Lecture 11 First-order Circuits (1) Hung-yi Lee. Dynamic Circuits Capacitor, Inductor (Chapter 5) Frequency Domain Time Domain (Chapter 6,7) S-Domain

Lecture 11First-order Circuits

(1)Hung-yi Lee

Dynamic Circuits

Capacitor, Inductor

(Chapter 5)

Frequency Domain

Time Domain

(Chapter 6,7)

S-Domain

(Chapter 11,13)

(Chapter 9)

Abstract

Textbook

• First-Order Circuits• Chapter 5.3, 9.1

First-Order Circuits

• Containing only one capacitor or inductor

The networks excluding capacitor or inductor only contains sources and resistors.

Can always be simplified by Thevenin or Norton Theorem

First-Order Circuits

sci

sci

RC:

RL:

First Order Circuits

sci

…0t

t

…0t

t

…1t

t2t

…t

0t…

titv scoc , voc and isc should be dynamic

(this lecture)

Unit Step Function

Perspective

• Differential Equation• Superposition• State

Perspective 1:Differential Equation

Zero-Input Response - RC

0V ttuv o

…0t

t


Find vc(t) and ic(t)

0tt Capacitor is open circuit

oc Vtv

0tic


0tt Capacitor is open circuit

0tvc

0tic



0tt

0tt

0tt 0tt ?

?



0tt 00 )( VtvC ic(t0) is unknown

Voltage on the capacitor should be continuous

00 )( VtvC


0tt

dt

tdvCti C

C

)()(

0)()(

tvdt

tdvRC C

C

tC Aetv )(

0 tt AeAeRC

Assume

01RCRC

1

00 )( VtvC ic(t0) is unknown00 )( VtvC


tC Aetv )(

RC

1

tRCAe1

0

10

VAet

RC 0

1

0

tRCeVA

001

0

11

0)(tt

RCt

RCt

RCC eVeeVtv

0

0 11

C)(

)(tt

RCo

ttRC

oC

C eR

V

dt

deV

dt

tdvCti

00 )( VtvC 00 )( VtvC


01

0)(tt

RCC eVtv

01

)(tt

RCoC e

R

Vti

Zero-Input Response - RC 0

1

0)(tt

RCC eVtv

RC

)(tvC

oV

oV

oVoV

01)( tt

oC eV

dt

tdv

oC V

dt

tdv

)( 0

oV

…0t

t 0

1

0)(tt

C eVtv

Zero-Input Response - RL

0ttuIi o

…0t

t

i

0I)(L ti

Zero-Input Response - RL

0L

R

0L I)(tt

eti

0RI-

)(L tv

0L

R

0)(tt

C eRItv

Zero-Input Response

01

0)(tt

RCC eVtv

0L

R

0L I)(tt

eti

R

LRC

0t-t

0Y)(

ety

Voltage of C,Current of L

Voltage,Current How fast?

…0t

t

Step Response - RC

…0t

t

0V ttuv o

0tt

Step Response - RC

• Solved by differential equation 0Vtv ttuo

0tt

0tt

0tvc

oc tv V

0tt 00 tvc

occ VtvtRi occ Vtvdt

tdvR C

00 tvc

Step Response - RC occ Vtvdt

tdvR C

tvtvtv FNc vN(t) is general solution

vF(t) is the solution of occ Vtvdt

tdvR C

vN(t) is the solution of 0C tvdt

tdvR c

c RCN )(

t

Aetv

oVtv F

00 tvc

vF(t) is special solution


tdvR C

tvtvtv FNc RCN )(

t

Aetv

oVtv F

00 tvc

o

t

c VAetv

RC 0RC0

o

t

VAe

RC0t

oeVA

0RC

1

0 1Vtt

c etv

Step Response

0RC

1

0 1Vtt

c etv

…0t

t

0L

R

0 1Itt

L eti

R

LRC

0t-t

0 1Y)( ety

Voltage of C,Current of L

Voltage,Current

How fast?

Step Response

0t-t

0 1Y)( ety

0Y

0Y

…0t

t

Step Response…

0tt

0t-t

0 1Y)( ety

0Y

0Y

0Y

10% time 90% time

Rise time

Step Response + Initial Condition

0tt

xc Vtv 0

00 tvc


tdvR C

tvtvtv FNc RCN )(

t

Aetv

oVtv F

xc tv V0

o

t

c VAetv

RC

xo

t

VAe VRC0

RC0

Vt

ox eVA

0

RC

1

0 VVV0

tt

xc etv

Perspective 2:Superposition

Step Response

• Solved by Superposition

0Vv ttut o

0Vv ttut o

01

0)(tt

C eVtv

RC

?)( tvC

Step Response

0

2

V

v

ttu

t

o

ot Vv1

Suppress v1, find vc2(t)

Suppress v2, find vc1(t)

tvtvtv ccc 21

…0t

t

…0t

t

…0t

t…

-=

Step Response

ot Vv1

0V

v

ttu

t

o

01

0)(tt

C eVtv

01

02 )(tt

C eVtv

01 )( VtvC

)()()( 21 tvtvtv CCC

01

0 1Vtt

e

0

2

V

v

ttu

t

o

Pulse Response

tv

oV

Solved by Superposition

tvtvdt

tdvR c

c C

Pulse Response

=

0 -oV

)(1 tvC …D

01

00)(

1

01 teV

ttv t

C

D1

D0)(

02 teV

ttv

Dt

C

oV

tv

oV

0

…)(2 tvC

D

Pulse Response

01

00)(

1

01 teV

ttv t

C

D1

D0)(

02 teV

ttv

Dt

C

)()()( 21 tvtvtv CCC

)(tvC

D

e1V00V

tD

ee

1V0

Pulse Response

)(tvC

D

e1V00V

tD

ee

1V0

DIf

x1ex (If x is small)

D

0V

t

eD

0V


xc Vtv 0

0

RC

1

0 VVV0

tt

xc etv

Violate Superposition?


xc Vtv 0

xV

The initial condition is automatically fulfilled.

Do not have to consider the initial condition anymore


xV xV

Zero-Input Response! Step Response (without initial condition)!

0RC

1

1 Vtt

xc etv

0RC

1

02 1Vtt

c etv

00 RC

1

0RC

1

21 1VVtttt

xccc eetvtvtv


Differential Equation Superposition

0

RC

1

0 VVV0

tt

x

c

e

tv

00 RC

1

0RC

1

1VVtttt

x

c

ee

tv

General solution

Special solution

Zero-input Response Step Response

The initial condition is considered in the general solution term.

The initial condition is automatically fulfilled.

Perspective 3:State

State

• The capacitor voltages and inductor currents constitute the state variables of any circuit. (P410)

If the circuit does not have any capacitor or inductor

The currents or voltages at time t do not depend on their values not at t.

Why? tRitv )(

State


With capacitor or inductor

0

00

0

V

tt

tttv

You can not explain the current or voltage at present unless considering the past.

diC

tvt

t1

11

State


dt

tdvCti

)()(

diC

tvtvt

t0

10

Capacitor voltages are States

If we know the voltage before at t0

We do not care about the current before t0

diC

tvt

1)(

……

State


diC

tvtvt

t0

10

State at t0 Source after t0

tvtvtv inputstate

State


tvtvtv inputstatec )(

xc Vtv 0

00 RC

1

0RC

1

1VVtttt

xc eetv

The response after t0

From state at t0

(Ignore input)From Input after t0

(Ignore state)

Response

y(t) = general solution + special solution

= natural response + forced response

= state response (zero input) + input response (zero state)

= =

y(t): voltage of capacitor or current of inductor

Zero-Input Response

Considering the circuit from t0:

No input after t0

State: vc(t0)=V0

Ignore everything before t0


0tvinput

Lead to tvstate

…0t

t

Zero-Input Response

State: vc(t0)=V0

01

0)(tt

state eVtv


0tvinput

01

0)(tt

statec eVtvtv

Zero-Input Response 0V ttuv o

…0t

t

Considering the circuit from t0-D:

Input after t0-D

State: vc(t0-D)=V0


0tt

Dt 0

Zero-Input Response


Dtt

state eVtv

0

1

0)( State: vc(t0-D)=V0

Input after t0-D

0tt

Dt 0

)(tvinput

0tDt 0

DtteV

01

0 1 DttD

ee

0

1V0

Example 9.4

0t 10 t 1t

V12

k6

V12

k4

Example 9.4

0t

00 v

00 v

00 v

Example 9.4 00 v

10 t

V12

k6No state response

0

10

tt

eVtv

6.0112

t

e

73.91 v

00 vState response:

Input response: Vvoc 12

Example 9.4 00 v

10 t

V12

k6

73.91 v

State response is zero

Example 9.4

1t

V12

k4

0

10

tt

eVtv

0tt

xeVtv

State response:

Input response:

00 v 73.91 v

Example 9.4

4.0

1

112t

etv

4.0

1

73.9

t

etv4.0

1

73.2112

t

e

00 v 73.91 v

V12

k4State response:

Input response:

1t

Example 9.4

4.0

1

73.2112

t

e

6.0112t

e

k

tv

dt

tdvFti

24100

Application:Touchscreen

Resistive Touchscreen電阻式觸控螢幕

http://www.analog.com/library/analogdialogue/archives/44-02/touch_screen.html

Capacitive Touchscreen 電容式觸控螢幕

http://www.eettaiwan.com/ART_8800583600_480702_TA_bc13e6c4.HTM

Before Touching

Finger is Touching

Homework

• 9.14

• 9.16

Homework - Stability

• The first-order circuit shown below is at steady state before the switch closes at t=0. This circuit contains a dependent source and so may be unstable. Find the capacitor voltage, v(t), for t>0.

Homework - Stability

• The gain of the dependent source below is B. What restrictions must be placed on the gain to ensure that the circuit is stable? Design this circuit to have a time constant of +20ms.

Thank you!

Homework

• 9.14

• 9.16

D

t

etv 2L 10

Dt 0

Dt D

Dt

etv 2L 93.3

65.8DC v

17.12DC v

81.83DC v

Stability

• The first-order circuit shown below is at steady state before the switch closes at t=0. This circuit contains a dependent source and so may be unstable. Find the capacitor voltage, v(t), for t>0.

201224t

etv V1

m2.0

m4.0 m1.0

m1.0

Stability

• The gain of the dependent source below is B. What restrictions must be placed on the gain to ensure that the circuit is stable? Design this circuit to have a time constant of +20ms.

2/3B 1B

Acknowledgement

• 感謝莊佾霖 (b02)• 指出投影片中 Equation 的錯誤

Documents

Lecture 11 First-order Circuits (1) Hung-yi Lee. Dynamic Circuits Capacitor, Inductor (Chapter 5) Frequency Domain Time Domain (Chapter 6,7) S-Domain