phin gia cc knh A v B.

B to gc thi gian iu khin tn s chuyn

mch 0t1: tn hiu t knh A c ni ti b K lch

ng, to thnh vt trn mn hin sng t1t2: tn hiu t knh B c ni ti b K

lch ng, to thnh vt trn mn hin sng Hai tn hiu hai knh c cng chu k T vc ng b vi nhau.

Dch chnh DC: dch chuyn tn hiu knh A(knh B) trn mn theo phng thng ngbng in p mt chiu

cc chu k tip theo: qu trnh lp li nhtrn. Tn s lp cao n mc m cc dngsng nhc hin ng thi.

c)

Dngsng

hin

Hnh 4.14

b)Phng php dng chuyn mch in t kiu ngt qung (Chop modeswitching):



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Trang 89

S dng tn s chuyn mch cao hn nhiu so vi ch lun phin. T1, T3, T5, T7,... tn hiu vo knh A c to

ra trn mn T2, T4, T6,... tn hiu vo knh B c to ra

trn mn Cc dng sng knh A v B c hin

hnh nh nhng ng t nt. Khi tn schuyn mch l cao tn khng th nhn ra

nhng cht nt fth nh: nh hin trn mn MHS gn nh lintc

fth ln; nfcm mfth : cc on ngt b lp do d huy ca ng v lu nh ca mt.

Ch : i vi tn hiu cao tn th kiu lunphin l tt nht, cn i vi tn hiu tn sthp th nn dng chuyn mch ngt qung Hnh 4.15

Chuyn mch in tphn ng

theo thi gian:



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Trang 90

theo thi gian: (a)

(b)

(c)

(d)

Hnh 4.17- Gin thi gianHnh 4.16 -S khi CM in t phn

ng theo thi gian

Mi knh tn hiu c cng thm mt lng in p 1 chiu E khc nhau

bi di t h h MHS t h i t



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Trang 91

cc ng biu din trn mn hnh MHS c tch ring tng ng,hnh (a). Sau tn hiu c a n mch ca, v ch qua c ca khi c tn

hiu m ca t b Pht sng chuyn mch.

Tn hiu m ca l cc xung vung c thi gian xut hin xen k v lnlt cho tng ca mt, hnh (b). Ti mi thi im chc duy nht 1 ca c m v cho tn hiu ca mt

knh i qua.

B tngcng cc tn hiu u ra cc ca, UY c dng xung vi bin tl vi gi tr ca cc tn hiu cn quan st ti thi im c xung m catng ng vi cc knh, hnh (c).

Sau khi khuch i Y, MHS c c hnh biu din tn hiu ca cc knhdi dng ng nt t, hnh (c).

MHS lm vic chng b vi chu k ca tn hiu cn quan st vkhng ng b vi tn hiu chuyn mch.

Dng nhng xung c rng rt nh (UZ) c to ra t mch vi phn tcc xung m ca a vo knh Z iu ch sng ca nh, hnh (d).

Chuyn mch in tphn ng theo mc:



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Trang 92

En

Hnh 4.18 -S khi CM in tphn ng theo mcHnh 4.19 -Gin thi gian

Phn hin th l ng truyn hnh lm lch bng ttrng

N l h t h i t t thi t hi k h



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Trang 93

Nguyn l hot ng: chuyn cc gi tr tc thi ca tn hiu cc knhthnh cc chui xung rt hp xut hin ti cc thi im m tu thucvo in p tn hiu nghin cu. Cc xung ny c a vo

khng ch sng ca ng hin hnh.

Mi knh tn hiu c cng thm mt lng in p 1 chiu E khcnhau, ri a n so snh vi tn hiu l xung rng ca a ti t b

K lch ng ca MHS (tn hiu qut dng). Mi khi URC = Uth, th u ra ca b so snh s xut hin mt xung

hp. Cc xung hp ny c cng vi nhau ri a vo khng ch sng ca ng hin hnh.

Ti thi im c xung, trn mn hnh xut hin mt chm sng trongkhi bnh thng th ti. Vt ca chm sng trn mn hnh biu dinhnh in p ca cc tn hiu cn quan st.



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Hnh 4.19.1 My hin sng

4.4. xil in t s:



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Trang 95

1. uim:

Duy tr nh ca tn hiu trn mn hnh vi khong thi gian khng

hn ch. Tc c c th thay i trong gii hn rng

C th xem li cc on hnh nh lu gi vi tc thp hn nhiu

Hnh nh tt hn, tng phn hn so vi loi xil tng t

Vn hnh n gin

S liu cn quan st di dng s c thc x l trong xil hoctruyn trc tip vo my tnh khi ghp xil vi my tnh.

2. Cu to v hotng:



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Trang 96

Chuyn mch S vtr 1: xil a nng thng thngChuyn mch S vtr 2: xil c nh s. in p cn quan st c a ti u vo Y, ti b ADC. Lc biu

khin gi 1 lnh ti u vo iu khin ca b ADCv khi ng qu trnhbin i. Kt qu l in p tn hiu c s ho. Khi kt thc qu trnh bini, bADCgi tn hiu kt thc ti biu khin.

Mi s nh phn c chuyn ti b nh v c nh v tr nh ring bit.

Hnh 4.20 My hin sng s n gin

Khi cn thit, mt lnh t b iu khin lm cho cc s nh

phn ny sp xp theo chui li theo th t xc nh v



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Trang 97

phn ny sp xp theo chui li theo th t xc nh vc a n b DAC

DACbin i cc gi tr nh phn thnh in p tng t

a qua b khuch i Y ti cp phin lm lch Y ca ng tiain t. Do b nh c lin tip qut nhiu ln trong 1 giy nn mn

hnh c sng lin tc v hin dng sng l hnh v cc imsng. Nhcim: di tn b hn ch (khong 1-10MHz) do tc

bin i ca bADC thp.

Hin nay, cc xil c nh s c di tn rng c pht trinnh ci t VXL, cc b bin i ADC c tc bin i nhanhhn

Khi nim chung

Chng 5.o tn s, khong thi gian & di pha


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Trang 98

Khi nim chung o tn s, s dng:

Cc mch in c tham s ph thuc tn s

Phng php so snh

Phng php s

o di pha, s dng: Phng php o khong thi gian Pha mt chth s

Phng php v dao ng

5.1. Khi nim chung

Tn s l s chu k ca 1 dao ng trong mt n v thi gian. ( )



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Trang 99

Tn s l s chu k ca 1 dao ng trong mt n v thi gian. Tn s gc: biu th tc bin i pha ca dao ng

l tn s gc tc thi v tn s tc thi

Quan h gia tn s v bc sng:

hay

Quan h gia chu k v tn s:

cim ca php o tn s:

l php o c chnh xc cao nht trong kthut o lng nh s pht trinvt bc ca vic ch to cc mu tn s c chnh xc v n nh cao. Lng trnh o rng (n 3.1011 Hz). Lng trnh o c phn thnh cc

di tn s khc nhau.

( )t

( )dt

dt

=

( ) ( )tft = 2 ( ) ( )tft ,

=c

f

f

c=

Tf

1=

Cc di tn s:

Di tn thp: < 16Hz Di tn s m thanh: 16 Hz < f < 20 KHz



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Trang 100

Di tn thp: < 16Hz Di tn s m thanh: 16 Hz < f < 20 KHz Di tn s siu m: 20 KHz < f < 200 KHz Di tn s cao: 200 KHz < f < 30 MHz

Di tn s siu cao: 30 MHz < f < 3000 MHz Di tn s quang hc: > 3GHzCc di tn s khc nhau c cc phng php o tn s khc nhau.Bao gm:

Nhm phng php o tn s bng cc mch in c tham s phthuc tn s Nhm phng php so snh Nhm phng php s

Php o tn s thngc sdng kim tra, hiu chun cc myto tn hiu o lng, cc my thu pht; xcnh tn s cng hngca cc mch dao ng; xc nh di thng ca b lc; kim tra lch tn s ca cc thit bang khai thc

5.2. o tn s bng cc mch in c tham s ph

thuc tn s Z Z



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Trang 101

thuc tn s5.2.1. Phng php cuDng cc cu o m iu kin cn bng ca cu

ph thuc vo tn s ngun in cung cp cho cu.* Mch cu tng qut: iu kin cn bng cu:

VD1:0.. 4231 == ABUZZZZ

iu kin cn bng cu:

+=

=

3

33

4231

1

..

CLjRZ

RRZR

x

Z1

Z3Z4

Z2

Hnh 5-1

3

Hnh 5-2

iu chun nhnh cng hng ni tip cho cng hng ti tn s cn o fx

(iu chnh C3).



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Trang 102

(iu chnh C ).

Khi

B ch th cn bng l vn mt chnh lu, vn mt in t.

Nhcim:

Kh o c tn s thp do kh ch to cun cm c L ln tn sthp.

Kh thc hin ch th 0 do c tc ng ca in t trng ln cuncm

4231 .. RRRR =33 RZ =

33 2

11

CLf

CL

x

x

x

xx =

=

VD2:iu kin cn bng cu:

'3'

1 1.RR



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Trang 103

Chn

+=+ 4

42

33

31 1

1

.

RjRR

CRj

RR

xx

4

3

3

4'

2

'

1

CC

RR

RR += 433

4

1 RRCC

x

x

=v

4343

12

CCRR

fxx ==

CCC == 43RRR == 43 v ta c:

Hnh 5-3

V

RC

fx

=

2

12

'

2

'

1 =R

R;

11

'

1 RVRR += 22'

2 RVRR +=

VR1,VR2 l phn in tr ca bin tr VR trn nhnh 1,2 tng ng

VD3: Cu T kp

iu kin cn bng cu:



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Trang 104

g

Khi :

Thang ca bin trR1 c khc trc tip theo n v tn s.

Phng php cu dng o tn s t vi chc Hz n vi trm Khz.Sai s: (0,5-1)%

=

=

12

2

21

2

1

2

21

2

2

2

RRC

CCR

x

x

12 2RR =12 2CC = v

112

1

CR

x =114

1

CR

fx

= Hnh 5-4

5.2.2. Phng php cng hng

Dng o tn s cao v siu cao

U(fx) Khi

ghp tn

hiu

Ch thCH

MchCH



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Trang 105

g

Nguyn tc chung: da vo nguyn lchn lc tn s ca mch cng hng.

Khi c bn ca tn s mt ny l mch cng hng.Mch ny c kch thch bng dao ng ly tngun c tn s cn o thng qua Khi ghp tn hiu.

Vic iu chnh thit lp trng thi cng hng nh dng Khi iu

chun. Hin tng cng hng c pht hin bng Khi ch th cng hng.Khi ny thng l Vnmt tch sng.

Tu theo di tn s m cu to ca mch cng hng khc nhau. C 3

loi mch cng hng: Mch cng hng c L, C tp trung Mch cng hng c L, C phn b Mch cng hng c L phn b, C tp trung.

iu

chun

Hnh 5-5

1. Tn smt cng hng c tham stp trung

+ C v L u l cc linh kin c thng stp trung B phn iu chnh cng hng chnh



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www.ptit.edu.vn GING VIN: KS. Mnh HB MN: KTT - KHOA KTT1

Trang 106

tp trung. B phn iu chnh cng hng chnhl t bin i C c thang khc theo n v tn s.

+ Ufx c ghp vo mch cng hng thng qua

cun ghp Lg.+ Mch ch th cng hn g l mch ghp h cmgia cundyL2 v L v c tch sng bng itv ch th bng c cu o t in.

+ Khi o ta a Ufx vo v iu chnh t C mchcng hng. Khi c cu o s ch th cc i.

Lg L

L2

Ufx C

D

T iuchnh

Ch th cnghng

+ Tn s mt loi ny thng dng trong di sng: 10 kHz 500 MHz.+ Sai s: (0,25-3)%

Hnh 5-6

LC

fx

=2

1

PV

2. Tn s mt c ng hng c tham sphn

bdng cp ng trc.



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Trang 107

P

Vg

Vlt

+ Cc ch ghp u gn v tr ni tt c nh sao cho cc v tr ny gnvi v tr bng sng khi c chiu di tng ng ltd=/2 ththit b chth s ch cc i.+ Khi dch chuyn pt tng vi di bng bi s nguyn ln /2 s t

cng hng c th xc nh bc sng bng cch ly 2 im cnghng ln cn l1=n/2; l2=(n-1) /2 l1-l2=/2+ Kt qu bc sng o c ca tn hiu siu cao tn x bi cng thc:

=2(l1-l2)

Hnh 5-7

+ y mch cng hng l 1 on cp ngtrc c ni tt 1 u, u kia c ni bng 1 pttng P c th dch chuyn dc trc bi h thngrng ca xon cckhc .+ vng ghp Vg a t/h vo, cn vng ghp Vghp t/h ra mch ch th cng hng.

+Bc sng (hoc tn s) c khc trc tip trn h thng iu chnh pttng.

+Tn s mt loi ny thng dng trong di sng t 3cm - 20cmD h h ht (kh 5000) i kh 0 5%



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Trang 108

+Do c h s phm cht cao (khong 5000) nn sai s ca n khong 0,5%.3. Tn smt cng hng c tham sphn bdngng dn sng

PVlt

G

+ ng dn sng c th l loi ng dn sng chnht hay ng dn sng trn.+ Piston P c th iu chnh dc theo ng bi hthng rng ca xon c c khc tn s.

Nng lng kch thch hc cng hng c ghpqua l hng G trn thnh c ni tt ca ng.+ Khi iu chnh piston P c ltd=n/2 th thit bch th s ch cc i.

+ Tn s mt vi hc cng hng thch hp vi disng nh hn 3cm.+ Do c h s phm cht cao (khong 30000) nnsai s ca n n h khong (0,010,05)%.

Hnh 5-8

5.3.o tn s bng phng php so snh

Phng php qut sin:MHS t ch khuch i



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Trang 109

- Khi : vi nY, nX nguyn dngnY : s giao im ca ng ct dc vi nhnX : s giao im ca ng ct ngang vi nh

- Tng qut: f X : tn s a vo knh lch ngang XfY : tn s a vo knh lch ng Y

X

Y

x

m

nn

ff =

X

Y

Y

X

n

n

f

f=

- MHS t ch khuch i.- in p c tn s cn o Ufx c a vo knhY, in p c tn s mu Ufm a vo knh X.

- Hnh nh nhn c trn mn l hnh Lixazu.Thay i fm sao cho trn mn nhn c hnhLixazu n nh nht.

nY=4, nX=2

Hnh 5-9

5.4.o tn s bng phng php s

L p2 hin i v thng dng nht o tn s



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Trang 110

u im:

chnh xc cao nhy ln Tc o ln, t ng ho hon ton trong qu trnh o Kt qu o hin th di dng s

Phn loi: Phng php xc nh nhiuchuk Phng php xc nh mt chuk

5.4.1. Phng phpxcnh nhiu chu k

Mchvo

To d ngxung

B mxung

fxU

dkU

xU dU

ngxa

xN

Kha


Bmxung

Mchvo

Todng

xung

KhoUfx

f T

Ux U

xung xoUk

Bmxung

Mchvo

Todng

xung

KhoUfx

f T

Ux U

xung xoUk


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Trang 111

t

t

t

xT

t

t{

a)

b)

c)

d)

fxU

xU

dkU

doU

xN

To xungchun

To xungiu khin

Gii m vchth

chU

Xung cht

Xun

hnh 5.10

hnh 5.11

Toxung

chun

Gii mv ch

th

Chia

tn

fch Uct Nx

xungcht

Toxung

iu

khin

Toxung

chun

Gii mv ch

th

Chia

tn

fch Uct Nx

xungcht

Toxung

iu

khin

5.4.1. Phng php xcnh nhiu chu k(tip)

a. S khi


Bmxung

Mch

Todng Kho

Ufx Ux U Bmxung

Mch

Todng

KhoUfx Ux U


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Trang 112

b. Chc nng cc khi:- Mch vo: thc hin tin x l nh phn p, lc nhiu... hoc bin i t/h

tun hon dng bt k u vo thnh hnh sin cng chu k vi t/h vo.- Mch to dng xung: bin i t/h hnh sin c chu k Tx thnh t/h xungnhn n cc tnh c chu k Tx.

Hnh 5-10: S khi my m tn theo phng php xc nh nhiu chu k

Toxungchun

xung

Gii mv ch

th

vodngxung

Chiatn

fch UctNx

xungcht

Toxung

iukhin

xung xoUk

Toxungchun

xung

Gii mv ch

th

vodngxung

Chiatn

fch UctNx

xungcht

Toxung

iukhin

xung xoUk

b. Chc nng cc khi: (tip)

- To xung chun: thng gm c b dao ng thch anh,



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Trang 113

cc b chia hay nhn tn s, v b to dng xung. B daong thch anh to ra cc xung t/g c chnh xc cao vi

tn s f0, xung chun ny c a qua b chia tn tora xung c tn s l fch = f0/n = 10k (Hz).- To xungiu khi n: nhn t/h Uch v to ra xung /k ng

mkho c rng t = Tch=10-k(s)- Mch gi i m v ch th: Gii m xung m c v avo cc c cu ch th s, c th l dng Led 7 on hoc

LCD ch th kt qu cn o.- B m: m cc xung u ra.

c. Nguyn l lm vic:

-Trong t/g c xung iu khinkho s c m, xung mkh k h th h h b

ta)fxU



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Trang 114

Hnh 5-11: Gin thi gian

qua kho kch thch cho bm xung.

-Gi s trong 1 chu k m t,m c Nx xung. S xung Nxny s c a qua mch giim v ch th hin th kt

qu l tn s cn o

t

t

xT

t

t{

b)

c)

d)

xU

dkU

doU

xN

Nu chn (Hz) th (Hz)kchf = 10

xx x

x

Nt = N T =

f

xx

Nf =

t

c h

ch

1 t = T =

f

xx x c h

ch

Nf = = N .f

Tvy

, vi

k

x xf = 10 .N

d.nh gi sai s:- Sai s ca xung /k (t) do sai s ca b to xung chun v b to/k



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xung /k gy ra.- Sai s lng t: sai s tuyt i l 1 xung m,

sai s tng i l 1/Nxfx tng Nx tng 1/Nx gim.fx gim Nx gim 1/Nx tng.

- Khi f x nh nh hng ca sai s lng t s ln trong trng hp

ny ta s chuyn sang phng php o xc nh 1 chu k.

5.4.2. Phng php xcnh mt chu k

fxU



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hnh 5.13

t

t

t

{

xT

T

t

t

xU

dkU

chU

doU

xt = nT(n = 1)

xN xung

Hnh 5-12

Toxung

mchun

B mxung

Giim v

ch ths

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

Uch U Nx

xung

xo

xung

cht

Hnh 5-12

Toxung

mchun

B mxung

Giim v

ch ths

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

Uch U Nx

xung

xo

xung

cht

Toxung

mchun

B mxung

Giim v

ch ths

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

Uch U Nx

xung

xo

xung

cht

a. S khi

5.4.2. Phng php xcnh mt chu k(tip)


Toxung B m

Giim v

Kh

Uch U NxToxung B m

Giim v

Kh

Uch U NxToxung B m

Giim v

Kh

Uch U Nx


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b. Chc nng cc khi:-Mch vo: Thc hin tin x l nh phn p, lc nhiu... hoc bin i tn

hiu tun hon dng bt k u vo thnh hnh sin cng chu k vi tnhiu vo .-Mch to dng xung: Bin i tn hiu hnh sin c chu k Tx thnh tnhiu xung nhn n cc tnh c chu k Tx.

Hnh 5-12

mchun

xung ch th

s

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

xung

xo

xung

cht

Hnh 5-12

mchun

xung ch th

s

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

xung

xo

xung

cht

mchun

xung ch th

s

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

xung

xo

xung

cht

b. Chc nng cc khi: (tip)

- To xung chun: To ra cc xung thi gian chun c



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g g g chnh xc cao vi tn s fch. Cc xung ny c s dng lm

xung m a ti kha.- To xungiu khi n: Nhn tn hiu t mch to dng xung vto ra xung iu khin ng mkho c rng t = nTx (s).- B m:m cc xung u ra, xung m ly t b to xungm chun.- Mch gii m v chth: Gii m xung m c v a vocc c cu ch th s, c th l dng Led 7 on hoc LCD

ch th kt qu cn o.

c. Nguyn l lm vic:- T/h Ufx a qua Mch vo ti Bto dng xung to ra xung nhn cchu k Tx. Xung ny s /k B to

fxU

t



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Trang 119

x g y dng xung/k to ra xung /k c rng t=nTx (VD: n=1)

- Trong t/gian c xung t, xung mchun Uch qua kho kch thch cho bm xung.- Gi s m c Nx xung th s

xung Nx ny s c a qua mchgii m v ch th t c kt qul tn s hoc chu k cn ot=Tx=Nx.Tch, vi Tch l chu k caxung m chun

fx=1/Tx = 1/NxTch = fch/Nx

Hnh 5-13 Gin thi gian

t

t

t

{

xT

T

t

xU

dkU

chU

doU

xt = nT(n = 1)

xN xungNu chn (Hz) th (Hz)

vi k = 0, 1, 2,...

k

chf = 10

ch

k

x

x

10f =

N

d.nh gi sai s:- Do sai s ca xung m

- Do sai s lng t (1/Nx)Kt hp 2 p2 o trn to ra 1 my



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p p ym tn c di tn o rng v chnhxc cao.

5.4.3. Gim nh sai s1 xungm

Cc phng php lm gim nh sai s 1 xung m:

- Tng tn s chun- Tng chu k xung tn hiu o q.Tx- Phng php m ni suy

t

xT

a)

b)

t

Phng php m ni suy

5.4.3. Gim nh sai s1

xungm (tip)



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-Tn hiu c dng hnh sin,c tn s fx cn o nh hnh5.14a. Tn hiu ny cthc hin bin i dngthnh chui xung nhn, ccng chu k Tx bi b todng xung nh hnh 5.14b.

t

t

t

t

1t

t

t

2 xt = nT

{

n

{

N

cT

t

c)

d)

e)

f)

g)

h)

-Mt tn hiu xung ca thnht c rng , c to ra

t b to ca logic (hnh5.14c). Tn hiu ny csinh ra l c lp vi chuixung trn.

5.4.3. Gim nh sai s1 xungm (tip)

Phng php m ni suy-Trong khong thi gian bng rng xung ca , s lng xung tn hiu cn o1t



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Trong khong thi gian bng rng xung ca , s lng xung tn hiu cn ou vo l n (hnh 5.14d), c m v ghi gi s liu li trong b nh.

Ta c:

Tr s l khc vi tr s fx cn o bi c s a i s xung m.Khixung tn hiu ca thhaic cha y xung tn hiu m Tx (hnh 5.14g),

v s lng xung m c l N, th cng c ghi gi li (hnh 5.14h).Ta c:

Do :

x

1

nf = t

xf

2 xt = n.T

1

cx

c x

n.fn.TN = =T f

x c

nf = f

N

5.4.3. Gim nh sai s1 xungm (tip)

nh gi sai sphng php m ni suy s



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5.5.o di pha

( )111 sin += tUu m

( )222 sin += tUu m



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Cc phng php:

phng php v dao ng

phng php bin i di pha thnh khong thi gian

phng php bin i di pha thnh in p; ...

( )222 m

21 =

5.5.1. o di pha bng pp o khong thigian

L phng php ph bin o pha



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T= 2 T=v

T

T= 00 360

T

T= 2 (rad) hay

Nguyn l:

+ Bin i cc in p c dng hnh sinthnh cc xung nhn tng ng vi ccthi im m in p bin i qua gi tr 0vi gi tr o hm cng du.

+ Khong thi gian gia 2 xung gn nhauca 2 in p o t l vi gc di pha cachng.

Hnh 5-14

Pha mt dng mch a hing b



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Cc in p hnh sin cn o di pha c a vo 2 u vo I v II in p hnh sin c bin i thnh cc xung vung nh Mch K hn ch v

a hing b, ri c a n Mch vi phn phn b. (Cc chu k dao ng bn thn ca ba hi c chn sao cho n ln hn

chu k ca in p o c tn s thp nht)

u ra ca Mch vi phn phn b l cc xung nhn, c a ti khng chhai b a hing b I v II.

u ra ca 2 ba hi ny c a ti mt mch tng hp, mch ny cng ho thi gian lch gia cc xung, cng l gc di pha ca 2 in p.

Hnh 5-15

- Mch vi phn phn b:

u ra ca n a ti u vo Ba hi ng b I ch cc xung



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nhn dng (hnh c) tng ngvi sn trc ca xung vungng th nht v cc xung nhnm (hnh d) tng ng vi snsau ca xung vung ng th 2

a ti B a hi ng b II chcc xung nhn dng (hnh d) cang th 2 v cc xung nhn m(hnh c) ng th nht

xc nh rng ca cc xunga ra (hnh , e)

mI

I000 180=T

TII m

=

20

Hnh 5-16

5.5.2. Pha mt chthsa.Chc nng cc khi:

Mch vo: thc hin tin x l tn hiu vo, lc nhiu. To dng xung: bin i tn hiu vo to ra cc xung o cc tnh c chu kT=chu k tn hiu vo Ux1, Ux2.



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x1, x2 Trigger: to ra xung vung c rng v chu k T chnh l nhUx1, Ux2(Ux1

c a vo u thit lp S ca Trigger, Ux2

c a vo u xo R caTrigger).

To xung m chun c chu k Tch . To xung o: chia tn s xung m chun to ra xung o c rng To.

Mchvo 1

Mchvo 2

To dngxung

To dngxung

Trigger

To xungchun

To xungo

Gii mv

ch th

B mxung

Kho

1

Kho

2

U1(t)

U2(t)

Ux1

Ux2

UT

Uch

U

o

xung cht

xung

xo

Nx

UUnx

Hnh 5.17- S khi ca Phamt s

Mchvo 1

Mchvo 2

To dngxung

To dngxung

Trigger

To xungchun

To xungo

Gii mv

ch th

B mxung

Kho

1

Kho

2

U1(t)

U2(t)

Ux1

Ux2

UT

Uch

U

o

xung cht

xung

xo

Nx

UUnx

Hnh 5.17- S khi ca Phamt s

b/ Nguyn l lm vic:

Xung UT t Trigger s iu khin ngm kho 1. Mi khi c xung, xung mUch t b to xung m chun s c


t

Ux1

t

Uth

U1U2

t

Ux1

t

Uth

U1U2

t

Ux1

t

Uth

U1U2

t

Ux1

t

Uth

U1U2


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Trang 129

ch g a qua kho 1 v u ra ca kho 1 lxung U

nxl 1 chui gm nhiu nhm

xung m v c a vo kho 2. Xung o Uiu khin ng m kho 2

trong thi gian c xung o To . Gi s c h nhm xung c a qua

kho 2 vo kch thch cho bm xung,tng s xung m c l Nx, s xung Nxny c a qua mch gii m v chth hin th kt qu l gc lch pha cn

o. Ta c gc lch pha gia 2 tn hiu U1(t)v U2(t) l

t

t

t

t

t

Nx xung

Ux2

UT

Uch

Unx

Uo

t

U

To

n xung

TT

Hnh 5.18 - Gin thi gian

t

t

t

t

t

Nx xung

Ux2

UT

Uch

Unx

Uo

t

U

To

n xung

TT

t

t

t

t

t

Nx xung

Ux2

UT

Uch

Unx

Uo

t

U

To

n xung

T

t

t

t

t

t

Nx xung

Ux2

UT

Uch

Unx

Uo

t

U

To

n xung

TT

Hnh 5.18 - Gin thi gian

,chnTT=

0360 .T

T

=

Nx

(n l s xung ca 1 nhm xung, Tch l chu k xung m chun).

T =h T



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Trang 130

x

do

ch NT

T .360=

hn x=To =h.T,

c. nh gi sai s:

Do sai s ca Tch. Do sai s lng t :

Sai s do khng ng nht ca knh 1, knh 2 l Khc phc:

+ a tn hiu U1(t) hoc U2(t) vo c 2 knh, gi s Phamt ch th gi tr l, ta c:

+ Qu trnh hiu chnh ny c th c thc hin nh b m xung thunnghch.

h

1n

1 ,

'' = do'

dodo =

+= '' do

5.5.3. o di pha bng phng php v dao ng1. Phng php dng qut tuyn tnh:

( )111 sin += tUu m

( )sin += tUu

U1

U2



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Trang 131

2. Phng php Lixazu:Gi thit o di pha ca t/hiu qua mt M4C. Phng php ny c th s dng

Oxilo 1 knh hoc 2 knh. Gi s ta s dng xil 2 knh, s nh hnh 5-20.

+iu chnh xil lm vic ch qut Lixazu:Chn chuyn mch X-Y

Vert.Mode CH2 = UCH2Knh Y

Source CH1 = UCH1Knh X

( )222 sin += tUu m

T

T

== .36021

T

T

Hnh 5.19

M4C

UV Ur

U1(t) U2(t)CH1 CH2

Hnh 5.20

M4C

UV Ur

U1(t) U2(t)CH1 CH2

M4C

UV Ur

U1(t) U2(t)CH1 CH2

Hnh 5.20

+ iu chnh h s lch pha nhn c dao ng Lixazu nmchnh gia v trong gii hn mn hnh.

Volts/div (CH1 v CH2)POS-Y (CH1)POS X

Ax

y

Y



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POS-XDao ng s c dng ng thng hoc ng Elip.

B

YMAX

XMAX

Hnh 5.21

+ Xc nh gc trung tm ca dao ng : acc chuyn mch kt ni u vo ca c 2 knh vv tr GND, trn mn hnh s l 1 im sng, dch

chuyn im sng v chnh gia mn hnh.+ a cc chuyn mch kt ni u vo v v tr AC, khi s nhnc dao ng c dng ng thng hoc Elip.

+ Xc nh gc lch pha:

=

===

maxmaxmaxmax

arcsinarcsinsinX

B

Y

A

X

B

Y

A

1. o dng in: mt chiu, xoay chiu

Chng 6.o dng in v in p


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2. o in p: mt chiu, xoay chiu

3. ng h vn nng: tng t, s

6.1.o dng in6.1.1. o dngin 1 chiu bng Ampe mt tin

Dng co: Ampe mt tin, c mc ni tip vi mch c dng in cn osao cho ti cc dng dng i vo v ti cc m dng i ra khi ampe mt.



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sao c o t cc d g d g o t cc d g a a pe t Yu cu: ni trRA nhm bo ampe mt nh hng rt t n n tr s dng

in cn o Ampe mt tin: lch ca kim t l thun vi dng in chy qua cun dy. o I ln mc in trsn vo mch o:

Io max = IA max + IS max

Ta c: IS max.RS = IA max.RA

S

SA

A

AS

S

A

A

S

R

RR

I

II

R

R

I

I +=

+=

max

maxmax

max

max

S

A

A

do

R

R

I

I+=1

max

max

max

max

A

do

I

In =; :h smrng thango

Hnh 6.1Hnh 6.1

Thay i RS bng cc gi tr khc nhau cc thang o khc nhauV d

1=

n

RR AS



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V d

Ampe mt nhiu thango

-Thay i v tr CM ( B, C, D)o c cc dng c tr s khc nhauCh : sdng cng tcng ri ctdng c khng b mt sn trnhdng qua qu ln gy hng

Hnh 6.2 - Ampe mt nhiuthang o n ginHnh 6.2 - Ampe mt nhiuthang o n gin

-Sn Ayrton: bo v cun dy ca khi b dng qu ln khi CM giacc sn-Phn tch:

CM B: RA // (R1 nt R2 nt R3)CM C: (R nt R ) // (R nt R )



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Trang 136

CM C: (RA nt R3) // (R1 nt R2)CM D: (R

A

nt R2

nt R3

) // R1

Hnh 6.3 Ampe mt nhiu thang o dng sn AyrtonHnh 6.3 Ampe mt nhiu thang o dng sn Ayrton

Sai sdo nhit:-Cun dy trong dng co TNCVC c qun bng dy ng mnh, v

in trca n c th thay i ng k theo nhit - I chy qua cun dynung nng nRcun dy thay i sai s php odng



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Trang 137

g-Khc phc: mc Rb bng Mangan hoc Constantan vi cun dy

(Mangan hoc Constantan c h sin trph thuc t0 bng 0)

nu Rb = 9 Rcun dy RA = Rb + Rcun dy = 10Rcun dy th khi Rcun dy thay

i 1% s khin cho RA thay i 0,1%RS cng c lm bng Mangan hoc Constantan trnh s thay i intrtheo t0

Hnh 6.4 - Mc in trb gim sais do nhit trong ampe mt


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Ch : dng qua ccu o khngc vt qu IA max- c cc thang o khc nhau cu to b bin dng vi cun th cp cnhiu u ra.



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Trang 139

Hnh 6.6 Ampe mt in t nhiu thang oHnh 6.6 Ampe mt in t nhiu thang o

6.2.o in p

6.2.1. cim & yu cu

- Php o d tin hnh, thc hin nhanh chng, chnh xc cao.

- Khong gi trin p cn o rng (vi V-vi trm KV), trong di tn s( i % H h h MH ) d i hi d t hi i



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g g p g ( ) gr

Cso Do Luong Dtu

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