Lecture 12 First-order Circuits (2) Hung-yi Lee. Outline Non-constant Sources for First-Order...

Lecture 12First-order Circuits

(2)Hung-yi Lee

Outline

• Non-constant Sources for First-Order Circuits (Chapter 5.3, 9.1)

sci

titv scoc ,

Outline

• Examples 5.12 and 5.11• Solved by Differential Equation• Solved by Superposition and State

Example 5.12

• RL circuit• R=4Ω, L=0.1H• Find i(t), t>0• i(t)=0, if t<0

ttv 280sin400 00 i

tvtRidt

tdiL tvti

dt

tdi 41.0

(t>0)

Example 5.12 – Differential Equation

tvtidt

tdi 41.0 00 i ttv 280sin400

iN(t): General solution (Natural Response)

tititi FN

iF(t): Special solution (Forced Response)Natural

Response:

tN Aeti

041.0 tt AeAe

40

tAe 40 A determined by initial condition

0A040 tAe

041.0 NN tidt

tdi

0ti NOT 0N ti


ttidt

tdi280sin40041.0 F

F

Table 5.3 (P222)

v(t)= iF(t)=0k 0K

tk1atek2

tktk sincos 43

01 KtK ateK2

tKtK sincos 43


tititi FN

iF(t): Special solution (Forced Response)Forced Response:

tvtidt

tdi 41.0 00 i ttv 280sin400



tititi FN

iF(t): Special solution (Forced Response)

ttti 280sinK280cosK 21F

t

tt

t

280sin400

280sinK280cosK4

cos280t280K280sinK2801.0

21

21

400428 21 KK

0428 12 KK

2,14K 21 K

tvtidt

tdi 41.0 00 i ttv 280sin400

ttidt

tdi280sin40041.0 F

F

Forced Response:



tititi FN


tN Aeti 40 ttti 280sin2280cos-14F

tteti t 280sin2280cos14-A 40

014-A 14A

tvtidt

tdi 41.0 00 i ttv 280sin400

0ti

xV14-A xA V14 xti V


tvtidt

tdi 41.0 00 i tetv 4010


tititi FN

iF(t): Special solution (Forced Response)Natural

Response:

tN Aeti 40 Independent to the sources

This circuit always has this term.

Change v(t)

041.0 NN tidt

tdi

tN Aeti 40

tetidt

tdi 40F

F 1041.0


?402

tF eKti

tvtidt

tdi 41.0 00 i tetv 4010


tititi FN

iF(t): Special solution (Forced Response)Forced Response:

If the form for iF(t) contains any term proportional to a component of the natural response, then that term must be multiplied by t. P224 - 225

tF teKti 40

2

tN Aeti 40


tvtidt

tdi 41.0 00 i tetv 4010


tititi FN


tF teKti 40

2

tttt eteKteKeK 40402

402

402 104401.0

101.0 2 K 1002 K

tF teti 40100

tetidt

tdi 40F

F 1041.0 Forced Response:

tN Aeti 40


tvtidt

tdi 41.0 00 i tetv 4010


tititi FN


tN Aeti 40

tF teti 40100

tt teeti 4040 100A

0A

tN Aeti 40

00 i

Differential Equation - Summary• List differential equation and find initial condition (from the

property of inductors and capacitors)• 1. Find general solution (natural response)• Exponential form: Ae-λt

• Find λ• 2. Find special solution (forced response)• Form: Consult Table 5.3 (P222) • If a term in special solution is proportional to

general solution, multiplying the term by t • Find the unknown constant

• 3. Add the general and special solution together, and then find A in the general solution by initial condition

Example 5.12 – Superposition + State• RL circuits• R=4Ω, L=0.1H• i(t)=0, if t<0

00 i

Consider the circuit from t=0

ttv 280sin400

4i

tti 280sin100

State is zero

No state term

Only input term

tititi inputstate

Find istate(t)

Example 5.12 – Superposition + State• Review: pulse response

tD

ee

1A

If D is small

Det

A

x1ex (If x is small)

A

R ti

ti

tiL

tiLA

t0

100

100

Example 5.12 – Superposition + State

tti 280sin100

……

The sin wav is composed of infinite tiny pulse!Find the response of each tiny pulse and sum them together.


Response of the pulse between time point t0-Δt and t0

Δt is small

tetitt

t

0

0A

tti 280sin100

0ttt 0 0tt-t0

0280sin100A t

tetitt

t

0

0 0100sin280t

0t 1t2t


The response of sin wave is the summation of all the pulse responses.

0t

1t2t

te

titt

t

0

0

0100sin280t

0t 1t2t


a

We do not have to care the pulse after time point a.

Let’s focus on the response of sin wave at time point a

Current Source(Input)

Current on Inductor

(Response)The function is zero at point a

a

… …

… …

…


Current on Inductor

(Response)

a0t 1t2t

3t

… … …

… …

Value at a:

tetitt

t

0

0 0100sin280t

Deta

0

0100sin280t

Deta

1

1100sin280t

Deta

2

2100sin280t


Current on Inductor

(Response)

a0t 1t2t

3t

… … …

… …Value at

a:

tetitt

t

0

0 0100sin280t

tet

aiat

t

ta

0

280sin100

t

280sin100

0

detat

t

ta


at

t

ta dtetai0

4040280sin100

at

t

ta dtet0

40280sin4000

at

t

ta dtete0

4040 280sin4000

t280sin100

0

det

aiat

t

ta

RL circuit R=4Ω, L=0.1H

R

LRC

40

1


at

t

ta dteteai0

4040 280sin4000

btbbtaba

edtebt

atat cossinsin

22

(P806)

at

a tte

e 022

4040 |280cos280280sin40

280404000

28028040

14000

280cos280280sin4028040

4000

2240

22

ae

aa

aae a 280sin2280cos14-41 40

We can always replace “a” with “t”.


tteti t 280sin2280cos14-41 40

tiN tiF

Example 5.12 – Superposition + State• From Differential Equation• If we have initial condition i(0)=Vx

• From Superposition State• Superposition (no state)

• State

tteti tx 280sin2280cos14-V41 40

tteti tinput 280sin2280cos14-41 40

txstate eVti 40 tititi inputstate


4i

tetv 4010 teti 405.2

RL circuit R=4Ω, L=0.1H Find i(t)• i(t)=0, if t<0


teti 405.2

0

Current Source(Input)

Current on Inductor

(Response)

Response of the pulse between time point t0-Δt and t0 teti

ttt

0

0A

tett

0-40t2.5e

Its contribution at point a:

a a

teaita

t

0

0

-40a2.5e


dteeaiat

t

tat

0

4040 405.2 dteeat

t

tat

0

4040100

dteat

t

a

0

40 1100 ae a40100

For the response of the pulse between time point t0-Δt and t0

Its contribution at point a is: teaita

t

0

0

-40a2.5e

tee

aiat

t

taa

0

405.2

40

1

at

t

taa

dtee

0

405.2

Homework

• 5.56• 5.60• 5.64

Thank you!

Answer

• 5.56: iF(t) =-10te^(-20t) - 3e^(-20t)• 5.60: vF(t) = 2 – 50te^(-25t)• 5.64: i(t) = 0.05e^(25t) + 0.02 – 0.07e^(-25t)

Lecture 12 First-order Circuits (2) Hung-yi Lee. Outline Non-constant Sources for First-Order...

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Transcript of Lecture 12 First-order Circuits (2) Hung-yi Lee. Outline Non-constant Sources for First-Order...