5/22/20141Chng 2PHNG TRNH VI PHN V NG DNGNi Dung1. Phng trnh vi phn cp 12. Phng trnh vi phn cp 23. ng dng phn gii mch in Mch DC (Direct Current) Mch bc nht RL, RC Mch bc hai RLC (ni tip, song song, tng qut) Mch AC (Alternating Current)5/22/20142MCH DCMch DC Bc NhtMch DC Bc HaiPhn Tch Mch DC C hai vn cn nh:1. Phn cc2. Th qua t v dng qua cun khng i tc thi nn 0+ = 0 ; 0+ = 0; trong t = 0: thi im va trc khi chuyn mch, t = 0+: thi im va sau chuyn mch; (vi gi thit rng thi im chuyn mch l t = 0)5/22/20143Mch DC Bc Nht L mch c c trng bi phng trnh vi phnbc nht (cp 1). Ta cn kho st 2 loi mch RC & RL Mch RC khng c ngun ngoi Mch RL khng c ngun ngoi Mch RC c ngun ngoi (p ng bc ca mch RC) Mch RL c ngun ngoi (p ng bc ca mch RL)MCH DC BC NHT KHNG C NGUN KCH THCH NGOIMch RCMch RL5/22/20144Mch DC Khng Ngun Mch khng c ngun, ta s kho st vi trnghp cc phn t C & L c tch tr nng lng. Nng lng tch tr ny s b tiu hao dn qua intr R.Mch RC Khng C Ngun Ngoi (1) Gi s ti thi im = 0, t c np (0) = 0 p dng KCL:

+

= 0

0

= 0

1

ln ln0 = 1

ln

0= 1

= 0

1

5/22/20145Mch RC Khng C Ngun Ngoi (2) = 0

1

p ng in th ca mch RC suy gim gi tr ban u ca ca in th qu t in theo hm m. p ng ny gi l p ng t nhin: Khi khng c ngun kch thch ngoi, p ng ny chph thuc vo c tnh ca t in (gi tr tch tr ban u & tnh cht vt l ca t in).Mch RC Khng C Ngun Ngoi (3) Thi hng (time-constant): l khong thi gian p ng suy gim theo h s 1/ (36,8%) ca gi trtch ly ban u. K hiu:

0

= 0

1 = 0.3680 = = 0

5/22/20146Mch RC Khng C Ngun Ngoi (4) V D: Cng tc trong mch in c ng trongmt thi gian di v m ra ti thi im = 0. Tm () khi 0.Mch RC Khng C Ngun Ngoi (5) Khi < 0, t in xem nh mch h - mch t

= 20 99 +3 = 15 Bi v th qua t khng i tc thi nn

0 = 0+ = 0 = 155/22/20147Mch RC Khng C Ngun Ngoi (6) Khi > 0, mch tr thnh mch RC khng c ngunkch thch ngoi v t c np. Mch t Thi hng: = t = 9 +1 . 20.103 = 0.2

=

0

= 15 0.2 = 155 Mch RL Khng C Ngun Ngoi (1) Gi s ti thi im = 0, cun cm tch (0) = 0 p dng KVL:

+ = 0

0

= 0

ln ln0 =

ln

0 =

= 0

5/22/20148Mch RL Khng C Ngun Ngoi (2) = 0

p ng dng in ca mch RL suy gim gi trban u ca dng in qua cun theo hm m. p ng ny gi l p ng t nhin: Khi khng c ngun kch thch ngoi, p ng ny chph thuc vo c tnh ca cun cm (gi tr tch trban u & tnh cht vt l ca cun cm).Mch RL Khng C Ngun Ngoi (3) Thi hng (time-constant): l khong thi gian p ng suy gim theo h s 1/ (36,8%) ca gi trtch ly ban u. K hiu:

0

= 0

1 = 0.3680 =

= 0

5/22/20149Mch RL Khng C Ngun Ngoi (4) V D: %Cng tc trong mch in c ng trongmt thi gian di v m ra ti thi im = 0. Tm khi 0.Mch RL Khng C Ngun Ngoi (5) Khi < 0, cun cm xem nh ni tt - mch t

1 =402 +3 = 8 = 12 1 12 +4= 6 Bi v dng qua cun khng i tc thi nn 0 = 0+ = 0 = 65/22/201410Mch RL Khng C Ngun Ngoi (6) Khi > 0, mch tr thnh mch RL khng c ngunkch thch ngoi v cun c np. Mch t Thi hng: =

t = 28 = 0.25 = 0

= 6 0.25 = 64 Bi TpBi 1: Cho

0 = 60. Tm

,

, 0 ti 05/22/201411Bi TpBi 2: Tm khi > 0MCH DC BC NHT CNGUN KCH THCH NGOIHm Bc n VMch RCMch RL5/22/201412Hm Bc n V Hm bc n v (unit step function): l hm chogi tr 0 khi < 0 v cho gi tr 1 khi > 0. Phng trnh ton hc: = 0, < 01, 0 Ngha Hm Bc Trong MchChng ta s dnghm bc biudin s thay i tngt ca ngun inth hoc ngun dngin khi a ngunny vo mch in.5/22/201413p ng Bc Ca Mch in p ng bc ca mch in: l p ng ca mchkhi n c kch thch bi mt hm bc (c th lngun th hoc ngun dng).p ng Bc Ca Mch RC (1)Gi s < 0, t tch 0 = 0KCL:

+

= 0 > 0:

=

; =

0

= 0

ln

0

=

5/22/201414p ng Bc Ca Mch RC (2) p ng ton phn =

0, < 0

+ 0

, > 0 p ng thnh phn = 0

p ng qu +

p ng n nh = 0

p ng t nhin +

1

p ng pp ng Bc Ca Mch RC (3)Khi > 0: =

+ 0

= () + (0) ()

(): l in th n nh ca t5/22/201415p ng Bc Ca Mch RC (4) V D: Xc nh () khi > 0, v tnh gi tr ca nti thi im = 1 v = 4p ng Bc Ca Mch RC (5) Khi < 0, 0 = 0 = 0+ =55+3. 24 = 15 Khi > 0, = = 4.103. 0,5.103 = 2 =

+ 0

= 30 + 15 30

2= 30 15

2 = 1: = 30 1512 = 20,9 = 4: = 30 1542 = 27,975/22/201416p ng Bc Ca Mch RL (1) p ng ton phn =

=

0, < 0

+ 0

, > 0 p ng thnh phn =

p ng n nh+ 0

p ng qu = 0

p ng t nhin+

1

p ng pp ng Bc Ca Mch RL (2)Khi > 0: =

+ 0

= () + (0) ()

(): l in th n nh ca t5/22/201417p ng Bc Ca Mch RL (3) V D: Xc nh () khi > 0p ng Bc Ca Mch RL (4) Khi < 0, 0 = 0 = 0+ = 102 = 5 Khi > 0, =

= 135 = 115

= () + 0 ()

= 2 + 5 2 15= 2 +3155/22/201418MCH DC BC HAI KHNG CNGUN KCH THCH NGOIRLC Ni tipRLC Song SongRLC Ni Tip Khng C Ngun Ngoi (1) Gi s ti < 0, t & cun d tr 0 = 0, 0 = 0 Khi > 0, p dng KVL: +

+ 1

= 0 2

2 +

+

= 0 5/22/201419RLC Ni Tip Khng C Ngun Ngoi (2) = 0 0 + 0

+0 = 0 0

= 1

0 +0 Nhc li, p ng t nhin ca mch bc nht cdng hm m =

RLC Ni Tip Khng C Ngun Ngoi (3)

2 +

+ 1

ptt 2 + +02 = 0 = 2

0 =1

lun cho 2 nghim, nn nghim ca c dng

q = 1

1

+2

2

5/22/201420RLC Ni Tip Khng C Ngun Ngoi (4) = 1

1

+2

2

> 0: 1 2

q = 1

1

+2

2

= 0: 1 = 2 =

q = 1 +2 < 0: 1,2 = = ; = 02 2

q = 1cos +2sinRLC Song Song Khng C Ngun Ngoi (1) Gi s ti < 0, t & cun d tr 0 = 0, 0 = 0 p dng KCL:

+1

+

= 0 2

2 + 1

+

= 0 5/22/201421RLC Song Song Khng C Ngun Ngoi (2)Phn tch tng t, ta cptt 2 + +02 = 0 =12

0 =1

lun cho 2 nghim, nn nghim ca c dng

q = 1

1

+2

2

RLC Song Song Khng C Ngun Ngoi (3) = 1

1

+2

2

> 0: 1 2

q = 1

1

+2

2

= 0: 1 = 2 =

q = 1 +2 < 0: 1,2 = = ; = 02 2

q = 1cos +2sin5/22/201422MCH DC BC HAI C NGUNKCH THCH NGOIRLC Ni tipRLC Song SongRLC Ni Tip C Ngun Ngoi (1) Khi > 0, KVL:

+ + =

=

2

2 +

+

=

Nhc li RLC nt khng c ngun ngoi

2

2 +

+

= 0C cngdng ptrVi phn5/22/201423RLC Ni Tip C Ngun Ngoi (2) p ng ton phn = o +q =

+q p ng qu q xem li phn mch bc 2 khng c ngun ngoiRLC Song Song C Ngun Ngoi (1) Khi > 0, KVL:

+ +

=

=

2

2 + 1

+

=

Nhc li RLC // khng c ngun ngoi

2

2 + 1

+

= 0C cngdng ptrVi phn5/22/201424RLC Song Song C Ngun Ngoi (2) p ng ton phn = o +q =

+q p ng qu q xem li phn mch bc 2 khng c ngun ngoiRLC Tng Qut Tm p ng ton phnca v khi > 05/22/201425AC Tn Hiu Dng SINE =

sin +

: bin : tn s gc : phaAC Bin Pha (Phasors) Bin Pha l mt s phc c trng cho bin v pha ca mt tn hiu dng SINE. =

cos +min thi gian =

bin pha Theo cng thc Euler:

= cos sin =

=

cos + =

=

t =

=

5/22/201426Cng Thc Lng Gicsin = sincos cos sincos = cos cos sinsinsin = sincos = cos sin

2= cos cos

2= sinAC Bin Pha (Phasors)Min thi gian Bin Pha

cos +

sin +

2

cos +

sin +

25/22/201427Tnh Cht Ca Bin Pha =

=

o hm

Tch phn

Chng Minh

=

sin + =

cos + +

2=

2

=

=

sin +

=

cos + +

2

=

2

=

5/22/201428V D Chuyn cc t/h sau thnh dng bin Pha = 6 cos 50

6 = 6

6 = 4 sin 30 +

5 = 4 710= 4 cos 30 +

5 +

2= 4 cos 30 +710V D Chuyn cc t/h sau thnh dng Sine = 3 + 4 = 5cos + 126= 5 1260 = 8

9 = 8 cos +718= 8

9

2= 87185/22/201429Bi Tp Xc nh dng () v () bng pp bin Phc = 1 +2

1 = 4 cos( + 300)

2 = 5 sin 200 = 1 +2

1 = 10 sin( 300)

2 = 20 cos +4504 + 8 3

= 50cos 2 + 7502

+ 5 + 10 = 50 cos 5 300Mi Quan H Gia Dng Th5/22/201430Tr Khng Tng ngMch AC Tm phng trnh 0 5/22/201431Ht Chng 2