| Transcriber: |
OOPS SJTU: 陆雯青, 章渊哲, 王芳华 |
| Brief Bio: |
Shanghai Jiao Tong University, China, luwq1984@gmail.com, mollyzyz@gmail.com, happygolucky.fang@gmail.com |
| Timecode: |
OOPS SJTU: 王芳华, 陆雯青 |
| Brief Bio: |
Shanghai Jiao Tong University, China, happygolucky.fang@gmail.com, luwq1984@gmail.com |
| Date finished: |
14, Nov, 2005 |
| Proofreader: |
OOPS SJTU: 郑晔鑫, 邹扬文 |
| Brief Bio: |
Shanghai Jiao Tong University, China, zhenyexin@citiz.net |
| Date finished: |
14, Nov, 2005 |
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Let's get started.
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Can you hear me about there?
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O.K. Let's get started.
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Before I begin,just a couple of announcements. Umm..
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B.B. is one of the students here,and he needs a note digger.
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It's a big positon.
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So if you are interested,you can stop by after class,
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and see him,he's sitting right here or there. O.K?
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Second,just a reminder that 6002 does have prerequisites.
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And prerequisites are 802 and 1803.
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O.K.
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So with that let me start of the usual..
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do a quick review of what we've done so far
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so we start out life,looking at the laws of physics,and maths equations and so on.
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and those may be too hard
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so we said let's make life easy for ourselves.
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So we choose to play in this playground in which we said we shall at here to the lump matter discipline
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O.K. the LMD,so we are in that playground,so this entire course
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and for that matter,large parts of EECS are within that playground
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O.K. Within which the lump matter discipline applies.
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So as soon as we made the jump into the playground,the LMD playground
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we could take maths equations and abstract them out into two very very simple rules. O.K.?
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The very simple rules are KVL and KCL.
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KVL simply said that I can sum the voltages in any loop in a circuit,
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and the result turns to be zero.
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Similarly,I can sum the currents that enter or exist
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any node and sum will also be zero.
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So what you can now do is if you feel like,you can go around and brag
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oh,I'll use maths equations in everyday life.
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yes,good staff.
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so a.. and the key is that this is really a calculation of maths equations
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within this playground that we were in
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so a.. I talk about the first method of circuit analysis in the last lecture,
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and that method simply took KVL for all the loops or KCL for all the nodes
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and wrote element vi relationships and together give you a big bunch of equations
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and you sat down and grouch through all the equations and you solve for branch voltages and currents.
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so a..We review the second method of circuit analysis and simply called circuit composition
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the basic idea behind this method was to learn some simple rules
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or how the resisters add and the conductors add and so on and so forth
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Look at the circuit and symplify the circuit by making serial simplifications
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when the resistance are unserial and so on and so forth
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and compose it till.. and ------on with it till the end up with the current voltages that we are looking for.
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This is intuite of the method,and so a section in charper 2,I believe,of the course notes
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discusses several examples using this method and attemps to make a little bit formal
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the intuite of approach that is applied in this method.
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O.K. So you then look at in node method.
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And the node method was simply a particular way of applying KVL and KCL.
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Node method,remember? We took a ground node,
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then we lable the nodes of the remaining voltages with respect to that ground
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then we wrote KCL for the each of the nodes
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O.K. We wrote KCL for the each of the nodes,remember..
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KVL was implicit in this expression that we used for the each of the node,
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each of the current that exiting each node
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ej was the node voltage,
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then ej minus ei if multiplied by the conductance Gi was the current that was going through
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one of those..I shall call it Gij,this is a conductance that connects nodes i and j
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O.K.that gives us the KVL that all of them fell into the same .
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uh,fell into the same system
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So these are three methods.
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Today,the node method by the way,is sort of the the world course of the 6002 industry
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and for that matter,for all of the circuits industry
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given an undoubt,apply the node method will be ok
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applys to linear circuits,nonlinear circuits,what have you.
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what i am going to do today,is go through two more methods,ok
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so notice that the first few lectures of the course,the first three lectures,
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simply comprise transiting you from the world of physics to the world of EECS
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and then,two lectures on giving you a bag of tracks
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we start you off this sort of tools-------
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and these five methods are your tools
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look at two methods today
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one method is called the method of superposition
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and the second method is called the Thevenin method
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and these method apply only to linear circuits
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ok,so look at the subset of circuits of linear,these two methods apply to only those circuits
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these are methods combined with an intuition,really enable you to solve very interesting circuits,very very quickly
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so let me do an example,and using a usuall node method
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and then,jumping to introducing the superposition method and Thevenin method using that same example
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so let me draw you an example circuit here
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so.. again and I'm using this example to..
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i will use this example to introduce the method of superposition and Thevenin method
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so what i am going to do is ...start off the usuall way
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and analyse the circuit using a method that you know now
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the node method
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and what i will do,is write down the node equations for this by applying the node method
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so..for the called node method,i choose a ground node
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i would choose this node
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it's go up with the voltage source connected to it
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and it also got many other edges,-----
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so i would choose as my ground node
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and i will label the other nodes with their voltages
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so this is an unknown,i label it as e
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i guess we just one unknown ...e
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and i know the voltage of this node
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and that is simply v
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since it's v above the...
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there's a voltage source between the ground node and that node
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ok,so what i can do next is that i can write down the node equations...for this node and then go from there
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so let me go ahead doing that
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so let me sum out the currunts going outside..going out
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so i have (see the picture)
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ok this is my node equation
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The first thing I want you to observe,
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ok,is that this equition is linear in V and I,
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what i mean by linear
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is that you don't see terms like vi,
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a product of vi,
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or v squared and things like that
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ok,it's ... some constants times v,
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plus some constants times i,
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equals some other constant.
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so,that's quite nice,
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so we can rearrange the terms in the following manner
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so move the known sources to the right-hand side
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and collect the coefficients of e on this side
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so i get one by R1 plus one by R2
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so stare at this for a moment
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and notice again here
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i have e,my unknown node voltage
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there are some conductance multiplier
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and that equals some function of v and
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summed up with some function of i
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and again notice that it's linear combination of v and i
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no one multiplication terms and so on and so forth
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this is a pretty standard form in which we'd represent the equations quite often
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and just label it
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this is often labeled G as a conductance matrix
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of course this is e,unknown node voltages
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and this is a linear sum of sources
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ok,so this is a really standard way that we could represent equations
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we did that last week as well or rather on Tuesday
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where i took a conductance matrix multiplied that
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by a column vector of unknown node voltages
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and equated that to some combination of...
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some linear combination of my source voltages
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the reason the circuit is linear
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is that i have only the linear elements in this circuit
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i don't have any non-linear elements
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and because of that
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i can rewrite this in the following manner
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i'm going to express e as a function of v and i
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and bring over to this side
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ok,so it's some function of i
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so I get R1 R2 divided by R1 plus R2
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and bring R1 R2 to this side
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that's what i get
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ok? so stare at this for a few seconds
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very common form
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my unknown node voltage
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is equal to these stuff on the right-hand side
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the stuff on the right-hand side has a term multiplying the source voltage v
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and some other term multiplying the current i
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and if i would put these things into symbolic form
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my unknown node voltage is some constant times V1 plus some constant times
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it's of the form,
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constant times the source current
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constant times the source voltage and so on
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the units of a's and b's are different
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because here are in this case,a has no units
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because V is voltage and so is e
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in this case b has a unit of resistance
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ok,so that b times i give me a voltage
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so stare at this equation for a few seconds
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and wish you help us build up some insight
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that will allow us to write down the answers
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almost by inspection
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ok,I'll show you a method now
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in a few minutes which will allow you to write down the answer e
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just by staring at the circuit without having to go through node equations and so on
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ok,the more and more method that I teach you
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the more you'll be able to do a lot of these completely by yourselves
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in this pretty clear example itself as a simple circuit
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but these method will be pratically useful when you have more complicated situations
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So,before I go on.
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Let me spend a few minutes upon ------- and linearing.
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So that's a linear circuit
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and this equation gives me the unknown voltage E as a linear sum of source voltages and source currents.
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The linearity implies two properties,
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the property homogeneity and also gives rise to the property of superposition.
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let's do homogeneity first
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So what it says is if-else circuit,sum circuit
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and I feed it some certain inputs A. Then let's say my output is As.
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If you are feeling hungry,think of this is a apples and the circuit could turn them to the apple sauce.ok?
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So what the homogeneity says is that what I can do is if I take each my apples,
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I instead of feeling in an entire apple what I will give it a three quarters of an apple.
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Say multiply all my inputs by some constant alpha three quarters
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What that is is that at the output of the circuit I will get one full of apple sauce
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I will get three quarters of the bottles of apple sauce
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so if I proportionally use of the inputs and this is a linear circuit
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Then so shall my output could use the same proportion
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So that's homogeneity,Next let's look at the superposition.
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The property of superposition are as the following
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Same circuit,if I feed it apples and I get apple sauce
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Ok,I take the same circuit and this time now if I feed the circuit a different sort of inputs
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I can say blueberries and let's say my output oops we do in this way
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So at the output I get a blueberry sauce,such as this
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So apple's apple sauce,blueberries could be blueberry sauce
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then what I am going to get...if I mix up the two
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So let's say I take my circuit,the same circuit with a set of inputs and it's an example of one output
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Let's mix up my inputs,and if i sum up my inputs in the following way
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Here I feed in A1 + B1,and here A2 + B2 and so on
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OK,then at the output I am going to get a mush of apple sauce and blueberry sauce
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all this says is that if I apply just apples,I get an apple sauce
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If I apply a blueberry,I got a blueberry sauce
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Then if I want to figure it out,how does the blender work?
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I fed it a combination of blueberries and apples,
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then for the purpose of understanding of the blender,all what is done was taken my two outputs
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and mix it together by myself and that's exactly what i get
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so ok if I sum up the inputs,my outputs will also be the sum of the outputs
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with the inputs apply by themselves.
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So let me take this here and ----------for a few seconds
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and I get something interesting out of it.
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So notice two inputs,two inputs,outputs
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In your notes I will give you another template for the next set of ------ I would make here
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so use next templates on page 3
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So what I will do here is something very simple
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Set one of the input to zero,and feed the voltage V1
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So lets feed the voltage V1 and set the other input zero,
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and let's say I get Y1 as an output
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And in this case,I set the first voltage to zero,and feed the different voltage V2 on the second input.
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And let's say the output is Y2
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This is just a particular application of the superposition principle I just outlined
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apply V1 set one input zero,I put V2 and set this original input zero
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then what we are going to find is that the answer was simply like this,
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just replace for As and Bs we just did.
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We will get V1 and zero here,and get zero and V2 here.
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OK,and as my outout,I am going to get exactly the sum Y1 + Y2.
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the simply particular application of superposition.
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Well,What I am saying is the following.
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If you look at this circuit here,effectively what I've done,
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is apply the voltage V1 on one input a voltage V2 at the other input.
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OK,V1 here,V2 here,and the output is Y1 + Y2
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what I'm saying is you look backwards now,
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What I’m saying is that the whole components of the output y1
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plus y2 could individually be derived in the following manner.
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I could get the component y1 by simply applying one of
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The voltages and setting the others to zero.
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00:20:23,210 --> 00:20:28,320
I can get the other component y2,by setting the other component to zero,
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00:20:28,320 --> 00:20:34,880
And applying the voltage v2 and to get y2,I sum them up,and that's my answer.
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00:20:34,880 --> 00:20:38,030
Ok? And this becomes a lot clearer of an example.
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00:20:38,030 --> 00:20:39,990
Ok,remember if I have a bunch of input of line two circuit,v1,v2 and so on,
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00:20:43,370 --> 00:20:48,270
And my output,and I get some output,then what' is to say is that,
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I can alternatively find out the answer by applying just one voltage.
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00:20:53,340 --> 00:20:56,210
setting the all the others to zero ，
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00:20:56,210 --> 00:20:59,540
Measuring the output by apply the second voltage,
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set all the input to zero,measure the output,
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00:21:02,230 --> 00:21:05,580
Ok? And sum up the apple source and blueberry source,and then we get the answer,ok?
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00:21:08,170 --> 00:21:14,440
Let’s do an example,and before we go in to that
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I'm talking about setting voltage sources and current sources to zero.
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00:21:17,860 --> 00:21:22,680
So,ah,first of all,what do I mean to set,
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the voltage source to zero,
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00:21:26,220 --> 00:21:33,820
Ah,this is the same as this.
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00:21:33,820 --> 00:21:39,050
Setting the voltage source to zero is simply displacing the voltage with a short
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and setting a current source,
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00:21:49,420 --> 00:22:03,620
Oops,setting current source to zero simply implies an open circuit.
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00:22:03,620 --> 00:22:06,510
Ok so when I say zero that source,
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00:22:06,510 --> 00:22:08,340
if it is a voltage source,short it,
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00:22:08,340 --> 00:22:16,520
If it is a current source,open it,yes
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I can take any two nodes in the world,right?
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00:22:18,990 --> 00:22:20,750
And measure the potential difference across them,
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00:22:20,750 --> 00:22:24,380
So there will be some potential difference,across these set by
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00:22:24,380 --> 00:22:27,510
The circuit that I haven't shown you on the site.
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00:22:27,510 --> 00:22:33,890
There will be other circuit that is controlling the voltage of these two nodes
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00:22:33,890 --> 00:22:36,040
the same is the short,the short the voltage is gonna be,
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00:22:36,040 --> 00:22:38,670
what's the V gonna be?
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00:22:38,670 --> 00:22:43,630
But the V is zero,ok?
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00:22:43,630 --> 00:22:52,870
so,that's the methods four,methods superposition
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00:22:52,870 --> 00:23:03,830
and this method says that the output of a circuit,
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00:23:03,830 --> 00:23:07,050
Again,remember focusing on linear circuits,
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00:23:07,050 --> 00:23:10,320
remember I haven't left the playground where LMV applies,
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00:23:10,320 --> 00:23:15,200
And within that playground I'm playing in the south goal area,in the south goal area,
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00:23:17,910 --> 00:23:21,140
In that subset of playground,circuits are linear,
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ok,so in that part of the playground,
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Superposition applies,because that circuit is linear.
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00:23:26,960 --> 00:23:43,450
So the output of a circuit is determined by summing of those responses to each source acting alone,
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Now in this statement here,this source stands for independent source
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00:24:01,360 --> 00:24:04,950
I haven't talked about independent source vs dependent sources,
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I will just talk about dependent sources a few weeks from today
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and you don't get confused for dependent sources,
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00:24:13,840 --> 00:24:19,260
You will be looking at section 3.3.3,of your course notes
280
00:24:19,260 --> 00:24:23,080
to see how superposition works with dependent sources,
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00:24:23,080 --> 00:24:25,550
Remember we haven't covered dependent sources yet,
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00:24:25,550 --> 00:24:35,080
we will be covering that about,ah,two weeks from now. Ok!
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00:24:35,080 --> 00:24:44,980
So let's go back to our example and apply the method of superposition to example,
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00:24:44,980 --> 00:24:52,510
So the methods says,sum up the outputs of each of the sub-circuit
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00:24:52,510 --> 00:24:55,950
where I'm applying one source acting alone,
286
00:24:55,950 --> 00:24:58,240
So let me just do this here,so let me start with this circuit
287
00:24:59,540 --> 00:25:04,520
And let me start with shutting "I" off,
288
00:25:04,520 --> 00:25:13,650
So I have a voltage V,
289
00:25:13,650 --> 00:25:18,500
And I've R2,and I'm shutting "I" off,
290
00:25:18,500 --> 00:25:21,810
Ok so I have displaced this with an open circuit,
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00:25:21,810 --> 00:25:30,280
So "I" is zero. Ok,so let me call the node voltage Ev to reflect
292
00:25:30,350 --> 00:25:36,020
That component of the node voltage,that arises due to V acting alone
293
00:25:36,880 --> 00:25:39,980
Ok and you should look at this pattern here,
294
00:25:39,980 --> 00:25:43,670
and very quickly be able to write the answer for patterns like this Voltage,
295
00:25:43,670 --> 00:25:46,510
The two resistors,let's call the resistor divider,
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00:25:46,510 --> 00:25:49,740
it appear again and again and again,
297
00:25:49,740 --> 00:25:59,300
and Ev is simply v times R2,divided by R1 plus R2,that’s still my ground node.
298
00:25:59,300 --> 00:26:04,620
Ok,so,the voltage here is simply this voltage,
299
00:26:04,620 --> 00:26:07,510
divided by the two resistances to give me the current,
300
00:26:07,510 --> 00:26:11,850
multiplied by R2 to give me the voltage across this R,
301
00:26:11,850 --> 00:26:13,270
remember this pattern,ok,
302
00:26:13,270 --> 00:26:19,200
you apply voltage divide pattern forty more times than any of the pattern you might imagine,
303
00:26:19,200 --> 00:26:25,570
ok so that's the v acting alone.
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00:26:25,570 --> 00:26:29,580
now let me do "I" acting alone,
305
00:26:29,580 --> 00:26:44,500
so for I acting alone,pointing up? yah,and what I want to do
306
00:26:44,500 --> 00:26:50,180
this time is to replace this with a short,
307
00:26:50,180 --> 00:26:57,410
to replace the voltage source for the short,and let me call this voltage Ei for the voltage,
308
00:26:57,410 --> 00:27:03,840
that is the component of the voltage due to the current I,
309
00:27:03,840 --> 00:27:08,820
and Ei,in this case,is simply given by another pattern here,
310
00:27:08,820 --> 00:27:14,360
the current across the pair of resistors is simply the effective resistants multiplied by the current,
311
00:27:14,360 --> 00:27:21,520
so it's I and effective resistance is R1,R2,
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00:27:21,520 --> 00:27:22,380
divided by R1 plus R2,
313
00:27:22,380 --> 00:27:27,650
that's Ei,that's component at that node due to current I,now
314
00:27:32,660 --> 00:27:35,670
so let us say that simply take these components,
315
00:27:35,670 --> 00:27:38,660
sum them up and then there you have the answer,
316
00:27:38,660 --> 00:27:43,200
so e is simply Ev plus Ei,the components should be e and I
317
00:27:44,150 --> 00:27:57,840
acting alone is simply v times R2 divided by R1 plus R2 plus R1,R2
318
00:27:57,840 --> 00:28:03,640
there we go,fortunately the fate has been kind to us,and
319
00:28:03,640 --> 00:28:07,040
answer is the same as we answer been in the node method.
320
00:28:07,040 --> 00:28:08,950
no surprise here,
321
00:28:08,950 --> 00:28:11,620
so this is actually incrediblely simple matter,
322
00:28:11,620 --> 00:28:14,310
so you can give a very complex circuit,
323
00:28:14,310 --> 00:28:17,060
ok what we've done here,you can take a very complex
324
00:28:17,060 --> 00:28:22,140
circuit and you can solve a very complex circuit by
325
00:28:22,140 --> 00:28:25,930
breaking it down into many simple individual sub-problems,
326
00:28:25,930 --> 00:28:30,090
ok you will do these in EECS time and time and time again,
327
00:28:30,090 --> 00:28:34,570
ok,whether there is a software system or hardware system,or what have you.
328
00:28:34,570 --> 00:28:36,500
You often have times building complicated systems
329
00:28:36,500 --> 00:28:40,240
,remember,do them on the site and when you put these things together,
330
00:28:40,240 --> 00:28:41,780
that's a large software system,
331
00:28:41,780 --> 00:28:44,390
you don't write whole piece of software studying main and you will down,you build a lot of little components
332
00:28:47,790 --> 00:28:49,640
and tied those components together,
333
00:28:49,640 --> 00:28:52,770
in the same manner here,you take a big circuit,
334
00:28:52,770 --> 00:28:59,070
and you find its behavior for each source acting alone,
335
00:28:59,070 --> 00:29:01,850
ok lots of little rinky simple little circuits,
336
00:29:01,850 --> 00:29:06,010
you will see examples in your homeworks where you will be given a big circuit or,
337
00:29:06,010 --> 00:29:10,000
because they set all the I to zero and the other v to zero,
338
00:29:10,000 --> 00:29:14,510
the whole circuit almost vanishes and the lecture left was little resistors will do,
339
00:29:14,510 --> 00:29:19,030
ok,so this is the very very powerful method,
340
00:29:19,030 --> 00:29:24,780
ok,I'd like to do a little demonstration for you.
341
00:29:24,780 --> 00:29:29,680
and what I will show you is the demo,
342
00:29:29,680 --> 00:29:32,820
is a vat of water,
343
00:29:32,820 --> 00:29:40,380
actually I’ll tell you what it is in a second,but assume it's salt water for now,
344
00:29:40,380 --> 00:29:48,240
ok I apply
345
00:29:48,240 --> 00:29:51,360
two voltages,in this case I want to apply,a
346
00:29:51,360 --> 00:29:59,210
sinusoid oh break it.
347
00:29:59,210 --> 00:30:08,730
and a triangular wave,and what I going to do,is measure the response at this site,
348
00:30:08,730 --> 00:30:11,010
and this is a vat of salt water,
349
00:30:11,010 --> 00:30:15,890
and I’ will tell you it behave like a linear system and ok
350
00:30:15,890 --> 00:30:19,400
you can view,if you view each little particle or each little
351
00:30:19,400 --> 00:30:25,100
cubit-centimeter or whatever of water. It behave like little resistor,
352
00:30:25,100 --> 00:30:30,070
ok so this big vat of water behave like big,the similar
353
00:30:30,070 --> 00:30:39,240
resistor of the following manner,
354
00:30:39,240 --> 00:30:42,990
and so on,ok,think of this,big,you know,mash of little
355
00:30:42,990 --> 00:30:46,880
resistors,but it's all resistors ok,it's a linear circuit,so I wanna
356
00:30:46,880 --> 00:30:49,730
apply two voltages a triangular and a sinusoid
357
00:30:49,730 --> 00:30:54,550
and we will observe the output. ok,and what we expect to
358
00:30:54,550 --> 00:30:57,730
see there?
359
00:30:57,730 --> 00:30:59,950
you see the superposition of the two,which is you see the
360
00:30:59,950 --> 00:31:10,900
sinusoid and you see the jag and triangular things articulating the sinusoid pattern,
361
00:31:10,900 --> 00:31:15,950
ok what I’m gonna do,right now,so,don't put any water yet,
362
00:31:15,950 --> 00:31:23,240
so this is a vat of ，nothing in it， it's all empty,
363
00:31:23,240 --> 00:31:34,870
ah,can we show the screen on that site? the oscilloscope screen,ok,
364
00:31:34,870 --> 00:31:38,760
there you go,so this is the screen of the oscillocscope now,
365
00:31:39,140 --> 00:31:42,740
so notice that I have a sinusoid,and I have a triangular
366
00:31:42,740 --> 00:31:47,310
wave,and the output is zero,and the reason is nothing in it,
367
00:31:47,310 --> 00:31:51,980
the vat is empty,so ah,previously when we talk about the
368
00:31:51,980 --> 00:31:55,220
course,I would get a salt water and pour salt water,
369
00:31:55,220 --> 00:31:59,870
then we discovered a much better source of water,that
370
00:31:59,870 --> 00:32:07,990
conduct electricity like one real mean fluid,cambridge water,
371
00:32:07,990 --> 00:32:13,180
it just works,works very pleasantly,it' s just conduct
372
00:32:13,180 --> 00:32:18,540
electricity like nothing or,and I was thinking using a child' s living
373
00:32:18,540 --> 00:32:20,680
water next time,and see what happens,
374
00:32:20,680 --> 00:32:23,440
although very probably we get some biological organisms doing strange things
375
00:32:24,750 --> 00:32:26,560
,go ahead,but
376
00:32:26,560 --> 00:32:30,800
ok so,our friendly demonstration expert Lorensa will
377
00:32:30,800 --> 00:32:37,830
pour some water into the vat,you should begin seeing as the output,being the superposition of
378
00:32:37,830 --> 00:32:47,140
the two,so as it pours,there you go,can you see that?
379
00:32:47,140 --> 00:32:53,120
so you do see the sinusoid articulation,and the jag waveform,and just
380
00:32:53,120 --> 00:32:55,870
to have some more fun
381
00:32:55,870 --> 00:33:07,110
,what I can do is to increase one of the voltages,and you will see
382
00:33:07,110 --> 00:33:10,870
now you know what will happen if I use child's living water.
383
00:33:10,870 --> 00:33:14,540
ok,so you know my output keeps increasing as I increase
384
00:33:14,540 --> 00:33:18,730
the corresponding wave form,
385
00:33:18,730 --> 00:33:27,780
ok ? ok I could do this,it's just fun,ok,so let me pause
386
00:33:27,780 --> 00:33:30,400
there,and go on to next topic,
387
00:33:30,400 --> 00:33:33,350
so that little demonstration shows you that,even
388
00:33:33,350 --> 00:33:39,860
something as simple as the physic entity,the vat of water
389
00:33:39,860 --> 00:33:42,500
behave like,the linear system,
390
00:33:42,500 --> 00:33:45,910
and we can bottle,like a linear system,as a set of resistors,
391
00:33:45,910 --> 00:33:48,870
I've been honest to you,right now,in the past 10 seconds,I
392
00:33:48,870 --> 00:33:51,960
introduce a new concept,
393
00:33:51,960 --> 00:33:56,060
ok it’s called the subliminal advertising,so what we do in
394
00:33:56,060 --> 00:33:59,720
double e a lot is modle real systems,
395
00:33:59,720 --> 00:34:04,010
ok,so often time if I want to look at the behavior of a vat-of
396
00:34:04,010 --> 00:34:07,580
water,I can model it as a set of resisters,for certain kind of activities,
397
00:34:08,460 --> 00:34:13,790
so just hold that part for some time later,in nuclear,
398
00:34:13,790 --> 00:34:24,210
ok,all right,next method for the superposition method,
399
00:34:24,210 --> 00:34:27,230
ok remember,it's methods like this,that make your life
400
00:34:27,230 --> 00:34:29,890
really really easy. If you find you have to
401
00:34:29,890 --> 00:34:32,810
do a grunge of homework or something,
402
00:34:32,810 --> 00:34:41,140
just step back and think superposition,ok,think Thevenin,or think composition rules,there must be a simpler way usually.
403
00:34:41,140 --> 00:34:51,330
ok,let's move to the next method,this is called the thevenin method.
404
00:34:51,330 --> 00:34:56,760
to derive this method let me start by applying
405
00:34:56,760 --> 00:35:02,990
superposition to some circuit,
406
00:35:02,990 --> 00:35:07,890
So let’s say,I have some arbitrary network N. Ok?
407
00:35:07,890 --> 00:35:12,270
As soon as a linear network. And the network has a whole bunch of good thing in it.
408
00:35:12,270 --> 00:35:19,990
It has a bunch of resisters.
409
00:35:19,990 --> 00:35:26,060
It has a bunch of voltage sources.
410
00:35:26,060 --> 00:35:30,970
And it has a bunch of current sources. Ok?
411
00:35:30,970 --> 00:35:35,070
Many current sources,many voltage sources,many resisters.
412
00:35:35,070 --> 00:35:44,520
Ok. Some jumbo or voltage sources,current sources and resisters. Ok?
413
00:35:44,520 --> 00:35:47,770
And I look at two nodes of the network.
414
00:35:47,770 --> 00:35:51,040
Ok. Here are two nodes in the network.
415
00:35:51,040 --> 00:35:53,670
Two points in the network where elements connect.
416
00:35:53,700 --> 00:35:55,300
I am looking at both the two nodes.
417
00:35:55,300 --> 00:35:58,040
And all I wanna do is following.
418
00:35:58,040 --> 00:36:01,520
Ok. I wanna figure out.
419
00:36:01,520 --> 00:36:02,820
If I take the rinky little current source and apply there.
420
00:36:09,720 --> 00:36:15,070
What I want to figure out is what is v and what is i .
421
00:36:15,070 --> 00:36:19,420
This is some mongo box out here,like a black box of resisters,
422
00:36:19,420 --> 00:36:22,180
voltage sources and current sources,do we need to count?
423
00:36:22,180 --> 00:36:25,480
I pick two nodes,apply current source.
424
00:36:25,480 --> 00:36:29,670
All I care about is the,er,what is the voltage.
425
00:36:29,670 --> 00:36:34,590
What is the voltage that I measure by applying here.
426
00:36:34,590 --> 00:36:38,040
notice the current here is i,because the current here is i.
427
00:36:38,040 --> 00:36:41,500
And I applied here. I want to measure what the voltage is.
428
00:36:41,500 --> 00:36:45,470
Ok. Now would a inside two obtained from superposition
429
00:36:45,470 --> 00:36:52,470
You should be able to jump up and step the form of the answer
430
00:36:52,470 --> 00:36:59,450
So by superposition,we know the following.
431
00:36:59,450 --> 00:37:06,780
Ok. We know that the effect of the circuit would be the same as some of
432
00:37:06,780 --> 00:37:09,050
the component we added out
433
00:37:09,050 --> 00:37:13,850
Some component,some component… A bunch of component added out
434
00:37:13,850 --> 00:37:19,100
Each component will be response of one sources acting alone.
435
00:37:19,100 --> 00:37:23,630
So if I can figure out the effect of one source acting alone. Ok?
436
00:37:23,630 --> 00:37:28,150
And put that down here. And do the same thing for all the sources.
437
00:37:28,150 --> 00:37:30,020
That’s what we will get.
438
00:37:30,020 --> 00:37:35,430
So,er… So for the source Vm. It is a linear circuit.
439
00:37:35,430 --> 00:37:40,660
So I know that my answer is gonna be… indeed,the final answer is gonna be Vm term
440
00:37:40,660 --> 00:37:44,720
And it is gonna be multiplied by some am term
441
00:37:44,720 --> 00:37:46,780
Ok. I know that it is a linear circuit.
442
00:37:46,780 --> 00:37:51,460
So I know the answer shall have a term Vm multiplied by some constant.
443
00:37:51,460 --> 00:37:54,470
Ooh,simple. I know that.
444
00:37:54,470 --> 00:37:57,060
Similarly the same is true for …er…
445
00:37:57,060 --> 00:38:01,140
This is the …er…the term Vm.
446
00:38:01,140 --> 00:38:03,900
What I can do is I can measure the just as a fact by setting all the
447
00:38:03,900 --> 00:38:06,190
other sources to zero.
448
00:38:06,190 --> 00:38:18,540
So I can set all the other current sources to zero and all voltage sources except for this one.
449
00:38:18,540 --> 00:38:21,910
Ok? I can get that answer.
450
00:38:21,910 --> 00:38:30,170
So similarly for every voltage source and I will get a term. Ok?
451
00:38:30,170 --> 00:38:33,190
So for every single source—m1,m2,m3 and so on.
452
00:38:33,190 --> 00:38:37,780
I wanna get such a term. Then I would sum up.
453
00:38:37,780 --> 00:38:43,750
Similarly,I am gonna get a term for In. Ok?
454
00:38:43,750 --> 00:38:46,510
And I know there will be a In term.
455
00:38:46,510 --> 00:38:52,900
And I know it's gonna be some constant data multiplying In?
456
00:38:52,900 --> 00:38:55,990
Ok. This is an example of ours here.
457
00:38:55,990 --> 00:38:57,070
And this is an example.
458
00:38:57,070 --> 00:39:04,260
Remember. a was this. Ok? And B was this.
459
00:39:04,260 --> 00:39:06,920
This constant here. Ok.
460
00:39:06,920 --> 00:39:09,430
Some constant B and some constant a
461
00:39:09,430 --> 00:39:12,660
And because I hold on to the current sources.
462
00:39:12,660 --> 00:39:16,410
Ok. There is gonna be such a term for each one of them.
463
00:39:16,410 --> 00:39:20,880
And each one of such terms Vm,In will be…
464
00:39:20,880 --> 00:39:23,720
The voltage I won’t see here.
465
00:39:23,720 --> 00:39:28,160
If I set all other Vm to zero,and I set all the other
466
00:39:30,540 --> 00:39:36,420
current sources except for that one to zero…
467
00:39:36,420 --> 00:39:40,050
What am I missing? Is that it?
468
00:39:40,050 --> 00:39:41,540
The response here.. v here.
469
00:39:41,540 --> 00:39:45,400
Am I missing anything here? Is that it?
470
00:39:45,400 --> 00:39:49,960
don't down it at once
471
00:39:49,960 --> 00:39:52,410
What am I missing?
472
00:39:52,410 --> 00:39:53,490
Current source I,exactly
473
00:39:53,490 --> 00:39:57,680
So if I have a current source I.then there is an effect of this current as well
474
00:39:58,870 --> 00:40:02,110
And so I ‘ve done an i there,too.
475
00:40:02,110 --> 00:40:05,340
There must be some constant multiple I.
476
00:40:05,340 --> 00:40:10,120
O.K,that constant is going to look like a resistor,right?
477
00:40:10,120 --> 00:40:14,730
Because this circuit contains current sources,voltage sources and resistors.
478
00:40:14,730 --> 00:40:21,020
If I short all my voltage sources,open all my current sources,what's lefting here?
479
00:40:21,020 --> 00:40:30,510
Just whole come to be all of Rs,just going to look like some resisters R. And that's what I get here.
480
00:40:30,510 --> 00:40:45,650
O.K,this is what V is to like,and that's the form,so let's take a look at these components.
481
00:40:45,650 --> 00:40:49,400
Let's focus on the easy part first,
482
00:40:49,400 --> 00:40:53,070
what is this look like? This component looks like an I,
483
00:40:53,070 --> 00:40:55,470
look like a current
484
00:40:55,470 --> 00:40:59,510
and some resistors. What's that resistors given by?
485
00:40:59,510 --> 00:41:04,150
Suppose they give you this network and this current source,and then ask to tell me R. How would you measure R?
486
00:41:07,210 --> 00:41:13,080
What you will do is open all the current sources,Short all the voltage sources,
487
00:41:13,080 --> 00:41:16,000
put a meter over there and measure there things R?
488
00:41:16,000 --> 00:41:18,470
That's,that's R.
489
00:41:18,470 --> 00:41:22,880
O.K,so then let's get to this term. What's about this term here?
490
00:41:22,880 --> 00:41:27,910
Can someone tell me the units of this term? The victim here?
491
00:41:27,910 --> 00:41:33,420
Voltage,O.K,this is the voltage,this is the voltage,RI is voltage,so,this behave like a voltage.
492
00:41:36,980 --> 00:41:45,060
O.K? And it behaves like some voltage V. So notice that,as far as this current I is concerned,
493
00:41:49,920 --> 00:42:00,310
the rest of the universe looks like a resistor and voltage sources behaving in some manner.
494
00:42:00,310 --> 00:42:09,800
O.K,let me just call it Vth,for now and you know while in a second.
495
00:42:09,800 --> 00:42:27,150
Or,the voltage has a form,some voltage plus Ri.
496
00:42:27,150 --> 00:42:44,190
So in other words,as far as this i is concerned,this whole network here,N,full of the nice,nice stuff,
497
00:42:44,190 --> 00:42:50,860
is distinguishable to this,I here,so my I is sitting up there injecting a current into two notes. O.K,I am I,
498
00:42:50,860 --> 00:43:05,120
O.K. I am looking at this,this network looks no different than a voltage source
499
00:43:05,120 --> 00:43:08,770
in series with an resistor R.
500
00:43:08,770 --> 00:43:14,950
O.K,notice that the equation for this simple circuit is this,
501
00:43:14,950 --> 00:43:30,340
so i is given by V minus Vth devided by R. O.K? Just remember.
502
00:43:30,340 --> 00:43:34,290
O.K,this circuit,in other words,
503
00:43:34,290 --> 00:43:38,300
I'd around sitting here,can not tell the difference
504
00:43:38,300 --> 00:43:44,070
if I'm measuring the voltage here between a circuit that looks Vth in series with the resistor,
505
00:43:44,070 --> 00:43:50,280
all this huge mess of voltage sources and current sources and so on.
506
00:43:50,280 --> 00:43:59,460
O.K? Now R,I would talk Vth and R. R is called the resistor of the network,
507
00:43:59,460 --> 00:44:04,930
as seen from the port,with all the sources short up. O.K?
508
00:44:04,930 --> 00:44:09,310
And similarly Vth,what is Vth?
509
00:44:09,310 --> 00:44:15,850
Vth is the open circuit voltage. In other words,if I apply the voltage here,
510
00:44:15,850 --> 00:44:21,770
look at this is the responce of all the current sources,and all the voltage sources acting together,
511
00:44:21,770 --> 00:44:27,810
O.K? So if I took this out,and simply measure my V here,
512
00:44:27,810 --> 00:44:32,650
is that i didn't exist right because this is the component I.
513
00:44:32,650 --> 00:44:36,640
So I open the I,and measure V,
514
00:44:36,640 --> 00:44:45,350
I will get that big current on the left hand side I,O.K,that what I means th.
515
00:44:45,350 --> 00:44:58,290
So that inspires the next method called Thevenin method.
516
00:44:58,290 --> 00:45:01,700
O.K,in this method,what I'm going to do is to take
517
00:45:01,700 --> 00:45:14,630
some circuit on the page nine with a mess of stuff,so a big mess of stuff.
518
00:45:16,050 --> 00:45:23,890
O.K,if I care,to look this impact on the something else from the outside,
519
00:45:23,890 --> 00:45:44,950
then,as far as the outside rule is concerned,this is indistinguishable from the circuit that looks like this.
520
00:45:44,950 --> 00:45:51,270
O.K? So what I can do is if I want to figure out what's happening here,then for the purpose of analysis,
521
00:45:51,270 --> 00:45:59,110
I can,this simple network carried out of Vth circuit becomes a thorough gate for this entire mess.
522
00:45:59,110 --> 00:46:03,390
So for the purpose to find out the behavior at this point,I can take this huge mess,
523
00:46:03,390 --> 00:46:08,090
and do replace it with this thevenin Vth,or this thevenin equivalent
524
00:46:08,090 --> 00:46:15,670
. O.K,this is called the thevenin equivalent of this big network.
525
00:46:15,670 --> 00:46:22,340
Let me give you an example that will make the method completely clear.
526
00:46:22,340 --> 00:46:29,850
Again,remember EECS,most of our lives about how can you make thing so simple as being analyzed by inspection.
527
00:46:29,850 --> 00:46:34,400
O.K,this is the method that takes you further down that path.
528
00:46:34,400 --> 00:46:43,950
So let me use a circuit that I have used before. A voltage V,
529
00:46:43,950 --> 00:46:49,430
R1,R2,
530
00:46:49,430 --> 00:46:54,810
this is the an R and fifty-five minutes so we have four
531
00:46:54,810 --> 00:47:04,390
minutes. So this is my circuit. And let's say all I care about is finding out i1.
532
00:47:04,390 --> 00:47:11,620
Look,that's all I care about,and what I'm going to do is I would box this up
533
00:47:11,620 --> 00:47:15,650
and see if I can replace that with an Thevenin equivalent.
534
00:47:15,650 --> 00:47:35,120
O.K? Some was boxed up.
535
00:47:35,120 --> 00:47:41,430
O.K,what I'm saying is that I'm going to box it up and replace it with a Thevenin equivalent.
536
00:47:41,430 --> 00:47:43,510
I don't know what Vth,R are at this
537
00:47:43,510 --> 00:47:47,860
point. I just call it Rth for fun. I don't know what these two value are?
538
00:47:49,150 --> 00:47:54,400
If I know these two values,then I can determine the nearly trivially as follows.
539
00:47:54,400 --> 00:48:01,140
I can get i1 as simply V minus Vth divided by R1 plus Rth.
540
00:48:06,650 --> 00:48:13,290
So if I know Vth,Rth,I can write down i1 by inspection in that manner.
541
00:48:13,290 --> 00:48:24,820
O.K.So,next,finally,how do I get Vth and Rth. You can get Rth by looking at this network
542
00:48:24,820 --> 00:48:28,740
and short up all the voltage sources and measuring the resistors there.
543
00:48:28,740 --> 00:48:43,850
So I short my voltage source,that's R1,that's,opps,wrong way.
544
00:48:43,850 --> 00:48:49,280
If you look this way,so look in this way,I open my,
545
00:48:49,280 --> 00:48:51,400
that's all I get.
546
00:48:51,400 --> 00:48:58,760
So what's Rth? Rth is simply R2.
547
00:48:58,760 --> 00:49:04,240
O.K,so I have
548
00:49:04,240 --> 00:49:05,950
open my current source.
549
00:49:05,950 --> 00:49:11,840
Similarly for Vth,remember all I want to do is look at,two nodes,stand
550
00:49:11,840 --> 00:49:16,380
back,put a what would be there,measure the voltage that's my opencircuit voltage.
551
00:49:16,380 --> 00:49:25,080
O.K,so what I do is I take the circuit and simply measure the voltage there,
552
00:49:25,080 --> 00:49:27,950
that's R2,
553
00:49:27,950 --> 00:49:30,800
that's my current capital I.
554
00:49:30,800 --> 00:49:36,380
O.K,and I simply want to measure the open voltage here which is one.
555
00:49:36,380 --> 00:49:42,510
And simply If I stand back and I can gingerly measure the voltage that was disturbing anything,I simply get IR2.
556
00:49:47,240 --> 00:50:00,940
O.K?So Vth is IR2,and Rth is R2,and here is the fomula for the current in this branch,
557
00:50:00,940 --> 00:50:08,870
when I replace,when i applied voltage source,and resistor R1,and resistor circuit here.
558
00:50:08,870 --> 00:50:14,640
O.K,so let's pause here for,let me summerize this for a seconds,
559
00:50:14,640 --> 00:50:17,540
have this circuit here,and I'd like to find i1,
560
00:50:17,540 --> 00:50:22,250
so what I say I do is to take this complicated mess,
561
00:50:22,250 --> 00:50:28,240
I said complicated mess and assume it is,and replace with the sistors,Rth,
562
00:50:28,240 --> 00:50:31,560
O.K? Got by turning off all of the sources.
563
00:50:31,560 --> 00:50:39,090
And voltages you see these,Vth which I get simply by pulling this thing out,
564
00:50:39,090 --> 00:50:46,140
taking my input this part out and simply measure the open circuit voltage out there,Vth.
565
00:50:46,140 --> 00:50:51,440
O.K? And then I do replace the whole network with this new network that called Thevenin network,
566
00:50:51,440 --> 00:50:56,610
and I got the answer at a second. A few homework problems left in this section as well.
Last Modified 11/15/05 10:59 PM
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