1. Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu
Engineering 43
Chp 5.[1-2] Superposition
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer

2. Additional Analysis Techniques
Chp5 Outline
Review Linearity
The Property Has Two Equivalent Definitions
Examination And Application Of Homogeneity
Apply Superposition
Review Some Implications Of The Superposition Property as Applied to Linear Circuits

3. Additional Analysis Techniques
Outline (cont.)
Develop Thevenin’s and Norton’s Theorems
These are two very powerful analysis tools that allow the Circuit Analyst
Focus On Specific Parts Of A Circuit
Hide Unnecessary Complexities
Maximum Power Transfer
A Very Useful Application of Thevenin’s and Norton’s theorems

4. Comments on Previous Methods
The GENERAL Methods Of NODE And LOOP Analysis Provide Powerful Tools To Determine The Behavior Of EVERY Component In A Circuit
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

5. Comments on Prev. Methods cont
However the Chp2 Methods Have Limited Applicability
Keep Them In The Solution “Tool Box” And Use Them When They Are More Convenient
Now Develop Additional Techniques That Simplify Analysis Of Some Ckts
These Techniques Expand On Concepts Previously Introduced
Linearity
Circuit Equivalence

6. Previous Equivalent Circuits
Series & Parallel Resistors
[Independent Srcs]
Vsrc’s in Series
Isrc’s in Parallel
Can’t have
V sources in Parallel
I sources in series
The Complementary Configs are Inconsistent with Source Definitions

7. Linearity Models Used So Far Are All Linear
Mathematically This Implies That They satisfy the principle of SUPERPOSITION
The Model T(u) is Linear IF AND ONLY IF
For All Possible
Input Pairs: u1 & u2
Scalars α1 & α2
AN Alternative, And Equivalent, Linearity & Superposition Definition
The Model T(u) is Linear IF AND ONLY IF It Exhibits
ADDITIVITY
HOMOGENEITY

8. Linearity cont. Linearity Characteristics NOTE
Additivity
NOTE
Technically, Linearity Can Never Be Verified Empirically On A System
But It Could Be Disproved By A SINGLE Counter Example.
It Can Be Verified Mathematically For The Models Used
Homogeneity
a.k.a. Scaling

9. Linearity cont.
Using Node Analysis For Resistive Circuits Yields Models Of The Form
The Model Can Be Made More Detailed
Where
A and B are Matrices
s Is A Vector Of All Independent Sources
For Ckt Analysis Use The Linearity Assumption To Develop Special Analysis Methods
Where
v Is A Vector Containing All The Node Voltages
f Is a Vector Containing Only independent Sources

10. Past Techniques  Case Study
Find Vo
Redraw the Ckt to Reveal Special Cases
After Untangling     
Solution Techniques Available? 

11. Case Study cont.
Loop Analysis for Vo
Node Analysis - -

12. Case Study cont. Series-Parallel Resistor-Combinations In other Words
By VOLTAGE Divider

13. Use Homogeneity Analysis
Find Vo by Scaling
If Vo is Given Then V1 Can Be Found By The Inverse Voltage Divider
Now Use VS As a 2nd Inverse Divider
Assume That The Answer Is KNOWN
How to Find The Input In A Very Easy Way ?
Then Solve for Vo

14. Homogeneity Analysis cont
The Procedure Can Be Made Entirely Algorithmic
Give to Vo Any Arbitrary Value (e.g., V’o = 1V )
Compute The Resulting Source Value and Call It V’s
Use linearity
The given value of the source (Vs) corresponds to
Then The Desired Output

15. Homogeneity Comment This is a Nice Tool For Special Problems
Normally Useful When
There Is Only One Source
Best Judgment Indicates That Solving The Problem BACKWARDS Is Actually Easier Than the Forward Solution-Path

16. Illustration  Homogeneity
Solve Using Homogeneity (Scaling)
Assume
V’out = V2 = 1volt
Then By Ohm’s Law

17. Illustration  Homogeneity cont
Solve Using Homogeneity
Scaling Factor
Again by Ohm’s Law
Using Homogeneity
Scale from Initial Assumption:
Then

18. Homogeneity Example Scaling Factor: k = 6ma/2ma = 3
Find Io Using Homogeneity
Given I = 6mA
For Convenience Assume Io = 1mA
Using Homogeneity
Scaling Factor: k = 6ma/2ma = 3
Io = 3mA by 6/2 scaling

19. Source Superposition
This Technique Is A Direct Application Of Linearity
Normally Useful When The Circuit Has Only A Few Sources

20. Illustration  Src Superposition
Consider a Circuit With Two Independent Sources: VS, IS
Now by Linearity
Calculated By Setting The CURRENT Source To ZERO (OPEN ckt) And Solving The Circuit
Calculated By Setting The VOLTAGE Source To ZERO (SHORT ckt) And Solving The Circuit

21. Illustration cont + By Linearity =
Circuit With Current Source Set To Zero
OPEN Ckt
Circuit with Voltage Source set to Zero
SHORT Ckt
By Linearity

22. Illustration cont + The Above Eqns Illustrate SUPERPOSITION =
This approach will be useful if solving the two, 1-Src circuits is simpler, or more convenient, than solving a circuit with two sources
We can have any combination of sources. And we can partition any way we find convenient
The Above Eqns Illustrate SUPERPOSITION

23. Example  Solve for i1 = Alternative for i1(t) By SuperPosition: +
Loop equations
Contribution of v1
Alternative for i1(t) By SuperPosition:
Contribution of v2
Once we know the “partial circuits” we need to be able to solve them in an efficient manner

24. Student Exercise Lets Turn on the Lights for 5-7 min
Students are invited to Analyze the following Ckt
Determine the Output Voltage Vo
List on board the solutions

25. Numerical Example Find Vo By SuperPosition Set to Zero The V-Src
i.e., SHORT it
Current division
Ohm’s law

26. Numerical Example cont.
Find Vo By SuperPosition
Set to Zero The I-Src
i.e., OPEN it
By V-Divider
Yields Voltage Divider
Finally, Add by SuperPosition

27. SuperPosition vs. OpAmp Ckts
Consider the Offset Inverting OpAmp Ckt
Spot TWO V-Srcs
Analyze One-by-One
Chunk-1
Chunk-2
Chunk-2 Redrawn for Clarity

28. SuperPosition vs. OpAmp Ckts
Chunk-1 → Basic Inverting Ckt
Add Chunks by Superposition
Chunk-2 → Basic NonInverting Ckt

29. WhiteBoard Work
Let’s Work Problem 5.21

30. Example  SuperPosition
Find Vo Using Source SuperPosition
Set to Zero The I-Src
i.e., OPEN it
Set to Zero The V-Src
i.e., SHORT it

31. Example cont Define V1 on the V-Src ckt
If V1 is known then V’o is obtained using the 6&2 Voltage-Divider
V1 can be obtained by series parallel reduction and divider

32. Numerical Example cont.2
Determine
Current I2 By Current Divider
V”o Using Ohm’s Law
When in Doubt REDRAW
Finally The SuperPosition Addition
The Current Division

33. Sample Problem Determine Io by Source SuperPosition
First Consider Only the Voltage Source
Yields
Second Consider Only the 3 mA I-Source
Yields Current Divider
Then

34. Sample Prob cont Determine Io by Source SuperPosition
By IO2 Current Divider
The Current will Return on the Path of LEAST Resistance; Thus
Third Consider 4mA Src
So by Source Superposition

35. Illustration Use Source Superposition to Determine Io
Open the Current Source
Next Short the V-Source
By Current Divider

36. Illustration cont Looks Odd & Confusing → REDRAW Now Use I-Divider2 2 2 2 2 1 3 1 3 2
Now Use I-Divider
Finally By Linearity

37. SuperPosition Of Plane-Polarized Light
All Done for Today
SuperPosition Of Plane-Polarized Light
Cyan = Red + Green
Red & Green Light-Waves are Polarized in Perpendicular Planes