Phep Bien Doi Laplace

_________________________________________Chng 10 Php bin i Laplace - 1

___________________________________________________________________________ Nguyn Trung Lp L THUYT

CHNG 10 PHP BIN I LAPLACE

DN NHP

PHP BIN I LAPLACE Php bin i Laplace

Php bin i Laplace ngc CC NH L C BN CA PHP BIN I LAPLACE

P DNG VO GII MCH CC PHNG PHP TRIN KHAI HM P(S)/Q(S)

Trin khai tng phn Cng thc Heaviside

NH L GI TR U V GI TR CUI nh l gi tr u nh l gi tr cui

MCH IN BIN I in tr Cun dy T in

_______________________________________________________________________________________________

10.1 DN NHP

Php bin i Laplace, mt cng c ton hc gip gii cc phng trnh vi phn, c s dng u tin bi Oliver Heaviside (1850-1925), mt k s ngi Anh, gii cc mch in.

So vi phng php c in, php bin i Laplace c nhng thun li sau: * Li gii y , gm p ng t nhin v p ng p, trong mt php ton. * Khng phi bn tm xc nh cc hng s tch phn. Do cc iu kin u c a vo phng trnh bin i, l phng trnh i s, nn trong li gii y cha cc hng s.

V phng php, php bin i Laplace tng t vi mt php bin i rt quen thuc: php tnh logarit (H 10.1) cho ta so snh s ca php tnh logarit v php bin i Laplace

Ly logarit Nhn chia trc tip Cng cc s Ly logarit ngc

Cc con s

Kt qu cc php tnh

logarit ca cc s

Tng logarit ca cc s

Pt vi tch phn

Pt sau Bin i

MCH



Bin i Laplace Php gii c in k u Php tnh i s k u Bin i Laplace ngc lnh vc thi gian Lnh vc tn s

(H 10.1)

lm cc php tnh nhn, chia, ly tha . . . ca cc con s bng php tnh logarit ta thc hin cc bc: 1. Ly logarit cc con s 2. Lm cc php ton cng, tr trn logarit ca cc con s 3. Ly logarit ngc c kt qu cui cng.

Thot nhn, vic lm c v nh phc tp hn nhng thc t, vi nhng bi ton c nhiu s m, ta s tit kim c rt nhiu thi gian v c th s dng cc bng lp sn (bng logarit) khi bin i. Hy th tnh 1,43560,123789 m khng dng logarit.

Trong bi ton gii phng trnh vi tch phn dng php bin i Laplace ta cng thc hin cc bc tng t: 1. Tnh cc bin i Laplace ca cc s hng trong phng trnh. Cc iu kin u c a vo 2. Thc hin cc php ton i s. 3. Ly bin i Laplace ngc c kt qu cui cng.

Ging nh php tnh logarit, cc bc 1 v 3 nh s dng cc bng lp sn chng ta c th gii quyt cc bi ton kh phc tp mt cch d dng v nhanh chng.

10.2 PHP BIN I LAPLACE

10.2.1 Php bin i Laplace

Hm f(t) xc nh vi mi t>0. Bin i Laplace ca f(t), c nh ngha == 0 stdtf(t).eF(s)[f(t)]L (10.1)

s c th l s thc hay s phc. Trong mch in s=+j Ton t L thay cho cm t 'bin i Laplace ca" iu kin f(t) c th bin i c l



Th d, vi hm f(t)=tn, dng qui tc Hospital, ngi ta chng minh c 00,etlim tn

t>=

Vi n=1, ta c

01dtt.e0

t >= ,2

Vi gi tr khc ca n, tch phn trn cng xc nh vi 0 C nhng hm dng khng tha iu kin (10.2) nhng trong thc t vi nhng

kch thch c dng nh trn th thng t tr bo ha sau mt khong thi gian no .

nate

Th d v(t)= >

0

0at

tt,Ktt0,e

2

v(t) trong iu kin ny tha (10.2)

Ta ni ton t L bin i hm f(t) trong lnh vc thi gian sang hm F(s) trong lnh vc tn s phc. Hai hm f(t) v F(s) lm thnh mt cp bin i

Th d 10.1 Tm bin i Laplace ca hm nc n v

u(t) =

0 ca mch (H 10.9a). Cho i(0)=4A v v(0)=8V

(a) (H 10.9) (b) Mch bin i cho bi (H 10.11b)

I(s)=2/ss3

8/s43)(2/s++

++

=3)2)(s3s(s3)-8)(s-(4s2s

2 ++++

=3)2)(s1)(s(s

24-6s4s2

++++

Trin khai I(s)

I(s)=3s

32s

201s

13++++

Suy ra, khi t>0 i(t)=-13e-t+20e-2t- 3e-3t A

Th d 10.13 Xc nh v(t) ca mch (H 10.10a). Cho i(0)=1A v v(0)=4V

(a) (b) (H 10.10) Vit phng trnh nt cho mch bin i (H 10.10b)

0244

24sV

s1

3sV

4V =+++

MCH



V(s)=4s

202s

164)2)(s(s

244s+++=++

v v(t)=-16e-2t+20e-4t V

10.5 CC PHNG PHP TRIN KHAI HM P(s)/Q(s)

Trong phn gii mch in bng php bin i Laplace, kt qu t c l mt hm theo s c dng P(s)/Q(s) , trong P(s) v Q(s) l cc a thc.

Nu P(s)/Q(s) c dng trong bng 1 th ta c ngay kt qu bin i Laplace ngc. Trong nhiu trng hp ta phi trin khai P(s)/Q(s) thnh tng cc hm n gin hn v c trong bng. Gi m v n l bc ca P(s) v Q(s) C 2 trng hp

* mn, c th trin khai ngay P(s)/Q(s) * m>n, ta phi thc hin php chia c

(s)Q(s)P

sA.....sAAQ(s)P(s)

1

1nmnm10 ++++= (10.18)

P1(s) v Q1(s) c bc bng nhau v ta c th trin khai P1(s)/Q1(s)

10.5.1. Trin khai tng phn Trng hp 1

Q(s)=0 c nghim thc phn bit s1 , s2, . . . sn.

n

n

2

2

1

1

s-sK

s-sK

s-sK

Q(s)P(s) +++= ..... (10.19)

Ki (i= 1, 2,. . . ., n) l cc hng s xc nh bi:

issQ(s)

P(s))s(sK ii

== (10.20)

Th d 10.14

Trin khai hm I(s)=23ss

1s2 ++

, xc nh i(t)=L -1[I(s)] Phng trnh s2+3s+2=0 c 2 nghim s1=-2 v s2=-1

I(s)= 23ss

1s2 ++

=1s

K2s

K 21+++

3Q(s)P(s)

2)(sK-s

1 =+== 2

-2Q(s)P(s)

1)(sK-s

2 =+== 1

I(s)= 1s

22s

3++

MCH



i(t)= 3e-2t-2e-t

Trng hp 2

Q(s)=0 c nghim a trng bc r

r2r ..... )s-(sK

)s-(sK

s-sK

)s-(sP(s)

Q(s)P(s)

i

r

i

2

i

1

i

+++== (10.21) xc nh K1, K2, . . . Kr, ta xt th d sau: Th d 10.15

Trin khai 21)(s2s

Q(s)P(s)

++=

21)(sK

1sK

Q(s)P(s) 21

+++= (1) Nhn 2 v phng trnh (1) vi (s+1)2

s+2=(s+1)K1+K2 (2) Cho s=-1, ta c K2=1

Nu ta cng lm nh vy xc nh K1 th s xut hin cc lng v nh xc nh K1, ly o hm theo s phng trnh (2)

1+0=K1+0 K1=1 Tm li

21)(s1

1s1

Q(s)P(s)

+++= V i(t) = e-t + te-t

Vi Q(s)=0 c nghim kp, mt hng s c xc nh nh o hm bc 1. Suy rng ra, nu Q(s)=0 c nghim a trng bc r, ta cn cc o hm t bc 1 n

bc r-1. Trng hp 3 Q(s)=0 c nghim phc lin hp s= j

)j-)(sj--(sP(s)

Q(s)P(s)

+= (10.22)

)j-(s*K

)j--(sK

Q(s)P(s)

++= (10.23) Cc hng s K xc nh bi

=+==

jAeQ(s)P(s)

)j(sKjs

,

V +==+=

jAeQ(s)P(s)

)j(sK*js

(10.24)

Th d 10.16

Trin khai I(s)= 54ss

1Q(s)P(s)

2 ++= Q(s)=0 c 2 nghim -2 j

MCH



I(s)= j)-2-(s

*Kj)2(s

KQ(s)P(s) +++=

==++==

0e21

21j

Q(s)P(s)j)2(sK

js

9j

2

==+=+=

0e21

21j

Q(s)P(s)j)2(sK*

js

9j

2

I(s)=j-2s

j1/2j2s

j1/2+++

i(t)= ]e[e21j )tj2()tj2( + = ]

2jee[e

tjt2t

j

Hay i(t)=e-2tsint A

10.5.2 Cng thc Heaviside

Tng qut ha cc bi ton trin khai hm I(s)=P(s)/Q(s), Heaviside a ra cng thc cho ta xc nh ngay hm i(t), bin i ngc ca I(s)

10.5.2.1 Q(s)=0 c n nghim phn bit

i(t)=L -1[I(s)] = L -1j

stn

1jj

ssQ(s)P(s)e)s(s]

Q(s)P(s)[

==

= (10.25)

Hoc

i(t) tsje)(sQ')P(sn

1j j

j=

= (10.26) Trong sj l nghim th j ca Q(s)=0

Th d 10.17 Gii li th d 10.14 bng cng thc Heaviside

I(s)=23ss

1s2 ++

, xc nh i(t)=L -1[I(s)] Phng trnh s2+3s+2=0 c 2 nghim s1=-2 v s2=-1

Q(s)= s2+3s+2 Q(s) = 2s+3 Ap dng cng thc (10.26)

i(t) te1)(Q'1)P(2te

2)(Q'2)P(

e)(sQ')P(s tsjn

1j j

j +

== =

i(t)= 3e-2t-2e-t A

10.5.2.2 Q(s)=0 c nghim a trng bc r

i(t)=L -1[I(s)] = L -1j

n-rj

n-r1nr

1n ssds)R(sd

1)!(nt

n)!-(r1]

Q(s)P(s)[ ==

=ts je (10.27)

MCH



sj l nghim a trng bc r r)) jj s(sQ(s)

P(s)R(s = (10.28) Th d 10.18

Gii li th d 10.15 bng cng thc Heaviside

I(s)= 21)(s2s

Q(s)P(s)

++=

Q(s)=0 c nghim kp, r=2, sj=-1 Ap dng cng thc (10.27)

Vi 2s1)(s1)(s2s)R(s 22j +=++

+=

1s2)(s1!t

0!1

ds2)d(s

0!t

1!1[e(t)

10t =+++= ;]i

V i(t) = e-t + te-t A

Th d 10.19 Cho mch in (H 10.11), t C tch in n V0=1V v kha K ng t=0. Xc nh

dng i(t)

0dtdtdLR

t =++ iii Ly bin i Laplace

L[sI(s)-i(0+)]+RI(s)+ Cs1 [I(s)+q(0+)]=0

Dng in qua cun dy lin tc nn i(0+)= i(0-)=0 q(0+): in tch ban u ca t:

s1

sV

Cs)q(0 o ==+

( du ca in tch u trn t ngc chiu in tch np bi dng i(t) khi chy qua mch)

Thay gi tr u vo, sp xp li

11)(s1

22ss1I(s) 22 ++=++=

i(t)=L -1[I(s)]=e-tsint.u(t)

Th d 10.20

Cho mch (H 10.12), kha K ng t=0 v mch khng tch tr nng lng ban u. Xc nh i2(t)

Vit pt vng cho mch

100u(t)1020dtd

211 =+ iii (1)

01020dtd

122 =+ iii (2)

MCH



Ly bin i Laplace, mch khng tch tr nng lng ban u:

(s+20)I1(s)-10I2(s)= s100 (3)

-10 I1(s)+ (s+20)I2(s)=0 (4) Gii h (3) v (4)

I2(s)= 300)40ss(s1000

20s101020s010s

10020s

2 ++=++

+

Trin khai I2(s)

30s1,67

10s5

s3,33(s)I 2 ++++=

i2(t)= 3,33-5e-10t+1,67e-30t

10.6 NH L GI TR U V GI TR CUI

10.6.1 nh l gi tr u

T php bin i ca o hm: Ldt

df(t) = sF(s)-f(0+)

Ly gii hn khi s [L

slim

dtdf(t)

] = [sF(s)-f(0+)] s

lim

m [Ls

limdt

df(t)]=

slim 0 dtedtdf(t) st =0

Vy [sF(s)-f(0+)]=0 s

lim

f(0+) l hng s nn f(0+)= sF(s) (10.29)

slim

(10.29) chnh l ni dung ca nh l gi tr u Ly trng hp th d 10.10, ta c:

I(s)=1/RCs1

R/CqV 0

+

i(0+)= sI(s)= s

limR

/CqV 0

10.6.2 nh l gi tr cui

MCH



T php bin i o hm: Ldt

df(t) = sF(s)-f(0+)

Ly gii hn khi s 0 [L

0slim

dtdf(t)

] = 0s

lim

0

dtedt

df(t) st = [sF(s)-f(0+)] 0s

lim

m 0s

lim

0

dtedt

df(t) st = = 0s

lim

+=0

)f(0-)f(df(t)

Vy f()-f(0+)= [sF(s)-f(0+)] 0s

lim

Hay f()= sF(s) (10.30) 0s

lim

(10.30) chnh l ni dung ca nh l gi tr cui, cho php xc nh gi tr hm f(t) trng thi thng trc.

Tuy nhin, (10.30) ch xc nh c khi nghim ca mu s ca sF(s) c phn thc m, nu khng f()= f(t) khng hin hu.

tlim

Th d, vi f(t)=sint th sin khng c gi tr xc nh (tng t cho e ). V vy (10.30) khng p dng c cho trng hp kch kch l hm sin. Ly li th d 10.13, xc nh dng in trong mch trng thi thng trc

I(s)= )R/Ls1

s1(

RV

+

i()= sI(s)= 0s

lim R

V)R/Lss(1

RV =+

i()=RV

BI TP

10.1 Mch (H P10.1). Kha K ng t=0 v mch khng tch tr nng lng ban u. Xc nh i(t) khi t> 0 10.2 Mch (H P10.2). Xc nh v(t) khi t> 0. Cho v(0)=10V

(H P10.1) (H P10.2)

10.3 Mch (H P10.3). Xc nh vo(t)

MCH



Cho vi(t) =

>0

(H P10.7)

10.8 Mch (H P10.8) t trng thi thng trc t=0. Xc nh v khi t>0

(H P10.8)

10.9 Mch (H P10.9) t trng thi thng trc t=0- Xc nh i khi t>0

MCH



(H P10.9)

10.10 Mch (H P10.10). Xc nh i(t) khi t>0. Cho v(0) = 4 V v i(0) = 2 A

(H P10.10)

MCH

( CHUONG 1010.1 DN NHP10.2 PHP BIN I LAPLACE10.2.1 Php bin i Laplace10.2.2 Php bin i Laplace ngc

10.3 CC NH L C BN CA PHP BIN I LAPLACE10.3.1 Bin i ca mt t hp tuyn tnh10.3.2 Bin i ca e-atf(t)10.3.3 Bi?n d?i c?a f(t-()u(t-()10.3.4 nh l kt hp (Convolution theorem)10.3.5 Bin i ca o hm10.3.6 Bin i ca tch phn10.3.7 Bin i ca tf(t)

10.4 P DNG VO GII MCH10.4.1 Gii phng trnh vi tch phn10.4.2 Mch in bin i

10.5 CC PHNG PHP TRIN KHAI HM P(s)/Q(s)10.5.1. Trin khai tng phn10.5.2 Cng thc Heaviside10.5.2.1 Q(s)=0 c n nghim phn bit10.5.2.2 Q(s)=0 c nghim a trng bc r

10.6 NH L GI TR U V GI TR CUI10.6.1 nh l gi tr u10.6.2 nh l gi tr cui

BI TP

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Phep Bien Doi Laplace