Let's get started.
Can you hear me back there? Loud and clear.
OK. Let's get started.
Before I begin, just a couple of announcements.
Brad Buren is one of our students here and he needs a
note-taker. It's a paid position.
So if you are interested you can stop by after class and see
him. He's sitting right here out
there, OK? Second, just a reminder that
6.002 does have prerequisites. And the prerequisites are 8.02
and 18.03. So with that let me start off
with the usual. Do a quick review of what we've
done so far. So we started out life looking
at the laws of physics and Maxwell's equations and so on.
And those were way too hard so we said let's make life easy for
ourselves. So we chose to play in this
playground in which we said we shall adhere to the lumped
matter discipline. OK?
The LMD. So we are in that playground.
So this entire course, and for that matter large parts
of EECS are within that playground, within which the
lumped matter discipline applies.
So as soon as we jumped into the playground,
the LMD playground, we could take Maxwell's
equations and abstract them out into two very,
very simple rules. And the very simple rules were
KVL and KCL. KVL simply said that I can sum
the voltages in any loop in a circuit and the result then
would be zero. Similarly, I can sum the
currents that enter or exit any node and the sum will also be
zero. So what you can now do is,
if you feel like, you can go around and brag.
Oh, yeah, we use Maxwell's equations in everyday life and,
yeah, it's good stuff. And the key is that this is
really an encapsulation of Maxwell's equations within this
playground that we are in. So I talked about the first
method of circuit analysis in the last lecture.
And that method simply took the, wrote KVL for all the
loops, wrote KCL for all the nodes and wrote element vi
relationships. And together gave you a big
bunch of equations. And you sat down and grunged
through the equations and you solved for branch voltages and
currents. So we reviewed a second method
of circuit analysis. And I'll simply call it circuit
composition. The basic idea behind this
method was to learn some simple rules of how resistors add and
conductances add and so on and so forth and look at a circuit
and simplify the circuit by making series simplifications
when the resistors are in series and so on and so forth,
and compose it and play around with it till we end up with the
current, the voltages that we are looking for.
This is the intuitive method. And so a section in Chapter 2,
I believe, of the course notes discusses several examples using
this method and attempts to make a little bit formal the
intuitive approach that is applied in this method.
So we then looked at the node method.
And the node method was simply a particular way of applying KVL
and KCL. Node method,
remember? We took a ground node.
Then we labeled the nodes of the remaining voltages with
respect to that ground. Then we wrote KCL for each of
the nodes. And when we wrote KCL for each
of the nodes, remember, KVL was implicit in
this expression that we used for each of the currents that were
exiting each node. So if Ej was a node voltage,
then Ej minus Ei multiplied by the conductance Gi was the
current that was going through one of those,
I should call it Gij. This is a conductance that
connects nodes i and j. That gave us the KVL that fed
into the same system. So these are three methods.
The node method, by the way, is sort of the
workhorse of the 6.002 industry. And for that matter for all of
the circuits industry. When in doubt,
apply the mode method, you'll be OK.
That applies to linear circuits, nonlinear circuits,
what have you. What I'm going to do today is
go through two more methods. So notice that the first few
lectures of this course, the first three lectures simply
comprise transitioning you from the world of physics to the
world of EECS. And then two lectures on giving
you a bag of tricks. So we start you off with the
sort of tools, your mallets and chisels and so
on and so forth. And these five methods are your
tools. We'll look at two methods
today. One method is called the method
of superposition and the second method is called the Thevenin
method. And these methods apply only to
linear circuits. So we look at the subset of
circuits that are linear, and these two methods apply to
only those circuits. These are methods that combined
with intuition really enables you to solve very interesting
circuits very, very quickly.
So let me do an example using a usual node method.
And then jump into introducing the superposition methods and
Thevenin methods using that same example.
So let me draw you an example circuit here.
So, again, I'm using this example, I will use this example
to introduce the method of superposition and the Thevenin
method. So what I'm going to do is
start off the usual way and analyze the circuit using a
method that you know now, the node method.
And what I'll do is write down the node equations for this by
applying the node method. So if you recall the node
method. I choose a ground node.
I'm going to choose this node. It's got both the voltage
source connected to it, and it's also got many other
edges impinging on it. So I'm going to choose that as
my ground node and I'm going to label the other nodes with their
voltages. So this is an unknown.
I'll label it as e. I guess we just have one
unknown e. And I know the voltage of this
node, and that is simply V. Since it's V,
there's a voltage source between the ground node and that
node. So what I can do next is that I
can write down the node equation for this node and then go from
there. So let me go ahead and do that.
So let me sum up the currents going outside,
going outwards. So I have e minus v divide by
R1, I have e minus zero divide by R2, and I have minus i equals
zero. This is a node equation.
The first thing I want you to observe is that this equation is
linear in V and i. What I mean by linear is that
you don't see terms like Vi or V-squared and things like that.
It's some constant times V plus some constant times i equals
some other constant. So that's quite nice.
So I'm going to rearrange the terms in the following manner.
I'll move the known sources to the right-hand side and collect
the coefficients of e on this side, so I get one by R1 plus
one by R2 over here.
So stare at this for a moment and notice again here I have e,
my unknown node voltage, there is some constant
multiplier, and that equals some function of V summed up with
some function of i. And, again, notice that this is
a linear combination of V and i. No multiplication terms and so
on and so forth. This is a pretty standard form
in which we will represent equations quite often.
And just to label it, this is often labeled G as the
conductance matrix. Of course this is e,
our unknown node voltages, and this is a linear sum of
sources. So this is a very standard way
that we will represent equations.
We did that last week as well, or rather on Tuesday where I
took a conductance matrix, multiplied that by a column
vector of unknown node voltages and equated that to some linear
combination of my source voltages.
The reason the circuit is linear is that I have only
linear elements in the circuit. I don't have any nonlinear
elements. And because of that I can
rewrite this in the following manner.
I'm just going to express e as a function of V and i and bring
it over to this side. So it's some function of i.
So I get R1 R2 divide by R1 plus R2.
And I bring R1 R2 to this side. That's what I get.
So stare at this for a few seconds, very common form.
My unknown node voltage is equal to this stuff on the
right-hand side. The stuff on the right-hand
side has a term multiplying the source voltage V and some other
term multiplying the current I. And if I were to put this in
sort of symbol-like form my unknown node voltage is some
constant times V1 plus some constant times,
is of the form constant times the source current,
constant times the source voltage and so on.
The units of As and Vs are different because in this case A
has no units because V is a voltage.
And so is e. In this case V has units of
resistance. So that V times i gives me a
voltage. So stare at this equation for a
few seconds and this should help us build up some insight that
will allow us to write down the answer almost by inspection.
I'm going to show you a method now, in a few minutes,
which will allow you to write down the answer e just by
starring at the circuit without having to go through node
equations and so on. The more and more methods I
teach you, the more you will be able to do a lot of this
completely by yourselves. In this particular example it's
a relatively simple circuit but these methods would be
particularly useful when you have more complicated
situations. But before I go on let me spend
a few minutes pontificating on linearity.
So that's a linear circuit. And this equation gives me the
unknown node voltage e as a linear sum of source voltages
and source currents. Linearity implies two
properties, the property of homogeneity and also gives vice
to the property of superposition.
Let's do homogeneity first. What this says is if I have a
circuit, some circuit and I feed it some sort of inputs,
A, then let's say my output is S.
If you're feeling hungry think of these as apples and the
circuit converts them into applesauce.
So what homogeneity says is that what I can do is if I take
each of my apples and instead of feeding it an entire apple what
if I give it three-quarters of an apple?
Say I multiple all my inputs by some constant alpha,
three-quarters. What that says is that at the
output instead of getting one full bottle of applesauce I'm
going to get three-quarters of a bottle of apple sauce.
So if I proportionately reduce all the inputs and if this is a
linear circuit then so shall my output be reduced in the same
proportion. So that's homogeneity.
Next, let's look at superposition.
The property of superposition says the following.
The same kind of circuit. If I feed it apples then I get
applesauce. I take the same circuit,
and this time around if I feed the circuit a different set of
inputs, say blueberries. And let's say my output,
oops, let me do it this way. So as my output I get blueberry
sauce, if such exists. So apples applesauce,
blueberries give me blueberry sauce.
Then what I'm going to get if I mix up the two,
so let's say I take my circuit, the same circuit with a set of
inputs and in this example one output.
Let's say I mix up my inputs and some of my inputs in the
following way, here I feed an A1 plus B1 and
here A2 plus B2 and so on then at the output I am going to get
a mush of apple sauce and blueberry sauce.
All this says is that if I apply just apples I get
applesauce. If I apply just blueberries I
get blueberry sauce. Then if I were to figure out
how this blender would have worked had I fed in the
combinations of apples and blueberries, then for the
purposes of understanding that blender all I could have done
was taken by two outputs and just mixed them up together
myself and that's exactly what I'd get.
So if I sum up the inputs my outputs would also be the sum of
the outputs with the inputs applied by themselves.
So let me take this here and munge around with hit for a few
seconds and get something interesting out of it.
So notice two inputs, two inputs, outputs.
In your notes I've given you another template for the next
set of scribbles I'm going to make here.
So use the next set of templates on page three.
What I'm going to do here is something very simple,
set one output to zero and feed a voltage V1.
So that's feed a voltage V1 and set the other output to zero.
And let's say I get Y1 as an output.
And in this case I set the first voltage to zero and feed a
different voltage V2 on the second input.
And let's say my output is Y2. This is just a particular
application of the superposition principle I just outlined.
Apply V1 set one output to zero.
Apply V2 set the original output to zero.
Then what I'm going to find is that the answer will simply look
like this, just replace for As and Bs what I just did and we
get V1 and zero here and we get zero and V2 here.
And as my output I'm going to get exactly the sum Y1 plus Y2.
This is simply a particular application of superposition
where what I'm saying is the following.
If you look at this circuit here effectively what have I
done? Effectively what I've done is
apply the voltage V1 on one input and a voltage V2 on the
other input. V1 here.
V2 here. And the output is Y1 plus Y2.
What I'm saying is look backwards now.
What I'm saying is that the whole components of the output
Y1 plus Y2 could individually be derived in the following manner.
I could get the component Y1 by simply applying one of the
voltages and setting the other to zero.
I can get the other component Y2 by setting yet another input
to zero and applying the voltage V2 to get Y2.
And sum then up and that's my answer.
This will become a lot clearer with an example.
Again, remember if I have a bunch of inputs applied to a
circuit, V1, V2 and so on, and I get some output then what
this is saying is that I can alternatively find out the
answer by applying just one voltage, setting all the others
to zero, measuring the output, apply a second voltage,
set all inputs to zero, measure the output and sum of
applesauce and blueberry sauce and there you get the answer.
Let's do an example. And before we go into that I
talked about setting voltage sources and current sources to
zero. First of all,
what does it mean to set a voltage source to zero?
This is the same as this. Setting a voltage source to
zero is simply replacing the voltage source with a short,
and setting a current source to zero simply implies an open
circuit. So when I say zero that source,
if it's a voltage source short it, if it's a current source
open it.
I can take any two nodes in the world and measure the potential
difference across them. So there may be some potential
difference across these set by the circuit that I haven't shown
you on this side. There might be some other
circuit that is controlling the voltage of these two nodes.
The same with the short. What's V going to be?
But there is a V. It's zero.
So that's method four, method of superposition.
And this method says that the output of a circuit --
Again, remember I'm focusing on linear circuits.
Remember, I have this playground where LMD applies.
And within that playground I'm playing in the south goal area.
In the south goal area, in that subset of the
playground circuits are linear. So in that part of the
playground superposition applies because there circuits are
linear. So the output of a circuit is
determined by summing up the responses to each source acting
alone.
Now, in this statement here this source stands for
independent source. I haven't talked about
independent versus dependent sources.
We'll talk about dependent sources a few weeks from today.
And just so you don't get confused, for dependent sources
you will be looking at Section 3.3.3 of your course notes to
see how superposition works with dependent sources.
But remember we haven't covered dependent sources yet.
We will be covering them about two weeks from now.
So let's go back to our example and apply the method of
superposition to an example. So the method says sum up the
outputs of each of the sub-circuits where I'm applying
one source acting alone. So let me just do this here.
Let me start with the circuit. And let me start with shutting
I off. So I have voltage V --
I have R2. And I'm shutting I off.
So I have replaced this with an open circuit.
So I is zero. Let me call the node voltage eV
to reflect that component of the node voltage that arises due to
V acting alone. And you should look at this
pattern here and very quickly be able to write the answer for
patterns like this voltage, the two resistors.
That's called a resistive divider.
It will appear again and again and again.
And eV is simply V times R2 divided by R1 plus R2.
That's still my ground node. So the voltage here is simply
this voltage divided by the two resistors to give you the
current multiplied by R2 to give you the voltage across this R.
Remember this pattern. You apply voltage divider
patterns probably more times than any other pattern that you
might imagine. So that's with the V acting
alone. Now, let me do I acting alone.
So for I acting alone --
And what I do this time around is replace this with a short,
replace the voltage source to the short.
And let me call this voltage eI for the component of the voltage
due to the current I. And eI, in this case,
is simply given by yet another pattern here,
the current across a pair or resistors is simply the
effective resistance multiplied by the current so it's i and the
effective resistance is R1, R2 or R1 plus R2.
That's eI. That's a component that node
due to the current I. Now, so the method says that.
Then take these components, sum them up and there you have
the answer. So E is simply ev plus ei.
The components of V and I acting alone,
just simply V times R2 divided by R1 plus R2 plus R1,
R2. There we go.
Fortunately, the fates have been kind to us
and the answer is the same as the answer we obtained with the
node method. No surprise here.
So this is actually an incredibly simple method.
So you can take a very complex circuit.
What have you really done here? You can take a very complex
circuit and you can solve a very complex circuit by breaking it
down into many simple individual sub problems.
You will do this in EECS time and time and time again.
Whether it's in software systems or hardware systems or
what have you, you're often times building
complicated systems. Remember doom on this side?
And the way and when you put these things together,
let's say a large software system, is you don't write the
whole piece of software starting main and grunge down.
You build a lot of little components and tie the
components together. In the same manner here you
take a big circuit and you find its behavior for each source
acting alone. Lots of little inky dinky
simple little circuits. And you will see examples in
your homework where you're given a big circuit or because it set
all the Is to zero and the other Vs to zero the whole circuit
almost vanishes and all that you're left with is a little
resistor or two. So this is the very,
very powerful method. I'd like to do a little
demonstration for you. And what I'm going to show you
is the demo is a vat of water. Actually, I'll tell you what it
is in a second. But assume it is salt water for
now. I'll apply two voltages.
In this case I'm going to apply a sinusoid.
That's not very good. A sinusoid and a triangular
wave. And what I'm going to do is
measure the response at this site.
Now, this is a vat of salt water.
And I'm going to tell you it behaves like a linear system.
If you view each little particle, or each little
cubic-centimeter or whatever of water, it'll behave like little
resistor. So this vat of salt water
behaves like big distributed resistor in the following
manner.
And so on. This of this big mesh of little
resistors, but it's all resistors.
It's a linear circuit. So I'm going to apply two
voltages, a triangular and a sinusoid, and we're going to
observe the output. And what do you expect to see
there? You will see the superposition
of the two, which is you'll see a sinusoid.
And then you'll see the jagged triangular thing articulating
the sinusoid pattern. What I'm going to do right now,
don't put any water yet. This is the vat of nothing
right now. It's all empty.
Can we show the screen on this side?
The oscilloscope screen?
OK. Oh, there you go.
So this is the screen of the oscilloscope now.
Notice that I have a sinusoid and I have a triangular wave and
the output is zero. And the reason is there is
nothing in this vat. It's empty.
So previously when I taught this course I would get
saltwater and pour saltwater. Then we discovered a much
better source of water that conducted electricity like one
real mean fluid. Cambridge water.
It just works very pleasantly. It just conducts electricity
like nothing at all. And I've been thinking of using
Charles River water next time and see what happens,
although there we'd probably get some biological organisms
doing strange things at you. But go ahead.
Our friendly demonstration expert, Lorenzo,
will pour some water into the vat.
And you should begin seeing the output being a superposition of
the two. So as he pours,
there you go, do you see that?
So you do see the sinusoidal articulation and the jagged wave
form. And just to have some more fun,
what I can do is increase one of the voltages.
And you'll see --
Now you know what would have happened if I had used Charles
River water. So my output keeps increasing
as I increase the corresponding wave form.
I could do this, this is fun.
So let me pause there and go onto the next topic.
So that little demonstration showed you that even something
as simple as this physical entity vat of water behaves like
a linear system, and we can model that linear
system as a set of resistors. Unbeknownst to you,
right now, in the past ten seconds I introduced a new
concept. It's called subliminal
advertising. So one of the things we do in
EE a lot is model real systems. So often times if I wanted to
look at the behavior of salt, behavior of a vat of water,
I can model it as a set of resistors for certain kinds of
activities. Just hold that thought for some
time later in your careers. All right.
That's method four, the superposition method.
Remember, it is methods like this that will make your life
really, really, really easy.
If you find that you are having to do a lot of grunging homework
or something, just step back and think
superposition, think Thevenin or think
composition rule. There must be a simpler way
usually. Let's do the next method.
This is called the Thevenin method.
To derive this method let me start by applying superposition
to some circuit. So let's say I have some
arbitrary network N. Assume it's a linear network
and the network has a whole bunch of goodies in it.
It has a bunch of resistors, it has a bunch of voltage
sources, and it has a bunch of current sources.
Many current sources. Many voltage sources.
Many resistors. Some jumbled voltage sources,
current sources and resistors. And I look at two nodes in this
network. Here are two nodes in the
network, two points in the network were elements connect.
I'm looking at those two nodes and all I want to do is the
following. I want to figure out if I take
a rinky-dinky little current source and apply it there,
all I want to figure out is what is V and what is I.
There is this mongo box out here, a black box of resistors,
voltage source and current sources, too many to count.
I pick two nodes, apply a current source,
and all I care about is what is the voltage that I will measure
by applying it here. Notice the current here will be
I because the current here is I. And I apply it here.
I want to measure what the voltage is.
Now, with the insight you've obtained from superposition,
you should be able to jump up and state the form of the
answer. So by superposition we know the
following. We know that the effect of the
circuit will be the same as the sum of components being added
up. Sum of component,
sum of component, a bunch of components added up.
Each component will be the response of one source acting
alone. So if I can figure out the
effect of one source acting alone and put that down here,
and do the same thing for all the sources, that's what I will
get. So for the source Vm it's a
linear circuit. So I know that my answer is
going to be, in the final answer is going to be a Vm term and
it's going to be multiplied by some alpha M term.
I know that. It's a linear circuit so I know
that the answer shall have a term Vm multiplied by some
constant. Simple, I know that.
Similarly, the same is true for, oh, this is the term Vm.
And what I can do is I can measure just this effect by
setting all the other sources to zero.
So I can set all the other current sources to zero and all
voltage sources, except for this one,
and I can get that answer. So, similarly,
for every voltage source I am going to get a term.
So for every single voltage source, M1, M2,
M3 and so on I'm going to get such a term and they're all
going to sum up. Similarly, I'm going to get a
term for In. And I know there will be an In
term, and I know it's going to be some constant beta
multiplying In. In this example of ours here,
in this example, remember alpha was this and
beta was this constant here. There's some constant beta,
some constant alpha. And because I have a whole
bunch of current sources there's going to be such a term for each
one of them. And each one of these terms,
Vm, In will be the voltage I would see here if I set all the
other Vms to zero and I set all the other current sources,
except for that one to zero. What am I missing?
Is that it? The response here,
V here. Am I missing anything here?
Is that it? Now, don't all yell at once.
What am I missing? Current source i,
exactly. So if I have a current source i
then there's an effect of this current as well.
And so I write down i there, too.
It's going to be some constant multiplying I.
And that constant is going to look like a resistor,
right, because this circuit contains current sources,
voltage sources and resistors. If I've shorted all my voltage
sources and opened all my current sources,
what's left in here? Just a whole caboodle full of
Rs. It's just going to look like
some resistance R. And that's what I get here.
So this is what V is going to look like and that's a form.
So let's take a look at these components.
Let's focus on the easy part first.
What does this look like? This component looks like an I,
it looks like a current and has some resistance.
What is that resistance given by?
Supposing I gave you this network and this currency source
and I asked you tell me R. How would you measure R?
What you would do is open all the current sources,
short all the voltage sources, put a ohmmeter in there and
measure the resistance R. That's R.
OK, so we understand this term. What about this term here?
Can someone tell me the units of this term here,
this big thing here? Voltage.
This is a voltage. This is a voltage.
iR is a voltage. So this does behave like a
voltage. And it behaves like some
voltage V. So notice that as far as this
current I is concerned the rest of the universe looks like a
resistor and a voltage source behaving in some manner.
And let me just call it Vth for now, and you'll know why in a
second.
The voltage has a form, some voltage plus Ri.
So, in other words, as far as this I is concerned
this whole network here N full of all the nice stuff is
indistinguishable to this I here.
So my I is sitting out there injecting a current into two
nodes. If I am i, I'm looking at this,
this network looks no different than a voltage source in series
with the resistor R. Notice that the equation for
this simple circuit is this, so I is given by V minus Vth
divided by R. Just remember.
It's a circuit. In other words,
Agarwal sitting here cannot tell the difference if I'm
measuring the voltage here between a circuit that looks
like a Vth in series to the resistor or this huge mess of
voltage sources and current sources and so on.
Now, we will talk about Vth and R.
R is called the resistance of the network as seen from the
port with all the sources shut off.
And similarly Vth, what is Vth?
Vth is the open circuit voltage.
In other words, if I apply the voltage here
this is the response of all the current sources and all the
voltage sources acting together. So it's as if I took this out
and simply measured my V here as if I didn't exist,
correct? Because this is the component
of i. So if I opened i and measured
V, I would get that big term on the left-hand side.
That's my Vth. So that inspires the next
method called the Thevenin method.
In this method what I'm going to do is take some circuit,
I'm on Page 9, with a mess of stuff.
It's a big mess of stuff. And if I care to look at its
impact on something else that I add from the outside then as far
as the outside world is concerned this is
indistinguishable from a circuit that looks like this.
So what I can do is if I want to figure out what's happening
here then, for the purpose of my analysis, this simple network
here with R and Vth becomes a surrogate for this entire mess.
So for the purpose of finding out the behavior at this point,
I can take this huge mess and replace it with its Thevenin
surrogate or Thevenin equivalent.
This is called the Thevenin equivalent of this big network.
Let me do an example that will make the method completely
clear. Again, remember in EECS,
most of our lives are about how can we make things so simple as
being able to be analyzed by inspection?
And so this is a method that takes you further down that
path. So let me use the same circuit
that I've been using before, my voltage V,
R1, R2. This is an R.
I'm 55 minutes fast so we have another three or four minutes.
So this is my circuit. And let's say all I care about
is finding out i1. That's all I care about.
And what I'm going to do is I'm going to box this up and see if
I can replace that with its Thevenin equivalent.
So I'm going to box that up.
What I'm saying is that I'm going to box it up and replace
it with this Thevenin equivalent.
I don't know what Vth and R are at this point.
I'm just calling it Rth for fun.
I don't know what these two values are, but if I knew what
these two values were I can determine I really trivially as
follows. I can get i1 as simply V minus
Vth divided by R1 plus Rth. So if I knew Vth and Rth,
I can write down i1 by inspection in that manner.
So next, finally, how do I get Vth and Rth?
You get Rth by looking at this network and shutting off all the
voltage sources and measuring the resistance there.
So I short my voltage source, that's R1.
Oops, wrong way. I need to look this way.
So looking this way, that's what I get.
So what's Rth? Rth is simply R2.
So I have opened my current source.
Similarly, for Vth, remember all I want to do is
look at the two nodes, step back, put a voltmeter
there, measure the voltage, that's my open circuit voltage.
So the way I do it is I take the circuit and simply measure
the voltage there. That's R2.
That's my current capital I. And I simply want to measure
the open circuit voltage here, which is what?
Just simply if I stand back and I kind of gingerly measure the
voltage here without disturbing anything, I simply get IR2.
So Vth is IR2 and Rth is R2 and here is the formula for the
current in this branch when I apply a voltage source and a
resistor R1 to this little circuit here.
OK, let's pause and let me summarize this in about ten
seconds. I had this circuit here.
I wanted to find out i1. So what I said I'd do is take
this complicated mess, well, it's not a complicated
mess but assume it is, and replace with it a
resistance Rth got by turning off all the sources.
And the voltage in series, Vth, which I get simply by
pulling this thing out, taking my input,
this part out and simply measuring the open circuit
voltage out there, Vth.
And then I replaced the whole network with this new network
that they call the Thevenin network, and voila,
I get the answer in a second.