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(1)

(2)

(3)

(4)

t t

TH x t

t

0 0

0

τ τ

τ

0

TH

τ

t

TH

TH t t

t

τ

τ τ

TH t t

τ τ

t

t0

t

t

TH x t t

t

0 0

0

τ τ

0

τ t

e /

*

o

(5)

t t

TH x

t t

t

0 0

0

τ τ

0

TH

t

t

x t TH

t t

0 0

τ 0 τ

− −

τ τ τ

x t x

t

t

t x t

TH t

t

0 0

τ 0 τ τ

t

t x t

TH t

t

0 0

0

τ τ

τ

τ τ

τ τ

0 0

0

t t

t TH t

t

## ( () )

τ 0

0

t t TH

TH

0

0 2

1

0

t t

τ

0 2

1

0

t t

τ

(6)

(7)

TH

vS +

RS a

b C

+ vc _

TH C

C

TH

0 ) 0 (

, =

= S C

S V v

v 假設

τ t S S

C

TH

，電容被充電

τ t

### te

dt C dvC

S S C

R v −v

0 : a KCL

− = +

S S C c

R v v

dt C dv

(8)

1. 電路只有獨立恆定電源

2. 對於關注的變數可以容易得到微分方程式。通常，使用基本的分析工具，如 KCL、 KVL. . .或戴維寧等效定律

3. 微分方程式的初始條件是已知的或者可以利用穩態分析得到

( )

1 2

t tO

O

ττττ

(9)

0 0 1

t t

τ τ

0 1 2

2 1

0 1

1

0

0 1 2

0

1

t

τ

τ

t

K e dt

dy

= 2

> ⇒ +

= , 0

)

(t K1 K2e t y

t τ

2

1

1 2 y(0 ) K

K = + −

0 0

1 0

1 a

y f dt dy a f a

y dt a

a dy + = ⇒ + =

### τ K

1

(10)

) 2 0 (

. 0 ),

(

VS

v t

t

v > 假設 =

0 ) ) (

( + =

dt t C dv R

V t

v S

2 / )

0 (

v =VS 初始條件

f dt y

dy + = τ

) (

, t

0

0 ,

) (

1 2 1

K v(t)

t e K K t

v

t

>

>

+

=

τ

τ

VS

dt v

dv = 0⇒ = 從微分方程式得到的 穩態值

VS

K =

1

) (

f K f

dt y

dy + = 1 = 假如數學模型為τ

1 2

2

1 (0)

) 0 (

0

K v

K K

K v

t

= +

= =

f v

K2 = (0)−

2 / 2

/ )

0

( VS K2 VS

v = ⇒ =−

0 , )

2 / ( )

(

: v t =VV eRC t >

t S

Vs

t v dt t

RC dv ( )+ ( ) =

R /

*

) 0 ( );

(

0 ,

) (

2 1

1

2 1

+

= +

=

>

+

=

x K

K x

K

t e K K

t x

t τ

(11)

### 求

− + vR

− + vL

(t) i KVL

) ( )

( t

dt Ldi t

Ri v

v

VS = R + L = +

0 ) 0 ) ( 0 ( ) 0 (

0 ) 0 (

0 + =



 +

=

⇒ −

=

⇒ −

< i

i i

i t

R t V

i dt t

di R

L S

= + ( ) )

( R

= L τ 步驟 2 穩態

R K V

i(∞) = 1 = S 步驟 3 初始條件

2

) 1

0

( K K

i + = +





=

LR

t

S e

R t V

i( ) 1

) 0 ( );

(

0 ,

) (

2 1

1

2 1

+

= +

=

>

+

=

x K

K x

K

t e K K

t x

t τ

1 2

t

τ

(12)

) ) (

( i t R

t IS = v + (t)

v

= ( )⇒ )

( t

dt Ldi t

v (t) i(t)

dt di R

IS = L +





=

LR

t

S e

I t

i( ) 1

R

= L τ 練習範例

1 2

t

τ

### t >

(13)

2

) ) (

( R

t t v

i =

V v

k V k

vC k (12) 4 (0 ) 4

6 3

) 3 0

( = ⇒ + =

= +

) 0 ) (

(

||

; ) 0 ) (

) ( (

2 1 2

1

= +

=

= +

+

P

P

R t t v

dt C dv

R R R R

t t v

dt C dv R

t v

=

= k k k RP 3 ||6 2

s F

C

RP =(2×103Ω)(100×10 6 ) =0.2

=

τ 步驟 1

], [ 4

)

(t = e0.2 V t >

v

t

0 ],

3 [ ) 4 (

:i t = e0.2 mA t >

t

, )

(t = K1+ K2e t >

i

t τ

### 當 t < 0

(14)

V t

v dt t

dv

O

O ( ) ( ) 6

5 .

0 + =

] [ 3 ) ( ) ( 5 . 0

12 ) ( 4 ) ( 2

A t

i dt t

di

t i dt t

di

= +

=

+ vO(t) = 2i(t)[V]

O

KVL(t>0) (t)

i

0 ) ( )

( )

( 3

1

1 + + + =

t R i t

dt Ldi t

i R VS

### τ

0 ,

)

(t = K1+ K2e t >

v

t

O τ

V v

t v dt t

dv

O O

O ( ) ( ) 6 ( ) 6

5 .

0 + = ⇒ ∞ =

) 1

( K

vO ∞ =

V K1 =6

) 0 ( );

(

0 ,

) (

2 1

1

2 1

+

= +

=

>

+

=

x K

K x

K

t e K K

t x

t τ

(15)

=

= 2||2 1 RTH

0 4 4

12+ 1 − =

I

I1

KVL

KVL VTH =VOC = 2I1 −4 =4[V] ]

[

1 4 A

I =

] 3[ ) 4 0 ( ) 0

( i A

iL − = + =

0 ,

)

(t = K1+ K2e t >

v

t

O τ

] 3[ ) 8 0 3 (

) 4 0

( v V

i + = ⇒ O + =

0 3 ,

3 5 )

(t = − e0.5 t >

i

t

a

b

0 ],

3 [ 6 10 )

(t = − e0.5 V t >

v

t O

L

3 6 10

3 8

2 2

2

1 + K = = − KK =

K

L

(16)

O

### 求

C R1

R2

0 )

( )

( 0 )

( 1 2

2 1

= +

⇒ + + =

+ C C c

C t v

dt C dv R R R

R t v

dt C dv

s F

C R

R ) (6 10 )(100 10 ) 0.6 ( 1+ 2 = × 3Ω × 6 =

=

) 3 (

) 1 4 (

2 ) 2

(t v t v t

vO C = C

= +

t

C τ K1 =0

− +

) 0

C(

v (12)V 9

= 6

] [ 8 8

) 0

( K1 K2 K2 V

vC + = = + ⇒ =

0 ],

[ 8

)

(t = e0.6 V t >

v

t C

0 ],

3 [ ) 8

(t = e0.6 V t >

v

t O

c

) 0 ( );

(

0 ,

) (

1 2 1

1

2 1

+

= +

=

>

+

=

i K K

v K

t e K K

t v

C

t

C τ

(17)

1

= ⇒ +181( ) 0

1 i t

dt Ldi

L

0 ) ( ) 9 (

1

1

1 t + i t = dt

di

) 0 ( );

(

0 ,

) (

1 2 1

1 1

2 1

1

+

= +

=

>

+

=

i K K

i K

t e K K

t i

t τ

9

= 1 τ

) 0

1( − i

V A

i 1

12 ) 12 0

1( =

= Ω

− 步驟 3

] [ 1 )

0 ( )

0

( 1 1 2 2

1 i K K K A

i − = + = + ⇒ =

1

19

9

t

t

### 解答

)

1(t i

− + vL

(18)

Circuit with resistances

and sources

Inductor or

Capacitor a

b

Representation of an arbitrary circuit with one storage element

VTH +

RTH

Inductor or

Capacitor a

b

VTH +

RTH a

b C

+ vc _

Case 1.1

Voltage across capacitor

VTH +

RTH a

b L

iL Case 1.2

Current through inductor

c

R

c

R

c

C

TH TH C

R

TH TH C

C

TH C

C

TH

− + vR

− + vL

TH L

R

L TH

R

L

L

TH L

TH

L

TH TH L

L

TH

SC

(19)

+

(t) iO

=0 t

>0 t

; (t) i Find O

TH TH O

O

TH

1

2

t O

τ

### 24

− +

>0 t

Thevenin for t>0

at inductor terminals a

b

TH

TH

TH

O O

2 0.3

1

2 0.3

t t

2

0.3

t

O 1

(20)

− +

) 0 ( ) 0

( − = O +

O i

i

<0 t

2

0.3

t O

2

0.3

t O

+

(t) iO

<0 t

Circuit for t<0

i1

2

3

1

1

3

1

2

2

1

2

3

3

1

3

2

3

C

C

O

v1

8 6 0

24 6

6 1

1 1

1 − = ⇒ =

+

+ v v v

v

6 6

) 24 0

( v1

iO + = +

(21)

+

-

O

O

C

O

C

TH C

C

TH

+

-

O

TH

TH

s F 0.3 10

* 100

* 10

*

3 36 =

=

C

C C

3 . 2 0 1

t

C K K e

v = +

2 1.5

1

2 0.3

t t

1

(22)

+

-

O

C

C

C

2

1

C

2

1

C

O C 微分方程式

(23)

iR iL

iC

= 0 + + +

iS iR iL iC

) ( );

( )

1 ( );

(

0

0

dt t C dv i

t i dx x L v

R i t

i v L C

t

t L

R = =

+ =

S L

t

t

i dt t

C dv t

i dx x L v

R

v + 1

### ∫

( ) + ( 0)+ ( ) =

0

S

2 2

− +vR

− +vC

=0 +

+ +

vS vR vC vL

) ( );

( )

1 (

; 0

0

dt t Ldi v

t v dx x C i

v Ri

v C L

t

t C

R = =

+ =

dt dv C

i dt Rdi dt

i

Ld 2 + + = S

2

S C

t

t

v dt t

Ldi t

v dx x C i

Ri + 1

### ∫

( ) + ( 0)+ ( ) =

0

101 學年度大學 學年度大學 學年度大學「 學年度大學 「 「繁星 「 繁星 繁星 繁星推薦 推薦 推薦 推薦」 」 」 」招生簡介 招生簡介

Normalization by the number of reads in the sample, or by calculating a Z score, should be performed on the reported read counts before comparisons among samples. For genes with

1 宙斯 Zeus Jupiter 天神之父，地上萬物的最高統治者，奧林匹 斯之主，諸神之神。. 2 希拉 Hera Juno

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Bìyè hòu, Chén Xiǎofēng céng sìchù xúnzhǎo gōngzuò, què wèinéng rúyuàn, yīncǐ.. gǎnjué

Hong Kong Institute of Vocational Education (IVE) Higher Diploma

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South Africa( 南非 南非 ) ) 簡介 簡介.  德班位於南非納塔爾省 德班位於南非納塔爾省 (Natal) (Natal) 南部之印度洋，與台