Bài tập dài đo lường
Transcript of Bài tập dài đo lường
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Bi tp ln o Lng TTCN Khoa in t
Cho mt trm bin p c cc tham s nh bng 1, bng 2.Bng 1: Tham s ca trm bin p
Ph ti Loi s Ch non ti cos28000KVA-35/10KV III U=Um; I=0.72Im; cos=0.75 0.92
Bng 2: Cp chnh xc ca cc thit b o v yu cu truyn dnCp chnh xc cc thit b S
knh chnh
xcPhng php
truyn dnTI TU V A KWh KVARh2 2 2 2 2 1.5 15 2 Thi gian
Phn I:Cu 1: V s o dng, p, cos, tn s, nng lng tc dng, nng lng
phn khng cho trm pha cao pS III: cng t tc dng 3 pha 3 phn t, cng t phn khng 3pha 2 phn t to gc lch pha 60o.
+ Cng t tc dng:.
AI,
.
AU.
BI ,.
BU.
CI ,.
CU+ Cng t phn khng:
.
AI,
.
ABU.
BI ,.
CAU
Chng minh:
+Cng t o nng lng tc dng
P3f = PA + PB + PC= UAIAcos(.
AU.
AI ) + UBIB cos(.
BU.
BI ) + UCIC cos(.
CU.
CI )
+Cng t o nng lng phn khngMq=K(ABA sin1 + BCA sin2 )
Trong :A=CAIA 2=(BCA)
B=CBIB 1=(ABA)BA=CBAUCA
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CA=CCAUCA1=210o-=>sin1=sin(30o- )2=150o-=>sin2=sin(30o+)
=>Mq=KIdUd[sin(30o+) + sin(30o-
)] =KIdUd2cos30osin= KIdUdsin=KQ3f (pcm)
Cu 2: Chn thang o cho cc thit btrn s 1, Chn thang o bin dng in.
Dng in nh mc pha s cp l:
( )28000
461,883. 3.35
dmdm
dm
SI A
U= = =
Mun o c dng in ny ta phi s dng my bin dngI
B vi yu
cu t ra: Iscm 461,88 (A); Itcm= 5 (A)
Vy chn BI c : KI =500
52, Chn thang o cho bin in p .
in p nh mc pha s cp ca my bin p l: UmBA=35 (KV)
Cho nn chn BU c h s bin in p: KU=35000
100=
35
0,1
3, Chn thang o cho ampemet.V dng in pha th cp ca BI nh mc l 5(A) nn chn ampemet c
thang o l: DA=5(A)
Dng ampemet in t:
Ampemet in t c ch to da trn c cu ch th in t.Ampemetny o c c dng mt chiu v dng xoay chiu
B phn chnh l mt cun dy dn, c th quay quanh mt trc, nm trongt trng ca mt nam chm vnh cu. Cun dy c gn vi mt kim ch gcquay trn mt thc hnh cung. Mt l xo xon ko cun v kim v v tr skhng khi khng c dng in
Nguyn l hot ng:Khi dng in mt chiu chy qua cun dy, dng in chu lc tc ng
ca t trng (do cc in tch chuyn ng bn trong dy dn chu lcLorentz) v b ko quay v mt pha, xon l xo, v quay kim. V tr ca u
kim trn thc o tng ng vi cng dng in qua cun dyy l mt thit b thc t o dng in c bn ngoi th trng:
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=> DA=5(A)
4, Chn thang o cho volmet.V in p pha th cp ca BU nh mc l 100(V) nn chn volmet c
thang o l: DV=100(V)
Cu to: Mch t gm li t bng tn silic c c tnh t tt v cun dycch in cao. C cu chnh Kim v 0 bng c kh. Cn du n nh kim bngnam chm vnh cu, bng nha baklit c cch in cao. V bng nhaABS. Ca s bng nha PC trong sut (hn ch chy). 2 ci v d cho viclp t Vn mt vo bng in.
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=>DV=100(V)
5, Chn thang o cho cosmet .V cosmet c cun dng mc vo th cp BI , cun p mc vo th cpBU nn ta chn cosmet c:
Imcosmet = 5(A)Umcosmet = 100(V)
Do gc lch pha gia in p v dng in khng ph thuc vo h sbin i ca BU v BI nn ta chn thang o ca cosmet l (0.5 1)
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Nguyn l hot ng:Cun tnh c mc ni tip vo pha A, hai cun dy ng c mc vi
hai in tr R v c t vo cc in p UAB v UAC
Gc quay ca c cu l:
Ch :
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+ Trong s ny cc cun p u ni tip vi in tr R nn khng
ph thuc tn s, hay 21
I
I = const.
+ Cun dng c th mc vo cc pha B, C ty .
6, Cng t o nng lng tc dng .V cng t o nng lng tc dngc cun dng mc vo th cp BI ,cun p mc vo th cp BU nn tachn:
Idm = 5(A)Udm = 100(V)=0,1(kV)
.3
W P ttddm pha dm
=
Vi Cos dm = 1, tdm = 30 ngy =720h
3.0,1.5.1.720 623,54( )W kWhtddm
=
Chn thang o cng t nng lng tc dng DWtd=625 KWh
7, Cng t o nng lng phn khng.V cng t o nng lng phn khng c cun dng mc vo B I , cun p
mc vo BU nn ta chn:Imctpk = 5(A)Umctpk = 100(V)=0,1(kV)
3 . 3. . .720pha dm dm dmW Q t U I Sinpkdm= = (ly Sin=1)
3.0,1.5.1.720 623,54Wpkdm
= AR(kV h)
Chn thang o ca cng t phn khng DWpk= 625 KVARh
8, Chn thang o cho tn s met.V thit b o tn s mc vo th
cp ca BU nn ta chn : Um = 100(V).Vi tn s ca mng in l 50HZ v t dao ng nn ta chn thang o trongkhong 45 55 HZ
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3. . . .W U I Cos t tddm dm dm dm=
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PhotoHnh nh
Cat.NoM s
TypePhn loi
RangeThang o
NG H O TN S - HIU MUNHEAN SINGAPORE Mt 96 x96 mm
CP-96FQP
Kim 220V ; 45 55Hz
ng h o tn s met cng hng
Nguyn l hot ng:Tn s mt cng hng gm mt nam chm in,to ra bi cun dy qun
trn li st t hnh ch U,mt ming thp nm trong t trng ca nam chmin,gn cht vo thanh l co l thp rung c tn s dao ng ring khcnhau. Tn s dao ng ring ca hai l thp k nhau hn km nhau l 0,25hoc 0,5Hz.in p ca tn hiu cn do tn s s c a vo cun dy canam chm in s to ra s dao ng ca tt c cc l thp.Tuy nhin l thpno c tn s dao ng ring bng tn s f th s dao ng cc i do cng
hng ring,cn cc thanh khc khng cng hng th khng dao ng cci.Nh vy ta s c kt qu ti tr s tng ng vi thanh rung cc i.
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Cu 3: Tnh s ch ca mi cng t trong thi gianmt thng, bit rng c80% thi gian my bin p lm vic ch nh mc , 20% thi gian my
bin p lm vic ch non ti. Xc nh cos TB ca ph ti.
1, Tnh s ch ca cng t o nng lng tc dng.a, S ch ca cng t tc dng trong ch nh mc:Thi gian nh mc:
Tm = 0,8.24.30 = 576(h)Im = IAm = IBm = ICm = 461,88(A)Um= 35(KV)cos m = 0,92 m = 23,074o
Ta c:
. .3 . .cos .,
dmIU AAW U I T AActtddm dmdmK K
U I
=
=35.0,1 461,88.5
3. . .0,92.57650035. 3
( )423,94 KWhb, S ch ca cng t tc dng trong ch non ti:
Thi gian non ti:Tnt = 0,2.24.30 = 144(h)Int = IAnt = IBnt = ICnt = 0,72.Idm =0,72.461,88 = 332,554(A)Unt = Udm = 35(KV)cos nt = 0,75 nt = 41,41o
Ta c:
. .
,cttdntW 3 . .cos .nt
nt
AAAA nt
U I
IUU I T
K K
=
=35.0,1 332,554.5
3. . .0, 75.14450035. 3
( )62,21 KWhc, S ch ca cng t tc dng trong mt thng.
( )cttd cttddm cttdntW W W 423, 94 62, 21 486,15 WhK= + = + =
2, Tnh s ch ca cng t o nng lng phn khng .a, S ch ca cng t phn khng trong ch nh mc.
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Thi gian nh mc:Tdm = 0,8.24.30 = 576(h)Idm = IAdm = IBdm = ICdm = 461,88(A)Udm= 35(KV)
cos dm = 0,92 sindm= sin(arccos dm) = 0,392
Ta c:
ctpkdm
35.0,1 461,88.5W 3. . . . 3. . .0,392.576
35 500
dm dmdm dm
U I
U ISin T h
K K= =
( )180,63 KVARh
b,S ch ca cng t phn khng trong ch non ti.Thi gian non ti:
Tnt = 0,2.24.30 = 144hInt = 0,61Idm = 0,72.461,88= 332,554(A)Unt = Udm = 35(KV)Cosnt = 0,75 sindm= sin(arccos dm) = 0,66
Ta c:
ctpknt35.0,1 332,554.5
W 3. . . . 3. . .0,66.14435 500
nt nt nt nt
U I
U ISin T
K K= =
( )54,743 KVARh=
c,S ch ca cng t pkn khng trong mt thng.
( )ctpkW 180, 63 54, 743 235, 373ctpkdm ctpknt W W KVARh= + = + =3, Xc nh costb ca ph ti.
Ta c :
( )2 2 2 2
486,15cos 0,9
486,15 (235,373)
cttdtb
cttd ctpk
W
W W = =
+ +
Cu 4: Tnh sai s tuyt i v sai s tng i ca cc php o trong haitrng hp ph ti.**C s l thuyt tnh sai s tng i , tuyt i
chnh xc l l tiu chun quan trong nht ca thit b o
xx dii =Trong :xi l kt qu o ca ln o th i
xd l gi tr ng ca i lng o i l sai lch ca ln o th i
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Sai s php o l sai lch gia gi tr thc v gi tr o c , sai sphp o ch c th xc nh 1 cch tng i v ta khng th bit c gi trthc ca i lng cn o
Sai s ca php c th biu din lm 2 dng
+Sai s tuyt i+Sai s tng iSai s tuyt i ca mt thit b o c nh ngha l gi tr ln nht ca
sai lch gy nn bi thit b trong khi o.]max[ i=
Tuy nhin sai s tuyt i cha nh gi c chnh xc v yu cu cacng ngh thit b o.Thng thng chnh xc ca mt php o hoc mtthit b o c nh gi bng sai s tng i
Sai s tng i c tnh vi mt php o
= vi x l gi tr i lng o
Sai s tng i c tnh vi mt thit b o
D=
vi D l khong gi tr nh nht ti gi tr ln nhtGi tr % gi l sai s tng i quy i dng xp xp cc thit b o
thnh cp chnh xcTheo quy nh hin hnh ca nh nc, cc dng c o c in c cp
chnh xc :0,05;0,1;0,2;0,5;1;1,5;2;2,5 v 4Thit b o s c cp chnh xc: 0,005;0,01;0,02;0,05;0,1;0,2;0,5;1Vy khi bit cp chnh xc ca mt thit b o ta c th xc nh c sai
s tng i quy i v suy ra sai s tng i ca thit b trong php o cth
X
D =
*Tnh ton sai s ca php o gin tip:
Gi s c 1 php o gin tip i lng y thng cc php o trc tip 1x , 2x
,.. nx . y=f( 1x , 2x ,.. nx ).
Ta c:
dy= 1 2. . ... . ny y y
dx dx dxx x x
+ + +
- Sai s tuyt i ca php o gin tip c nh gi:
y= 2 2 2 21 211 2
( . ) ( . ) ... ( . ) ( . )n
n k
kn k
y y y yx x x x
x x x x=
+ + + =
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1x , 2x ,.. nx : Sai s tuyt i ca php o cc i lng trc tip 1x , 2x ,..
nx .
- Sai s tng i ca php o gin tip c nh gi:
y =y
y
=
2 2 2 2 2 2 2 2 21 21 2
1 2
( ) .( ) ( ) .( ) ... ( ) .( ) ...nn
n
xx xy y yx x x
y x y x y x
+ + + = + + +
1x , 2x ,.. nx : Sai s tng i ca php o trc tip cc i lng 1x ,
2x ,..
n
x .
-Bng tnh sai s ca hm y= 1x . 2x
Hm y Sai s tuyt i y Sai s tng i y = y/y
1x . 2x2 2 2 2
1 2 2 1( ) ( )x x x x +
2 21 21 2
( ) ( )x x
x x
+
1, Sai s ca cc php o ch nh mc.a, Sai s ca php o dng in.
Dng in qua cc pha A,B,C ( ba pha nh nhau ) c xc nh.Idm = KI.IAdm
Vi IAm l dng m ampemet o c ch nh mc.
( )461,88
.5 4, 62500
dmAdm
I
II A
K= = =
Ta c:Sai s tuyt i ca ampemet l:
( )ADI AAAdm 1,05.100
2. ===
Sai s tuyt i ca bin dng in l:
2 500
. . 2100 5
I KI IK K = = =
Vy sai s tuyt i ca php o dng in ch nh mc l :
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( ) ( ) 22 .. AdmIAdmIdm IKIKI +=
( ) ( )
( )
2 22.4,62 100.0,1
13,62 A
= +
Sai s tng i ca php o dng in ch nh mc l:
%13,62
.100% .100% 2, 95%461,88dm
dmI
dm
I
I
= = =
b, Sai s ca php o in p.Sai s tuyt i ca bin in p:
2 35. . 7
100 0,1U KU U
K K = = =
Sai s tuyt i ca volmet :
( )VDU VVV 2100.100
2. ===
Kt qu o in p dy :
( )KVUKU VUd 35100.1,0
35. ===
Vy sai s tuyt i ca php o in p dy l:
( ) ( ) 22 .. VUVUd UKUKU +=
( ) ( )
( )
2 27.100 350.2
989,95 V
= +
Sai s tng i ca php o in p dy:
%989,95
.100% .100% 2,83%35000d
dU
d
U
U
= =
Sai s tuyt i ca php o in p pha:
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( ) 22
.3
. VUV
Uf UKU
KU +
=
( )
( )
2
21007. 350.23
808,29 V
= + =
Sai s tng i ca php o in p pha :
%
808,29.100% . 3.100% 4%
35000ff
U
f
U
U
= =
c, Sai s ca php o nng lng tc dng.Nng lng tc dng ca ph ti ch nh mc l:
( )tddm cttddm35 500
W . .W . .423,94 14837900 Wh0,1 5
U IK K K= = =
Sai s tuyt i ca cng t tc dng ch nh mc:
cttddm WW .cttd td D =
Ta c : ( )W 625tdD KWh=
( )cttddm2
W .625 12,5100
KWh = =
Vy sai s tuyt i ca php o nng lng tc dng ch nh mcl :
( ) ( ) ( )
222
tddm ......W cttddmIUcttddmIUcttddmIU WKKWKKWKK ++=
( ) ( ) ( )2 2 2
7.100.423,94 350.2.423,94 350.100.12,5= + +
( )606248,1927 KWh=
Sai s tng i ca php o nng lng tc dng l :
tddmW
606248,1927.100% .100% 4%
14837900
tddm
tddm
W
W
= = =
d, Sai s ca php o nng lng phn khng .
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Nng lng phn khng ca ph ti ch nh mc l :
( )pkdm ctpkdmW . .W 350.100.180, 63 6322050 AR U IK K KV h= = = Sai s tuyt i ca cng t phn khng :
ctpkdm WW .
ctpk pk
D
=Ta c:( )W 625 AR pkD KV h=
( )1, 5
.625 9, 375 AR 100
ctpkdmW KV h = =
Vy sai s tuyt i ca php o nng lng phn khng l :
( ) ( ) ( )222pkdm ......W ctpkdmIUctpkdmIUctpkdmIU WKKWKKWKK ++=
( ) ( ) ( )2 2 2
7.100.180, 63 350.2.180, 63 350.100.9,375= + +
( )373685, 2534 ARKV h=
Sai s tng i ca php o nng lng phn khng l:
pkdmW
373685,2534.100% .100% 5,9%
6322050
pkdm
pkdm
W
W
= =
2, Sai s ca cc php o ch non ti.a, Sai s ca php o dng in
Dng in ch non ti c xc nh :Int = KI.IAnt
C : IAnt = 0,72.IAdm=0,72.4,62=3,3264
Sai s tuyt i ca ampek :
( )AII AdmAnt 1,0==
Vy sai s tuyt i ca php o dng in ch non ti l :
( ) ( )22 .. AntIAntInt IKIKI +=
( ) ( )
( )
2 22.3,3264 100.0,1
12,01 A
= +
=
Sai s tng i ca php o dng in ch non ti l :
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%12,01
.100% .100% 3, 61%332,554nt
ntI
nt
I
I
= = =
b, Sai s ca php o nng lng tc dng .Nng lng tc dng ca ph ti ch non ti l:
( )tdnt cttdnt35 500
W . .W . .62, 21 2177350 Wh0,1 5
U IK K K= = =
Sai s tuyt i ca cng t tc dng ch non ti :
( )cttdnt cttdntW W 12,5 KWh = =Vy sai s tuyt i ca php o nng lng tc dng ch non ti l
( ) ( ) ( ) 222tdnt ......W cttdntIUcttdntIUcttdntIU WKKWKKWKK ++=
( ) ( ) ( )2 2 2
7.100.62, 21 350.2.62, 21 350.100.12,5= + +
( )441813, 2325KWh=
Sai s tng i ca php o nng lng tc dng ch non ti:
tdntW
441813, 2325.100% .100% 20,3%
2177350
tdnt
tdnt
W
W
= = =
c, Sai s ca php o nng lng phn khng.Nng lng phn khng ca ph ti ch non ti l :
( )pknt ctpkntW . .W 350.100.54, 743 1916005 AR U IK K KV h= = =
Sai s tuyt i ca cng t phn khng ch non ti :
ctpknt WW .ctpk pk D =Ta c :
( )W 625pkD KVarh= ( )9,375 AR ctpknt ctpkdmW W KV h = =
Vy sai s tuyt i ca php o nng lng phn khng ch non ti l :
( ) ( ) ( )222pknt ......W ctpkntIUctpkntIUctpkntIU WKKWKKWKK ++=
( ) ( ) ( )2 2 2
7.100.54, 743 350.2.54, 743 350.100.9,375= + +
( )332570,1065 ARKV h=
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Sai s tng i ca php o nng lng phn khng ch non ti l:
pkntW
332570,1065.100% .100% 17,36%
1916005
pknt
pknt
W
W
= = =
Phn II: o lng v bo v cho trm ngi ta cn thu thpcc tn hiu in v khng in. Hy v s truyn dn chotrm.
II.1. Thu thp tn hiu :II.1.1.M hnh
Ta s dng m hnh ni tip songII.1.2. S lng knh: 15
- Knh 1 (S1) : Tn hiu dng pha A : IA
- Knh 2 (S2) : Tn hiu dng pha B : IB
- Knh 3 (S3) : Tn hiu dng pha C : IC
- Knh 4 (S4) : Tn hiu p pha A : UA
- Knh 5 (S5) : Tn hiu p pha B : UB
- Knh 6 (S6) : Tn hiu p pha C : UC- Knh 7 (S7) : Tn hiu p dy AB : UAB
- Knh 8 (S8) : Tn hiu p dy BC : UBC
- Knh 9 (S9) : Tn hiu p dy CA : UCA
- Knh 10(S10) : Tn hiu hin th h s cos
- Knh 11(S11) : Tn hiu hin th tn s pha A
- Knh 12(S12) : Tn hiu hin th tn s pha B
- Knh 13(S13) : Tn hiu hin th tn s pha C- Knh 14(S14) : Tn hiu cng sut tc dng
- Knh 15(S15) : Tn hiu cng sut phn khng
II.1.3. Chn cc Mux
Trong thc t nhiu i lng o l cc i lng khng in, v d nh:
p sut, nhit V vy c th truyn ti cc thng tin, o c v kim
tra c ta cn phi c cc thit b chuyn i chun ha trong o lng. u
vo l cc tn hiu khng in, u ra l cc tn hiu in vi chun p 0-10V,
chun dng 4-20mA. Do s lng knh lin lc ln, v th s dng cc thit
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b i ni (MUX) thc hin phn knh lin lc, tin cho vic truyn tn
hiu v kim tra sa cha khi c s c. gim sai s ca cc b i ni,
trong s ta b tr cc b i ni theo nhm v phn theo tng, nh m
tc x l thng tin nhanh hn.- Tn hiu dng : Mux1 (2 bit)
- Tn hiu p : Mux2 (3 bit)
- Cc tn hiu cn li : Mux3(3 bit)
- Mux (2bit)
II.1.4.Lp bng trng thi
II.1.4.1.Bng trng thi ca Mux1 (2bit)
- Dng in pha A : Chn D0
- Dng in pha B : Chn D1
- Dng in pha C : Chn D2
- Chn a ch : A1, A2
Chn a chChnclock
u ra
A1 A2 C1 Q1
0 0 1 D0
0 1 1 D1
1 0 1 D2
x x 0 0
=>Q1=C1(D0. 1 2A .A +D1. 1 2A .A +D2 1 2A .A )
II.1.4.2.Bng trng thi ca Mux2 (3bit)- in p pha A : chn D3
- in p pha B : chn D4
- in p pha C : chn D5
- in p dy UAB : Chn D6
- in p dy UBC : Chn D7
- in p dy UCA : Chn D8
- Chn a ch : A3, A4, A5
GVHD: Phm Ph Thim - 17 - SVTH: Mai Vn Tuyn
-
7/30/2019 Bi tp di o lng
18/20
Bi tp ln o Lng TTCN Khoa in t
Chn a chChnclock u ra
A3 A4 A5 C2 Q2
0 0 0 1 D3
0 0 1 1 D4
0 1 0 1 D5
0 1 1 1 D6
1 0 0 1 D7
1 0 1 1 D8
x x x 0 0
x x x 0 0
=>Q2=C2(D3. 53 4A .A .A +D4. 53 4A .A .A +D5. 53 4A .A .A +D6. 53 4A .A .A +D7.
53 4A .A .A
+D8. 53 4A .A .A
)
II.1.4.3.Bng trng thi ca Mux3 (3bit)- Tn hiu hin th h s cos : chn D9
- Tn hiu hin th tn s pha A : chn D10
- Tn hiu hin th tn s pha B : chn D11
- Tn hiu hin th tn s pha C : chn D12
- Tn hiu cng sut tc dng : chn D13- Tn hiu cng sut phn khng : chn D14
- Chn a ch : A6, A7, A8
Chn a ch Chnclock u ra
GVHD: Phm Ph Thim - 18 - SVTH: Mai Vn Tuyn
-
7/30/2019 Bi tp di o lng
19/20
Bi tp ln o Lng TTCN Khoa in t
A6 A7 A8 C3 Q3
0 0 0 1 D9
0 0 1 1 D10
0 1 0 1 D11
0 1 1 1 D12
1 0 0 1 D13
1 0 1 1 D14
x x x 0 0
x x x 0 0
=>Q3=C3(D9. 76 8A .A .A +D10. 76 8A .A .A +D11. 76 8A .A .A +D12. 76 8A .A .A +D13.
76 8A .A .A +D14. 76 8A .A .A )
II.1.4.4. Bng trng thi ca Mux ( 2bit )
Chn a chChnclock u ra
A9 A10 C Q
0 0 1 Q1
0 1 1 Q2
1 0 1 Q3
x x 0 0
GVHD: Phm Ph Thim - 19 - SVTH: Mai Vn Tuyn
-
7/30/2019 Bi tp di o lng
20/20
Bi tp ln o Lng TTCN Khoa in t
9 10 1 9 10 2 9 10 3Q C (A .A .Q +A .A .Q +A .A .Q =
II.1.4.5. Bng trng thi tng hp cc Mux
Cc chn a ch Mux1 Mux2 Mux3 Mux
A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 C1 Q1 C2 Q2 C3 Q3 C Q
0 0 x x x x x x 0 0 1 0D 0 0 0 0 1 1Q
0 1 x x x x x x 0 0 1 1D 0 0 0 0 1 1Q
1 0 x x x x x x 0 0 1 2D 0 0 0 0 1 1Q
x x 0 0 0 x x x 0 1 0 0 1 3D 0 0 1 2Q
x x 0 0 1 x x x 0 1 0 0 1 4D 0 0 1 2Q
x x 0 1 0 x x x 0 1 0 0 1 5D 0 0 1 2Q
x x 0 1 1 x x x 0 1 0 0 1 6D 0 0 1 2Q
x x 1 0 0 x x x 0 1 0 0 1 7D 0 0 1 2Q
x x 1 0 1 x x x 0 1 0 0 1 8D 0 0 1 2Q
x x x x x 0 0 0 1 0 0 0 0 0 1 9D 1 3Q
x x x x x 0 0 1 1 0 0 0 0 0 1 10D 1 3Q
x x x x x 0 1 0 1 0 0 0 0 0 1 11D 1 3Q
x x x x x 0 1 1 1 0 0 0 0 0 1 12D 1 3Q
x x x x x 1 0 0 1 0 0 0 0 0 1 13D 1 3Q
x x x x x 1 0 1 1 0 0 0 0 0 1 14D 1 3Q
II.2. S truyn dn cho trm: