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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

WhiteBoard Work Let’s Work Problem 5.21

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

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

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

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

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

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

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

SuperPosition Of Plane-Polarized Light All Done for Today SuperPosition Of Plane-Polarized Light http://www.enzim.hu/~szia/cddemo/edemo4.htm Cyan = Red + Green Red & Green Light-Waves are Polarized in Perpendicular Planes