Chapter 4 Transients

Chapter 4Transients

2008.9

Electrical Engineering and Electronics II Electrical Engineering and Electronics II

Scott

•Main Contents

1. Solve first-order RC or RL circuits.

2. Understand the concepts of transient response and steady-state response.

3. Relate the transient response of first-order

circuits to the time constant.

4. Solve RLC circuits in dc steady-state

conditions.

Introduction

Initial state and DC Steady State

First-order RC Circuits

First-order RL Circuits

Summary

•Main Contents

t

E

Cu New steady statetransient

C

Old steady statesteady state

K R

E+

_ CuSwitch K is closed

4.1 Introduction

New steady steady statestate

R

Us+

_ Cu

Conception of steady state and transient state

When t=0 ， uc(0)=0

When t=∞, uc(∞)=Us

Old steady state

Why the transient response happens?

No transient

I

Resistance circuit

t = 0

E R

+

_

I

K

•Resistor is a energy-consumption element, current is

proportional to voltage, no transient response will happen

even if changing source

Energy can not change instantly because of accumulating or decaying period.

CW Charging or discharging Cu Change

gradually

Electric field energy ）（ 2

2

1CCuWc

E

K R

+

_ CuC

E

t

Cu

Magnetic field energy ）（ 2

2

1LL LiW

LWLi Change

gradually

K R

E+

_

t=0iL

t

LiE/R

Energy can not change instantly because of accumulating or decaying period.

Transients

•The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching.

•By writing circuit equations, we obtain integrodifferential equations.

The causes of transients:

1. Energy storage elements -inductors and capacitors

change gradually；

2.Changing circuit, such as switching source.

LC iu ,

4.2 Initial state and steady state

Assume changing circuit when t=0, then t=0– is end point of old steady state; t=0+ is the start point of transient state.

)0()0(

)0()0(

CC

LL

WW

WW

)0()0(

)0()0(

CC

LL

uu

iiFrom t=0–to t=0+,iL 、 uC

change continuously.

t=0tt=0

-

t=0+

The law of changing circuit

DC Steady State Response

•The steps in determining the forced response or steady state response for RLC circuits with dc sources are:

1. Replace capacitances with open circuits.

2. Replace inductances with short circuits.

3. Solve the remaining circuit.

Example 4.1 Find steady-state values of vx and ix in this circuit for t>>0.

Answer: vx =5V, ix = 1A t>>0

Exercise 4.3 Find steady-state values of labeled currents and voltages for t>>0.

Answer: va =50V, ia = 2A

i1 = 2A, i2=1A, i3=1A

How to get initial valueExercise 1: Assuming old circuit is in DC steady state

before switch K is closed. how to get uC(0+),iR(0+)?

iR

R14k

12V

K

t=08kR2

2FuC

Solution:

When t=0-, capacitor is considered as open circuit, we get equivalent circuit. R1

4k

12V uC(0–)8k

t=0-

8(0 ) 12 8V

4 8Cu

R14k

12V uC(0–)8k

iR

R14k

12V

K

t=08kR2 2F

uC

8(0 ) 12 8V

4 8Cu

Vuu CC 8)0()0(

substituting voltage source for uC(0+)

iR(0+)

8kR2

+

– u

C(0+)

t=0 ＋2

(0 ) 8(0 ) 1m A

8C

R

ui

R

How to get initial value

•Exercise 2: Given by Exercise 2: Given by RR11=4Ω, =4Ω, RR22=6Ω, =6Ω, RR33=3Ω, =3Ω, CC=0.1µF, =0.1µF,

LL=1mH, =1mH, UUSS=36V, switch S is closed for a long time. =36V, switch S is closed for a long time.

Open the switch S wOpen the switch S whenhen t=0, how to get the initial values t=0, how to get the initial values

of all elements?of all elements?

How to get initial value

Equivalent circuit of First-order circuit

Two parts: one (equivalent) capacitor or inductor; a two terminal network with resistance and sources.

N L N Cor

First-order circuit

Only one (equivalent) capacitor or inductor is included in a linear circuit.

4.3 First-order RC Circuits

According to Thevenin Law

N L N Cor

RU LuL

iL

+

-

RU CuC

iC

+

-

4.3 First-order RC Circuits

Differential equation of first-order RC circuit

R

U LuL

iL

+

-

R

U CuC

iC

+

-

Uuu CR

Uudt

duRC C

C

Uuu LR

Udt

tdiLtRi L

L )(

)(

R

Uti

dt

tdi

R

LL

L )()(

)0( tS

R

C

Ci

Cu

SU

2

1Ru

0)0( Cu

Solution:

0

Cu Ru i

0

t)(tf

0 0 0

0 SU

SU

R

U S

0 0

First-order RC Circuits•Example: to find the transient response after changing circuit when t=0.

)0( tS

R

C

Ci

Cu

SU

2

1Ru

SCR Uuu

dt

duCiRiu C

R

SCC Uudt

duRC

0)0( Cu 0)0( Cu 0)0()0( CC uu

First-order RC Circuits

)0( tS

R

C

Ci

Cu

SU

2

1Ru

"'CCC uuu

stC Aeu '

"Cu

——homogeneous solution

——particular solution

SCC Uudt

duRC

First-order RC Circuits

homogeneous solution

)0( tS

R

C

Ci

Cu

SU

2

1Ru

SCC Uudt

duRC

01 RCs

RCs

1

tRC

C Aeu1

First-order RC Circuits

)0( tS

R

C

Ci

Cu

SU

2

1Ru

SCC Uudt

duRC

Therefore

SCC Uuu )("

Then, the final solution is

Sst

CCC UAeuuu "'

Particular solution

First-order RC Circuits

The solution of differential equation

Sst

CCC UAeuuu "'

Substituting the initial condition：

0)0( 0"' Ss

CCC UAeuuu

SCC UuuA )()0(

tRC

SS

tRC

CCCC

eUU

euuutu1

1

)]()0([)()(

First-order RC Circuits

The solution of differential equation

RC ——Time constant

t

CCCC euuutu

)]()0([)()(

)(Cu

)0( Cu

——Steady state value

——Initial value

First-order RC Circuits

Solution of other parameters

( ) ( ) (0 )

( ) [ (0 ) ( )]

t t

R S C S R

t

R R R

u t U u t U e u e

u u u e

( )

( ) (0 )

( ) [ (0 ) ( )]

t tSR

t

Uu ti t e i e

R R

i i i e

Three elements method

Three elements: 1.steady state value f(∞); 2.time constant τ; 3. initial value f(0+).

Formula of Three element method ：

t

effftf

)]()0([)()(

f(∞)——steady state value

τ——time constant

f(0+)——initial value

τ=RC ——time constant of RC circuitτ= ？？ —— time constant of RL circuit

4.3 First-order RL Circuits

4.3 First-order RL Circuits

Time constant

τ=RC

τ=L/R

Uudt

duRC C

C

R

Uti

dt

tdi

R

LL

L )()(

RU LuL

iL

+

-

RU CuC

iC

+

-

4.3 First-order RL Circuits

• Time constant reflects the length of transient period.

t 2 3 4 5 6 7

e-t/ 36.8% 13.5% 5% 1.8% 0.3% 0.25% 0.09%

•After about five time constants, the transient response is over.

•After one time constants, the transient response is equal to 36.8 percent of its initial value.

The curves versus time

SU

)(tuC

)(tuR

R

U S

0t

)()( titu

)(ti

SU632.0

)0(368.0 RU

2

The initial slop intersects the final value at one time constant.

Mounting curve

Decaying curve

• Time constant reflects the length of transient period.

•Three element method

Initial value: t=0-→t=0+ f(0+) Steady state value: t =∞ f(∞) Time constant : τ=RC τ=L/R Substituting three elements

Draw the curve versus time

t

effftf

)]()0([)()(

Steps

Limited Condition:

1) first-order circuit2) DC source

•Example 4.2 Find voltage of v(t) and current i(t) in this circuit for t>0.

Answer:( ) 2 2 (A), ( ) 100 (V)

0.12(ms)

50

t t

i t e v t e

L

R

( ) 2 2 (A), ( ) 100 (V)

0.12(ms)

50

t t

i t e v t e

L

R

•Example 4.3 Find voltage of v(t) and current i(t) in this circuit for t>0.

Answer: 1 1

2

( ) , ( )t t

s SV LVi t e v t e

R R

L

R

1 1

2

( ) , ( )t t

s SV LVi t e v t e

R R

L

R

•Exercise 4.5 Find voltage of v(t) and current iR(t) , iL(t) in this circuit for t>0, assume that iL(0)=0.

Answer:

)(2.0

)(20)(),(22)(),(2)(

s

VetvAetiAetitt

L

t

R

•Exercise 4.5 Find voltage of v(t) and current i(t), v(t) in this circuit for t>0, assume that the switch has been closed for a very long time prior to t=0.

Answer:

1, 0 1, 0( ) ( ) 5( )

0.5 0.5 , 0 100 , 0t t

t ti t v t ms

e t e t

P4.8 P4.18 P4.26 P4.30

•Homework 4