TRNG AI HOC S PHAM KY THUAT TP. HCM KHOA IE N BO MON. C S KY THUAT IEN ------------0----------- BIEN SOAN: ThS. LE TH THANH HOA NG BAI GIANG. MACH IEN II TP. HCM Thang 12 / 2007 K 1 k 1 C+_ k 2 k 22R1RX(P) ) P ( X1) P ( YLI NI U MCH IN l mt mn hc c s quan trng i vi sinh vin khi k thut ni chung v sinh vin ngnh in ni ring. c th tip tc nghin cu chuyn su v lnhvc in th sinh vin phi nm vngnhng kin thc trongmn hc MCH IN.Ngoiramnhcnylcnlmncschosinhvinhctipccmn chuyn ngnh khc nh mn iu Khin T ng, My in, L Thuyt Tn HiuMch in II ny bao gm ba chng :Chng I: Phn tch mch trong min thi gianChng II: Phn tch mch trong min tn s Chng III : Mch khng tuyn tnh Chng IV. ng dy di Quyn sch ny tc gi trnh by cc phng php phn tch mch c km theo ccvdcthvccbitpcsontheotngccchnglthuyt,gip ngi hc c th gii v ng dng vo cc mn hc c lin quan. Tc gi vit bi ging ny vi s c gng su tm cc ti liu trong v ngoi nc, vi s ng gp tn tnh ca cc ng nghip trong v ngoi b mn, cng vi kinh nghim ging dy mn hc ny trong nhiu nm. Tuy nhin y cng l ln u tin bin son bi ging mch in II nn khng th trnh khi nhng thiu st. Ti rt mong s nggp kin ca cc ng nghip, ca cc em sinhvinv cc bnc quan tm n bi ging ny. Xin chn thnh cm n. TP. HCM thng 12 nm 2007. Th vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vnMC LC Trang CHNG : PHN TCH MCH TRONG MIN THI GIAN (QU TRNH QU ) .......................................................................... 1 I.1. KHI NIM ............................................................................................................. 1 I.2. P DNG PHNG TRNH VI PHN GII BI TON QU (PHNG PHP TCH PHN KINH IN) ................................................................ 1 I.2.1. Gii bi ton vi iu kin ban u bng 0 ....................................................... 1 I.2.2. Gii bi ton vi iu kin u khc 0 .............................................................. 6 a. Mch c cun dy ......................................................................................... 6 b. Mch c t .................................................................................................... 8 I.3. P DNG PHNG PHP TON T LAPLACE GII BI TON QU .. 12 I.3.1. Mt s kin thc c bn bin i Laplace .................................................. 12 I.3.2. nh lut Kirchhoff dng ton t .................................................................... 16 I.3.3. S ton t Laplace ..................................................................................... 17 I.3.4. Thut ton tnh qu trnh qu bng phng php ton t ........................... 17 I.3.5. Mt s v d v cc bi ton qu vi cc iu kin ban u bng 0 ............ 17 I.3.6. Cc bi ton qu vi cc iu kin ban u khc 0 .................................... 21 BI TP CHNG I ................................................................................................. 27 CHNG II: PHN TCH MCH TRONG MIN TN S ................................. 36 II.1. NH NGHA HM TRUYN T .................................................................... 36 II.2. BIU DIN TH CA HM TRUYN ......................................................... 40 II.2.1. c tuyn logarit - tn s logarit ................................................................... 40 II.2.2. c tuyn bin - tn s logarit .................................................................. 41 II.2.3. c tuyn pha tn s Logarit ......................................................................... 45 BI TP CHNG II................................................................................................ 48 CHNG III: MCH PHI TUYN .......................................................................... 51 III.1. CC PHN T KHNG TUYN TNH ............................................................ 51 III.1.1. in tr phi tuyn ........................................................................................ 51 III.1.2. in cm phi tuyn (cun dy phi tuyn) ..................................................... 51 III.1.3. in dung phi tuyn..................................................................................... 52 III.2. CC THNG S C TRNG CA CC PHN T PHI TUYN ................ 53 III.2.1. in tr tnh v in tr ng ...................................................................... 53 Th vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vnIII.2.2. in cm tnh v in cm ng .................................................................. 53 III.2.3. in dung tnh v in dung ng ............................................................... 54 III.3. CC PHNG PHP PHN TCH MCH KTT .............................................. 54 III.3.1. Phng php th ...................................................................................... 54 III.3.2. Phng php d ........................................................................................... 55 III.3.3. Phng php gii tch .................................................................................. 57 III.4. CCH GHP NI CC PHN T KTT ............................................................ 61 III.4.1. Mc ni tip cc phn t KTT ..................................................................... 61 III.4.2. Mc song song ............................................................................................. 62 III.4.3. Cch ni cc phn t KTT vi ngun tc ng ............................................ 63 III.4.4. Mch KTT dng mt chiu .......................................................................... 64 III.5. BI TP CHNG III (Mc III.4) ..................................................................... 67 III.6. CHUI FOURIER ............................................................................................... 69 III.6.1. Chui Fourier lng gic ............................................................................. 69 III.5.2. Chui Fourier dng phc ............................................................................. 70 III.7. BI TP CHNG III (Mc III.6) .................................................................. 76 CHNG IV. NG DY DI ............................................................................. 78 IV.1. CC THNG S N V CA NG DY DI ......................................... 78 IV.1.1. nh ngha ......................................................................................................... 78 IV.1.2. Phng trnh ng dy di v nghim............................................................ 79 IV.1.3. Nghim ca phng trnh ng dy di vi tc ng sin ................................. 80 IV.1.4. Cc quan h nng lng trn ng dy di ...................................................... 83 IV.2. BI TP CHNG IV ....................................................................................... 84 IV.3. QU TRN NG DY DI .................................................................. 86 IV.3.1. Phng trnh ton t ca DD .......................................................................... 86 IV.3.2. ng in p vo ng dy h mch cui ....................................................... 86 IV.3.3. ng in p vo ng dy ti in tr ........................................................... 88 IV.3.4. th Zig Zac (gin bounce) ..................................................................... 89 TI LIU THAM KHO Th vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 1 CHNG I: PHN TCH MCH TRONG MIN THI GIAN (QU TRNH QU ) I.1. KHI NIM Qu trnh qu l qu trnh bin i dng in ban u thnh gi tr xc lp. Xt mch in nh hnh v (1.1): Trong : K l kha dng ng m mch in. Trc khi kha K ng i = 0 gi l gi tr ban u. Kha K ng trong mt thi gian di th dng in t n gi tr xc lp l i = RE Qu trnh bin i t gi tr ban u n gi tr xc lp c gi l qu trnh qu . I.2.PDNGPHNGTRNHVIPHNGIIBITONQU (PHNG PHP TCH PHN KINH IN) I.2.1. Gii bi ton vi iu kin ban u bng 0 V d 1: Cho mch in nh hnh v (1.2): Ti t = 0 ng kho K li. Tm cng dng in i(t) chy trong mch in. Li gii Khi kha K ng li: uR + uL = E (1.1.1) M: uR = iR E K R i(t) L H nh (1.1) E K R i(t) L Hnh (1.2) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 2 dtdiL uL =thay vo pt(1.1) ta c: EdtdiL iR = + (1.1.2) Vy ta phi gii phng trnh vi phn tm i(t). Gi s i l nghim ca phng trnh: i = it do + ixc lp(1.1.3) ixclp:ldngintrongmchsaukhing(hocm)khoKsau mt thi gian di. Trong mi mch in c th c mt gi tr xc lp. itdo:lnghimcaphngtrnhviphncvphibngkhng (phng trnh thun nht). (Thnh phn t do ca in p v dng in ph thuc vo nng lng tch ly trong mch v cc thng s mch, n khng ph thuc vo hnh dng ca ngun tc ng) t itd = keSt Trong : k: hng s S: s phc t: thi gian iR + Ldtdi = 0 (1.1.4) Thay vo: keStR + Ldt) d(kest= 0 0 LS) (R keSt= + nghim itd = 0 ( 0 keSt= ) R + LS = 0 LRS = LRttdke i= M: ixc lp = RE Vy:tLRkeREi(t)+ = Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 3 Xc nh k: Da vo iu kin ban u ca bi ton i(0+)= 0 Ti t = 0: 0 keREi(0)o= + = k = RE||.|

\| = = tLRtLRe 1REeREREi(t) (A) Vy: Ti t = 0 i = 0 Ti t = i = RE t RL = : hng s thi gian i(t) = ||.|

\|te 1RE Khi t = 3 th i~ ixc lp (96%) Thi gian qu l thi gian dng in i t gi tr ban u n gi tr xc lp. Cha ngngng K t t0- t0+ i(0-)i(0+) i t 0 RE Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 4 V d 2: Cho mch in nh hnh v (1.3): Yu cu: Ti t = 0 ng kha K, tm uc(t). Li gii Khi ng kha K: uR + uc = E(1.2.1) M: uc + RCdtduC = 0(1.2.2) y l phng trnh vi phn. Gii phng trnh vi phn trn tm uc(t). t: uc = uc t do + uc xc lp (1.2.3) ucxclp:linpxclptrntmtthigiandisaukhing(hoc m) kha K. uc xc lp = E (khi t c np y) uc t do: l nghim ca phng trnh vi phn c v phi bng khng. uc + RCdtduC = 0(1.2.4) t: uc t do = keSt Vy: 0dt) RCd(kekeStSt= +Trong : k: hng s S: s phc t: thi gian keSt + RCS.keSt = 0 keSt(1 + RCS) = 0 Do keSt = 0 nn: (1 + RCS) = 0 S = RC1uR = iRthay vo(1.2.1) dtduC iC=C E uc(t) i(t)K RHnh (1.3) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 5 Phng trnh trn l phng trnh c trng uc t do = kRCte u(t) = E + kRCte Xc nh k: Da vo iu kin ban u ca bi ton:uc(0) = 0Ti t = 0: uc(0) = E + ke0 = 0 k = E ||.|

\| = RCtce 1 E (t) ut= RC: hng s thi gian ca mch (n v s) Vy: uc(t) = E(1 te) -khi t = 0 uc(t) = 0 -khi t = uc(t) = E Theo bi ta tm i(t) i = CdtduC = dt) E.e d(ECRCt = RCteRCCE = RCteRE i(t) = teREvi t = RC -Ti t = 0 i = RE -Ti t = i = 0 RE t i 0 E 0tucChuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 6 I.2.2. Gii bi ton vi iu kin u khc 0 a. Mch c cun dy Cho mch in nh hnh v (1.4) Ti t = 0, m kha K. Xc nh i(0+). iukinbotontthng:Tngtthngmcvngtrongmtvngknlin tc ti thi im ng m: (0) = (0+)(1.1) -Ti t0 (0) -Ti t0+ (0+) T thng = L.i L.i(0) = L.i(0+)(1.2) Ti t0-: (0) = L1.i(0) iL1(0-) = RE iL2(0-) = 0 Ti t0+: (0+) = L1.i(0+) + L2.i(0+) = (L1 + L2).i(0+) M: (0) = (0+) L1.i(0) = (L1 + L2).i(0+) Vy 2 11L LREL) i(0+=+(1.3) L2KEi(t) R L1Hnh (1.4) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 7 V d p dng: Cho mch in nh hnh v (1.5) Ti t = 0 m K, tm i(t). Li gii Trc khi m K:3A412RE) i(0 = = = Ti t0+: A43L L) i(0 L) i(02 11=+=+ Khi m K: iR + (L1 + L2)dtdi = E : phng trnh vi phn Gii phng trnh vi phn t i = itd + ixl ixl =3RE=(A) itd l nghim ca phng trnh vi phn c v phi bng 0 iR + (L1 + L2)dtdi = 0 t itd = keSt

keStR + (L1 + L2)dt) d(keSt = 0 keSt[R + (L1 + L2)S] = 0 Do keSt = 0 nn R + (L1 + L2)S = 0 S =2 1L LR+ itd = tL LR2 1ke+ L1 = 1H i(t) 4 L2 = 3H KE = 12V Hnh (1.5) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 8 i(t) = 3 + tL LR2 1ke+ Xc nh k: i (0+) = 3 + keo = 43 k = 49Vy i(t) = 3 te49vi t = RL L2 1 + tqu = 3s dng in t gi tr n nh. Khi m kha K dng in tng ln 3A (gi tr ixl) b. Mch c t Cho mch in nh hnh v (1.6) Ti t = 0 ng kha K. Tm uc(t). Li gii Trc khi ng K: uc1(0) = E uc2(0) = 0 Ti t(0+):i 43 3 t 0 Lc m K K E R C1C2 uc(t) a Hnh (1.6) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 9 uc1(0+) = uc2(0+) = uc(0+) iukinbotonintch:intchti1 nh(nt)lintctithiimng m: q(0+) = q(0)(1.4) in tch ti a t(0) t(0): q(0) = C1.uc1(0) = C1.E t(0+): q(0+) = C1.uc1(0+) + C2.uc2(0+) = (C1 + C2).Uc(0+) q(0+) = q(0) (C1 + C2).Uc(0+) = C1.E uc(0+) = 2 11C CE C+ V d p dng:Cho mch in nh hnh v (1.7): Ti t = 0 ng K, tm uc(t). Li gii+ Tm iu kin ban u: uc(0+) =2 11C CE C+=320412110 .21=+(V) + Khi ng K li ta c: uR + uc = E Vi C = C1 + C2; uR = iR = RCdtduc RCdtduc + uc = E : phng trnh vi phn Gii phng trnh vi phn tm uc Ta t: uc(t) = uctd + ucxl Vi ucxl = E (in p sau khi ng kha K thi gian di) Tm uctd bng cch cho v phi ca phng trnh vi phn bng 0 K E 2O C1 C2 1 2 1 4 FF Hnh (1.7) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 10 RCdtduc + uc = 0 t uctd = keSt thay vo phng trnh ta c: 0dt) RCd(kekeStSt= + Trong : k: hng s S: s phc t: thi gian keSt + RCS.keSt = 0 keSt(1 +RCS) = 0 Do keSt = 0 nn:(1 +RCS) = 0 S = RC1Phng trnh trn l phng trnh c trng. Ta c uc(t) = E + kRCte Xc nh k: Da vo iu kin ban u ca bi ton. uc1(0) = E ;uc2(0) = 0 uc(t) = E + kRCte Ti t = 0 uc(0+) = E + ke0 = 10 + ke0 =320 k = 310 t = RC: hng s thi gian ca mch (n v s) t = RC = 2|.|

\|+4121 = 23 Vy uc(t) = 10 32te310 (V) uc

10V 320t 0 Lc ng K Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 11 V d: Cho mch in nh hnh v (1.8)

Cho e(t) = 10cos(10t + 450). Khi K ang ng v tr 1, ti t = 0 ng K sang v tr 2. Tm i(t). Li gii Trc khi ng K sang (2) ta c: i(0) = 21RE=(A) Khi va ng sang (2) i(0+) i(0+) = 21 (A)(do L.i(0) = L.i(0+), khng gy t bin v ch c 1 cun dy) Khi ng K sang (2) iR + Ldtdi = e = 10cos(10t + 450) t i = itd + ixl ixl: dng in xc lp l dng in khi ng in mt thi gian di. Ta c s tng ng: Tng tr phc ton mch: 045 2 10 10 j 10 Z Z = + = 2145 2 1045 10ZEI00XL=ZZ= = ixl = 21cos10t Xc nh itd ta gii phng trnh vi phn: 10O j10045 10 E Z = xlI 10O 1H e(t) 1 2 K 5V Hnh (1.8) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 12 iR + Ldtdi = 0 itd = ktLRe = ke10t i(t) = ke10t + 21cos10t Xc nh k: Da vo iu kin ban u ca bi ton i(0+) = ke0 + 21cos0 = 21 k = 0,207 Vy i(t) = 0,207e10t + 21cos10t I.3.PDNGPHNGPHPTONTLAPLACEGIIBITON QU Phng php tch phn kinh in nghin cu mc trn c u im l cho thy rhintngvtlcadnginvinpqunhngkhngtindngcho cc mch phc tp v vy vic gii trc tip phng trnh vi phn s kh khn, khi bc ca phng trnh vi phn cao. Phng php ton t c u im l ch, n cho php i s ha phng trnh vi tch phn, vi cc iu kin u c t ng a vo phng trnh i s, do kt qu nhn c s nhanh hn trong trng hp gii trc tip. I.3.1. Mt s kin thc c bn bin i Laplace Gif(t)lhmgc,binthintheothigiantvtabinithnhhmF(p). F(p) c gi l hm nh; p: s phc. Biu thc (1.5) dng xc nh nh ca mt hm f(t).L [f(t)]= 0( ) ( )ptF p f t e dt=}(1.5) Trong P l s phc: p = o + je Cc tnh cht c bn ca bin i Laplace l: nh ca o hm gc: L [f(t)] = F(p) = }0ptdt e ) t ( fdtd(1.6) Dng cng thc tch phn phn on ta c: }0ptdt e ) t ( f= f(t)0Pte+ p}0ptdt e ) t ( f= p.F(P) f(0)(1.7) nh ca o hm gc bng hm nh nhn vi p. L P) P ( Fdt ) t ( f0=((

}(1.8) nh ca tch phn hm gc bng hm nh chia cho p. Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 13 NhhaitnhchtquantrngcabiniLaplacetachuynphngtrnhvitch phn theo hm gc thnh phng trnh i s vi nh l F(p). BNG BIN I LAPLACE Hm gc f(t)Hm nh F(p) 1 1p te o 1p o + ( )11te oo( )1p p o + .tt e o ( )21p o + coset 2 2e + PP sinet 2 2ee+ P t 21p tn 1 nP! n+ 1 22 11( )t te eo oo o 1 21( )( ) p p o o + + 1 21 21 21( )t te eo oo oo o 1 2( )( )pp p o o + + n tt e o 1!; 0,1, 2...( )nnnp o+=+ 211 (1 )tt e ooo ( + 21( ) p p o + 21( 1)te t oo+ 21( ) p p o + (1 )tt e oo2( )pp o + sinte toe 2 2( ) peo e + + coste toe 2 2( )ppoo e++ + Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 14 21(1 cos ) t ee2 21( ) p p e + sin t t e2 2 22( )pp ee + cos t t e2 22 2 2( )ppee+ 1 2 2 12 21 2sin sin t t e e e ee e 1 22 2 2 21 2( )( ) p pe ee e + + 1 1 2 22 21 2sin sin t t e e e ee e 22 2 2 21 2( )( )pp p e e + + 2 12 21 2cos cos t t e ee e 2 2 2 21 2( )( )pp p e e + + 2 21 1 2 22 21 2cos cos t t e e e ee e 32 2 2 21 2( )( )pp p e e + + sin tte arctgpe NgclinubithmnhF(P)=(p) P(p) P21tacthtmchmgctheocng thc sau: ==n1 KpKtK 2K 1e) (p P') (p Pf(t) Trong '2P (PK) l o hm ca a thc P2(p) ti im P = PK Sau y l mt s v d cch tm hm gc: V d 1: Cho hm nhF(p) = ( )( )41 2 p p + + Hy tm hm gc f(t). Li gii Khi gp hm phc tp ta dng phng php phn tch: Bc 1: Phn tch ( )( )41 2 1 2A Bp p P P= ++ + + + Tm A: nhn 2 v cho (P+1) ( ) 1 42 2B PAp P += ++ + Cho P = 1 A = 4 Tm B: nhn 2 v cho (P + 2) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 15 ( ) 2 41 1PA Bp P+= ++ + Cho P = 2 B = 4 Bc 2: Tra bng 2( ) 4. 4t tf t e e = Cch 2: Ta c th tm A v B bng cch ly gii hn A = = + ) P ( F ). P ( limP114241=+ PlimP B = = + ) P ( F ). P ( limP22414lim2 =+ PP V d 2:

( )8( )2F PP P=+ Hy tm hm gc f(t). Li gii Bc 1: Phn tch ( )82 2A BP P P P= ++ + Tm A: Nhn 2 v cho p 8 .2 2B PAP P = ++ + Cho p = 0 A = 4 Tm B: Nhn 2 v cho p + 2 ( ) 2 8 PA Bp P+ = +Cho p = 2 B = 4 Bc 2: Tra bng f(t) = 4 4e4t Cch 2: ta c th tm A v B bng cch ly gii hn A =) P ( F . P limP 0 =428lim0=+PP B = = + ) P ( F ). P ( limP2248lim2 = PP V d 3: ( )( )24( )1 2F PP P=+ + Hy tm hm gc f(t). Li gii Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 16 Bc 1: Phn tch ( )( ) ( )2 241 21 2 2A B CP PP P P= + ++ ++ + + Tm A: nhn 2 v cho (P+1) ( )( ) ( )( )2 21 1 422 2B P C PAPP P+ += + +++ + Cho P = 1 A = 4 Tm C: nhn 2 v cho (P + 2)2 1) C(P 2) 1)(P B(P 2) A(P 42 2+ + + + + + = Cho P = 2 4 = C ( 2 + 1) C = 4 Tm B: nhn 2 v cho (P + 2)2 ( )( )( )22421 1A PB P Cp P + = + + ++ + o hm P theo 2 v: ( )( )( )( )2 22 .... 41 1A PBp P+= ++ + Gi tr () khng cn quan tm Cho p = 2 B = 4 Bc 2: Tra bng f(t) = 4.et 4.e2t 4t.e2t Cch 2: ta c th tm A, B, v C bng cch ly gii hn A == + ) P ( F ). P ( limP114) 2 (4lim21=+ PP C = ) P ( F . ) P ( limP222 + =414lim2 =+ PP Tm B bng cch nhn 2 v ca phng trnh cho (p + 2)2, sau ly o hm 2 v ca phng trnh v cho p = 2, ta c: B = 4. I.3.2. nh lut Kirchhoff dng ton t nh lut Kirchhoff 1 T biu thc = 0 i = 0 ) P ( I (1.9) nh lut Kirchhoff 2 Cho mch vng kn gm R - L - C ni tip t vo in p u ta c: ) 0 ( u idtC1dtdiL Ri uct0+ + + =} Chuyn sang bin i Laplace ta c: Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 17 L.i(0)p(0) upC1PL R I(p) U(p)c +((

+ + = (1.10) T ta suy ra: I(P) = PC1PL R) 0 ( LiP) 0 ( u) P ( Uc+ ++ Cng thc trn tng ng vi s ton t ca hnh (1.9) di y: Trong : L.i(0) v P) 0 ( UCc trng cho iu kin u ca bi ton. I.3.3. S ton t Laplace I.3.4. Thut ton tnh qu trnh qu bng phng php ton t Bc 1: Xc nh cc iu kin ban u Bc 2: Lp s ton t, gii s ton t theo cc phng php bit tm I(p). Bc 3: Dng bin i Laplace ngc tm hm gc i(t). I.3.5. Mt s v d v cc bi ton qu vi cc iu kin ban u bng 0 Bi 1: Cho mch in nh hnh v (1.10) i(t) C i s ha I(p) CP1 i s ha i(t)LI(p)Lp i s ha R i(t) I(p) R U(p) R pL PC1 L.i(0) I(p) PUC) 0 (Hnh (1.9) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 18 Ti t = 0 ng kho K, tm i(t). Li gii Bc 1: Xc nh iu kin ban u Theo bi ti t = 0 ng kha K tm i(t). Trc khi kha K ng thmch in h. V th cc iu kin ban u u bng khng. Bc 2: Bin i cc thng s Trc khi mun gii mt bi ton qu trnh qu ta phi bin i cc thng s vdngLaplacevishamchin(tclamchinvstng ng di dng Laplace). S tng ng Laplace: Bc 3: Tnh ton cc gi tr theo bin i Laplace Ta c: Tng tr ca mch in l nh sau: 8( ) 24 4P PZ P+= + =Cng dng in chy qua mch: 10( ) 40( )8( ) ( 8)4U PPI PPZ P P P= = =++ Bc 4: Phn tch 40( 8) 8A BP P P P= ++ += F(P) Tm A v B bng cch ly gii hn A =) P ( F . P limP 0 =5840lim0=+PP B == + ) P ( F ). P ( limP88540lim8 = PP I(P) 2 10 P P 4 10V K 2 i(t) 1 4 H Hnh (1.10) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 19 Vy:40 5 5( )( 8) 8I PP P P P= =+ + 8 8( ) 5 5 5(1 )t ti t e e = = (A) Thi gian qu l: Bi 2: Cho mch in nh hnh v (1.11) Yu cu: Ti t = 0 ng kha K, tm i(t) qua R v uc(t) t trn hai u t in. Li gii Bc 1: Xc nh iu kin ban u Ti t = 0 ng kha K. Do trc khi kha K ng th mch in trn h. V vy cc iu kin ban u bng 0. Bc2:ishamchin(tclamchinvstngngdi dng Laplace) 12 ) t ( uc= V P12) P ( U =C = 21F C(p) = P2 S tng ng: Bc 3: Tnh ton cc gi tr theo bin i Laplace Ta c: Tng tr ca mch i(t) 5 t I(p) 4 12 p2pUc(p) 12V uc(t) F2 1Ki(t) 4Hnh (1.11) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 20 2 4 2 2(2 1)( ) 4P PZ PP P P+ += + = =Cng dng in chy trong mch: Z(p)U(p)I(P) = 21p32 4p12p1) 2(2pp12I(p)+=+=+= Vy 12( ) 3ti t e A= Thi gian qu : t = 3t = 6s Tm uc(t): Ta c: in p t trn hai u t in2 12 2( ) ( )4 2Uc P I PP P P= = + 24 61(4 2)( )2P PP P= =++ Bc 4: Phn tch61( )2P P + =12A BPP++= F(p) Tm A v B bng cch ly gii hn A = = + ) P ( F ). P ( limP2121126lim21 = PP B = =) P ( F . P limP 012216lim0=+PP Vy A = 12; B = 12 12 12( )12Uc tPP= + 1 12 2( ) 12 12 12(1 )t tUc t e e = = (V) t 3 0 0tuc12Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 21 I.3.6. Cc bi ton qu vi cc iu kin ban u khc 0 -f(t) F(p)-) f(0 p.F(p)dtdf(t) -i(t)I(p) -) i(0 p.I(p)dtdi(t) - LdtdiLLp.I(p) L.iL(0) a. Cun dy uL = LdtdiLUL(P) = Lp.I(p) L.iL(0) b. i vi t in in p ban u trn t: Bi 1: Cho mch in nh hnh v (1.12) Yu cu: Ti t = 0 m kha K, tm cng dng in i(t) chy trong mch in. L iL(0-)LLiL(0-) Lpp) 0-( ucCp 1L C0-) ( uc+_C0-) ( ucE = 60Vi(t)K 7 5 H21Hnh (1.12) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 22 Li gii Bc 1: Xc nh iu kin ban u Ti t = 0 m kha K, do trc t = 0 th mch in ang hot ng. Vy ta phi xc nh iu kin ban u: + Xc nh dng in i qua cun dy trc khi kha K m ra: 12560) (0 iL= =(A) Bc 2: Bin i cc thng si s hamch in (tc l bin imch in v s tng ng di dng Laplace) u(t) = 60 VU(p) = 60P L = 21HL.p = 2P S tng ng: Bc 3: Tnh ton cc thng s theo Laplace60( )(5 7) 6260 60 6612( 10)( )24(24 )5 72 2PI PPPPP PI PP PP P+ + = ++++ = = =+++ + Bc 4: Phn tch 12( 10)( 24) 24P A BP P P P+= ++ += F(p) Tm A v B bng cch ly gii hn A = =) P ( F . P limP 0524) 10 ( 12lim0=++PPP B = = + ) P ( F ). P ( limP24247) 10 ( 12lim24=+ PPP Vy:4 2 p7p5) 4 2 p(p) 0 1 12(p++ =++ L I(p) 7p 60 p5 2V 6 _+L Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 23 (A) 7e 5 i(t)t 4 2 + = Cho t = 0 i = 12 (A) t = i = 5 (A) Bi 2: Cho mch in nh hnh v (1.13) Yu cu: Ti t = 0 ng kha K, tm cng dng in i(t) chy trong mch in? Li gii Bc 1: Xc nh iu kin ban u Ti t= 0 ngkha K, do trc t= 0 thmch in ang hot ng. Vvy ta phi xc nh iu kin ban u. Cng dng in chy qua mch khi kha K cha ng li: 51260) (0 iL= =(A) Bc 2: Bin i cc thng s i s ha mch in (a v mch in tng ng di dng Laplace) p60U(p) 60 u(t) = = L = 21H L.p = 2P 25215 ).L (0 i ) (0 UL L= = = (V) 12 ti50 7 K 1 60Vi(t) 5H 2Hnh (1.13) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 24 Mch in tng ng di dng Laplace: Bc 3: Tnh ton cc thng s theo Laplace25p602p5 I(p) + = |.|

\| +10) p(p24) 5(p2p 102p5p 1202P525p60I(P)++=++=++= Bc 4: Phn tch F(p) = 5( 24)( 10) 10P A BP P P P+= + + +

Tm A v B bng cch ly gii hn A = =) P ( F . P limP 01210) 24 ( 5lim0=++PPP B = = + ) P ( F ). P ( limP10107) 24 ( 5lim10 =+ PPP Vy: 10 P7P1210) P(P24) 5(P+ =++ (A) 7e 12 i(t)10t = I(p)52p _+V25L.iL(0-) p600 ti12 5Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 25 Bi 3: Cho mch in nh hnh v (1.14) Ti t = 0 m kha K, tm cng dng in i(t) chy trong mch v in p uc(t) t ln hai u t in. Li gii Bc 1: Xc nh iu kin ban u Ti t = 0 m kha K do trc t = 0 th kha K ng, v vy ta phi xc nh iu kin ban u: 32 212) i(0 =+=(A) uc(0) = i(0).2 = 6 (V) Bc2:ishamchin(binimchinvstngngdi dng Laplace) p12U(p) 12 u(t) = = p4C(p) F41C = =S tng ng: Bc 3: Tnh ton cc thng s theo Laplace p6p12p42 I(p) =||.|

\| +p6p4 2pI(p) =||.|

\| +2I(p)p 126p4 p12V 2 2 F41 i(t)K uc(t) Hnh (1.14) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 26 2 p34 2p6I(p)+=+= Vy: 2( ) 3ti t e=(A) Tm uc(t):

3 4 6 12 6( )( 2) ( 2)Uc PP P P P P P= + = ++ + Bc 4: Phn tch F(p) = 12( 2) 2A BP P P P= ++ + Tm A v B bng cch ly gii hnA = =) P ( F . P limP 06212lim0=+PP B = = + ) P ( F ). P ( limP22612lim2 = PP Vy: 12 6 6 6 6 12 6( 2) 2 2 P P P P P P P P+ = + = + + + ) e 6(2 6e 12 (t) u2t 2tc = = (V) Bi 4: Cho mch in nh hnh v (1.15) Yu cu: Ti t = 0 ng kha K, tm cng dng in iR(t) chy trong mch in. Li gii Bc 1: Xc nh iu kin ban u Tit=0ngkhaK,dotrct= 0thkhaKm.Vvytaphixcnh iu kin ban u. i(0) = 1020 = 2(A) uc(0) = 2.3 = 6(V) in p trn t in bng in p trn in tr 3 t 6 12 uc iR 20V 1F101 i(t) 6 3KHnh (1.15) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 27 Bc 2: khi ng kha k ta c S ton t Laplace

1 10( )10C F C PP= = Bc 3: Tnh ton cc thng s theo Laplace Da vo phng trnh li gii 10 6( )( )63( )2 105I P ZP PPI PPPP+ == =++ vi Z = 23 63 . 6=+ Vy cng dng in chy qua in tr 3O: 5 p25).9 (p3.696I(p) (p) IR+=+= = (A) 2e (t) i5tR= BI TP CHNG I Bi 1.1: Cho mch in nh hnh v (1.16) Yu cu: Ti t = 0 ng kha K tm cng dng in i1(t) chy trn in tr 12. p s: Cng dng in chy trn in tr 12 l i1(t) = 2 (A) 63 10p6p IR(P) 412 i1(t)8HK24V i(t)Hnh (1.16) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 28 Bi 1.2: Cho mch in nh hnh v (1.17) Ti t = 0 ng kha K, tm cng dng in i(t) chy trong mch. p s: t35e348 i(t) =(A) Bi 1.3: Cho mch in nh hnh v (1.18) Yu cu: Ti thi im t = 0 tm uc(t) vi V = 2tet (v). p s: uc(t) = 4e-3t 4e-2t + 4t.e-2t (v) Bi 1.4: Cho mch in nh hnh v (1.19) Yu cu: Ti t = 0 ng kha K, tm cng dng in i(t) chy trong mch. Cho bit: u(t) = 30e0,5t (V) K100Vi(t) 10 55i1(t)5H Hnh (1.17) Ki(t) 1F 212 v = 2te-t uc(t)+_ Hnh (1.18) Ki(t) 3 u(t) = 30e-0,5t6H Hnh (1.19) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 29 p s: t215t.e i(t)=(A) Bi 1.5: Cho mch in nh hnh v (1.20) Yu cu: Ti t = 0 ng kha K, tm in p uc(t) t trn t in. p s: uc(t) = 2e2t 2e3t = 2(e2t e3t) (V) Bi 1.6: Cho mch in nh hnh v (1.21) Yu cu: Ti t = 0 ng kha K, hy tm in p t trn in tr R = 2. p s: 340(t) uR=(V) Bi 1.7: Cho mch in nh hnh v (1.22) Ki(t) 1 F2 1 2 v = e-2tuc(t)+_Hnh (1.20) 3 3 5R = 21 3 K60V Hnh (1.21) 10VKi(t)2F2 1 2 2uc(t)+_Hnh (1.22) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 30 Yu cu: Ti t = 0 ng kha K, tm uc(t). p s: uc(t) = 5 5t32e= 5(1 t32e) (V) Bi 1.8: Cho mch in nh hnh v (1.23) Yu cu: Ti t = 0 m kha K, tm in p uR(t) t ln in tr R = 75 . p s: uR(t) = 150e10t (V) Bi 1.9: Cho mch in nh hnh v (1.24) Yu cu: Ti t = 0 m kha K, tm in p uR(t) trn in tr R = 8 . p s: uR(t) = 12e3t (V) uR(t) K150V 150 10H50 75 Hnh (1.23) 32V 2 12 8 2H 12KiR(t)uR(t) Hnh (1.24) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 31 Bi 1.10: Cho mch in nh hnh v (1.25) Yu cu: Ti t = 0 m kha K, tm iR(t). p s: 2tRe81(t) i=(A) Bi 1.11: Cho mch in nh hnh v (1.26) Yu cu:Ti t = 0 m kha K, tm cng dng in i(t) chy trong mch. p s: i(t) = 3e2t + 6t.e2t (A) Bi 1.12: Cho mch in nh hnh v (1.27) Yu cu:Ti t = 0 m kha K, tm cng dng in i(t) chy trong mch. p s: i(t) = 4 + e8t (A) 12V5K630iR(t)3 116F Hnh (1.25) 12V 4F 4 1K i(t)1HHnh (1.26) 2 H21 4 4H4 1 K24V i(t)Hnh (1.27) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 32 Bi 1.13: Cho mch in nh hnh v (1.28) Yu cu:Ti t = 0 m kha K, tm in p uR(t) t trn in tr 2 . p s: t101Re38(t) u=(V) Bi 1.14: Cho mch in nh hnh v (1.29) Yu cu:Ti t = 0 m kha K. Xc nh v v dng dng in iR(t). p s: iR(t) = 2,5et (A) Bi 1.15: Cho mch in nh hnh v (1.30) Yu cu:Ti t = 0 m kha K. Xc nh v v dng dng in iR(t) v in p uC(t). p s: iR(t) = 2,5et (A) v uc(t) = 10et (V) 100V 151FK8 4 3 2uRHnh (1.28) e(t) = 20sin(t +900) (V) 2 21F K 4 iR(t) 4H Hnh (1.29) e(t) = 20sin(t +900) (V) 2 K 2H 41F 4 iR(t) uc(t) Hnh (1.30) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 33 Bi 1.16: Cho mch in nh hnh v (1.31) Yu cu: Ti t = 0 kha K chuyn t v tr 1 2. Xc nh v v dng dng in i(t). p s: i(t) = 5e5t (A) Bi 1.17: Cho mch in nh hnh v (1.32) Ti t = 0 m K. Xc nh v v dng dng in ic(t) v in p uc(t). p s: ic(t) = 2,5e2t (A) ; uc(t) = 10e2t (V) Bi 1.18: Cho mch in nh hnh v (1.33) j(t) = 20cos10t (A) 5 1 5 iL(t) 21H K2 i(t) Hnh (1.31) 22 e(t) = 20cos4t (V) RR K F81C =ic(t) uc(t) Hnh (1.32) Hnh (1.33) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 34 Ti t = 0, kha K chuyn t v tr 1 sang v tr 2. Hy xc nh v v dng sng ca dng in i1(t), i2(t), i3(t), bit e(t) = 2E0coset, e = LR, E0 > 0. p s: i1(t) = tLReRERE0 0 =) 1 (0tLReRE (A) i2(t) = tLReRERE20 02 2=) 1 (220tLReRE (A) i3(t) = i1(t) + i2(t) =RE230 tLReRE0tLReRE202= )2123(20tLRtLRe eRE (A) Bi 1.19: Cho mch in nh hnh v (3.34) Hy xc nh v v dng dng in i(t) trong mch trnkhi < t < + , nu ti t = 0 m kho K. Bit rng: e(t) = Ecoset; E > 0 v e = RC LR 1=p s: i(t) = tLRe tLRRE ). 1 (3(A)hay i(t) = RE3(1 et)e et (A) Bi 1.20: Cho mch in nh hnh v (1.35) Hnh (1.34) 100V50V K 25O 0,01H i(t) 12 Hnh (1.35) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng I. Phn tch mch trong min thi gian 35 Yu cu: Tit=0chuynkhaKtvtr1sangvtr2.Tmcngdngini(t) chy trong mch. p s: 2500t6e 4 i(t) =(A) Chuong ITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 36 CHNG II: PHN TCH MCH TRONG MIN TN S Hm truyn t TrongmcI.3taninvicpdngphngphptontphntchqu trnh qu trongmch TTD. Nhvyvi tt c cc phng php hc, ta c th xc nh c tt c cc dng in v in p trn cc phn t mch, mi trng thi camch.Trongthctikhingitakhngquantmntonbmch,mch chnmtbphnno.Trongtrnghpnhvyngitatmramtcch khcmtmch,trongchchnccilngmtacntmvquanh canvinguntcng.Mchtrongtrnghpnycxtvikhinimtc ng - p ng (hay l nhn qu), cng ng nghavi khi nim truyn t Vo- Ra. II.1. NH NGHA HM TRUYN T Gi thit rng, ti t = 0 mch c tc ng bi ngun p hay ngun dng (k hiu l hm x(t), v i lng cn xt l dng hoc p u ra k hiu l y(t)). Vi x(t) v y(t) xut hin trn cc cc ca mch (Hnh v II.1.a, b, c). Khi iu kin u bng 0, hm truyn t c nh ngha nh sau: W(p) = X(p)Y(p) Trong : Y(p) = L[y(t)] X(p) = L[x(t)] Hm truyn t l mt hm c trng cho cc tnh cht ca mch, mt khi bit W(P) ta c th tm c p ng ca mch i vi mt tc ng bt k theo biu thc sau: Y(p) = W(p).X(p) y(t) = L1[Y(p)] quanhgiax(t)vy(t)lntr,thiukinquantrngliukinu phi bng 0. Mch TTD x(t)y(t) Hnh II.1.a Hai cc i(t) u1(t) Hnh II.1.b Bn cc i1(t) u1(t) Hnh II.1.c i2(t) u2(t) Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 37 Hm truyn ca 2 cc l tr khng hay dn np ty theo cc i lng vo ra c chn l dng hay p. Khi x(t) = u(t)v y(t)= i(t), th hm truyn ca 2 cc s l dn np. W(p) = U(p)I(p) = Y(p) Khi x(t) = i(t) v y(t) = u(t), th hm truyn ca 2 cc s l tr khng: W(p) = I(p)U(p) = Z(p) (Ch thch: T hm truyn t hay truyn t thng c dng cho mng hai ca (4 cc) v n mang ngha truyn t tn hiu. Khi dng cho 2 cc, n ch c ngha l tr khng hay dn np ca 2 cc ). V d1: Cho mch in nh hnh v (2.1) u1(t): tn hiu vo ca mch (x(t)) u2(t): tn hiu ra ca mch (y(t)) Tnh hm truyn W(p) = X(p)Y(p) Li gii Bc 1: a mch v s ton t Laplace Ta c: X(p) = U1(p) Y(p) = U2(p) Bc 2: Xc nh hm truyn t p: U2(p) = U1(p).CPRCP11+ RCp 1 U1(p)U2(p)Ru1(t)Cu2(t)Hnh (2.1) Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 38 W(p) = CP1RCP1) P ( U) P ( U12+== RCP 11+ V d 2: Cho mch in nh hnh v (2.2) Tnh hm truyn t p W(p). Li gii Bc 1: a mch v s ton t Laplace Ta c:X(p) = U1(p) Y(p) = U2(p) Bc 2: Xc nh hm truyn t p W(P) = CP1R RCP1R) P ( U) P ( U2 1212+ ++== CP ) R R ( 1CP R 12 12+ ++ Vy W(P) = P 10 1P 10 134++ k 9 R1 = k 1 R2 =CP1) P ( U1) P ( U2R1 = 9kR2 = 1kC1 = 0,1F u1(t) u2(t)Hnh (2.2) Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 39 V d 3: Cho mch in nh hnh v (2.3) Tnh hm truyn W(p). Li giiBc 1: a mch v s ton t Laplace

Bc 2: Xc nh hm truyn t p ) P ( U .CP1RCP1RRR) P ( U111222++=W(p) = 1 2 2 11 212R R CP R R) 1 CP R ( R) P ( U) P ( U+ ++= V d 4: Cho mch in nh hnh v (2.4) Tnh hm truyn W(p) 1R2R) P ( U1) P ( U2CP11R2R ) t ( u1) t ( u2CHnh (2.3) 1R2R ) t ( u1) t ( u2CHnh (2.4) Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 40 Li giiBc 1: a mch v s ton t Laplace Bc 2: Xc nh hm truyn t p W(p) = CP1RCP1RRCP1RCP1R) P ( U) P ( U2212212+++== 1 CP RRR1 CP RR22122+++ W(p) = 1 2 2 12R R CP R RR+ + II.2. BIU DIN TH CA HM TRUYN II.2.1. c tuyn logarit - tn s logarit Trong thc t ngi ta thng quan tm n c tuyn bin W(je); bi v n d o lng v n cho ta bit nhiu tnh cht ca mch i vi tn s. Khi nim v Bel v Decibelbel B decibel dB 1b = 10db L n v o mc tng gim ca tn hiu vaoraPPlg [b] 1b {Pr = 10 PV} 10vaoraPPlg [db] + 10db Pr = 10 PV Pvo Pra 1R2R ) P ( U1) P ( U2CP1Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 41 + 20db Pr = 100 PV 0db Pr = PV 10db Pr = 10PV 20db Pr = 100PV 2VrVrUUPP||.|

\|= 10lgVrPP = 10lg2VrUU||.|

\| db = 20lgVrUU (db) Thng thng c tuyn tn s c vit di dng:W(p) = P T + 11 hay W(je) = Tj 11+ Trong : p = jeTje: s phcModun W(je) Argumen (e) II.2.2. c tuyn bin - tn s logarit (Gin Bode) V d ta kho st s bin thin ca hm truyn: W(je) = Tj 11+ 20lgW(je) = 20lg Tj 11+ = 20lg1 20lgTje +1(dB) - Khi e T1 Te >> 1 Tje +1 ~ Te Vy 20lgW(je)~ 20lgTe( 20db/dec)Gii thch: -dec decade (10 ln tn s) -( 20db/dec) gim 20db khi tn s tng 10 ln-Ti e0 20lgTe = 20lgTe0 = xdb -Ti e = 10e0 20lgTe = 20lgT.10.e0 = 20lgT.e0 20lg10 = x 20db Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 42 c tuyn bin tn s logarit: V d1: Cho hm truyn: W(p) = P T 1K+vi K, T: hng s p = je. Hy v c tuyn bin - tn s logaritLi gii:Ta c: W(je) = Tj 1K+

20lgW(je) = 20lg Tj 1K+ = 20lgK 20lgTje +1 - Khi e T1 Te >> 1 Tje +1 ~ Te Vy 20lgW(je)~ 20lgK 20lgTe( 20db/dec) 0 T1 T10 20db 20db/dec e db 20lgK 0 T1 T10 20db 20db/dec e db Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 43 CC BI TP V DV d 1: Cho mch in nh hnh v (2.5) Tnh W(p); V c tuyn bin - tn s logarit (gin Bode): 20lgW(je) Tm li gi tr C tn hiu vo tn s 105 khng b suy gim. Li gii Bc 1: a mch v s ton t Laplace Bc 2: Xc nh hm truyn t p W(P) = CP1RCP1) P ( U) P ( U12+== RCP 11+ = P 10 11P 10 . 10 114 7 3 +=+ W(je) = 1 ) j ( 1014+Vi p = je Bc 3: V c tuyn bin - tn s logarit (gin Bode)20lgW(je) = 20lg104 (je) +1- Khi e T1 Te >> 1 Tje +1 ~ Te 20lgW(je) = 20lgTe (dB) ( 20 dB/dec) RCP1) P ( U1) P ( U21Ku1(t)C = 0,1F u2(t)Hnh (2.5) Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 44 c tuyn bin tn s logarit: Ta c: RC1T1C= => 105 C < 3 5 510 . 101R 101== 108 F V d 2: Cho hm truyn: W(p) = K(Tp + 1) Vi K, T: hng s; p = j.e V c tuyn bin - tn s logarit (gin Bode). Li gii Ta c: 20lgW(je) = 20lgK(Tje +1) = 20lgK + 20lg(Tje +1) - Khi e T1 Te >> 1 Tje +1 ~ Te 20lgW(je) = 20lgK + 20lgTe (dB)(20 dB/dec) V d 3: Cho hm truyn: W(p) = 1 P T) 1 P T ( K12++ Vi K, T1, T2: hng s; T1 > T2. W(je)= 1112++j T) j T ( K V c tuyn bin - tn s logarit (gin Bode) dB 20lgK e + 20dB/dec T1 0 410T1= 20db/dec e db Di thng T10 20db Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 45 dB 20lgK 1T1 1T10 e 2T1 20db/dec Li gii Ta c: 20lgW(je) =20lgK + 20lg(T2je+1) 20lg(T1je+1) - Khi e > 1 Tje +1 ~ Tje W(je) = KTje = 2t 20lgK db 0 T1 e - 20db/dec () 0 T1 e 2t e 4 t2t 103 20db/dec db e 103 1K u1(t)C = 1F u2(t)Hnh (2.6) Chuong IITh vien H SPKT TP.HCM - http://www.thuvienspkt.edu.vn Chng II. Phn tch mch trong min tn s 47 V d 3: Cho hm truyn W(p) = 1 P T) 1 P T ( K12++ Vi K, T1, T2: hng s; T1 > T2 W(je)= 1112++j T) j T ( K V c tuyn pha - tn s logarit: (e) Li gii - Khi e