no.11 chapter 6 first-order circuit 一阶电路 1.rc and rl circuits 2.initial conditions...

NO.11

Chapter 6

First-Order Circuit

一阶电路

1. RC and RL Circuits

2. Initial Conditions

3. First-order Circuit Zero-input Response

4. First-order Circuit Zero-state Response

5. First-order Circuit Complete Response

6. Three quantities in finding the response

7. Applications

Items:

Introduction

换路：即电路变化

C

t = 0

Us

R K +

–uC

RC Circuits

t = 0

+

–uR

Us

R KRL

R Circuits

+

–uR

Us

RRL

Before K is switched oni = 0 , uC = 0

i = 0 , uC= Us

(I) Dynamic circuit

i

+

–uC

Us

RC

Introduction to dynamic circuits

Steady state analysis

K+

–uC

Us

RC

i t = 0

K is switched on long time

K+

–uC

Us

RC

i

Initial state Transient

state

Steady state

t1

USuc

t0

? iR

U S

Response of a Circuit• Transient response of an RL or RC circuit

is– Behavior when voltage or current source are suddenly

applied to or removed from the circuit due to switching.

– Temporary behaviorDynamic Circuit ： circuit which containing energy-store elements (L, C).

过渡过程

(II) Why the circuit produce transient process

1. The circuit containing L and C elements

• Inductor current cannot change instantaneously

• Capacitor voltage cannot change instantaneously

2. voltage or current source are suddenly applied to or removed from the circuit

switching

动态电路及过渡过程动态电路及过渡过程

动态电路 dynamic circuit 特点：当改变原来的工作状态时会有过渡过程。（瞬态）

原稳态开关动作

接入 ( 去掉 ) 电源

t

换路

过渡过程新稳态

Transient state

研究动态电路基本规律的意义：认识、掌握过渡过程 ( 暂态过程 transient state) 的现象和规律。

动态电路的基本规律反映出一般动态系统（机械、自动控制系统等）的普遍规律。

FORMULATING RC AND RL CIRCUIT EQUATIONS

A first-order circuit is characterized by a first-order differential equation.

k k

0)()()(

01 ttxtyadt

tdya

6-1 Initial conditions of the circuits6-1 Initial conditions of the circuits1.Initial conditionsInitial conditions ：： The values of voltages The values of voltages and currents and the derivatives at t = 0+:and currents and the derivatives at t = 0+:

dt

di

dt

duuiiu LLCc

)0()0()0()0()0()0(

、、、、、

0 t

0 0

确定微分常数时须利用电路初始条件 f(0+)

just prior to switching just after switching

Switching at t=0

2. Switching rule 换路规则： At the instant of switching, if the current flowing through the capacitance is finite, the charges remain unchanged; if the voltage across the inductance is finite, the current flowing through the inductance remains unchanged.

uC(0+)=uC(0-)

iL(0+)=iL(0-)

Note ： Only the voltage on a capacitor and the current through an inductor cannot chang abruptly 。

3. 电路中其他初始电压、电流的一般求解方法：

具体求法是：画出 t=0+ 电路：在该电路中若 uC (0+)= uC (0-)= US ，电容用一个电压源 US 代替，若 uC (0+)= 0 则电容用 a short circuit 代替。若 iL(0+)= iL(0-)=IS ，电感用一个电流源 IS 代替，若 iL(0+)= 0 则电感作 an open circuit 。

由 t=0+ 电路来求得

电阻电路

S(t=0)

si Ri Ci Li

+

-Ru

+

-Cu

+

-Lu

t

0-

0+

Li CuCi RisiLu Ru

5A 10V0 5A00 10V

5A 10V 00-10A-10V

LcRs iiii 10

15A

Example 1

Assume Close the switch S when t=0. Find i1(0+) 、 i2 (0+) 、i3 (0+) 、 uc (0+) 、 uL (0+).

example2

)0()0( Lc iu = = 0 t=0+ circuit

.0)0()0( Lc iu

求初始值的一般步骤：(1) 根据 t=0- 时的电路，求出 uC(0-) 及 iL(0-) 。

(2) 由换路定则 : uC(0+) = uC(0-) , iL(0+) = iL(0-)

。(3) 作出 t=0+ 时的等效电路，并在图上标出各

待求量。(4) 由 t=0+ 等效电路，求出各待求量的初始值

。

§6-2 Zero-input response of first-order RC circuits

一阶 RC 电路的零输入响应

Assume that at time t=0, shut the key K, and uc(0-)=U0

Find ： uc(t) ， ic(t) （ t≥0 ）

(t=0)

Capacitor Discharging

即：放电过程 discharge

Solution:

Zero input response ： The behavior (in terms of the voltages and currents) of the circuit itself, with no external sources of excitation.

Solution: -uR+uc=0By KVL:

∵uR=i R, dt

duCi C

0 CC u

dt

duRC

Eq.(7-1) is a homogeneous equation because the right side is zero.

Eq.(7-1)

A solution in the form of an exponential　　　 uc=Kest 　　 t≥0 　　 Eq.(7-2)

where K and s are constants to be determined.

characteristic equation

Substituting the trial solution into Eq.(7-1) yields OR

Eq.(7-3)

0)1(

0

CsRKe

KeCKseR

Tst

ststT

RTCs+1=0

CRs

T

1

CTRt

KeuC

a single root of the characteristic equation

zero -input response of the RC circuit:

∵uc(0+)=U0 00)0( UKKeu

The time constant

时间常数 τ=RC

0 , 0t

R C

Cu U e t

the zero-input response:

τ 小：过渡过程短τ 大：过渡过程长

It is customary to assume that it takes 3 3 ～～ 5 5 for the circuit to reach the new steady state.

0.368U0

k (t=0)

(t=0)uc(t)

U0

CR

t

C TeR

U

dt

duCi

0t≥0

i

(t=0)

Note: The key to working with a source-free RC circuit is finding:

(1)The initial voltage u(0)=U0 across the capacitor;

(2)The time constant τ;

Example 7.1

Let uc(0-)=15V. Find uC ,ux , and ix f

or t>0.

Vev tc

5.215 Aei tx

5.275.0

uC ux

R0

R

L

U0

uL

1

2

S(t=0)

i

§6-3Zero-input response of first-order RL circuits

Assume :

00 tiRdt

diL

RU

L Ii 0)0(

0)0()0( Iii LL

)0( tS 2

Find : iL(t) , t≥0

00

0)0( IAAeIit

t

LLR

From the initial conditions ：

Let i = Aept

L

Rp

tLR

Aei

Then (Lp+R)ept=0 Lp+R=0

0 iRdt

diL

tLR

eIti 0)(Hence ： t

eI 0=

R

L

Then ：

t

eIti

0)(t

eRIiRuR 0

t

eRIdt

tdiLuL

0

)(

R0

R

L

U0

uL

1 2

S(t=0)

i

Curve ：

tO

iL, uL

-RI0

I0

iL(t)

uL(t)

§6-2 and §6-3 Zero-input Response

(t=0)

Capacitor Discharging

uc(0-)=U0

00

teUu CRt

C

K(t=0)i

L uL

R

0)0( IiL

00

teIi LRt

L

τ=RC GLR

L

为什么ＲＣ的 τ 与Ｒ成正比，ＬＣ的却成反比？

•因为 RC 的 i0=U0/R, 此时Ｒ的耗能为 U20

／Ｒ ,与Ｒ成反比，即说明Ｒ越大，过渡时间持续越长， τ 与Ｒ成正比；

•而 LC 的 i0= Ｉ 0 , 此时Ｒ的耗能为Ｉ 20

Ｒ ,与Ｒ成正比，即说明Ｒ越大，过渡时间持续越短， τ 与Ｒ成反比．

K(t=0)i

L uL

R

0)0( IiL

R0

R

L

U0

uL

1

2

S(t=0)

i

Zero-input response of first-order RL circuits

Assume :

RU

L Ii 0)0(

0)0()0( Iii LL

)0( tS 2

Find : iL(t) , uR(t), t≥0

Multisim

第９周周一期中考试

范围 : 1-6 章

方式 : 闭卷 ( 英文试题 )

例 7.3.2 下图是一台 300kw 汽轮发电机的励磁回路。已知R=0.189Ω, L=0.398H, U=35V ，电压表量程为 50V ，内阻 RV=5kΩ 。 t=0 时开关 S 打开（设 S 打开前电路已稳定） .

求： 1 、 i(0+)

2 、 τ （ t≥0 ） 3 、 i 和 uv （ t≥0 ）4 、 S 刚断开时的 uv(0+)

Multisim

τ=L/R=79.6μs ,

Aeti t12560185)( kVetu tV

12560926)(

D

_

+i(0-)=i(0+)=185A, 解 :

t=0

§6-4 First-order Circuit Complete Response

全响应

Us

sCC Utudt

tduRC )(

)(

For t≥0

If uc(0)=0, it is Zero-State Response. 零状态响应

uc(t)

t=0i

1. Complete Response of a RC Circuit

How to find the complete response of the RC circuit ?How to find the complete response of the RC circuit ?

If uc(0) ≠ 0, it is Complete Response. 全响应

Us

sCC Utudt

tduRC )(

)(

For t≥0uc(t)

t=0i

divide solution v(t) into two components:

)()()( tututu FNC

The homogeneous solution (natural response) is the general solution of Eq.1 when the input is set to zero.

00)()(

ttudt

tduRC N

N

0)( / tKetu RCtN

total particularhomogeneous

The particular solution( forced response) :seek a particular solution of the equation

SFF Utudt

tduRC )(

)(t≥0

Now combining the forced and natural responses, we obtain

0)()()( / tUKetututu S

RCtFNC

Setting UF(t)=US meets this condition.

Us

Zero-State Response : uc(0)=0

uc(t)

t=0i

0)()()( / tUKetututu SRCt

FNC

∴K= -US

0)1()( // teUUeUtu tSS

RCtSC

using the initial condition:

O

US

t

uC(t)

i(t)

uC(t) , i(t)

R

U S

0)1()( / teUtu tSC

0)(

)( / teR

U

dt

tduCti tSC

即：充电过程

charge

dteRdteRdtiWt

st

s

RU

RU

R

0

2

00

222

)(

CsS WCUe

CURC

t

2

21

0

22

2

不论 R 、 C 如何，电源充电能量的一半被 R 吸收，一半转换为电容的电场能量，充电效率为 50 ％。

Us u(t)

t=0i

0)()()( / tUKetututu SRCt

FNC

∴K= U0 -US

Complete Response : uc(0)=U0

The complete response of the RC circuit:

0)()( /0 tUeUUtu S

RCtSC

uc(0)=U0

0)()( /0 tUeUUtu S

RCtSC

Complete Response :

Step response of first-order RC circuit

U0

US

Note: 0)()( /0 tUeUUtu S

RCtSC

0)1()( //0 teUeUtu RCt

SRCt

C

全响应 = 零输入响应 + 零状态响应

Zero-input response

Zero-State Response

The superposition principleuc(t)

US

t=0

uc(0)=U0

The RL circuit is the dual of the RC circuit:

0)()(

tItiRdt

tLdiS

LIS

The complete response of the RL circuit is

0)()( /0 tIeIIti S

GLtS

0)()(: /0 tUeUUtucircuitRC S

RCtSC

2. Complete Response of a RL Circuit

+

_uc(t)

30Ω

20Ω

t=0

KIS=1A

0.5F

Example1 :Example1 : The switch is opened at t=0. uc (0) =5V, Find u C (t) ． (t≥0)

ic

方法一：（经典法）求解微分方程( )

25 ( ) 20cc

du tu t

dt

Example1 :Example1 : The switch is opened at t=0. uc (0) =5V, Find u C (t) ． (t≥0)

t≥0 时的电路

方法二：先利用戴维南等效，

再套全响应公式

ic

Solution:

05)0()('

teeututt

cc

sRC 255.0)2030(

0)1(20)(''

tVetut

c

01520)()()( 04.0''' tVetututu tccc

Zero-input:

Zero-state:

Complete response:

3 、 Step function 阶跃函数

1

The unit step function单位阶跃激励 : ε(t)

t0

01

00)(

t

tt

The unit step function ε(t) is 0 for negative values of t and 1 for positive values of t.

延迟单位阶跃信号

0

00 ,0

,1)

tt

tttt－（

We use the step function to represent an abrupt change in voltage or current, for example, the voltage

0

00)(

ttU

tttu

A

A

May be expressed :

)()( 0ttUtu AA

UA ε(t-t0)

a

b

Equivalent circuit:a

b

UA=

t=t0

30Ω

+

_uc(t)20Ω

t=0

KIS

1A 0.5F

Example 1:Example 1: The switch is opened at t=0. uc (0) =5V, Find u C (t) , (t≥0)

Solution:

)(5)0()(' teeututt

cc

)()1(20)('' tetut

c

)()1520()()()( 04.0''' tetututu tccc

Zero-input:

Zero-state:

Complete response:

Cuus

Example 2:Example 2:

Find u C (t) , for t≥0.

Solution :

方法一 : 分段分析

to

us

10V

1S

uc (0) =0V, ,1: Assume

Cuus

st 10

to

us

10V

1S

Cu

t≥1s

UC(1)

uc (0) =0V,

,1: Assume

10 t

1t Veuu CC 32.6)1(10)1()1( 1

Vetu tc

)1(32.6)(

1) 用分段函数表示

Veu tC )1(10

方法二 : 用阶跃函数表示

)]1()([10)( tttus

Cuus

to

us10V

1S

to

to

)(10 t

)1(10 t

u ´

u "

Cu

to

u10V

1Sus

CCC uuu '''

)]1()([10)( tttus

According to the superposition principle:

Vteu tC )()1(10' Vteu t

C )1(]1[10 )1(''

s(t=0)

)(tuC

Determine the expressions for uC (t) and i(t). (t≥0)

Example ：

i(t)

0.1F

50/7V

10/7Ω

uC

t≥0

Vuc 5)0(

Cu

i

0.1F

s(t=0)

0,)7

505(

7

50)( 7 tVetu t

c

0tA,)(

5

)()(

dt

tduC

tuti CC

0,)7

105.2(

7

10 7 tAe t

s(t=0)

)(tuC

The question is changed to ‘Determine i(t)’. (t≥0)

Example ：

i(t)

0.1F

Is there another

way?

The complete response of a first-order circuits depends on three quantities:

1. The initial value of state variable(U0+ or I0+)

2. The final value of state variable(US or IS)

3. The time constant (RC or GL)

0)()( /0 teIIIti GLt

SSL

0)()( /0 teUUUtu RCt

SSC

f(0+)f(∞)

0)()0()()( / teffftf t

三要素法

Note:

(1) Get f(0+) --- use 0+ equivalent circuit

(2) Get f()---use equivalent circuit

(3) Get τ---calculate the equivalent resistance R, τ=RC or L/ R

Then,t

effftf

))()0(()()(

Note: method of “three quantities” can be applied in step response on any branch of First-order circuit.

6. Method of “three quantities” (method 3)

Solution ： Using the method of “three quantities”

(1) uc(0+) , i(0+)

Vuu CC 5)0()0(

s(t=0)

)(tuC

Determine the expressions for uC (t) and i(t). (t≥0)Example ：

i(t)

0.1F

)0()0( ii ?×

Find i(0+):s(t=0)

Cu

i

0.1F

Vuu CC 5)0()0(

i(0+):

Equivalent circuit at t=0+

5VAi 5.2

2

510)0(

s(t=0)

Cu

(2) uc(∞) , i(∞)

V

uC

7

50

1052

5)(

10V

2Ω

5Ω uC (∞)

i(∞)

Ai7

10

52

10)(

s(t=0)

Cu

(3) τ (t≥0)R=2//5=10/7Ω

SRC7

11.0

7

10

R

2Ω

5Ω

4 、 t

CCCC euuuu

)]()0([)(

Ve t7)7

505(

7

50

s(t=0)

Cu

i

(t≥0)

VuC 5)0( VuC 7

50)( S

7

1

s(t=0)

Cu

t

eiiii

)]()0([)( Ae t7)7

105.2(

7

10

i

(t≥0)

Ai 5.2)0( Ai7

10)( S

7

1

How to get initial value f(0+) ?

1. the capacitor voltage and inductor current are always continuous in some conditions. ( 换路定则 )

Vc(0+)=Vc(0-); IL(0+)=IL(0-)

2. ---use 0+ equivalent circuit .

C: substituted by voltage source;

L: substituted by current source.

3. Find f(0+) in the above DC circuit.

Note:

How to get final value f(∞) ?

How to get time constant τ?

The key point is to get the equivalent resistance R.R is the Thévenin equivalent resistance “seen” by the inductor (or the capacitor)

Use ∞ equivalent circuit(stead state) to get f(∞).

C: open circuit;

L: short circuit.

t

effftf

))()0(()()(

Example 2:Example 2: The switch is closed at t=0. iL (0)

=2A, Find iL (t) , u L (t) , i (t) (t≥0)

Solution ：“ Three quantities”

(1) Find f(0+ ):

220

)0(

10

10)0()

10

1

10

1(

L

L

uu AiVuL 3

5)0(

3

20)0(

When t→∞

(2) Find f(∞ ):

AiAiu LL 1)(1)(0)(

(3)Find ： R

a

b外加电源法

3

200 i

uR s

R

L3.0

20

32

AiAiVu LL 2)0(3

5)0(

3

20)0(

AiAiu LL 1)(1)(0)(

(4)According to:

0)()0()()(

teffftft

s3.0

§6-5 Applications

Differential Circuit and Integral Circuit

1 、 Differential Circuit

o t

ui /V

10

t1 t2

tP

o t

-10

uC

uO /V

10

<<tP

微分电路的条件(1) << tP ；(2) uR as output

RCui uO

uC

i

dt

tduRC

dt

tduRCiRtu iC )()(

)(0

Example:

2. Integral Circuit

o t

ui /V

10

t1 t2

to

u O /V

tP

>>tP

C

R

ui uO

uR

i

积分电路的条件(1) >> tP ；(2) uC as output

t

i

t

R

t

duRC

duRC

diC

tu

00

00

)(1

)(1

)(1

)(

§6-6§6-6 一阶电路的冲击响应一阶电路的冲击响应（不讲）（不讲）

作业 11 ：

《电路》 :

P150 6-1

《 Fundamentals of Electric Circuits 》 :

P284 7.5

P286 7.20

作业 12 ：


P288-289

7.36 7.43 《电路》P155 6-23

作业 13: 《电路》 P152-155 6-8 6-21


P291 7.55

no.11 chapter 6 first-order circuit 一阶电路 1.rc and rl circuits 2.initial conditions...

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