2/5/20151Chng 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)2/5/20152MCH 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)2/5/20153Mch 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 RL2/5/20154Mch 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

2/5/20155Mch RC Khng C Ngun Ngoi (2) = 0

1

p ng in th ca mch RC suy gim gi tr ban u 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

2/5/20156Mch 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 = 152/5/20157Mch 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

2/5/20158Mch 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

2/5/20159Mch 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 = 62/5/201510Mch 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 02/5/201511Bi TpBi 2: Tm khi > 0MCH DC BC NHT CNGUN KCH THCH NGOIHm Bc n VMch RCMch RL2/5/201512Hm 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.2/5/201513p 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

=

2/5/201514p 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 t2/5/201515p 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 = 27,972/5/201516p 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 t2/5/201517p 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 +315