ADDITIONAL ANALYSIS TECHNIQUES

LEARNING GOALS

REVIEW LINEARITYThe property has two equivalent definitions. We show and application of homogeneity

APPLY SUPERPOSITIONWe discuss some implications of the superposition property inlinear circuits

DEVELOP THEVENIN’S AND NORTON’S THEOREMSThese are two very powerful analysis tools that allow us tofocus on parts of a circuit and hide away unnecessary complexities

MAXIMUM POWER TRANSFERThis is a very useful application of Thevenin’s and Norton’s theorems

The techniques developed in chapter 2; i.e., combination series/parallel, voltage divider and current divider are special techniques that are more efficient than the general methods, but have a limited applicability. It is to our advantage to keep them in our repertoire and use them when they are more efficient.In this section we develop additional techniques that simplify

the analysis of some circuits. In fact these techniques expand on concepts that we have already introduced: linearity and circuit equivalence

THE METHODS OF NODE AND LOOP ANALYSIS PROVIDE POWERFUL TOOLS TODETERMINE THE BEHAVIOR OF EVERY COMPONENT IN A CIRCUIT

SOME EQUIVALENT CIRCUITSALREADY USED

LINEARITYTHE MODELS USED ARE ALL LINEAR.MATHEMATICALLY THIS IMPLIES THAT THEYSATISFY THE PRINCIPLE OF SUPERPOSITION

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22112211

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,

)(

scalars possible all and

pairsinput possible allfor

IFF LINEAR ISMODEL THE

uu

TuTuuuT

Tuy

AN ALTERNATIVE, AND EQUIVALENT, DEFINITION OF LINEARITY SPLITS THESUPERPOSITION PRINCIPLE IN TWO.

yhomogeneit 2.

additivity

IFF LINEAR IS MODEL THE

uTuuT

uuTuTuuuT

Tuy

,,)(

,,)(.1 212121

NOTICE THAT, TECHNICALLY, LINEARITY CANNEVER BE VERIFIED EMPIRICALLY ON A SYSTEM.BUT IT COULD BE DISPROVED BY A SINGLECOUNTER EXAMPLE.IT CAN BE VERIFIED MATHEMATICALLY FOR THEMODELS USED.

USING NODE ANALYSIS FOR RESISTIVECIRCUITS ONE OBTAINS MODELS OF THEFORM fAv

v IS A VECTOR CONTAINING ALL THE NODEVOLTAGES AND f IS A VECTOR DEPENDINGONLY ON THE INDEPENDENT SOURCES. IN FACT THE MODEL CAN BE MADE MOREDETAILED AS FOLLOWS

BsAv

HERE, A,B, ARE MATRICES AND s IS A VECTOROF ALL INDEPENDENT SOURCES

FOR CIRCUIT ANALYSIS WE CAN USE THELINEARITY ASSUMPTION TO DEVELOPSPECIAL ANALYSIS TECHNIQUES

FIRST WE REVIEW THE TECHNIQUESCURRENTLY AVAILABLE

A CASE STUDY TO REVIEW PAST TECHNIQUES

Redrawing the circuit may help usin recognizing special casesO VDETERMINE

SOLUTION TECHNIQUES AVAILABLE??

USING HOMOGENEITY

Assume that the answer is known. Can we Compute the input in a very easy way ?!!

1V

If Vo is given then V1 can be computed using an inverse voltage divider.

02

211 V

R

RRV

… And Vs using a second voltage divider

02

2141

4 VR

RR

R

RRV

R

RRV

EQ

EQ

EQ

EQS

Solve now for the variable Vo

The procedure can be made entirely algorithmic

1. Give to Vo any arbitrary value (e.g., V’o =1 )

2. Compute the resulting source value and call it V’_s

3. Use linearity. kkVkVVV SS ,'0''

0'

4. The given value of the source (V_s) corresponds to

'S

S

V

Vk

Hence the desired output value is

'0'

'00 VV

VkVV

S

S

This is a nice little toolfor special problems.Normally when there isonly one source and when in our judgementsolving the problembackwards is actuallyeasier

EQR

SOLVE USING HOMOGENEITY

][12 VVVout ASSUME

1I

OV

NOW USE HOMOGENEITY

][2][12

][1][6

VVVV

VVVV

outO

outO

mA1OI ASSUME

][31 VV

][5.0 mA

][5.1 mASV

][62][5.1 1 VVkmAVS

][5.0 mAmA2

____6

12

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O

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YHOMOGENEIT USE

mAIIO 6 USE Y.HOMOGENEIT USING COMPUTE

LEARNING EXTENSION

Source Superposition

This technique is a direct application of linearity.

It is normally useful when the circuit has onlya few sources.

c i r c u i t

+ -

VS

IS

+

VL

_

IL

Due to Linearity

V aV a IL S S 1 2

V L1 Can be computed by setting the current

source to zero and solving the circuit

V L2 Can be computed by setting the voltage

source to zero and solving the circuit

FOR CLARITY WE SHOW A CIRCUIT WITH ONLY TWO SOURCES

SVBY ONCONTRIBUTI1LV

SIBY ONCONTRIBUTI2LV

Circuit with currentsource set to zero(OPEN)

1LI

1LV

Circuit with voltage sourceset to zero (SHORT CIRCUITED)

2LI

2LV

SOURCE SUPERPOSITION

= +

The approach will be useful if solving the two circuits is simpler, or more convenient, than solving a circuit with two sources

Due to the linearity of the models we must have2121LLLLLL VVVIII Principle of Source Superposition

We can have any combination of sources. And we can partition any way we find convenient

LEARNING EXAMPLE 1i

=

][6||33 kReq

eq

eq

R

vi

kR

2"2

][)3||3(6

+

Loop equations

Contribution of v1

Contribution of v2

WE WISH TO COMPUTE THE CURRENT

Once we know the “partial circuits”

we need to be able to solve them in

an efficient manner

LEARNING EXAMPLE

We set to zero the voltage source

Current division

Ohm’s law

Now we set to zero the current sourceVoltage Divider

+-

V 0"6 k

3k

3V ][6"0

'00 VVVV

][2 V

ionsuperposit source using Compute 0V

LEARNING EXAMPLE

Set to zero voltage source

We must be able to solve each circuit in a very efficient manner!!!

Set to zero current source

1V

If V1 is known then V’o is obtained using a voltage divider

V1 can be obtained by series parallel reduction and divider

2I

The current I2 can be obtained using a current dividerand V”o using Ohm’s law

2k||4k

2k

6k

I2+

V "0

_

2m A

+-

2k

4k||8k+

V1_

)6(3/82

3/81 V

+

V 1_

6k

2k

+

V '0

_

][7

18

26

61

' VVkk

kVO

WHEN IN DOUBT… REDRAW!

mAkkkk

kkkI )2(

)4||2(62

)4||2(22

"'

2" 6

OOO

O

VVV

kIV

ionsuperposit source using Compute 0V

Sample Problem

1. Consider only the voltage source

mAI 5.101

3. Consider only the 4mA source

2. Consider only the 3mA source

Current divider

mAI 5.102

003 I

mAIIII 30302010 Using source superposition

IONSUPERPOSIT SOURCE USING COMPUTE 0I

1I

1

1

2

2

3

3

211 I

IO

Linearity