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Solving the Second Order Systems

Continuing with the simple RLC circuit(4) Make the assumption that solutions are of the exponential form:

( ) ( )stexpAti =

where A and s are constants of integration.Then substituting into the differential equation

01

2

2

=++ iCdt

diR

dt

idL

( ) ( )( ) 0stexpA

C

1

dt

stexpdAR

dt

stexpAdL

2

2

=++

( ) ( ) ( ) 0stexpC

AstexpRsAstexpALs2 =++

Dividing out the exponential for the characteristic equation

0C

1RsLs2 =++

Also called the Homogeneous equation

Thus quadratic equation and has generally two solutions.

There are 3 types of solutions

Each type produces very different circuit behaviour

Note that some solutions involve complex numbers.

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General solution of the Second Order Systems

Consider the characteristic equations as a quadraticRecall that for a quadratic equation:

0cbxax2 =++

The solution has two roots

a2

ac4bbx

2 =

Thus for the characteristic equation

0C

1RsLs2 =++

or rewriting this

0LC

1s

L

Rs2 =++

Thus

LC1c

LRb1a ===

The general solution is:

LCL

R

L

Rs

1

22

2

=

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General solution Second Order Systems Cont'd

Second order equations have two solutionsUsually define

LCL

R

L

Rs

1

22

2

1

+=

LCL

R

L

Rs

1

22

2

2

=

The type of solutions depends on the value these solutions

The type of solution is set by the Descriminant

=

LCL

RD

1

2

2

RecallL

Ris the time constant of the resistor inductor circuit

Clearly the descriminate can be either positive, zero, or negative

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3 solutions of the Second Order Systems

What the Descriminant represents is about energy flows

= LCL

RD 12

2

How fast is energy transferred from the L to the C

How fast is energy lost to the resistor

There are three cases set by the descriminant

D > 0 : roots real and unequal

In electronics called the overdamped case

D = 0 : roots real and equal

In electronics the critically damped case

D < 0 : roots complex and unequal

In electronics: the underdamped case: very important

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Second Order Solutions

Second order equations are all about the energy flow

Consider the spring case

The spring and the mass have energy storageThe damping pot losses the energy

The critical factor is how fast is energy lost

In Overdamped the energy is lost very fastThe block just moves to the rest pointCritically Dampedthe loss rate is smaller

Just enough for one movement up and downFor Underdampedspring moves up and down

Energy is transferred from the mass to the spring and back again

Loss rate is smaller than the time for transfer

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Complex Numbers

Imaginary numbers necessary for second order solutions

Imaginary number j

1=j

Note: in math imaginary number is called iBut i means current in electronics so we use jComplex numbers involve Real and Imaginary parts

WW jIRW +=

r

May designate this in a vector coordinate form:( )

WW IRW ,=

r

Example:

( ) 2)Im(1Re

)2,1(21

==

=+=

WW

jWr

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Complex Numbers Plotted

In electronics plot on X-Y axis

X axis real, Y axis is imaginary

A vector represents the imaginary number has lengthVector has a magnitude M

Vector is at some angle to (theta) the real axisThen the real and imaginary parts are

( ) ( )cosRe MRWalW==

( ) ( )sinIm MIWaginaryW==

( ) ( )[ ] sincos jMW +=r

The magnitude

( ) ( )22WW

IRWWMag +== rr

The angle

=

W

W

R

Iarctan

Thus can give the vector in polar coordinates

( ) == MMW ,r

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Complex Numbers and Exponentials

Polar coordinates are connected to complex numbers in expConsider an exponential of a complex number

( ) ww jIRMW +== ,

r

This is given by the Euler relationship

( ) ( ) ( )] sincosexp jj +=

( ) ( ) ( )[ ] sincosexp jMjMW +==r

This relationship is very important for electronicsUsed in second order circuits all the time.

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Overdamped RLC

This is a very common caseIn RLC series circuits this is the large resistor

Energy loss in the resistor much greater than energy transfersTwo real roots to the characteristic equation

LCL

R

L

Rs

1

22

2

=

The solution is a double exponential decay

( ) ( ) ( )tsAtsAti 2211 expexp += As R is increased damping increasesGet the overdamped case

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Overdamped RLC Energy Flows

For the example case L = 5 mH, C = 2 F, R = 200 Also assume C is charged to 10 V at t = 0

Initially energy starts in the Capacitor CSome energy transfers to the Inductor LC looses charge much faster than L gains currentSo energy starts to rise in L but only to a limited levelThen energy is removed from both by the resistor

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Overdamped RLC Initial Conditions

For the overdamped case the ss are real & different

( ) ( ) ( )tsAtsAti2211

expexp +=

To solve the constants A need the initial conditionsFor second order need two conditionsThus both initial current & its derivativeThis varies from circuit to circuitIn the case of a charged C switched into the circuitSince L acts as an open initially then i(0) = 0, thus

( ) 210 AAi += Thus

12 AA =

Again since the inductor acts open at time zero & i(0)=0Thus voltage drop across the resistor is zero

( ) ( ) ( )dt

diLVV LC 000 ==

( ) ( )L

V

dt

dic

00=

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Overdamped RLC Full Solution

Now using substituting A equation into the exp equation

( ) ( ) ( )tsAtsAti2111

expexp =

To solve the constants A with the derivative initial condition

( )( ) ( )[ ]tsstssA

dt

tdi22111

expexp =

Now applying the initial condition derivative

( )

( ) ( )[ ] [ ]

( )L

V

ssAssAdt

tdic

0

0exp0exp

0211211

===

=

Now solving the equations

( )[ ]LssV

A C

21

1

0

=

( ) ( )

[ ]

( ) ( )[ ]tstsLss

Vti C

21

21

expexp0

=

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Overdamped RLC Circuit Example

For the example case L = 5 mH, C = 2 F, R = 200 Solving for the roots first what are the discriminate values

sec501

sec102005.02

200

2

14 ===

=

L

R

( )28

6sec10

102005.0

11 =

=

LC

( )[ ] 288242

sec103101021

2

==

=

LCL

RD

Thus gives

138

2

1sec1068.2103

005.02

2001

22

=+=

+=

LCL

R

L

Rs

148

2

2sec1073.3103

005.02

2001

22

=

=

=

LCL

R

L

Rs

Thus( )

[ ] [ ] A

Lss

VA C

2

44

21

11077.5

005.01087.51087.1

100 =+

==

( ) ( ) Ati 42 1073.3exp10368.2exp1077.5 =

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Critical Damped RLC

If we decrease the damping (resistance) energy loss decreasesChange the exponential decay until discriminant=0

01

2

2

=

=

LCL

RD

LCL

R 1

2

2

=

This is the point called critical damping

Difficult to achieve: only small change in R moves from this point

Small temperature change will cause that to occurEnergy transfer from C to L is now smaller than loss in R

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Critical Damped RLC Solutions

The characteristic equation has two identical solutions

0LC

1sL

Rs2

=++

L

Rss

221 ==

This is a special case, with special solution

( ) [ ]

+=

L

RttAAti

2

exp21

Why this solution?

Given in the study differential equations in math

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Critical Damped RLC Two Solutions

The critically damped equation has two solutiosn

( ) [ ]

+= L

RttAAti 2exp

21

Solution has two possible behaviours depending on A values

The characteristic equation has two identical solutions

Get only one oscillation (transaction)

Or get slow approach to final value

Difference depends on initial conditions only

If starts with i(t=0) = 0, slow approach to rest value

If start with i(t=0) 0, get one oscillation then rest valueThus same circuit will have different solutions

Depending on the initial current conditions & energy storage

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Critical Damped RLC Example i(0)=0

Keeping L & C the same

If R is increased 100 ohm get critical damping

Here L = 5 mH, C = 2 F, R = 100 Also assume C is charged to 10 V at t = 0Also assume C is charged to 10 V at t = 0 but i(t)=0This is the no oscillation case

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Critical Damped RLC, i(t=0) = 0 Energy Plot

Energy is transferred from Capacitor to Inductor

Inductor energy is later and higher than overdamped case

Then both energies decline

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Critical Damped RLC i(0)=0 Solution

We define the Damping Decay Constant

14

sec10005.02

100

2

=== L

R

The damping constant gives how fast energy is decayingThe basic Critical Damping equation is

( ) [ ]

+=

L

RttAAti

2exp21

Solving for A constants from the initial conditionsSince current is at t=0 is zero then

( ) [ ] 1212

0exp000 A

L

RAAti =

+===

( )

=

L

RttAti

2exp2

Again since the inductor acts open at time zero

( ) ( )L

V

dt

di c 00 =

Now applying this to the equation

( )222

2

01

2

0exp

21

2exp

0A

L

R

L

RA

L

Rt

L

RtA

dt

tdi=

=

=

=

Thus at time t=0 then ( )A

L

VA c 2000

005.0

1002 ===

( ) [ ] ( )At10expt2000L2

RtexptAti

42 =

=

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Critical Damped RLC, i(t=0) = 0 Voltage Plot

Capacitor voltage starts at 10 V and declines

Inductor voltage starts at 10 V then reverses & declines

Resistor voltage starts at zero, rises to peak above Vcthen declines

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Critical Damped RLC, i(t=0) = 0 Current Plot

Current rises to peak due to A2t term

Then current declines near exponential with

( ) [ ] ( )At10expt2000L2

RtexptAti 42 =

=

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Critical Damped RLC, i(t=0) 0

Now consider the case when i(t=0) is non zero

The practical case is when capacitor is uncharged

But inductor has current flowing in it

The equation has both constants nonzero

( ) [ ]

+=

L

RttAAti

2exp21

Again the example have L = 5 mH, C = 2 F, R = 100 Now assume L initially carries 100 mA at t = 0, thus

( ) mA100AI0ti 10 ====

Since C acts as a short at time t=0 thus VC(t=0) = 0Then the only voltage drop is across the resistance

( )0

00 =

=+

dt

tdiLRI

( )LRI

dttdi 00 ==

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Critical Damped RLC, i(t=0) 0 Equation

Now for the derivative of the equation

( ) [ ]

++= L

RtLRtAAA

dttdi

2exp

2212

For the initial conditions

( )[ ]

L

RIA

L

RAA

L

R

L

RAA

dt

tdi

222

0exp

2

0 021212 =

=

+=

=

Relating this to the resistance

( )L

RI

L

RIA

dt

tdi 002

2

0==

=

AL

RIA 1000

005.02

1.0100

2

02 =

==

Thus the critically damped i(t=0)0 current equation is

( ) [ ] [ ] ( )At10expt10001.0L2

RtexptAAti 4

21 =+=

Thus the current will reverse direction at t=0.1 msec.

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Critical Damped RLC, i(t=0) 0 Current Plot

Current reaches zero at t=0.1 msec

Reverses, reaches peak at t=0.2 msec then declines

( ) [ ] [ ] ( )At10expt10001.0L2

RtexptAAti

421 =

+=

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Critical Damped RLC, i(t=0) 0 Energy Plot

Inductor energy falls to minimum at t=0.1 msec

Energy is transferred from L to C and back to L

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Critical Damped RLC, i(t=0) 0 Energy Plot Expanded

Capacitor energy reaches max when ELis minimum

Then Inductor energy rises again

Both energies decay

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Critical Damped RLC, i(t=0) 0 Voltage Plot

Inductor voltage starts at 10 V and declines

Resistor voltage starts at -10 V and rises to zero

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Critical Damped RLC, i(t=0) 0 Voltage Plot Expanded

Resistor voltage reaches zero at t=0.1 msec

Capacitor voltage reaches max also at t=0.1 msec

Inductor voltage falls to zero at t=0.2 msec then goes negative

Reason VLdepends on direction of Current derivative

Resistor voltage changes to positive, reaches peak then declines