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Let's get started.

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Can you hear me back there?
Loud and clear.

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OK.
Let's get started.

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Before I begin,
just a couple of announcements.

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Brad Buren is one of our
students here and he needs a

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note-taker.
It's a paid position.

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So if you are interested you
can stop by after class and see

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him.
He's sitting right here out

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there, OK?
Second, just a reminder that

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6.002 does have prerequisites.
And the prerequisites are 8.02

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and 18.03.
So with that let me start off

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with the usual.
Do a quick review of what we've

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done so far.
So we started out life looking

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at the laws of physics and
Maxwell's equations and so on.

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And those were way too hard so
we said let's make life easy for

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ourselves.
So we chose to play in this

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playground in which we said we
shall adhere to the lumped

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matter discipline.
OK?

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The LMD.
So we are in that playground.

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So this entire course,
and for that matter large parts

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of EECS are within that
playground, within which the

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lumped matter discipline
applies.

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So as soon as we jumped into
the playground,

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the LMD playground,
we could take Maxwell's

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equations and abstract them out
into two very,

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very simple rules.
And the very simple rules were

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KVL and KCL.
KVL simply said that I can sum

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the voltages in any loop in a
circuit and the result then

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would be zero.
Similarly, I can sum the

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currents that enter or exit any
node and the sum will also be

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zero.
So what you can now do is,

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if you feel like,
you can go around and brag.

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Oh, yeah, we use Maxwell's
equations in everyday life and,

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yeah, it's good stuff.
And the key is that this is

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really an encapsulation of
Maxwell's equations within this

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playground that we are in.
So I talked about the first

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method of circuit analysis in
the last lecture.

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And that method simply took
the, wrote KVL for all the

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loops, wrote KCL for all the
nodes and wrote element vi

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relationships.
And together gave you a big

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bunch of equations.
And you sat down and grunged

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through the equations and you
solved for branch voltages and

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currents.
So we reviewed a second method

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of circuit analysis.
And I'll simply call it circuit

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composition.
The basic idea behind this

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method was to learn some simple
rules of how resistors add and

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conductances add and so on and
so forth and look at a circuit

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and simplify the circuit by
making series simplifications

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when the resistors are in series
and so on and so forth,

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and compose it and play around
with it till we end up with the

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current, the voltages that we
are looking for.

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This is the intuitive method.
And so a section in Chapter 2,

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I believe, of the course notes
discusses several examples using

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this method and attempts to make
a little bit formal the

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intuitive approach that is
applied in this method.

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So we then looked at the node
method.

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And the node method was simply
a particular way of applying KVL

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and KCL.
Node method,

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remember?
We took a ground node.

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Then we labeled the nodes of
the remaining voltages with

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respect to that ground.
Then we wrote KCL for each of

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the nodes.
And when we wrote KCL for each

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of the nodes,
remember, KVL was implicit in

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this expression that we used for
each of the currents that were

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exiting each node.
So if Ej was a node voltage,

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then Ej minus Ei multiplied by
the conductance Gi was the

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current that was going through
one of those,

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I should call it Gij.
This is a conductance that

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connects nodes i and j.
That gave us the KVL that fed

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into the same system.
So these are three methods.

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The node method,
by the way, is sort of the

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workhorse of the 6.002 industry.
And for that matter for all of

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the circuits industry.
When in doubt,

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apply the mode method,
you'll be OK.

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That applies to linear
circuits, nonlinear circuits,

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what have you.
What I'm going to do today is

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go through two more methods.
So notice that the first few

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lectures of this course,
the first three lectures simply

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comprise transitioning you from
the world of physics to the

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world of EECS.
And then two lectures on giving

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you a bag of tricks.
So we start you off with the

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sort of tools,
your mallets and chisels and so

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on and so forth.
And these five methods are your

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tools.
We'll look at two methods

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today.
One method is called the method

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of superposition and the second
method is called the Thevenin

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method.
And these methods apply only to

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linear circuits.
So we look at the subset of

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circuits that are linear,
and these two methods apply to

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only those circuits.
These are methods that combined

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with intuition really enables
you to solve very interesting

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circuits very,
very quickly.

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So let me do an example using a
usual node method.

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And then jump into introducing
the superposition methods and

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Thevenin methods using that same
example.

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So let me draw you an example
circuit here.

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So, again, I'm using this
example, I will use this example

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to introduce the method of
superposition and the Thevenin

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method.
So what I'm going to do is

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start off the usual way and
analyze the circuit using a

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method that you know now,
the node method.

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And what I'll do is write down
the node equations for this by

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applying the node method.
So if you recall the node

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method.
I choose a ground node.

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I'm going to choose this node.
It's got both the voltage

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source connected to it,
and it's also got many other

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edges impinging on it.
So I'm going to choose that as

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my ground node and I'm going to
label the other nodes with their

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voltages.
So this is an unknown.

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I'll label it as e.
I guess we just have one

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unknown e.
And I know the voltage of this

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node, and that is simply V.
Since it's V,

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there's a voltage source
between the ground node and that

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node.
So what I can do next is that I

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can write down the node equation
for this node and then go from

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there.
So let me go ahead and do that.

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So let me sum up the currents
going outside,

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going outwards.
So I have e minus v divide by

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R1, I have e minus zero divide
by R2, and I have minus i equals

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zero.
This is a node equation.

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The first thing I want you to
observe is that this equation is

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linear in V and i.
What I mean by linear is that

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you don't see terms like Vi or
V-squared and things like that.

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It's some constant times V plus
some constant times i equals

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some other constant.
So that's quite nice.

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So I'm going to rearrange the
terms in the following manner.

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I'll move the known sources to
the right-hand side and collect

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the coefficients of e on this
side, so I get one by R1 plus

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one by R2 over here.

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So stare at this for a moment
and notice again here I have e,

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my unknown node voltage,
there is some constant

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multiplier, and that equals some
function of V summed up with

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some function of i.
And, again, notice that this is

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a linear combination of V and i.
No multiplication terms and so

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on and so forth.
This is a pretty standard form

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in which we will represent
equations quite often.

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And just to label it,
this is often labeled G as the

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conductance matrix.
Of course this is e,

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our unknown node voltages,
and this is a linear sum of

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sources.
So this is a very standard way

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that we will represent
equations.

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We did that last week as well,
or rather on Tuesday where I

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took a conductance matrix,
multiplied that by a column

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vector of unknown node voltages
and equated that to some linear

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combination of my source
voltages.

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The reason the circuit is
linear is that I have only

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linear elements in the circuit.
I don't have any nonlinear

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elements.
And because of that I can

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rewrite this in the following
manner.

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I'm just going to express e as
a function of V and i and bring

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it over to this side.
So it's some function of i.

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So I get R1 R2 divide by R1
plus R2.

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And I bring R1 R2 to this side.
That's what I get.

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So stare at this for a few
seconds, very common form.

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My unknown node voltage is
equal to this stuff on the

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right-hand side.
The stuff on the right-hand

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side has a term multiplying the
source voltage V and some other

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term multiplying the current I.
And if I were to put this in

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sort of symbol-like form my
unknown node voltage is some

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constant times V1 plus some
constant times,

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is of the form constant times
the source current,

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constant times the source
voltage and so on.

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The units of As and Vs are
different because in this case A

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has no units because V is a
voltage.

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And so is e.
In this case V has units of

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resistance.
So that V times i gives me a

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voltage.
So stare at this equation for a

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few seconds and this should help
us build up some insight that

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will allow us to write down the
answer almost by inspection.

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I'm going to show you a method
now, in a few minutes,

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which will allow you to write
down the answer e just by

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starring at the circuit without
having to go through node

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equations and so on.
The more and more methods I

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teach you, the more you will be
able to do a lot of this

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completely by yourselves.
In this particular example it's

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a relatively simple circuit but
these methods would be

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particularly useful when you
have more complicated

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situations.
But before I go on let me spend

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a few minutes pontificating on
linearity.

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So that's a linear circuit.
And this equation gives me the

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unknown node voltage e as a
linear sum of source voltages

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and source currents.
Linearity implies two

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properties, the property of
homogeneity and also gives vice

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to the property of
superposition.

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Let's do homogeneity first.
What this says is if I have a

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circuit, some circuit and I feed
it some sort of inputs,

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A, then let's say my output is
S.

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If you're feeling hungry think
of these as apples and the

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circuit converts them into
applesauce.

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So what homogeneity says is
that what I can do is if I take

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each of my apples and instead of
feeding it an entire apple what

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if I give it three-quarters of
an apple?

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Say I multiple all my inputs by
some constant alpha,

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three-quarters.
What that says is that at the

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output instead of getting one
full bottle of applesauce I'm

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going to get three-quarters of a
bottle of apple sauce.

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So if I proportionately reduce
all the inputs and if this is a

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linear circuit then so shall my
output be reduced in the same

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proportion.
So that's homogeneity.

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Next, let's look at
superposition.

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The property of superposition
says the following.

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The same kind of circuit.
If I feed it apples then I get

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applesauce.
I take the same circuit,

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and this time around if I feed
the circuit a different set of

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inputs, say blueberries.
And let's say my output,

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oops, let me do it this way.
So as my output I get blueberry

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sauce, if such exists.
So apples applesauce,

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blueberries give me blueberry
sauce.

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Then what I'm 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

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inputs and in this example one
output.

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Let's say I mix up my inputs
and some of my inputs in the

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following way,
here I feed an A1 plus B1 and

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here A2 plus B2 and so on then
at the output I am going to get

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a mush of apple sauce and
blueberry sauce.

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All this says is that if I
apply just apples I get

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applesauce.
If I apply just blueberries I

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get blueberry sauce.
Then if I were to figure out

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how this blender would have
worked had I fed in the

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combinations of apples and
blueberries, then for the

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purposes of understanding that
blender all I could have done

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was taken by two outputs and
just mixed them up together

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myself and that's exactly what
I'd get.

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So if I sum up the inputs my
outputs would also be the sum of

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the outputs with the inputs
applied by themselves.

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So let me take this here and
munge around with hit for a few

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seconds and 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've given you
another template for the next

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set of scribbles I'm going to
make here.

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So use the next set of
templates on page three.

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What I'm going to do here is
something very simple,

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set one output to zero and feed
a voltage V1.

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So that's feed a voltage V1 and
set the other output to 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 a

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different voltage V2 on the
second input.

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And let's say my output is Y2.
This is just a particular

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application of the superposition
principle I just outlined.

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Apply V1 set one output to
zero.

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Apply V2 set the original
output to zero.

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Then what I'm going to find is
that the answer will simply look

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like this, just replace for As
and Bs what I just did and we

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get V1 and zero here and we get
zero and V2 here.

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And as my output I'm going to
get exactly the sum Y1 plus Y2.

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This is simply a particular
application of superposition

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where what I'm saying is the
following.

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If you look at this circuit
here effectively what have I

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done?
Effectively what I've done is

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apply the voltage V1 on one
input and a voltage V2 on the

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other input.
V1 here.

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V2 here.
And the output is Y1 plus Y2.

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What I'm saying is look
backwards now.

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What I'm saying is that the
whole components of the output

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Y1 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 the

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voltages and setting the other
to zero.

262
00:20:21 --> 00:20:25
I can get the other component
Y2 by setting yet another input

263
00:20:25 --> 00:20:30
to zero and applying the voltage
V2 to get Y2.

264
00:20:30 --> 00:20:32
And sum then up and that's my
answer.

265
00:20:32 --> 00:20:36
This will become a lot clearer
with an example.

266
00:20:36 --> 00:20:40
Again, remember if I have a
bunch of inputs applied to a

267
00:20:40 --> 00:20:44
circuit, V1, V2 and so on,
and I get some output then what

268
00:20:44 --> 00:20:48
this is saying is that I can
alternatively find out the

269
00:20:48 --> 00:20:53
answer by applying just one
voltage, setting all the others

270
00:20:53 --> 00:20:57
to zero, measuring the output,
apply a second voltage,

271
00:20:57 --> 00:21:01
set all inputs to zero,
measure the output and sum of

272
00:21:01 --> 00:21:07
applesauce and blueberry sauce
and there you get the answer.

273
00:21:07 --> 00:21:13
Let's do an example.
And before we go into that I

274
00:21:13 --> 00:21:20
talked about setting voltage
sources and current sources to

275
00:21:20 --> 00:21:22
zero.
First of all,

276
00:21:22 --> 00:21:28
what does it mean to set a
voltage source to zero?

277
00:21:28 --> 00:21:38
This is the same as this.
Setting a voltage source to

278
00:21:38 --> 00:21:49
zero is simply replacing the
voltage source with a short,

279
00:21:49 --> 00:22:00
and setting a current source to
zero simply implies an open

280
00:22:00 --> 00:22:05
circuit.
So when I say zero that source,

281
00:22:05 --> 00:22:09
if it's a voltage source short
it, if it's a current source

282
00:22:09 --> 00:22:10
open it.

283
00:22:10 --> 00:22:16


284
00:22:16 --> 00:22:19
I can take any two nodes in the
world and measure the potential

285
00:22:19 --> 00:22:22
difference across them.
So there may be some potential

286
00:22:22 --> 00:22:26
difference across these set by
the circuit that I haven't shown

287
00:22:26 --> 00:22:29
you on this side.
There might be some other

288
00:22:29 --> 00:22:33
circuit that is controlling the
voltage of these two nodes.

289
00:22:33 --> 00:22:38
The same with the short.
What's V going to be?

290
00:22:38 --> 00:22:42
But there is a V.
It's zero.

291
00:22:42 --> 00:22:48
So that's method four,
method of superposition.

292
00:22:48 --> 00:22:56
And this method says that the
output of a circuit --

293
00:22:56 --> 00:23:03


294
00:23:03 --> 00:23:06
Again, remember I'm focusing on
linear circuits.

295
00:23:06 --> 00:23:10
Remember, I have this
playground where LMD applies.

296
00:23:10 --> 00:23:14
And within that playground I'm
playing in the south goal area.

297
00:23:14 --> 00:23:17
In the south goal area,
in that subset of the

298
00:23:17 --> 00:23:21
playground circuits are linear.
So in that part of the

299
00:23:21 --> 00:23:25
playground superposition applies
because there circuits are

300
00:23:25 --> 00:23:32
linear.
So the output of a circuit is

301
00:23:32 --> 00:23:42
determined by summing up the
responses to each source acting

302
00:23:42 --> 00:23:44
alone.

303
00:23:44 --> 00:23:54


304
00:23:54 --> 00:23:58
Now, in this statement here
this source stands for

305
00:23:58 --> 00:24:01
independent source.
I haven't talked about

306
00:24:01 --> 00:24:04
independent versus dependent
sources.

307
00:24:04 --> 00:24:08
We'll talk about dependent
sources a few weeks from today.

308
00:24:08 --> 00:24:12
And just so you don't get
confused, for dependent sources

309
00:24:12 --> 00:24:17
you will be looking at Section
3.3.3 of your course notes to

310
00:24:17 --> 00:24:21
see how superposition works with
dependent sources.

311
00:24:21 --> 00:24:25
But remember we haven't covered
dependent sources yet.

312
00:24:25 --> 00:24:30
We will be covering them about
two weeks from now.

313
00:24:30 --> 00:24:36
So let's go back to our example
and apply the method of

314
00:24:36 --> 00:24:43
superposition to an example.
So the method says sum up the

315
00:24:43 --> 00:24:49
outputs of each of the
sub-circuits where I'm applying

316
00:24:49 --> 00:24:55
one source acting alone.
So let me just do this here.

317
00:24:55 --> 00:25:03
Let me start with the circuit.
And let me start with shutting

318
00:25:03 --> 00:25:06
I off.
So I have voltage V --

319
00:25:06 --> 00:25:13


320
00:25:13 --> 00:25:16
I have R2.
And I'm shutting I off.

321
00:25:16 --> 00:25:21
So I have replaced this with an
open circuit.

322
00:25:21 --> 00:25:25
So I is zero.
Let me call the node voltage eV

323
00:25:25 --> 00:25:32
to reflect that component of the
node voltage that arises due to

324
00:25:32 --> 00:25:37
V acting alone.
And you should look at this

325
00:25:37 --> 00:25:42
pattern here and very quickly be
able to write the answer for

326
00:25:42 --> 00:25:46
patterns like this voltage,
the two resistors.

327
00:25:46 --> 00:25:49
That's called a resistive
divider.

328
00:25:49 --> 00:25:52
It will appear again and again
and again.

329
00:25:52 --> 00:25:56
And eV is simply V times R2
divided by R1 plus R2.

330
00:25:56 --> 00:26:02
That's still my ground node.
So the voltage here is simply

331
00:26:02 --> 00:26:07
this voltage divided by the two
resistors to give you the

332
00:26:07 --> 00:26:12
current multiplied by R2 to give
you the voltage across this R.

333
00:26:12 --> 00:26:16
Remember this pattern.
You apply voltage divider

334
00:26:16 --> 00:26:21
patterns probably more times
than any other pattern that you

335
00:26:21 --> 00:26:24
might imagine.
So that's with the V acting

336
00:26:24 --> 00:26:27
alone.
Now, let me do I acting alone.

337
00:26:27 --> 00:26:31
So for I acting alone --

338
00:26:31 --> 00:26:43


339
00:26:43 --> 00:26:49
And what I do this time around
is replace this with a short,

340
00:26:49 --> 00:26:53
replace the voltage source to
the short.

341
00:26:53 --> 00:27:00
And let me call this voltage eI
for the component of the voltage

342
00:27:00 --> 00:27:04
due to the current I.
And eI, in this case,

343
00:27:04 --> 00:27:08
is simply given by yet another
pattern here,

344
00:27:08 --> 00:27:12
the current across a pair or
resistors is simply the

345
00:27:12 --> 00:27:17
effective resistance multiplied
by the current so it's i and the

346
00:27:17 --> 00:27:21
effective resistance is R1,
R2 or R1 plus R2.

347
00:27:21 --> 00:27:24
That's eI.
That's a component that node

348
00:27:24 --> 00:27:31
due to the current I.
Now, so the method says that.

349
00:27:31 --> 00:27:38
Then take these components,
sum them up and there you have

350
00:27:38 --> 00:27:44
the answer.
So E is simply ev plus ei.

351
00:27:44 --> 00:27:49
The components of V and I
acting alone,

352
00:27:49 --> 00:27:56
just simply V times R2 divided
by R1 plus R2 plus R1,

353
00:27:56 --> 00:28:00
R2.
There we go.

354
00:28:00 --> 00:28:02
Fortunately,
the fates have been kind to us

355
00:28:02 --> 00:28:06
and the answer is the same as
the answer we obtained with the

356
00:28:06 --> 00:28:08
node method.
No surprise here.

357
00:28:08 --> 00:28:11
So this is actually an
incredibly simple method.

358
00:28:11 --> 00:28:14
So you can take a very complex
circuit.

359
00:28:14 --> 00:28:18
What have you really done here?
You can take a very complex

360
00:28:18 --> 00:28:22
circuit and you can solve a very
complex circuit by breaking it

361
00:28:22 --> 00:28:25
down into many simple individual
sub problems.

362
00:28:25 --> 00:28:30
You will do this in EECS time
and time and time again.

363
00:28:30 --> 00:28:33
Whether it's in software
systems or hardware systems or

364
00:28:33 --> 00:28:35
what have you,
you're often times building

365
00:28:35 --> 00:28:38
complicated systems.
Remember doom on this side?

366
00:28:38 --> 00:28:41
And the way and when you put
these things together,

367
00:28:41 --> 00:28:44
let's say a large software
system, is you don't write the

368
00:28:44 --> 00:28:47
whole piece of software starting
main and grunge down.

369
00:28:47 --> 00:28:50
You build a lot of little
components and tie the

370
00:28:50 --> 00:28:53
components together.
In the same manner here you

371
00:28:53 --> 00:28:57
take a big circuit and you find
its behavior for each source

372
00:28:57 --> 00:29:00
acting alone.
Lots of little inky dinky

373
00:29:00 --> 00:29:04
simple little circuits.
And you will see examples in

374
00:29:04 --> 00:29:09
your homework where you're given
a big circuit or because it set

375
00:29:09 --> 00:29:14
all the Is to zero and the other
Vs to zero the whole circuit

376
00:29:14 --> 00:29:18
almost vanishes and all that
you're left with is a little

377
00:29:18 --> 00:29:21
resistor or two.
So this is the very,

378
00:29:21 --> 00:29:24
very powerful method.
I'd like to do a little

379
00:29:24 --> 00:29:28
demonstration for you.
And what I'm going to show you

380
00:29:28 --> 00:29:36
is the demo is a vat of water.
Actually, I'll tell you what it

381
00:29:36 --> 00:29:42
is in a second.
But assume it is salt water for

382
00:29:42 --> 00:29:46
now.
I'll apply two voltages.

383
00:29:46 --> 00:29:52
In this case I'm going to apply
a sinusoid.

384
00:29:52 --> 00:29:59
That's not very good.
A sinusoid and a triangular

385
00:29:59 --> 00:30:03
wave.
And what I'm going to do is

386
00:30:03 --> 00:30:06
measure the response at this
site.

387
00:30:06 --> 00:30:08
Now, this is a vat of salt
water.

388
00:30:08 --> 00:30:13
And I'm going to tell you it
behaves like a linear system.

389
00:30:13 --> 00:30:17
If you view each little
particle, or each little

390
00:30:17 --> 00:30:22
cubic-centimeter or whatever of
water, it'll behave like little

391
00:30:22 --> 00:30:25
resistor.
So this vat of salt water

392
00:30:25 --> 00:30:30
behaves like big distributed
resistor in the following

393
00:30:30 --> 00:30:32
manner.

394
00:30:32 --> 00:30:39


395
00:30:39 --> 00:30:41
And so on.
This of this big mesh of little

396
00:30:41 --> 00:30:44
resistors, but it's all
resistors.

397
00:30:44 --> 00:30:47
It's a linear circuit.
So I'm going to apply two

398
00:30:47 --> 00:30:51
voltages, a triangular and a
sinusoid, and we're going to

399
00:30:51 --> 00:30:54
observe the output.
And what do you expect to see

400
00:30:54 --> 00:30:57
there?
You will see the superposition

401
00:30:57 --> 00:31:01
of the two, which is you'll see
a sinusoid.

402
00:31:01 --> 00:31:07
And then you'll see the jagged
triangular thing articulating

403
00:31:07 --> 00:31:13
the sinusoid pattern.
What I'm going to do right now,

404
00:31:13 --> 00:31:18
don't put any water yet.
This is the vat of nothing

405
00:31:18 --> 00:31:21
right now.
It's all empty.

406
00:31:21 --> 00:31:25
Can we show the screen on this
side?

407
00:31:25 --> 00:31:29
The oscilloscope screen?

408
00:31:29 --> 00:31:34


409
00:31:34 --> 00:31:35
OK.
Oh, there you go.

410
00:31:35 --> 00:31:38
So this is the screen of the
oscilloscope now.

411
00:31:38 --> 00:31:43
Notice that I have a sinusoid
and I have a triangular wave and

412
00:31:43 --> 00:31:46
the output is zero.
And the reason is there is

413
00:31:46 --> 00:31:49
nothing in this vat.
It's empty.

414
00:31:49 --> 00:31:53
So previously when I taught
this course I would get

415
00:31:53 --> 00:31:57
saltwater and pour saltwater.
Then we discovered a much

416
00:31:57 --> 00:32:01
better source of water that
conducted electricity like one

417
00:32:01 --> 00:32:05
real mean fluid.
Cambridge water.

418
00:32:05 --> 00:32:10
It just works very pleasantly.
It just conducts electricity

419
00:32:10 --> 00:32:14
like nothing at all.
And I've been thinking of using

420
00:32:14 --> 00:32:18
Charles River water next time
and see what happens,

421
00:32:18 --> 00:32:23
although there we'd probably
get some biological organisms

422
00:32:23 --> 00:32:26
doing strange things at you.
But go ahead.

423
00:32:26 --> 00:32:30
Our friendly demonstration
expert, Lorenzo,

424
00:32:30 --> 00:32:34
will pour some water into the
vat.

425
00:32:34 --> 00:32:39
And you should begin seeing the
output being a superposition of

426
00:32:39 --> 00:32:42
the two.
So as he pours,

427
00:32:42 --> 00:32:44
there you go,
do you see that?

428
00:32:44 --> 00:32:50
So you do see the sinusoidal
articulation and the jagged wave

429
00:32:50 --> 00:32:53
form.
And just to have some more fun,

430
00:32:53 --> 00:32:58
what I can do is increase one
of the voltages.

431
00:32:58 --> 00:33:01
And you'll see --

432
00:33:01 --> 00:33:06


433
00:33:06 --> 00:33:10
Now you know what would have
happened if I had used Charles

434
00:33:10 --> 00:33:13
River water.
So my output keeps increasing

435
00:33:13 --> 00:33:17
as I increase the corresponding
wave form.

436
00:33:17 --> 00:33:24


437
00:33:24 --> 00:33:26
I could do this,
this is fun.

438
00:33:26 --> 00:33:30
So let me pause there and go
onto the next topic.

439
00:33:30 --> 00:33:34
So that little demonstration
showed you that even something

440
00:33:34 --> 00:33:39
as simple as this physical
entity vat of water behaves like

441
00:33:39 --> 00:33:42
a linear system,
and we can model that linear

442
00:33:42 --> 00:33:46
system as a set of resistors.
Unbeknownst to you,

443
00:33:46 --> 00:33:50
right now, in the past ten
seconds I introduced a new

444
00:33:50 --> 00:33:53
concept.
It's called subliminal

445
00:33:53 --> 00:33:56
advertising.
So one of the things we do in

446
00:33:56 --> 00:34:02
EE a lot is model real systems.
So often times if I wanted to

447
00:34:02 --> 00:34:06
look at the behavior of salt,
behavior of a vat of water,

448
00:34:06 --> 00:34:11
I can model it as a set of
resistors for certain kinds of

449
00:34:11 --> 00:34:14
activities.
Just hold that thought for some

450
00:34:14 --> 00:34:17
time later in your careers.
All right.

451
00:34:17 --> 00:34:21
That's method four,
the superposition method.

452
00:34:21 --> 00:34:25
Remember, it is methods like
this that will make your life

453
00:34:25 --> 00:34:29
really, really,
really easy.

454
00:34:29 --> 00:34:34
If you find that you are having
to do a lot of grunging homework

455
00:34:34 --> 00:34:38
or something,
just step back and think

456
00:34:38 --> 00:34:41
superposition,
think Thevenin or think

457
00:34:41 --> 00:34:45
composition rule.
There must be a simpler way

458
00:34:45 --> 00:34:48
usually.
Let's do the next method.

459
00:34:48 --> 00:34:51
This is called the Thevenin
method.

460
00:34:51 --> 00:34:57
To derive this method let me
start by applying superposition

461
00:34:57 --> 00:35:03
to some circuit.
So let's say I have some

462
00:35:03 --> 00:35:09
arbitrary network N.
Assume it's a linear network

463
00:35:09 --> 00:35:15
and the network has a whole
bunch of goodies in it.

464
00:35:15 --> 00:35:22
It has a bunch of resistors,
it has a bunch of voltage

465
00:35:22 --> 00:35:30
sources, and it has a bunch of
current sources.

466
00:35:30 --> 00:35:33
Many current sources.
Many voltage sources.

467
00:35:33 --> 00:35:37
Many resistors.
Some jumbled voltage sources,

468
00:35:37 --> 00:35:43
current sources and resistors.
And I look at two nodes in this

469
00:35:43 --> 00:35:46
network.
Here are two nodes in the

470
00:35:46 --> 00:35:50
network, two points in the
network were elements connect.

471
00:35:50 --> 00:35:56
I'm looking at those two nodes
and all I want to do is the

472
00:35:56 --> 00:36:00
following.
I want to figure out if I take

473
00:36:00 --> 00:36:05
a rinky-dinky little current
source and apply it there,

474
00:36:05 --> 00:36:10
all I want to figure out is
what is V and what is I.

475
00:36:10 --> 00:36:15
There is this mongo box out
here, a black box of resistors,

476
00:36:15 --> 00:36:20
voltage source and current
sources, too many to count.

477
00:36:20 --> 00:36:24
I pick two nodes,
apply a current source,

478
00:36:24 --> 00:36:29
and all I care about is what is
the voltage that I will measure

479
00:36:29 --> 00:36:35
by applying it here.
Notice the current here will be

480
00:36:35 --> 00:36:40
I because the current here is I.
And I apply it here.

481
00:36:40 --> 00:36:43
I want to measure what the
voltage is.

482
00:36:43 --> 00:36:48
Now, with the insight you've
obtained from superposition,

483
00:36:48 --> 00:36:53
you should be able to jump up
and state the form of the

484
00:36:53 --> 00:36:57
answer.
So by superposition we know the

485
00:36:57 --> 00:37:01
following.
We know that the effect of the

486
00:37:01 --> 00:37:06
circuit will be the same as the
sum of components being added

487
00:37:06 --> 00:37:07
up.
Sum of component,

488
00:37:07 --> 00:37:11
sum of component,
a bunch of components added up.

489
00:37:11 --> 00:37:16
Each component will be the
response of one source acting

490
00:37:16 --> 00:37:18
alone.
So if I can figure out the

491
00:37:18 --> 00:37:23
effect of one source acting
alone and put that down here,

492
00:37:23 --> 00:37:28
and do the same thing for all
the sources, that's what I will

493
00:37:28 --> 00:37:32
get.
So for the source Vm it's a

494
00:37:32 --> 00:37:35
linear circuit.
So I know that my answer is

495
00:37:35 --> 00:37:40
going to be, in the final answer
is going to be a Vm term and

496
00:37:40 --> 00:37:44
it's going to be multiplied by
some alpha M term.

497
00:37:44 --> 00:37:47
I know that.
It's a linear circuit so I know

498
00:37:47 --> 00:37:52
that the answer shall have a
term Vm multiplied by some

499
00:37:52 --> 00:37:54
constant.
Simple, I know that.

500
00:37:54 --> 00:38:00
Similarly, the same is true
for, oh, this is the term Vm.

501
00:38:00 --> 00:38:05
And what I can do is I can
measure just this effect by

502
00:38:05 --> 00:38:09
setting all the other sources to
zero.

503
00:38:09 --> 00:38:15
So I can set all the other
current sources to zero and all

504
00:38:15 --> 00:38:19
voltage sources,
except for this one,

505
00:38:19 --> 00:38:23
and I can get that answer.
So, similarly,

506
00:38:23 --> 00:38:30
for every voltage source I am
going to get a term.

507
00:38:30 --> 00:38:34
So for every single voltage
source, M1, M2,

508
00:38:34 --> 00:38:39
M3 and so on I'm going to get
such a term and they're all

509
00:38:39 --> 00:38:44
going to sum up.
Similarly, I'm going to get a

510
00:38:44 --> 00:38:48
term for In.
And I know there will be an In

511
00:38:48 --> 00:38:53
term, and I know it's going to
be some constant beta

512
00:38:53 --> 00:38:57
multiplying In.
In this example of ours here,

513
00:38:57 --> 00:39:01
in this example,
remember alpha was this and

514
00:39:01 --> 00:39:08
beta was this constant here.
There's some constant beta,

515
00:39:08 --> 00:39:11
some constant alpha.
And because I have a whole

516
00:39:11 --> 00:39:16
bunch of current sources there's
going to be such a term for each

517
00:39:16 --> 00:39:19
one of them.
And each one of these terms,

518
00:39:19 --> 00:39:24
Vm, In will be the voltage I
would see here if I set all the

519
00:39:24 --> 00:39:28
other Vms to zero and I set all
the other current sources,

520
00:39:28 --> 00:39:33
except for that one to zero.
What am I missing?

521
00:39:33 --> 00:39:36
Is that it?
The response here,

522
00:39:36 --> 00:39:39
V here.
Am I missing anything here?

523
00:39:39 --> 00:39:43
Is that it?
Now, don't all yell at once.

524
00:39:43 --> 00:39:47
What am I missing?
Current source i,

525
00:39:47 --> 00:39:50
exactly.
So if I have a current source i

526
00:39:50 --> 00:39:55
then there's an effect of this
current as well.

527
00:39:55 --> 00:40:00
And so I write down i there,
too.

528
00:40:00 --> 00:40:03
It's going to be some constant
multiplying I.

529
00:40:03 --> 00:40:07
And that constant is going to
look like a resistor,

530
00:40:07 --> 00:40:11
right, because this circuit
contains current sources,

531
00:40:11 --> 00:40:15
voltage sources and resistors.
If I've shorted all my voltage

532
00:40:15 --> 00:40:19
sources and opened all my
current sources,

533
00:40:19 --> 00:40:22
what's left in here?
Just a whole caboodle full of

534
00:40:22 --> 00:40:25
Rs.
It's just going to look like

535
00:40:25 --> 00:40:30
some resistance R.
And that's what I get here.

536
00:40:30 --> 00:40:35
So this is what V is going to
look like and that's a form.

537
00:40:35 --> 00:40:40
So let's take a look at these
components.

538
00:40:40 --> 00:40:45


539
00:40:45 --> 00:40:47
Let's focus on the easy part
first.

540
00:40:47 --> 00:40:51
What does this look like?
This component looks like an I,

541
00:40:51 --> 00:40:54
it looks like a current and has
some resistance.

542
00:40:54 --> 00:40:57
What is that resistance given
by?

543
00:40:57 --> 00:41:01
Supposing I gave you this
network and this currency source

544
00:41:01 --> 00:41:05
and I asked you tell me R.
How would you measure R?

545
00:41:05 --> 00:41:09
What you would do is open all
the current sources,

546
00:41:09 --> 00:41:14
short all the voltage sources,
put a ohmmeter in there and

547
00:41:14 --> 00:41:17
measure the resistance R.
That's R.

548
00:41:17 --> 00:41:21
OK, so we understand this term.
What about this term here?

549
00:41:21 --> 00:41:25
Can someone tell me the units
of this term here,

550
00:41:25 --> 00:41:27
this big thing here?
Voltage.

551
00:41:27 --> 00:41:32
This is a voltage.
This is a voltage.

552
00:41:32 --> 00:41:37
iR is a voltage.
So this does behave like a

553
00:41:37 --> 00:41:40
voltage.
And it behaves like some

554
00:41:40 --> 00:41:45
voltage V.
So notice that as far as this

555
00:41:45 --> 00:41:52
current I is concerned the rest
of the universe looks like a

556
00:41:52 --> 00:42:00
resistor and a voltage source
behaving in some manner.

557
00:42:00 --> 00:42:04
And let me just call it Vth for
now, and you'll know why in a

558
00:42:04 --> 00:42:05
second.

559
00:42:05 --> 00:42:21


560
00:42:21 --> 00:42:26
The voltage has a form,
some voltage plus Ri.

561
00:42:26 --> 00:42:32
So, in other words,
as far as this I is concerned

562
00:42:32 --> 00:42:39
this whole network here N full
of all the nice stuff is

563
00:42:39 --> 00:42:43
indistinguishable to this I
here.

564
00:42:43 --> 00:42:50
So my I is sitting out there
injecting a current into two

565
00:42:50 --> 00:42:55
nodes.
If I am i, I'm looking at this,

566
00:42:55 --> 00:43:03
this network looks no different
than a voltage source in series

567
00:43:03 --> 00:43:11
with the resistor R.
Notice that the equation for

568
00:43:11 --> 00:43:18
this simple circuit is this,
so I is given by V minus Vth

569
00:43:18 --> 00:43:22
divided by R.
Just remember.

570
00:43:22 --> 00:43:30


571
00:43:30 --> 00:43:32
It's a circuit.
In other words,

572
00:43:32 --> 00:43:37
Agarwal sitting here cannot
tell the difference if I'm

573
00:43:37 --> 00:43:42
measuring the voltage here
between a circuit that looks

574
00:43:42 --> 00:43:47
like a Vth in series to the
resistor or this huge mess of

575
00:43:47 --> 00:43:51
voltage sources and current
sources and so on.

576
00:43:51 --> 00:43:54
Now, we will talk about Vth and
R.

577
00:43:54 --> 00:43:59
R is called the resistance of
the network as seen from the

578
00:43:59 --> 00:44:04
port with all the sources shut
off.

579
00:44:04 --> 00:44:06
And similarly Vth,
what is Vth?

580
00:44:06 --> 00:44:09
Vth is the open circuit
voltage.

581
00:44:09 --> 00:44:12
In other words,
if I apply the voltage here

582
00:44:12 --> 00:44:17
this is the response of all the
current sources and all the

583
00:44:17 --> 00:44:23
voltage sources acting together.
So it's as if I took this out

584
00:44:23 --> 00:44:27
and simply measured my V here as
if I didn't exist,

585
00:44:27 --> 00:44:31
correct?
Because this is the component

586
00:44:31 --> 00:44:35
of i.
So if I opened i and measured

587
00:44:35 --> 00:44:40
V, I would get that big term on
the left-hand side.

588
00:44:40 --> 00:44:43
That's my Vth.
So that inspires the next

589
00:44:43 --> 00:44:48
method called the Thevenin
method.

590
00:44:48 --> 00:44:58


591
00:44:58 --> 00:45:04
In this method what I'm going
to do is take some circuit,

592
00:45:04 --> 00:45:07
I'm on Page 9,
with a mess of stuff.

593
00:45:07 --> 00:45:13
It's a big mess of stuff.
And if I care to look at its

594
00:45:13 --> 00:45:20
impact on something else that I
add from the outside then as far

595
00:45:20 --> 00:45:25
as the outside world is
concerned this is

596
00:45:25 --> 00:45:32
indistinguishable from a circuit
that looks like this.

597
00:45:32 --> 00:45:45


598
00:45:45 --> 00:45:49
So what I can do is if I want
to figure out what's happening

599
00:45:49 --> 00:45:54
here then, for the purpose of my
analysis, this simple network

600
00:45:54 --> 00:45:59
here with R and Vth becomes a
surrogate for this entire mess.

601
00:45:59 --> 00:46:03
So for the purpose of finding
out the behavior at this point,

602
00:46:03 --> 00:46:07
I can take this huge mess and
replace it with its Thevenin

603
00:46:07 --> 00:46:10
surrogate or Thevenin
equivalent.

604
00:46:10 --> 00:46:14
This is called the Thevenin
equivalent of this big network.

605
00:46:14 --> 00:46:18
Let me do an example that will
make the method completely

606
00:46:18 --> 00:46:21
clear.
Again, remember in EECS,

607
00:46:21 --> 00:46:25
most of our lives are about how
can we make things so simple as

608
00:46:25 --> 00:46:30
being able to be analyzed by
inspection?

609
00:46:30 --> 00:46:36
And so this is a method that
takes you further down that

610
00:46:36 --> 00:46:40
path.
So let me use the same circuit

611
00:46:40 --> 00:46:45
that I've been using before,
my voltage V,

612
00:46:45 --> 00:46:48
R1, R2.
This is an R.

613
00:46:48 --> 00:46:55
I'm 55 minutes fast so we have
another three or four minutes.

614
00:46:55 --> 00:47:01
So this is my circuit.
And let's say all I care about

615
00:47:01 --> 00:47:05
is finding out i1.
That's all I care about.

616
00:47:05 --> 00:47:10
And what I'm going to do is I'm
going to box this up and see if

617
00:47:10 --> 00:47:14
I can replace that with its
Thevenin equivalent.

618
00:47:14 --> 00:47:17
So I'm going to box that up.

619
00:47:17 --> 00:47:35


620
00:47:35 --> 00:47:39
What I'm saying is that I'm
going to box it up and replace

621
00:47:39 --> 00:47:42
it with this Thevenin
equivalent.

622
00:47:42 --> 00:47:45
I don't know what Vth and R are
at this point.

623
00:47:45 --> 00:47:48
I'm just calling it Rth for
fun.

624
00:47:48 --> 00:47:52
I don't know what these two
values are, but if I knew what

625
00:47:52 --> 00:47:57
these two values were I can
determine I really trivially as

626
00:47:57 --> 00:48:00
follows.
I can get i1 as simply V minus

627
00:48:00 --> 00:48:06
Vth divided by R1 plus Rth.
So if I knew Vth and Rth,

628
00:48:06 --> 00:48:11
I can write down i1 by
inspection in that manner.

629
00:48:11 --> 00:48:16
So next, finally,
how do I get Vth and Rth?

630
00:48:16 --> 00:48:22
You get Rth by looking at this
network and shutting off all the

631
00:48:22 --> 00:48:29
voltage sources and measuring
the resistance there.

632
00:48:29 --> 00:48:36
So I short my voltage source,
that's R1.

633
00:48:36 --> 00:48:43
Oops, wrong way.
I need to look this way.

634
00:48:43 --> 00:48:51
So looking this way,
that's what I get.

635
00:48:51 --> 00:48:56
So what's Rth?
Rth is simply R2.

636
00:48:56 --> 00:49:05
So I have opened my current
source.

637
00:49:05 --> 00:49:09
Similarly, for Vth,
remember all I want to do is

638
00:49:09 --> 00:49:13
look at the two nodes,
step back, put a voltmeter

639
00:49:13 --> 00:49:18
there, measure the voltage,
that's my open circuit voltage.

640
00:49:18 --> 00:49:24
So the way I do it is I take
the circuit and simply measure

641
00:49:24 --> 00:49:26
the voltage there.
That's R2.

642
00:49:26 --> 00:49:32
That's my current capital I.
And I simply want to measure

643
00:49:32 --> 00:49:37
the open circuit voltage here,
which is what?

644
00:49:37 --> 00:49:44
Just simply if I stand back and
I kind of gingerly measure the

645
00:49:44 --> 00:49:50
voltage here without disturbing
anything, I simply get IR2.

646
00:49:50 --> 00:49:56
So Vth is IR2 and Rth is R2 and
here is the formula for the

647
00:49:56 --> 00:50:02
current in this branch when I
apply a voltage source and a

648
00:50:02 --> 00:50:08
resistor R1 to this little
circuit here.

649
00:50:08 --> 00:50:12
OK, let's pause and let me
summarize this in about ten

650
00:50:12 --> 00:50:14
seconds.
I had this circuit here.

651
00:50:14 --> 00:50:18
I wanted to find out i1.
So what I said I'd do is take

652
00:50:18 --> 00:50:22
this complicated mess,
well, it's not a complicated

653
00:50:22 --> 00:50:26
mess but assume it is,
and replace with it a

654
00:50:26 --> 00:50:31
resistance Rth got by turning
off all the sources.

655
00:50:31 --> 00:50:35
And the voltage in series,
Vth, which I get simply by

656
00:50:35 --> 00:50:38
pulling this thing out,
taking my input,

657
00:50:38 --> 00:50:42
this part out and simply
measuring the open circuit

658
00:50:42 --> 00:50:43
voltage out there,
Vth.

659
00:50:43 --> 00:50:48
And then I replaced the whole
network with this new network

660
00:50:48 --> 00:50:51
that they call the Thevenin
network, and voila,

661
00:50:51 --> 50:54
I get the answer in a second.