電路學 - [第五章] 一階rc/rl電路
TRANSCRIPT
RC/RL
Department of Electronic EngineeringNational Taipei University of Technology
••• RC RL
•
Department of Electronic Engineering, NTUT2/32
i
+
−C
+
−
v
dvi C
dt=
L
vL−+
i
L
div L
dt=
( ) ( ) ( )0
0
1 t
ti t v t dt i t
L= +∫( ) ( ) ( )
00
1 t
tv t i t dt v t
C= +∫
Department of Electronic Engineering, NTUT3/32
RC RL
RCv(t)+
−
ic iR
+
−v t( ) L R
−
+vR
i t( )
KCL 0c Ri i+ =
( ) ( ) ( ) ( )10 0
dv t v t dv tC v t
dt R dt RC+ = ⇒ + =
( ) ( ) ( ) ( )0 0di t di t R
L Ri t i tdt dt L
+ = ⇒ + =
KVL
( ) ( ) ( )dy tay t f t
dt+ =
( ) ( ) ( )dy tay t f t
dt+ =
• RC RL
Department of Electronic Engineering, NTUT4/32
RC RL ( )
Vs C
+
−v
vR+ −
it = 0 Rt =0 iL
+ −vL
RVs
vs(t) C
+
−v
vR+ −
iR iL
+ −vL
Rvs(t)
( ) 0f t ≠
Department of Electronic Engineering, NTUT5/32
• RC RL
y v(t) i(t) af(t) f(t)
•
c y(0)
t
c e-at
eat
( ) ( ) ( )dy tay t f t
dt+ =
( ) ( ) ( )at at atdy te e ay t e f t
dt+ =
( ) ( ) ( ) ( ) ( ) ( )at at at at atdy t dy tde y t e e ay t e ay t e f t
dt dt dt
= + = + =
( )( ) ( ) ( )
at
at at atd e y t
dt e f t dt e y t e f t dt cdt
= ⇒ = +∫ ∫ ∫
( ) ( )at at aty t e e f t dt ce− −= +∫
Department of Electronic Engineering, NTUT6/32
•(Complete response)
( ) ( )at at aty t e e f t dt ce− −= +∫
( ) ( )at at aty t e e f t dt ce− −= +∫
yp (Particular solution)(Forced response)
(Steady-state response)
(Homogeneous solution)
(Natural response) t
(Transient response)
t
A Be τ−
= +
f ny y= +
( )at atfy A e e f t dt−= = ∫
tat
ny Be ceτ− −= =
1
aτ =
Department of Electronic Engineering, NTUT7/32
• f(t)
::::
::::
(Natural response)
f (t) = 0
f (t) = b
( ) ( ) ( ) ( ) ( )at at atdy tay t f t y t e e f t dt ce
dt− −+ = ⇒ = +∫
( ) ( )0at aty t ce y e− −= =
( ) atby t ce
a−= + ( )0
by c
a= + ( )0
bc y
a= −
( ) ( )0 atb by t y e
a a− = + −
:::: (y (0) = 0) f (t) = b
( ) atby t ce
a−= + ( )0 0y =
bc
a= −
( ) atb by t e
a a−= −
Department of Electronic Engineering, NTUT8/32
1
• 5-1(a) f (t) = 0 y(0) = 10 ( )(b) f (t) = 5 y(0) = 10 ( )(c) f (t) = 5 y(0) = 0 ( = )
(a)
(b)
(c)
( ) ( ) ( )5dy t
y t f tdt
+ =
( ) ( )at at aty t e e f t dt ce− −= +∫
( ) 5ty t ce−=
( )0 10y c= =
( ) 510 ty t e−∴ =
( ) 5 5 5 55 1t t t ty t e e dt ce ce− − −= × × + = +∫( )0 10 1 9y c c= = + ⇒ =
( ) 51 9 ty t e−∴ = +
( ) 5 5 5 55 1t t t ty t e e dt ce ce− − −= × × + = +∫
( )0 0 1 1y c c= = + ⇒ = −
( ) 51 ty t e−∴ = −
Department of Electronic Engineering, NTUT9/32
(I)
•
• RC (a)
(a) RC
4 Ω 6 Ω
2 Ω 7 Ω
C = 1 F 150 V
+ −v
t =05 Ω
+
−
vc(t)
(b) t = 0
t = 0 t = 0
t = 0
(b)
vc ( )0− 150 V
+ −v+
−4 Ω 6 Ω
2 Ω 7 Ω 5 Ω
Department of Electronic Engineering, NTUT10/32
(II)
Rth =10 Ω 1 F
+
−vc t( )
(c) t > 0
( )4 2 / /6 7 10thR = + + = Ω
( ) 100 150 100V
10 5cv − = × = +
t 0
vc 100V vc
(c)( ) ( )0 0 100Vc cv v+ −= =
Department of Electronic Engineering, NTUT11/32
RC (I)
• t > 0
KCL
( ) t > 0
( ) 00cv V− =
0c Ri i+ =
( ) ( )0c cdv t v t
Cdt R
+ =
( ) ( )10C
C
dv tv t
dt RC+ = ( )
t
RCcv t ce
−=
( ) 00cv V− = ( ) 00cv c V− = =
( ) ( ) 00 0c cv v v+ −= =
( ) 0
t
RCcv t V e
−= ( ) ( ) 0
tc RC
c
dv t Vi t C e
dt R
−= = − ( ) ( ) 0
tc RC
R
v t Vi t e
R R
−= =
RCvc(t)+
−
ic iR
Department of Electronic Engineering, NTUT12/32
RC (II)
• RC
(a)
K t
R C
RC ( , )
• RC
(Time constant)
v(t) i(t)
(b)
K
0 t
f t( )
(a)
(b)
( )t
f t Ke τ−
= t → ∞
( ) ( ) 0dy t
ay tdt
+ =
RCτ =
R v i= C q v=
( ) 0f t →
τ 1
a
τ τ
τ
V0
vc(t)
0t
0.368V0
1 Kτ =2 2Kτ =
3 3Kτ =
K 2K 3K1τ 2τ 3τ
Department of Electronic Engineering, NTUT13/32
2
• (a) t = 0 t = 0 t 0 vc(t) , v(t) ic(t)
(c) t > 0
τ = RC
(a) RC
4 Ω 6 Ω
2 Ω 7 Ω
C = 1 F 150 V
+ +v
t =05 Ω
+
−
vc(t) Rth =10 Ω 1 F
+
−vc t( )
(c) t > 0
( ) ( ) ( )0 0 100 Vc cv v+ −= =
( ) ( )0.110100 100 Vt
tcv t e e
− −= =
( ) ( ) ( ) ( )0.1 0.17 7100 70 V
7 6 // 2 4 10t t
cv t v t e e− −= = × =+ +
( ) ( ) ( ) ( )0.1 0.1100 0.1 10 Ac t tc
dv ti t C e e
dt− −= = × − = −
Department of Electronic Engineering, NTUT14/32
3 ( )
• v(0) = 4 (V) t > 0 i
KCL
t > 0
+−
18 F 3 Ω2i (V)
+
−v
i
6 Ω
2 10
6 8 3
v i dv v
dt
− + + =1
8
dv dvi C
dt dt= =
114 0 6 0
6 8 3
dvv
dv v dvdt vdt dt
−+ + = ⇒ + =
( ) ( ) ( )60 4 Vt
tv t v e eτ− −= =
( ) ( ) ( )6 614 6 3 A
8t tdv
i t C e edt
− −= = × × − = −
Department of Electronic Engineering, NTUT15/32
RC (I)
• v(0–)=V0
t = 0
• t > 0 v(t)
(b)
( )
( )
RC K( V0–Vs) (
RC )
( ) ( )0
t
RCs sv t V V V e
−= + −
sV
( )0
t
RCsV V e
−−
(a)
Vs
0 t
v(t)
( )0
t
RCsV V e
−−( )0 sV V−
(b)0 t
V0
Vs
v(t)
( )0
t
RCs sV V V e
−+ −
Vs C+
−v
vR+ −
it = 0 R
Department of Electronic Engineering, NTUT16/32
RC (II)
• RC
1. RC
2. ( )
3. ( ) τ
t
e τ−
×
Department of Electronic Engineering, NTUT17/32
4
• t = 0 v(0 ) = V0
t 0 v(t)
v(t)
t = 0+
KCL t 0
I
+
−vC
t =0a
b
R
iR iC
dv v dv v IC I
dt R dt RC C+ = ⇒ + =
( )t t t t
RC RC RC RCI
v t e e dt ce RI ceC
− − − −= + = +∫
( ) ( ) 00 0v v v+ −= = ( ) 00v RI c V+ = + = 0c V RI= −
( ) ( )0 , 0t
RCv t RI V RI e t−
= + − >
Department of Electronic Engineering, NTUT18/32
5
• (a) v(0 ) = 15 (V) t > 0 v(t)
24 Ω 13F
+
−v40V
8 Ωi a
b(a) (b)
40V
a
bRth = 6 Ω
8 Ω
24 Ω
A. ( (b))
30(V) abVoc
( )2440 30 V
24 8× =
+
B. (c) 8 24
68 24thR
×= = Ω+
16 2 sec
3thR Cτ = × = × =
( ) 0.5215 30 15t
te e− −− = −
C. t > 0 ( ) ( )0.530 15 Vtv t e−= −
13F30V
a
b(c)
6 Ω
Department of Electronic Engineering, NTUT19/32
RC
• RCv(0) = 0 t > 0
• RC
Voc = ( )
Rth =
( ) ( )1t
tRCs s sv t V V e V e τ− −= − = −
( ) ( )1 tht R Cocv t V e−= −
a te−
Vs C
+
−v
vR+ −
it =0 R
Department of Electronic Engineering, NTUT20/32
6
• v(0) = 0 t 0 v(t) i(t)
(b)
(a)
+−36 V 12 Ω
+
−v t( )
6 Ω
18
F
i t( )
+−Voc 24= V
a
b
+
−v t( )
Rth = 4 Ω
(b)
18
F
( )1236 24 V
12 6ocV = × =+
6 / /12 4thR = = Ω
1 14 sec
8 2thR Cτ = = × =
( ) ( )( )224 1 Vtv t e−= −
( ) ( ) ( )236 6=2+4e Ati t v t −= −
Department of Electronic Engineering, NTUT21/32
•I0 ( )
• RL
RL
0di dt = 0Lv L di dt= =
Department of Electronic Engineering, NTUT22/32
RL
• RL i (0 ) = I0
t > 0
• RL τ
L/R τ ( L )
i(0+) = i(0 ) = I0 t > 0
i(0 ) = I0
+
−v t( ) L R
−
+vR
i t( )( ) ( ) ( ) ( )0 0di t di t R
L Ri t i tdt dt L
+ = ⇒ + =
( )R
tLi t ce
−=
( ) 00i c I− = =
( ) 0
Rt
Li t I e−
= ( ) 0
Rt
LL
div t L I Re
dt
−= = − ( ) 0
Rt
LRv t iR I Re
−= =
( )t
f t Ke τ−
=
Department of Electronic Engineering, NTUT23/32
7
• 1 ( 1 ) t = 0
2 t > 0 v(t) i(t) 2 Ω
t =0+
−30V
2+
−v t( )
3 Ωa
b
i t( ) 2 Ω1 H
3 Ω1
i 1H158 Ω
a
b
15
845
4
t
ab
div e
dt
−= = − ( )
15 15
8 82 45 2 9
V3 2 4 5 2
t t
abv t t e e− −
= × = − × = −+
(3)
1
6A
2 a-b
( )0
30 306 A
2 3 5i I= = = =
+
( )( )
3 2 3 15
3 2 3 8thR× +
= = Ω+ +
( )15
86t
i t e−
= 8 sec
15th
L
Rτ = =
(1)
(2)
t > 0
Department of Electronic Engineering, NTUT24/32
8 ( )
•L/R i(0) = 2 A
v6A
a
b
2 Ω
32H
4 Ω
i
+ −v
34 2
2 6ab
di vv i i
dt = + = × −
3
2
di div L
dt dt= =
33 24 22 6
didi dti idt
+ = × −
6 0di
idt
+ = 62 Ati e−=
1
6τ = L
Rτ =
Department of Electronic Engineering, NTUT25/32
RL (I)
• RL
• i(0 ) = I0 t = 0
KVL t > 0
t = 0+
t > 0
t =0 iL
+ −vL
RVs
ss
di di R VL Ri V i
dt dt L L+ = ⇒ + =
( )R R R R
t t t ts sL L L L
V Vi t e e dt ce ce
L R
− − −= + = +∫
( ) ( ) 00 0i i I+ −= =
( ) 0 00 s sV Vi c I c I
R R+ = + = ⇒ = −
( ) 0
Rt
s s LV V
i t I eR R
− = + −
ss
VI
R= ( ) ( )0
Rt
Ls si t I I I e
−= + −
Department of Electronic Engineering, NTUT26/32
RL (II)
• i(t) Vs / R
•
• RL
RL
– × τ
( ) ( )0
Rt
Ls si t I I I e
−= + −
0
Rt
s LV
I eR
− −
t
e τ−
0
Rt
s LV
I eR
− −
VssV
R
sV R
Department of Electronic Engineering, NTUT27/32
9
• i(0 ) = 2 A t > 0 i(t)
A. t > 0 a-b
( )
B.
a-b
+−
3 Ω
6 Ω 3 Ω2 H
b
a
it =0
6 Ω
36V+−6 V
2 Ωa
2 H
b
i t( )
3 336 6(V)
6 3 3 3oc abV v = = × = + + ( )6 / /6 3 / /3 2thR = + = Ω
21
2th
L
Rτ = = = ( ) ( ) ( )3 2 3 3 At ti t e e− −= + − = −
( ) ( )36 63 A
6 6 / /3 3 6× =
+ +( )6 / /6 3 / /3 2+ = Ω ( ) ( ) ( )3 2 3 3 At ti t e e− −= + − = −
Department of Electronic Engineering, NTUT28/32
•
K K = 1 u(t) (Unit step function)
us(t) (a) t = 0 0 1
• t – t0 t
R (b) u(t – t0) t0 t0
u t tK t
( ) = <= >
0 00
t t0R t t0
,,
u t t0( )− = <= >
0
1
0 t
(a)
(b)
( )u t
1
0 t0t
( )0u t t−
Department of Electronic Engineering, NTUT29/32
10
• i(t)
KCL
B. t > 0 i t i i Aef nt( ) = + = + −3 2di
dt i+ =2 6
A. t < 0
i(0) = 0 A = 0 i(t) = 0
didt i+ =2 0 i t Ae t( ) = −2
C. t i(t) = 0 , t < 0
= 3 – 3e-2t , t > 0 i t e u tt( ) ( ) ( )( )= − −3 1 2 A
i(0) = 0 = 3 + A = 0 , A = -3 i t e t( ) ( )( )= − −3 1 2 A
+−6u t( ) V
2 Ω
1H
i
Department of Electronic Engineering, NTUT30/32
11
• i(t)
A. u(t) (b)
B. –u(t – 2) (c)
i2(t) i2 t( ) = – 0.5 [1 –e-2(t – 2)] , t > 2
i1 t( ) = 0.5 (1 –e-2t)(A) , 0 < t < 2i t( ) =
i1 t( ) += i2 t( ) = – 0.5 [1 –e-2(t – 2)] (A) , t > 20.5 (1 –e-2t)
i1(0) = 0 A = 0 i1(t) = 0
t 0 didt i+ =2 0 i t Ae t( ) = −2
t 0 i1 t i1 i1 Aef nt( ) = + = + −0.5 2
di1dt i1+ =2 0
i1(0) = 0.5 + A = 0 A = - 0.5⇒
0.5 (1 –e-2t) (A) , t > 0i1 t( ) =
+−[u(t)−u(t−2)] V
2 Ω i
1H+
−v t( )
u(t) i1(t) –u(t – 2)
i2(t) i(t) = i1(t) + i2(t)
+−
2 Ω
1H
(c)
i2(t)
−u(t−2) V
(a)
+− u t( )
2 Ω
1H
(b)
i1(t)
Department of Electronic Engineering, NTUT31/32
• RC RL
yf yn
• yf ( ) ( )yf yf yn y(0)
B yn = Be –at
( ) ( ) ( )dy tay t f t
dt+ =
1
aτ
( )t
aty t A Be A Beτ− −= + = +
( ) ( )0 atf f f ny t y y y e y y− = + − = +
( ) ( )0 0 fB y A y y= − = −
Department of Electronic Engineering, NTUT32/32