Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-43_Lec-07-2_Step-by-Step_Response.ppt1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

Chp 7Chp 7Step-by-StepStep-by-Step

Pulse ResponsePulse Response



11stst Order Ckts: Step-by-Step Order Ckts: Step-by-Step

This Approach Relies On The Known Form Of The Solution But Finds The ODE Parameters Using Basic Circuit Analysis Tools

This Method Eliminates the Need For The Determination Of The Differential Equation Model

Most Useful When Variable of Interest is NOT vC or iL



Basic ConceptBasic Concept Recall The form of the ODE Solution for a Ckt

w/ One E-Storage Element and a Constant Driving Ckt

0,21 teKKtx

t

Where

• K1 The final Condition for the Variable of Interest

– Can Be determined by Analyzing the Ciruit in Steady State; i.e., t→

• x(0+) The Initial Condition for the Variable– Provides the Second Eqn for Calculating K2

Ckt Time Constant– Determine By Finding RTH Across the Storage Element



The General ApproachThe General Approach

With This Analysis Find• Time Constant using RTH

• Final Condition using vTH

Circuitw ith

resistancesand

sources

I nductororCapacitor

a

b

R epresentation of an arbitrarycircuit w ith one storage elem ent

Thevenin VTH

R TH

I nductororCapacitor

a

b

THCC

TH vvdt

dvCR

NTH

THL

L

TH

IR

vi

dt

di

R

L

FC

Obtain The Voltage Across The Capacitor or The Current Through The Inductor



The Steps: 1-4The Steps: 1-4 STEP 1. Assume

The Form Of The Solution

• STEP 3: Draw The Circuit At t = 0+0+

• The CAPACITORCAPACITOR Acts As a VOLTAGE SOURCEVOLTAGE SOURCE

• The INDUCTORINDUCTOR Acts As a CURRENT SOURCECURRENT SOURCE

• Determine The VARIABLE of INTEREST At t=0+

Determine x()• STEP 4: Draw The Circuit a

Loooong Time After Switching to Determine The Variable In Steady State

Determine x(0+)• STEP 2: Draw The Circuit

In Steady State just PRIOR To Switching And Determine Capacitor-Voltage Or Inductor-Current

0;

0,

211

21

xKKxK

teKKtxt



The Steps: 5-6The Steps: 5-6 STEP 5: determine

the time constant With These 3-Parameters

Write the Solution For the Variable of interest using The Assumed Solution

STEP 6: Determine The Constants K1 & K2,

inductor one withcircuit

capacitor one withcircuit

TH

TH

R

L

CR

)()0(

)0(

)(

2

21

1

xxK

xKK

xK

texxxtx )()0()()(

Step-By-Step DOWNside• Do NOT have ODE So Can

NOT easily Check Solution– Can usually chk the

FINAL Condition

• RTH Determined at Cap/Ind Connection Terminals



Step-By-Step: Inductor ExampleStep-By-Step: Inductor Example

Note That vO is NOT Directly Related to The Storage Element• → Use Step-by-Step

STEP-1: The Form of the Soln

STEP-2: Initial inductor current (L is Short to DC)

2

2 4

V6

0t

H2

Ov

t

O eKKtv

21)(

2

2 4

V6

0t

)0( Li

AiL 3)0(

For the Circuit Below Find vO for t>0



Inductor Example cont.Inductor Example cont. STEP 3: Determine

output at 0+• By Inductor Physics

At t=0+, Replace The L with a 3A Current Src

Note That at t=0+• The 6V Source is

DISCONNECTED from the Ckt Elements– No Connection on

Supply Side

• Single Loop Ckt

Aii LL 3)0()0(

2

2 4

V6

0t

_

)0(

Ov

A3

)0( Oi

Vv

kAv

Ai

kiv

O

O

O

OO

6)0(

23)0(

3)0(But

2)0()0(



Inductor Example cont.2Inductor Example cont.2 STEP 4: Find Output In

Steady State After The Switching• By Inductor Physics In

Steady State– L is SHORT to DC

Recall at t=0+The 6V Source is DISconnected from the Ckt Elements• The Ckt Has

NO Power Source

• Over A long Time All the Energy Stored by The Inductor Will be Dissipated as HEAT by The Resistors, Hence

0)()( OO iv2

2 4

V6

0t

_

)(

Ov



Inductor Example cont.3Inductor Example cont.3 STEP 5: Find Time

Constant After Switch• Find RTH With Respect to

the L Terminals

Then The Time Constant,

2

2 4

V6

0t

THR

RTH by Series Calc8THR

SS

AV

ASVH

R

L

TH

25.08

2

8

2

8

2



Inductor Example cont.4Inductor Example cont.4 STEP 6: Find The

Solution

Then The Solution

Alternatively use x = v in:

2

2 4

H2

V6

0t

_

)(tvO

VvKK O 6)0(21

0)(1 OvK

0;6)( 25.0

teVtv s

t

O

0;6)(4

teVtvt

s

rads

O

texxxtx )()0()()(

tHz

st

Vetv

eVtv4

4/1

6)(

060)(



WhiteBoard WorkWhiteBoard Work

1 k 9V t = 0 6 k

50 F 2 k 2 k V0 ( t)

+

-

12 k

Let’s Work This 1st Order Cap Problem• Power Source DISengaged



Pulse ResponsePulse Response Consider The Response Of Circuits To A

Special Class Of SINGULARITY functions

01

00)(

t

ttu VOLTAGE STEP

CURRENT STEP TIME SHIFTED STEP



Pulse ConstructionPulse Construction Pulse = Sum of Steps Examples

)]01.0()([10)( tutumAti

)]2()1([10)( tutuVtv

01

00)(

t

ttu



PieceWise Transient RepsonsePieceWise Transient Repsonse Non-Zero Initial

Condition (std ODE)

The Response is Shifted From the Time Origin by an Amount t0

For CONSTANT fTH, The Time-Shifted Exponential Solution

This expression will hold on ANY interval where the sources are CONSTANT

The values of the constants may be different and must be evaluated for each interval

The values at the END of one interval will serve as INITIAL conditions for the NEXT interval

00 )(; xtxfxdt

dxTH

021 ;)(0

tteKKtxtt



PieceWise ExamplePieceWise Example

+-

k10

F20

Ov

a

b

V12

)(st5.0

12

)(tvs

)(tvS

Piecewise constant sourceOtt

O eKKtv

21)(

The Switch is Initially At a. At Time t=0 It Moves To b, and At t=0.5 it moves back to a.

Find vO(t) for t>0

On Each Interval Where The Source is Constant The Response Will Be of the Form



PieceWise Example contPieceWise Example cont For 0t<0.5 (Switch at b)

• t0 = 0

• Assume Solution

Find Parameters And Piece-1 Solution

Now Piece-2 (Switch at a)• t0=0.5S

+-

k10

F20

Ov

a

b

V12t

O eKKtv

'2

'1)(

'2

'1][12)0( KKVv

'10)( KvO

5.00,12)( 2.0

tVetv s

t

O

sFk 2.0)20)(10(

VVevv s

s

OO 985.012)5.0()5.0( 2.0

5.0

'

)5.0("2

"1)(

t

O eKKtv

"985.0)5.0( 2"1 KKvO

"112)( KvO

015.1112985.0"2 K

steVVtv s

st

O 5.0,015.1112)( 2.0

5.0



First Order Pulse-Response Transient Solution

0.0

2.0

4.0

6.0

8.0

10.0

12.0

-0.1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3

Time (s)

vO (

V)

file = Engr44_Lec_06-1_Last_example_Fall03..xls

s

t

O eVtv 2.012)(

s

st

O eVVtv 2.0

5.0

015.1112)(

PieceWise EndPoints MUST Match



WhiteBoard WorkWhiteBoard Work

t=0

R 16V

R3R 2

+VC ( t)-

R 4

i0 (t )

C

R1= R2=R3= R4= 2 k C= 0.2 m F

F or t= 0 -

Let’s Work This 1st Order Cap Problem• R1→4 = 2 kΩ

• Power Source ENgaged– IF we Have Time

Documents

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege