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All right.
Good morning,
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all.
So we take another big step
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forward today and get onto a new
plane of understanding,
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if you will.
In the last week and a half,
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our focus was on the storage
element or storage elements
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called inductors and capacitors.
And capacitors stored change
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and inductors essentially stored
energy in the field,
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the magnetic flux.
And the state variable for an
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inductor was the current while
that for a capacitor was the
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capacitor voltage.
We also looked at circuits
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containing a single storage
element, we looked at RC
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circuits and we also looked at
circuits containing a single
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inductor.
And this was a single inductor
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with a resistor and a current
source or a voltage source and
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so on.
What we are going to do today
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is do what are called
"second-order systems".
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So they are on the next plane
now.
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And with this second-order of
systems, they are characterized
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by circuits containing two
independent storage elements.
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They could be an inductor and a
capacitor or two independent
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capacitors.
And you will see towards the
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end what I mean by two
independent capacitors.
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If I have two capacitors in
parallel, they can be
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represented as a single
equivalent capacitor so that
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doesn't count.
It has to be two independent
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energy storage elements and
resistors and voltage sources
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and so on.
And what we end up getting is
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what is called "second-order
dynamics".
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And much as first order
circuits were represented using
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first order differential
equations, this kind you end up
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getting second-order
differential equations.
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Before we go into this,
I would like to start
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motivating this and give you one
example of why this is important
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to study.
There are many,
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many examples but I will give
you one.
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What I would like to do is draw
your attention to our good old
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inverter driving a second
inverter.
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The same circuit that we used
to motivate RC studies,
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one inverter driving another.
So let me draw the circuit.
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Here is one inverter.
This is, let's say,
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5 volts and this is,
let's say, 2 kilo ohms.
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And I connect the output of
this inverter to a second
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inverter.
And what we saw in the last few
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lectures was that in this
specific example there was a
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parasitic capacitor or a
capacitor associated with the
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gate of this MOSFET.
And that could be modeled by
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sticking a capacitor CGS between
the gate of the MOSFET and
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ground.
And we saw that the waveforms
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here, if I had some kind of step
here.
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Let's say, for example,
a step that went from high to
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low.
Then out here I would have a
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transition that instead of going
up rapidly like this would
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transition a little bit more
slowly.
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And this transition was
characterized by an RC time
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constant.
And this is what gave rise to a
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delay in the eventual output.
So that is what we saw
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previously, single energy
storage element.
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Today what we are going to do
is we are going to look at the
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same circuit,
the exact same circuit,
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and have some fun with it.
What we are going to say is
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look, this thing is pretty slow,
so what I would like to do is
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-- why don't we go ahead and put
that up.
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What we are going to see is
that the yellow waveform is the
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waveform at the input here.
And the green waveform here is
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the waveform at this
intermediate node.
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And notice that this waveform
here is characterized by the
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slowly rising characteristics
that are typical of an RC
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circuit.
There are some other
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weirdnesses and so on going on
here like a little bump and
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stuff like that.
You can ignore all of that for
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now.
It happens because of certain
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other very subtle circuit
effects that you won't be
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dealing with,
called Miller effects and so on
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that you won't be dealing with
in 6.002.
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So focus then on this part
here.
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It is pretty slow.
And because of that slow
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rising, I get a very slow
transition and I get some delay
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in my inverter.
So you say ah-ha,
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we learned about this in 6.002,
I can make it go faster.
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How can you make the circuit go
faster?
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What could you do?
This is rising very slowly.
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How can you make it go faster?
Anybody?
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You have multiple choices,
actually.
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What are your choices here?
Pardon.
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Decrease the time constant.
And how would you decrease the
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time constant?
The capacitance is connected to
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this MOSFET gate here.
I didn't want it in the first
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place but it is there,
I cannot help it,
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so I can decrease the
resistance.
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Good.
Let me go ahead and do that.
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What I will do is I am going to
knock this sucker out and stick
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in a new resistance that is say
50 ohms, a much smaller
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resistance.
That should speed things up,
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right?
That should make things go much
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faster because this is a smaller
time constant because R is
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smaller, correct?
OK, let's go do it.
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And let's see if we get what we
expect.
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I have a little switch here.
And using that switch,
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I am going to switch in this
little resistance.
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Whoa, what on earth is
happening out there?
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This is so much fun.
What I did is I switched in a
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small resister here to decrease
the time constant,
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but it looks like I got a whole
bunch of crapola that I did not
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bargain for.
This is certainly very fast,
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it goes up really fast,
but I am not sure where it is
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going, though.
Let's stare at that a little
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while longer.
Let me expand the time scale
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for you.
Look at this.
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Instead of a nice little smooth
thing going up.
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I get something that looks like
this.
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It looks something like a
sinusoid.
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It looks sinusoidal,
but then it is a sinusoid that
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kind of gives up and kind of
gets tired and kind of goes
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away.
Right?
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It kind of dies out.
So nothing that you have
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learned so far has prepared you
for this.
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And, trust me,
when I first did some circuit
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designs myself a long,
long time ago I got nailed by
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that.
I looked at my circuit,
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and what ended up happening was
I was noticing these sharp lines
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at all my transitions.
When I looked at my scope,
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I expected to see nice little
square waves but I saw these
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little nasty spikes sitting out
there.
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And then when I stared at it
more carefully,
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those spikes were really
sinusoids that seemed to kind of
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get tired and kind of go away.
So those are nasty,
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those are real and they happen
all the time.
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And what we will do today is
try to get into that and
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understand why that is the case.
We will understand how to
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design that away.
And that is a real problem,
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by the way.
And the reason that is a real
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problem is the following.
Look at this.
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Look down here.
Because this intermediate
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voltage is meandering all over
the countryside here,
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at this particular point the
intermediate voltage dips quite
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low.
And because it dips quite low
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look at the output.
The output has a bump here.
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And it is quite possible for
this output bump to now go into
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the forbidden region.
Or worse.
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If this swing here was higher,
this could have actually gone
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onto a one, so I would have
gotten a false one pulse here.
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Instead of having a nice one to
zero transition,
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I would have gotten a one to
zero, oh, back to one,
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oh, back to zero and then back
down to zero.
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So this is nasty stuff,
really, really nasty stuff.
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What we will do is understand
why that is the case today and
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see if we can explain it.
What is going on here?
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What is really going on here is
take a look at this circuit
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here.
I will take a look at this path
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here.
So this is your VS voltage
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source.
Path kind of goes like this and
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around.
It turns out that this circuit
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is a loop here.
And when there is current flow,
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going down to basic physics you
remember that I also enclose
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some amount.
So there is a current flowing
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in a loop.
And because of that there is an
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effective inductance here.
And, in fact,
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any current flowing through a
wire above a ground plane,
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for that matter,
can be characterized by the
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inductance.
So I can model that by sticking
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a little inductor here.
So my real circuit is not
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exactly a resistor and a
capacitor, but my real circuit
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is an inductor as well that
comes into play because of this
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wire.
Every wire, when there is a
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current flow,
has an inductance associated
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with it.
And because of that the real
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circuit is resistor,
inductor and capacitor.
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So I end up with two storage
elements now,
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and the dynamics of that are
very different from that with a
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single storage element.
That is just a bit of
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motivation for why our study of
inductors is important.
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And I can draw a quick circuit
here.
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If you look at the circuit,
start from ground,
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the voltage VS and there is a
resistor here.
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And then I have an inductor and
then I have a capacitor.
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So it is a voltage source,
resistor, inductor and
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capacitor.
For this whole week we will be
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looking at circuits like this.
Today what I would like to do
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is start very simple,
start with the simplest
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possible form of this so that
you can begin building up your
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insight and then go into more
complicated cases.
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Today what I will do is simply
begin with a case where I don't
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have a resistor here and simply
study a voltage source,
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an inductor and a capacitor and
understand what the voltage
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looks like out here.
So we look at the dynamics of a
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little system like this.
Before we go on,
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I want to caution you about
something.
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It is just happenstance that I
have introduced for you
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capacitors based on the
parasitic capacitance here and
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inductance based on parasitic
inductance.
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I would hate to leave you with
the impression that inductors
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and capacitors are "bad".
Because when you think of a
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parasitic, you know,
parasites.
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These are parasitic.
You didn't expect them there,
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didn't expect this here and we
got the weird behavior.
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So parasitics have a bad
connotation to them.
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I do not want to leave you with
a bad taste in your mouth about
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capacitors and inductors that
these are just bad things.
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We just have to deal with them
and deal with second-order
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differential equations and all
that stuff because they're just
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bad stuff and we just have to
deal with them.
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I don't want you to end up
going through life hating
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capacitors and inductors.
Just because of my choice of
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examples, it just happened to be
introducing them as capacitors.
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I want to point out that these
are fundamental lumped elements
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in their own right.
They are very,
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incredibly important and useful
circuits where we designed
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capacitors and inductors because
we want to have them in there.
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There are many circuits that we
will look at where we really
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want the inductor in there.
We will design an inductor by
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wrapping wire around in a coil
and get bigger inductances and
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so.
Just remember that this can be
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parasitic in some cases,
but in many cases it's good,
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inductors are good,
so just stick with that
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thought.
These are mostly good so don't
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go around hating them.
All right.
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Let's go on and analyze a basic
circuit like this.
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And what I would like to cover
in the next hour are the
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foundations of something like
that.
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I will take you through the
foundations so you understand
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how it works.
And, as always,
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what I am going to end up with
is build up the foundations,
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help you understand why we got
where we were and then help you
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build intuition.
And then show you a really,
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really simple intuitive way of
doing things in terms of how
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experts do it.
And the real cool thing about
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EECS is that the way experts do
things, things are really,
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really very simple in the end.
But you need to build up some
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intuition to get there.
So our circuit looks like this
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in terms of my two storage
elements.
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I have a voltage vI,
inductor L, capacitor C and I
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am going to look at the voltage
across the capacitor and my
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current through the capacitor.
So v(t) is the voltage across
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the capacitor and my current is
the current through this loop
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here, which is the same as the
current through the capacitor or
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the current through the
inductor.
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And we are going to proceed in
exactly the same manner as we
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did for first order differential
equations, write the equations
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down and just boom,
boom, boom, boom,
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go down the same sets of steps
but just get to some place
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different.
We are going to start by
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writing a node equation for this
node here.
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That's the only node for which
I have an unknown voltage.
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The node here is vI,
so I need to find this,
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there's just one unknown node
voltage.
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And I am going to need some
element laws.
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For the capacitor I know the iV
relation is given by the i for
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the capacitor is Cdv/dt.
And just to show the capacitor
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I am just calling it dvc/dt.
Similarly, for an inductor,
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L, the voltage across the
inductor is given by Ldi/dt.
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So this is the vI relation for
the capacitor,
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the vI relation for an
inductor.
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It also suits us to write this
in an integral form.
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So if I integrate both sides of
this equation and I bring L down
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to this side,
I end up getting something like
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this, 1/L minus infinity to t,
VLdt, and that is simply iL.
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I am just simply replacing this
with an integral form.
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So this is a VI relationship
for the inductor and this is for
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the capacitor.
So let me now go ahead and
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apply the node method for my
circuit here.
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00:17:16 --> 00:17:21
Here, for the node method,
I have to equate the currents
265
00:17:21 --> 00:17:26
coming into the node or sum the
currents coming into the node
266
00:17:26 --> 00:17:32
and equate that to zero.
And while I do that I simply
267
00:17:32 --> 00:17:37
replace the currents by the
corresponding voltages using the
268
00:17:37 --> 00:17:40
element laws.
So what do I get?
269
00:17:40 --> 00:17:45
I get the current going in here
to the inductor is equal to the
270
00:17:45 --> 00:17:48
current going through the
capacitor.
271
00:17:48 --> 00:17:51
What is the current going the
capacitor?
272
00:17:51 --> 00:17:55
In terms of its v relationship
it is Cdv/dt.
273
00:17:55 --> 00:17:59
And the current going to the
inductor is given by this
274
00:17:59 --> 00:18:03
relation here,
which is simply 1/L minus
275
00:18:03 --> 00:18:08
infinity to t.
The voltage across the
276
00:18:08 --> 00:18:14
capacitor is simply (vI-v)dt.
I have just written down the
277
00:18:14 --> 00:18:17
node quotation for this node
here.
278
00:18:17 --> 00:18:23
Now I will just apply a bit of
math and simplify it and get the
279
00:18:23 --> 00:18:28
resulting equation.
What I can do is simply
280
00:18:28 --> 00:18:33
differentiate with respect to t
here.
281
00:18:33 --> 00:18:41
And get this to be Cd^2v/dt^2,
the second derivative of v.
282
00:18:41 --> 00:18:47
And here what I end up getting
is 1/L(vI-v).
283
00:18:47 --> 00:18:54
So I just differentiated the
whole thing by d/dt here.
284
00:18:54 --> 00:19:02
And then I just move L up here.
I bring d^2v/dt^2 out here.
285
00:19:02 --> 00:19:08
And then I get a minus v here,
and that will be equal to,
286
00:19:08 --> 00:19:12
oh, I'm sorry.
Let me leave this here.
287
00:19:12 --> 00:19:18
Bring the minus v to this side
so it becomes a plus and leave
288
00:19:18 --> 00:19:22
vI on this side.
So I end up getting
289
00:19:22 --> 00:19:25
LCd^2v/dt^2.
I bring L up here.
290
00:19:25 --> 00:19:30
And then I take v to the other
side.
291
00:19:30 --> 00:19:33
Plus v and leave vI here so I
get vI.
292
00:19:33 --> 00:19:37
That is second order
differential equation that
293
00:19:37 --> 00:19:40
governs the characteristics of
the voltage, v.
294
00:19:40 --> 00:19:45
So much as the voltage across
the capacitor was a state
295
00:19:45 --> 00:19:51
variable in our RC circuits or
the current through the inductor
296
00:19:51 --> 00:19:55
was a state variable in our RL
circuits, out here both the
297
00:19:55 --> 00:20:01
current through the inductor and
the voltage across the capacitor
298
00:20:01 --> 00:20:06
are my two state variables.
And so here I have a
299
00:20:06 --> 00:20:09
second-order equation in my
voltage, v.
300
00:20:09 --> 00:20:12
Again, going through the
foundations here,
301
00:20:12 --> 00:20:15
I am now going to go through a
bunch of math.
302
00:20:15 --> 00:20:18
Up to here it was circuit
analysis, and now I am just
303
00:20:18 --> 00:20:22
going to do math.
For the next three or four
304
00:20:22 --> 00:20:25
blackboards just math.
You can solve this second-order
305
00:20:25 --> 00:20:30
differential equation any which
way you want.
306
00:20:30 --> 00:20:33
But just to keep things as
simple as possible,
307
00:20:33 --> 00:20:36
in 6.002 I solve all the
differential equations,
308
00:20:36 --> 00:20:40
it turns out we are fortunate
enough we can do that,
309
00:20:40 --> 00:20:43
using the exact same method
again and again and again,
310
00:20:43 --> 00:20:48
the same thing can be applied.
And the method that we use to
311
00:20:48 --> 00:20:51
solve it is the method of
homogenous and particular
312
00:20:51 --> 00:20:54
solutions.
So the first step we are going
313
00:20:54 --> 00:20:58
to find the particular solution,
vP.
314
00:20:58 --> 00:21:03
Second step we find the
homogenous solution,
315
00:21:03 --> 00:21:07
vH.
And the third step we are going
316
00:21:07 --> 00:21:14
to find the total solution as
the sum of, v is simply the
317
00:21:14 --> 00:21:21
particular plus the homogenous
solution and then solve for
318
00:21:21 --> 00:21:27
constants based on the initial
conditions and the applied
319
00:21:27 --> 00:21:32
voltage.
So let's write down initial
320
00:21:32 --> 00:21:34
conditions.
Let's assume,
321
00:21:34 --> 00:21:37
for simplicity,
that my initial conditions are
322
00:21:37 --> 00:21:42
simply the voltage across the
capacitor is zero to begin and
323
00:21:42 --> 00:21:47
the current through my inductor
is also zero as I begin life.
324
00:21:47 --> 00:21:51
Now, this is what is called
"zero state".
325
00:21:51 --> 00:21:54
v and i are both zero,
and so the response of my
326
00:21:54 --> 00:21:58
circuit for some input is going
to be called ZSR.
327
00:21:58 --> 00:22:05
You've probably heard this term
in one of your recitations.
328
00:22:05 --> 00:22:10
So zero state response simply
says I start with my circuit at
329
00:22:10 --> 00:22:14
rest and looks at how it behaves
for some given input.
330
00:22:14 --> 00:22:18
That is a little term you may
end up using.
331
00:22:18 --> 00:22:22
My input next.
I am going to use the following
332
00:22:22 --> 00:22:25
input.
vI of t is going to be a step,
333
00:22:25 --> 00:22:31
is going to look like this.
My input is at t=0 v is going
334
00:22:31 --> 00:22:35
from zero to some voltage VI and
then stay at that voltage.
335
00:22:35 --> 00:22:37
It is going to be a step.
Kaboom.
336
00:22:37 --> 00:22:41
And you can see why I am going
with this set of variables,
337
00:22:41 --> 00:22:45
because I want make this
situation as close as possible
338
00:22:45 --> 00:22:48
to the funny behavior we
observed there.
339
00:22:48 --> 00:22:52
Remember we had a step,
and because of the step we had
340
00:22:52 --> 00:22:56
some behavior at that node?
So I will try to bring you as
341
00:22:56 --> 00:23:00
close to that.
In tomorrow's lecture,
342
00:23:00 --> 00:23:04
I am going to close the loop
around that and derive for you
343
00:23:04 --> 00:23:07
exactly the behavior we saw on
the scope.
344
00:23:07 --> 00:23:11
And to get there I am going to
be try to be as close as
345
00:23:11 --> 00:23:15
possible to the constants and
other parameters in the demo.
346
00:23:15 --> 00:23:19
So VI is a step and zero state.
Just in terms of notation,
347
00:23:19 --> 00:23:22
this kind of a step input
occurs pretty frequently.
348
00:23:22 --> 00:23:25
And we just have a special
notation for it.
349
00:23:25 --> 00:23:30
We simply call it VI is the
final value here.
350
00:23:30 --> 00:23:33
And we call it u(t).
So VIu(t), u(t) simply
351
00:23:33 --> 00:23:39
represents a step at time t=0,
steps from zero volts to VI.
352
00:23:39 --> 00:23:44
That is just a little more
notation that will come in handy
353
00:23:44 --> 00:23:47
at some point.
More math now.
354
00:23:47 --> 00:23:50
Three steps,
particular solution,
355
00:23:50 --> 00:23:54
homogenous solution,
total solution/constants.
356
00:23:54 --> 00:24:00
This is almost like a mantra
here, like a chorus.
357
00:24:00 --> 00:24:03
Homogenous solution we compute
using a four-step method.
358
00:24:03 --> 00:24:06
And four-step method for
homogenous solutions,
359
00:24:06 --> 00:24:09
it turns out that it happens to
be that way for all the
360
00:24:09 --> 00:24:12
equations we will see in our
course.
361
00:24:12 --> 00:24:16
The first step would be assume
a solution of the form Ae^st.
362
00:24:16 --> 00:24:19
Exactly as with RCs.
If you close your eyes and do
363
00:24:19 --> 00:24:23
exactly what you did for RCs you
will get to where you want to
364
00:24:23 --> 00:24:25
be.
You assume a solution of the
365
00:24:25 --> 00:24:27
form Ae^st.
Substitute that into your
366
00:24:27 --> 00:24:31
homogenous equation.
Obtain the characteristic
367
00:24:31 --> 00:24:34
equation.
Solve for the roots.
368
00:24:34 --> 00:24:37
And then write down your
homogenous solution.
369
00:24:37 --> 00:24:42
Same sort of steps again and
again and again until you get
370
00:24:42 --> 00:24:45
bored to tears.
Particular solution.
371
00:24:45 --> 00:24:49
For the particular solution,
I simply need to find a
372
00:24:49 --> 00:24:53
solution, any solution,
if not the most general one but
373
00:24:53 --> 00:24:57
any solution that satisfies the
particular equation which
374
00:24:57 --> 00:25:03
satisfies that equation.
LCd^2vP/dt^2+vP=VI.
375
00:25:03 --> 00:25:09
My input is a step and I am
going to look for the solution
376
00:25:09 --> 00:25:15
for time t greater than zero.
Notice that for time t less
377
00:25:15 --> 00:25:20
than or equal to zero,
v is going to be zero.
378
00:25:20 --> 00:25:26
So I am looking for a solution
greater than t=0.
379
00:25:26 --> 00:25:34
Here, if I substitute vP=VI,
that is a particular solution.
380
00:25:34 --> 00:25:39
Because if I substitute VI here
this goes to zero and then I get
381
00:25:39 --> 00:25:43
VI=VI, so this works.
I promised you this was going
382
00:25:43 --> 00:25:47
to be simple.
You cannot get any simpler than
383
00:25:47 --> 00:25:49
that.
I have done my first step.
384
00:25:49 --> 00:25:52
I found the particular
solution.
385
00:25:52 --> 00:25:57
And VI is a good enough
particular solution so I will
386
00:25:57 --> 00:26:03
use it, I will take it.
As my second step I am going to
387
00:26:03 --> 00:26:08
find vH or the solution to the
homogenous equation.
388
00:26:08 --> 00:26:15
And the homogenous equation is
simply that equation with drive
389
00:26:15 --> 00:26:18
set to zero.
What I get here is
390
00:26:18 --> 00:26:23
LCd^2vH/dt^2+vH=0.
That is my homogenous equation.
391
00:26:23 --> 00:26:28
I simply set the drive to be
zero.
392
00:26:28 --> 00:26:32
And to find the solution here,
I go through my four-step
393
00:26:32 --> 00:26:34
method.
Again, in 6.002 following the
394
00:26:34 --> 00:26:38
kind of Occam's principle,
we just show you the absolute
395
00:26:38 --> 00:26:41
minimum necessary to get to
where you want.
396
00:26:41 --> 00:26:45
The absolute minimum necessary
is it turns out that we can
397
00:26:45 --> 00:26:50
solve all our differential
equations that we use here by
398
00:26:50 --> 00:26:54
using the methods of homogenous
and particular solutions.
399
00:26:54 --> 00:26:58
And every homogenous solution
can be solved by a four-step
400
00:26:58 --> 00:27:04
method.
That is about as minimal as it
401
00:27:04 --> 00:27:09
can get.
So no extraneous stuff there.
402
00:27:09 --> 00:27:13
The four-step method,
four steps.
403
00:27:13 --> 00:27:21
The first step is assume a
solution of the form vH=Ae^st.
404
00:27:21 --> 00:27:28
What I have noticed is that
students starting out are
405
00:27:28 --> 00:27:35
usually scared of differential
equations.
406
00:27:35 --> 00:27:36
I know I was when I was a
student.
407
00:27:36 --> 00:27:40
And the trick with differential
equations is that it is all a
408
00:27:40 --> 00:27:42
matter of psych.
Just because you see some
409
00:27:42 --> 00:27:46
squigglies and squagglies and a
bunch of math and so on you say
410
00:27:46 --> 00:27:49
oh, that must be hard.
But differential equations are
411
00:27:49 --> 00:27:52
actually the simplest thing
there is because in a large
412
00:27:52 --> 00:27:55
majority of cases the way you
solve them is you assume you
413
00:27:55 --> 00:27:59
know the answer,
someone tells you the answer.
414
00:27:59 --> 00:28:02
And then all you are left to do
is shove the answer into the
415
00:28:02 --> 00:28:05
equation and find out the
constants that makes it the
416
00:28:05 --> 00:28:07
answer.
Just a matter of psych.
417
00:28:07 --> 00:28:09
Psych yourselves that this
stuff is easy,
418
00:28:09 --> 00:28:11
because I am telling you what
the solution is.
419
00:28:11 --> 00:28:14
All you have to do is
substitute and verify.
420
00:28:14 --> 00:28:17
If you think about differential
equations that way or a large
421
00:28:17 --> 00:28:20
majority of them,
it really is very simple if you
422
00:28:20 --> 00:28:22
can just get past the squigglies
here.
423
00:28:22 --> 00:28:26
Just get past the squigglies
and then just simply stick in
424
00:28:26 --> 00:28:31
some simple stuff and it works.
I mean it just cannot get any
425
00:28:31 --> 00:28:34
easier.
I cannot think of any other
426
00:28:34 --> 00:28:39
field where the way you find a
solution is assume you know the
427
00:28:39 --> 00:28:43
solution and stick it in.
It has never made any sense to
428
00:28:43 --> 00:28:48
me but that is how it is.
So we assume the solution to
429
00:28:48 --> 00:28:51
the form Ae^st,
you stick it in there,
430
00:28:51 --> 00:28:55
and you have to find out the A
and s that make it so.
431
00:28:55 --> 00:28:58
It cannot get any simpler than
that.
432
00:28:58 --> 00:29:02
Let's stick the sucker in here
and see what we can get.
433
00:29:02 --> 00:29:07
Substitute Ae^st here I get
LCA, and second derivative,
434
00:29:07 --> 00:29:12
so it's s^2 e^st.
And Ae^st on this one here.
435
00:29:12 --> 00:29:17
And that equals zero.
And then let me just solve for
436
00:29:17 --> 00:29:22
whatever I can find.
Assuming I don't take the
437
00:29:22 --> 00:29:26
trivial case A=0,
I cancel these guys out.
438
00:29:26 --> 00:29:32
And what I am left with is
simply LCs^2+1=0.
439
00:29:32 --> 00:29:36
In other words,
what I end up getting is B,
440
00:29:36 --> 00:29:38
s^2=-1/LC.
My first step was,
441
00:29:38 --> 00:29:43
I am giving you solutions,
stick them in there,
442
00:29:43 --> 00:29:48
assume a solution of this form.
Second step is get the
443
00:29:48 --> 00:29:53
characteristic equation.
And the way you get the
444
00:29:53 --> 00:29:59
characteristic equation is that
you simply stick this guy in
445
00:29:59 --> 00:30:04
there.
And what you end up getting is
446
00:30:04 --> 00:30:09
some equation in s^2.
Do you remember what you got
447
00:30:09 --> 00:30:13
for first order circuits?
What s was?
448
00:30:13 --> 00:30:17
What is s?
For first order circuits,
449
00:30:17 --> 00:30:21
what did you get as a
characteristic equation?
450
00:30:21 --> 00:30:24
s+1/RC=0.
The same thing.
451
00:30:24 --> 00:30:30
Just remember to blindly apply
the steps.
452
00:30:30 --> 00:30:33
It will lead you to the answer.
This is called the
453
00:30:33 --> 00:30:37
"characteristic equation".
This is incredibly important.
454
00:30:37 --> 00:30:41
You will see in about a couple
weeks from now that once you
455
00:30:41 --> 00:30:44
write the characteristic
equation down for a circuit,
456
00:30:44 --> 00:30:48
it tells you all there is to
know about the circuit.
457
00:30:48 --> 00:30:51
And often times you can stop
solving right here.
458
00:30:51 --> 00:30:54
To experienced circuit
designers this tells me
459
00:30:54 --> 00:30:59
everything there is to know.
This is really key.
460
00:30:59 --> 00:31:02
That's why it's called a
characteristic equation.
461
00:31:02 --> 00:31:05
I believe in problem number
three of the homework that will
462
00:31:05 --> 00:31:09
be coming out this week,
that is exactly what you are
463
00:31:09 --> 00:31:11
going to do.
I am going to give you a
464
00:31:11 --> 00:31:15
circuit, ask you to get to the
characteristic equation quickly
465
00:31:15 --> 00:31:18
and then from there intuit the
solution.
466
00:31:18 --> 00:31:21
Write the characteristic
equation and then just intuit
467
00:31:21 --> 00:31:25
solution, it's that simple.
So, step A, assume a solution
468
00:31:25 --> 00:31:27
of the form, step B,
write the characteristic
469
00:31:27 --> 00:31:33
equation down.
And let me just simplify that a
470
00:31:33 --> 00:31:37
little bit.
I go ahead and find my roots.
471
00:31:37 --> 00:31:42
And my roots here,
remember that j is the square
472
00:31:42 --> 00:31:47
root of minus one.
And so what I end up getting
473
00:31:47 --> 00:31:53
is, my two roots here are,
plus j square root of 1/LC and
474
00:31:53 --> 00:31:59
minus j square root of 1/LC.
Two roots.
475
00:31:59 --> 00:32:03
And just as a shorthand
notation, much like I had a
476
00:32:03 --> 00:32:08
shorthand notation for RC,
what was my shorthand notation
477
00:32:08 --> 00:32:09
for RC?
Tau.
478
00:32:09 --> 00:32:15
Just as tau was big in first
order, we have a corresponding
479
00:32:15 --> 00:32:20
thing that is big in second
order and that is omega nought.
480
00:32:20 --> 00:32:24
Omega nought is simply square
root 1/LC.
481
00:32:24 --> 00:32:28
Just as tau was RC,
omega nought is a shorthand
482
00:32:28 --> 00:32:33
here.
And so s is simply plus or
483
00:32:33 --> 00:32:40
minus j omega nought.
Notice that in this equation
484
00:32:40 --> 00:32:48
here, if you take the square
root of LC there that has units
485
00:32:48 --> 00:32:56
of time, so one divided by that
has units of frequency.
486
00:32:56 --> 00:33:04
Notice that this guy is a
frequency in radians.
487
00:33:04 --> 00:33:09
I end up getting my roots of
the homogenous equation,
488
00:33:09 --> 00:33:13
and that is my third step.
And as my fourth step,
489
00:33:13 --> 00:33:18
I simply write down the
homogenous solution as
490
00:33:18 --> 00:33:23
substituting s with its roots
and writing the most general
491
00:33:23 --> 00:33:29
possible form of the solution,
and that would be A1e^(j omega
492
00:33:29 --> 00:33:34
nought t)+A2e^(-j omega nought
t).
493
00:33:34 --> 00:33:36
Done.
Some constant times this
494
00:33:36 --> 00:33:38
solution plus some other
constant times,
495
00:33:38 --> 00:33:41
the other solution.
Plus zero omega nought.
496
00:33:41 --> 00:33:44
Remember it comes from here,
Ae^st.
497
00:33:44 --> 00:33:48
I assume the solution of this
form, so my solution in this
498
00:33:48 --> 00:33:52
most general case would be s
being j omega nought in one
499
00:33:52 --> 00:33:55
case, minus j omega nought in
the other case,
500
00:33:55 --> 00:34:00
and I sum the two to get the
most general solution.
501
00:34:00 --> 00:34:05
502
00:34:05 --> 00:34:10
So blasting ahead.
I now have my homogenous
503
00:34:10 --> 00:34:14
solution.
And as my third step of
504
00:34:14 --> 00:34:21
solution to differential
equations I write down the total
505
00:34:21 --> 00:34:27
solution, v=vP+vH,
particular plus the homogenous
506
00:34:27 --> 00:34:32
solutions.
And v=VI, was my particular
507
00:34:32 --> 00:34:38
solution, +A1e^(j omega nought
t)+A2e^(-j omega nought t) is my
508
00:34:38 --> 00:34:41
complete solution.
The final step,
509
00:34:41 --> 00:34:46
write down the total solution
and find the constants from the
510
00:34:46 --> 00:34:51
initial conditions.
To find the constants from the
511
00:34:51 --> 00:34:54
initial conditions,
let's start with,
512
00:34:54 --> 00:34:59
the voltage is zero to begin
with.
513
00:34:59 --> 00:35:03
This equation governs the
characteristics of v,
514
00:35:03 --> 00:35:08
so I need to find the initial
conditions.
515
00:35:08 --> 00:35:12
First of all,
I know that know that v(0)=0.
516
00:35:12 --> 00:35:17
From there I substitute t=0.
And so this goes to one,
517
00:35:17 --> 00:35:21
this goes to one,
and I end up getting
518
00:35:21 --> 00:35:25
0=VI+A1+A2.
That is my first expression.
519
00:35:25 --> 00:35:31
And then I am also given that
i(0)=0.
520
00:35:31 --> 00:35:37
And so I can get that as well.
How do I get i?
521
00:35:37 --> 00:35:41
This is v.
I know that i=Cdv/dt,
522
00:35:41 --> 00:35:47
so I can get i by simply
multiplying by C and
523
00:35:47 --> 00:35:52
differentiating this with
respect to t.
524
00:35:52 --> 00:36:00
I get C, this guy vanishes so I
get d/dt of this.
525
00:36:00 --> 00:36:07
So it is CA1(j omega nought)
e^(j omega nought t)+CA2(-j
526
00:36:07 --> 00:36:12
omega nought)e^(-j omega nought
t).
527
00:36:12 --> 00:36:21
From here I am given that that
is zero, and so therefore this
528
00:36:21 --> 00:36:26
guy becomes a one,
this guy becomes a one,
529
00:36:26 --> 00:36:34
j omega nought,
j omega nought cancel out.
530
00:36:34 --> 00:36:43
What I end up getting is A1=A2.
From the second initial
531
00:36:43 --> 00:36:50
condition I get A1=A2.
From these two,
532
00:36:50 --> 00:36:58
if I substitute here for A2,
I get VI + 2A1 = 0,
533
00:36:58 --> 00:37:05
or A1=-VI/2.
That is also equal to A2.
534
00:37:05 --> 00:37:13
Therefore, my total solution
now can be written in terms of
535
00:37:13 --> 00:37:19
the actual values of the
constants I have obtained.
536
00:37:19 --> 00:37:24
I get VI-VI/2.
So A1 and A2 are equal.
537
00:37:24 --> 00:37:33
I just pull them outside.
I pull VI-2 outside and I stick
538
00:37:33 --> 00:37:38
these two guys in parenthesis
in.
539
00:37:38 --> 00:37:46
Again, I promised you no more
circuits from here on until the
540
00:37:46 --> 00:37:52
very last board or something
like that.
541
00:37:52 --> 00:37:57
It is all math,
so not much else happening
542
00:37:57 --> 00:38:01
there.
More math.
543
00:38:01 --> 00:38:06
If you would like,
I could skip all the way to the
544
00:38:06 --> 00:38:13
end and show you the answer.
But I just love to write
545
00:38:13 --> 00:38:19
equations on the board so let me
just go through that.
546
00:38:19 --> 00:38:25
I am going to simplify this a
little further here.
547
00:38:25 --> 00:38:31
And we should remember this
form by the Euler relation,
548
00:38:31 --> 00:38:38
ejx=cos x+j sin x.
And by the same token,
549
00:38:38 --> 00:38:47
(e^jx + e^-jx)/2=cos x.
You all should know this from
550
00:38:47 --> 00:38:54
the Euler relation.
So were are using this guy
551
00:38:54 --> 00:39:04
here, ej^x + e^-jx=2cos x.
And so this one is 2 cosine of
552
00:39:04 --> 00:39:08
omega nought t,
2 and 2 cancel out,
553
00:39:08 --> 00:39:16
and what I am left with is
v(t)=VI-VI cos( omega nought t).
554
00:39:16 --> 00:39:24
And the current is Cdv/dt,
which is simply CVI sin( omega
555
00:39:24 --> 00:39:30
nought t).
Just remember that omega nought
556
00:39:30 --> 00:39:36
is the square root of 1/LC.
We are done.
557
00:39:36 --> 00:39:44
In fact, I did not give that
answer the importance that was
558
00:39:44 --> 00:39:48
due so let me just draw.
559
00:39:48 --> 00:39:57
560
00:39:57 --> 00:40:00
There.
That is better.
561
00:40:00 --> 00:40:02
Enough math.
In a nutshell,
562
00:40:02 --> 00:40:06
what did we do.
We wrote the node method,
563
00:40:06 --> 00:40:10
it's a very simple circuit,
to write down the equation
564
00:40:10 --> 00:40:15
governing that circuit.
And then we grunged through a
565
00:40:15 --> 00:40:18
bunch of math.
Not a whole lot here.
566
00:40:18 --> 00:40:23
It is pretty simple.
And ended up with a relation
567
00:40:23 --> 00:40:28
that says the voltage across the
capacitor for a step input,
568
00:40:28 --> 00:40:33
assuming zero state,
is a constant VI-VI cos omega
569
00:40:33 --> 00:40:36
t.
Notice that even though I have
570
00:40:36 --> 00:40:39
a step input,
the circuit dynamics are such
571
00:40:39 --> 00:40:42
that I get a cosine in there.
You can begin to see where
572
00:40:42 --> 00:40:44
these cosines are coming from
now.
573
00:40:44 --> 00:40:47
They come in here.
And if you recall the example I
574
00:40:47 --> 00:40:50
showed you earlier of the
inverter circuit,
575
00:40:50 --> 00:40:52
remember there was a cosine
that decayed,
576
00:40:52 --> 00:40:56
that was sort of losing energy
and kind of dying out?
577
00:40:56 --> 00:41:00
So you can see where the
cosines are coming from.
578
00:41:00 --> 00:41:11
And just to draw you a little
sketch here.
579
00:41:11 --> 00:41:26
Let me draw v and i for you and
let me plot omega t,
580
00:41:26 --> 00:41:35
pi/2, pi and so on.
Let me plot VI.
581
00:41:35 --> 00:41:41
When time t=0,
VI=0, cosine omega t is one,
582
00:41:41 --> 00:41:48
and so VI-VI=0.
That is simply a cosine that
583
00:41:48 --> 00:41:56
starts out at zero here,
and at pi I get cosine omega t
584
00:41:56 --> 00:42:02
is minus one,
so I get plus VI on the other
585
00:42:02 --> 00:42:08
side.
So I end up at +2VI.
586
00:42:08 --> 00:42:13
At this point the voltage is
here.
587
00:42:13 --> 00:42:19
And notice that this guy looks
like this.
588
00:42:19 --> 00:42:27
It is a cosine that is
translated up so that its mean
589
00:42:27 --> 00:42:37
value is not zero but VI.
It is just a translation up of
590
00:42:37 --> 00:42:42
a cosine.
Similarly, in this case for the
591
00:42:42 --> 00:42:48
current it is a sinusoidal
characteristic.
592
00:42:48 --> 00:42:56
And it looks something like
this where the peak is given by
593
00:42:56 --> 00:43:00
CVI, oh, I messed up.
594
00:43:00 --> 00:43:08
595
00:43:08 --> 00:43:14
When I differentiated this is
missed the omega nought out
596
00:43:14 --> 00:43:15
there.
597
00:43:15 --> 00:43:25
598
00:43:25 --> 00:43:30
What I would like to do now --
This is the form of the output
599
00:43:30 --> 00:43:33
for a step input.
What I would like to do next is
600
00:43:33 --> 00:43:36
show you a demo.
But before I show you a demo,
601
00:43:36 --> 00:43:40
I always found it strange that
I have a step input and then I
602
00:43:40 --> 00:43:44
have two little elements,
how can I get a sine coming out
603
00:43:44 --> 00:43:47
of the output?
I would like to get some
604
00:43:47 --> 00:43:50
intuition as to why things
behave the way they are.
605
00:43:50 --> 00:43:54
I could go and pray to find
out, but let me just give you
606
00:43:54 --> 00:43:58
some very basic insight as to
why this behaves the way it
607
00:43:58 --> 00:44:02
does.
Let me draw the circuit for you
608
00:44:02 --> 00:44:05
here.
And this is my inductor L and
609
00:44:05 --> 00:44:08
capacitance C.
Remember this is v.
610
00:44:08 --> 00:44:13
Let me just walk you through
what is happening there and get
611
00:44:13 --> 00:44:17
you to understand this.
Now, you have seen sines occur
612
00:44:17 --> 00:44:20
before.
If you go and write down the
613
00:44:20 --> 00:44:24
equation of motion of a
pendulum, you know,
614
00:44:24 --> 00:44:28
you have a pendulum,
you move it to one side,
615
00:44:28 --> 00:44:31
let go.
It is also governed by
616
00:44:31 --> 00:44:35
sinusoidal characteristics.
And you will find that the
617
00:44:35 --> 00:44:39
equation governing its motion is
very much of the same form,
618
00:44:39 --> 00:44:43
and you get the sinusoid where
you have energy that is sloshing
619
00:44:43 --> 00:44:47
back and forth between maximum
potential energy to maximum
620
00:44:47 --> 00:44:51
kinetic energy and zero
potential energy back to maximum
621
00:44:51 --> 00:44:53
potential energy,
zero kinetic.
622
00:44:53 --> 00:44:56
So it is energy sloshing back
and forth.
623
00:44:56 --> 00:45:01
The same way here.
Capacitors and inductors store
624
00:45:01 --> 00:45:04
energy.
Let's walk through and see what
625
00:45:04 --> 00:45:06
happens.
I start off with both of them
626
00:45:06 --> 00:45:09
having the stage zero,
zero current,
627
00:45:09 --> 00:45:11
zero voltage.
I apply a step here.
628
00:45:11 --> 00:45:14
Boom, the step comes
instanteously to VI.
629
00:45:14 --> 00:45:18
I notice that the capacitor
voltage cannot change instantly
630
00:45:18 --> 00:45:22
unless there is an infinite
pulse of a sort,
631
00:45:22 --> 00:45:24
so this guy cannot change
instantly.
632
00:45:24 --> 00:45:29
And so its voltage starts off
being zero.
633
00:45:29 --> 00:45:32
So the entire voltage here,
KVL must be true no matter
634
00:45:32 --> 00:45:34
what.
They are absolutely fundamental
635
00:45:34 --> 00:45:36
principles from Maxwell's
equations.
636
00:45:36 --> 00:45:38
KVL must hold,
which means that the entire
637
00:45:38 --> 00:45:40
voltage VI must appear across
the inductor.
638
00:45:40 --> 00:45:44
I put a big voltage across the
inductor and its current begins
639
00:45:44 --> 00:45:45
to build up.
There you go.
640
00:45:45 --> 00:45:49
A voltage across the inductor,
its current begins to build up.
641
00:45:49 --> 00:45:52
As its current begins to build
up that current must flow
642
00:45:52 --> 00:45:54
through the capacitor,
too.
643
00:45:54 --> 00:45:57
And as current flows through a
capacitor it is depositing
644
00:45:57 --> 00:46:02
charge into the capacitor.
As the capacitor begins to get
645
00:46:02 --> 00:46:07
charge deposited on it,
its voltage begins to rise.
646
00:46:07 --> 00:46:12
Let's see what happens here.
Its voltage keeps rising.
647
00:46:12 --> 00:46:16
At some point,
the voltage across the
648
00:46:16 --> 00:46:21
capacitor is equal to VI.
But then VI equals this VI
649
00:46:21 --> 00:46:24
here.
So when the two become VI,
650
00:46:24 --> 00:46:29
the inductor has zero volts
across it.
651
00:46:29 --> 00:46:32
So there is no longer a
potential difference that is
652
00:46:32 --> 00:46:35
increasing the current in that
direction.
653
00:46:35 --> 00:46:37
At that point,
at pi divided by 2,
654
00:46:37 --> 00:46:42
I have some current going into
the inductor so there is no
655
00:46:42 --> 00:46:46
longer a pressure that is
forcing more current through the
656
00:46:46 --> 00:46:49
inductor because this voltage
reaches VI.
657
00:46:49 --> 00:46:53
But remember capacitors like to
sit around holding voltages.
658
00:46:53 --> 00:46:57
Just remember that demo.
That rinky-dink capacitor sat
659
00:46:57 --> 00:47:01
there stubbornly holding its
voltage.
660
00:47:01 --> 00:47:03
And it had a huge spark towards
the end.
661
00:47:03 --> 00:47:05
It just sat there holding its
voltage.
662
00:47:05 --> 00:47:09
In the same manner,
inductors love to sit around
663
00:47:09 --> 00:47:12
holding a current.
They will do whatever they can
664
00:47:12 --> 00:47:14
to keep the current going
through them.
665
00:47:14 --> 00:47:17
It has got the current going
through.
666
00:47:17 --> 00:47:19
And few forces on earth can
change that.
667
00:47:19 --> 00:47:22
And so therefore,
even though the capacitor
668
00:47:22 --> 00:47:26
voltage is VI and the voltage
drop across the inductor is
669
00:47:26 --> 00:47:30
zero, it still keeps supplying a
current.
670
00:47:30 --> 00:47:32
It has got the current.
It's got inertia.
671
00:47:32 --> 00:47:34
It keeps going.
It is like a runaway train.
672
00:47:34 --> 00:47:37
You may not be pushing the
train from the back,
673
00:47:37 --> 00:47:41
but once it is running it has
got kinetic energy and is going
674
00:47:41 --> 00:47:43
to run no matter what for a
least some more time,
675
00:47:43 --> 00:47:46
even if you take away the force
on the train.
676
00:47:46 --> 00:47:49
So I have taken away the force
on the punching more current
677
00:47:49 --> 00:47:52
through, but it has kinetic
energy.
678
00:47:52 --> 00:47:55
It has current flowing through
it so it continues to supply a
679
00:47:55 --> 00:47:57
current.
Because it continues to supply
680
00:47:57 --> 00:48:02
the current the capacitor
voltage keeps increasing.
681
00:48:02 --> 00:48:05
This is a subtle insight which
is absolutely spectacular that
682
00:48:05 --> 00:48:09
with zero volts across it,
it still keeps pumping that
683
00:48:09 --> 00:48:11
current.
Capacitor voltage has gone up.
684
00:48:11 --> 00:48:14
And guess what?
The voltage on this side is
685
00:48:14 --> 00:48:17
higher now but this guy is still
pumping a current.
686
00:48:17 --> 00:48:20
Man, I have been born to do
this, you know,
687
00:48:20 --> 00:48:24
I shall pump a current.
However, because the voltage
688
00:48:24 --> 00:48:29
has now gone up here gradually
the current begins to diminish.
689
00:48:29 --> 00:48:33
So the capacitor is concerned.
You pump a current into me,
690
00:48:33 --> 00:48:36
my voltage goes up.
At some point,
691
00:48:36 --> 00:48:39
like a runaway train,
it comes to a halt.
692
00:48:39 --> 00:48:44
The current through the
capacitor drains and now goes to
693
00:48:44 --> 00:48:47
zero and the capacitor voltage
reaches 2VI.
694
00:48:47 --> 00:48:50
So this is at 2VI now and this
is at VI.
695
00:48:50 --> 00:48:53
Now the situation is not in
equilibrium.
696
00:48:53 --> 00:48:57
At this point there is zero
current through it,
697
00:48:57 --> 00:49:01
but guess what?
I have a VI pumping in this
698
00:49:01 --> 00:49:05
direction now.
I have the same VI punching in
699
00:49:05 --> 00:49:07
this direction.
So guess what?
700
00:49:07 --> 00:49:11
Its current must now build up
in this direction and its
701
00:49:11 --> 00:49:14
current begins to build up in
that direction.
702
00:49:14 --> 00:49:18
That begins to discharge the
capacitor and the capacitor then
703
00:49:18 --> 00:49:22
goes on to a negative,
or the current goes down to a
704
00:49:22 --> 00:49:26
maximum negative current,
and this process continues.
705
00:49:26 --> 00:49:30
What you are seeing here is
energy.
706
00:49:30 --> 00:49:33
It is sloshing back and forth
between the two,
707
00:49:33 --> 00:49:37
and that is kind of a key.
I will just quickly put up a
708
00:49:37 --> 00:49:40
demo that you can watch as you
are walking out.
709
00:49:40 --> 00:49:44
With a step input,
notice the green is the voltage
710
00:49:44 --> 00:49:48
across the capacitor and the
orange is the current through
711
00:49:48 --> 49:51
the capacitor.