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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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00:15:21,000 --> 00:15:26,000
the capacitor and my current is
the current through this loop

235
00:15:26,000 --> 00:15:31,000
here, which is the same as the
current through the capacitor or

236
00:15:31,000 --> 00:15:35,000
the current through the
inductor.

237
00:15:35,000 --> 00:15:38,000
And we are going to proceed in
exactly the same manner as we

238
00:15:38,000 --> 00:15:41,000
did for first order differential
equations, write the equations

239
00:15:41,000 --> 00:15:43,000
down and just boom,
boom, boom, boom,

240
00:15:43,000 --> 00:15:47,000
go down the same sets of steps
but just get to some place

241
00:15:47,000 --> 00:15:48,000
different.
We are going to start by

242
00:15:48,000 --> 00:15:51,000
writing a node equation for this
node here.

243
00:15:51,000 --> 00:15:54,000
That's the only node for which
I have an unknown voltage.

244
00:15:54,000 --> 00:15:56,000
The node here is vI,
so I need to find this,

245
00:15:56,000 --> 00:16:00,000
there's just one unknown node
voltage.

246
00:16:00,000 --> 00:16:04,000
And I am going to need some
element laws.

247
00:16:04,000 --> 00:16:11,000
For the capacitor I know the iV
relation is given by the i for

248
00:16:11,000 --> 00:16:17,000
the capacitor is Cdv/dt.
And just to show the capacitor

249
00:16:17,000 --> 00:16:23,000
I am just calling it dvc/dt.
Similarly, for an inductor,

250
00:16:23,000 --> 00:16:30,000
L, the voltage across the
inductor is given by Ldi/dt.

251
00:16:30,000 --> 00:16:34,000
So this is the vI relation for
the capacitor,

252
00:16:34,000 --> 00:16:37,000
the vI relation for an
inductor.

253
00:16:37,000 --> 00:16:42,000
It also suits us to write this
in an integral form.

254
00:16:42,000 --> 00:16:48,000
So if I integrate both sides of
this equation and I bring L down

255
00:16:48,000 --> 00:16:53,000
to this side,
I end up getting something like

256
00:16:53,000 --> 00:17:00,000
this, 1/L minus infinity to t,
VLdt, and that is simply iL.

257
00:17:00,000 --> 00:17:04,000
I am just simply replacing this
with an integral form.

258
00:17:04,000 --> 00:17:09,000
So this is a VI relationship
for the inductor and this is for

259
00:17:09,000 --> 00:17:13,000
the capacitor.
So let me now go ahead and

260
00:17:13,000 --> 00:17:16,000
apply the node method for my
circuit here.

261
00:17:16,000 --> 00:17:21,000
Here, for the node method,
I have to equate the currents

262
00:17:21,000 --> 00:17:26,000
coming into the node or sum the
currents coming into the node

263
00:17:26,000 --> 00:17:32,000
and equate that to zero.
And while I do that I simply

264
00:17:32,000 --> 00:17:37,000
replace the currents by the
corresponding voltages using the

265
00:17:37,000 --> 00:17:40,000
element laws.
So what do I get?

266
00:17:40,000 --> 00:17:45,000
I get the current going in here
to the inductor is equal to the

267
00:17:45,000 --> 00:17:48,000
current going through the
capacitor.

268
00:17:48,000 --> 00:17:51,000
What is the current going the
capacitor?

269
00:17:51,000 --> 00:17:55,000
In terms of its v relationship
it is Cdv/dt.

270
00:17:55,000 --> 00:17:59,000
And the current going to the
inductor is given by this

271
00:17:59,000 --> 00:18:03,000
relation here,
which is simply 1/L minus

272
00:18:03,000 --> 00:18:08,000
infinity to t.
The voltage across the

273
00:18:08,000 --> 00:18:14,000
capacitor is simply (vI-v)dt.
I have just written down the

274
00:18:14,000 --> 00:18:17,000
node quotation for this node
here.

275
00:18:17,000 --> 00:18:23,000
Now I will just apply a bit of
math and simplify it and get the

276
00:18:23,000 --> 00:18:28,000
resulting equation.
What I can do is simply

277
00:18:28,000 --> 00:18:33,000
differentiate with respect to t
here.

278
00:18:33,000 --> 00:18:41,000
And get this to be Cd^2v/dt^2,
the second derivative of v.

279
00:18:41,000 --> 00:18:47,000
And here what I end up getting
is 1/L(vI-v).

280
00:18:47,000 --> 00:18:54,000
So I just differentiated the
whole thing by d/dt here.

281
00:18:54,000 --> 00:19:02,000
And then I just move L up here.
I bring d^2v/dt^2 out here.

282
00:19:02,000 --> 00:19:08,000
And then I get a minus v here,
and that will be equal to,

283
00:19:08,000 --> 00:19:12,000
oh, I'm sorry.
Let me leave this here.

284
00:19:12,000 --> 00:19:18,000
Bring the minus v to this side
so it becomes a plus and leave

285
00:19:18,000 --> 00:19:22,000
vI on this side.
So I end up getting

286
00:19:22,000 --> 00:19:25,000
LCd^2v/dt^2.
I bring L up here.

287
00:19:25,000 --> 00:19:30,000
And then I take v to the other
side.

288
00:19:30,000 --> 00:19:33,000
Plus v and leave vI here so I
get vI.

289
00:19:33,000 --> 00:19:37,000
That is second order
differential equation that

290
00:19:37,000 --> 00:19:40,000
governs the characteristics of
the voltage, v.

291
00:19:40,000 --> 00:19:45,000
So much as the voltage across
the capacitor was a state

292
00:19:45,000 --> 00:19:51,000
variable in our RC circuits or
the current through the inductor

293
00:19:51,000 --> 00:19:55,000
was a state variable in our RL
circuits, out here both the

294
00:19:55,000 --> 00:20:01,000
current through the inductor and
the voltage across the capacitor

295
00:20:01,000 --> 00:20:06,000
are my two state variables.
And so here I have a

296
00:20:06,000 --> 00:20:09,000
second-order equation in my
voltage, v.

297
00:20:09,000 --> 00:20:12,000
Again, going through the
foundations here,

298
00:20:12,000 --> 00:20:15,000
I am now going to go through a
bunch of math.

299
00:20:15,000 --> 00:20:18,000
Up to here it was circuit
analysis, and now I am just

300
00:20:18,000 --> 00:20:22,000
going to do math.
For the next three or four

301
00:20:22,000 --> 00:20:25,000
blackboards just math.
You can solve this second-order

302
00:20:25,000 --> 00:20:30,000
differential equation any which
way you want.

303
00:20:30,000 --> 00:20:33,000
But just to keep things as
simple as possible,

304
00:20:33,000 --> 00:20:36,000
in 6.002 I solve all the
differential equations,

305
00:20:36,000 --> 00:20:40,000
it turns out we are fortunate
enough we can do that,

306
00:20:40,000 --> 00:20:43,000
using the exact same method
again and again and again,

307
00:20:43,000 --> 00:20:48,000
the same thing can be applied.
And the method that we use to

308
00:20:48,000 --> 00:20:51,000
solve it is the method of
homogenous and particular

309
00:20:51,000 --> 00:20:54,000
solutions.
So the first step we are going

310
00:20:54,000 --> 00:20:58,000
to find the particular solution,
vP.

311
00:20:58,000 --> 00:21:03,000
Second step we find the
homogenous solution,

312
00:21:03,000 --> 00:21:07,000
vH.
And the third step we are going

313
00:21:07,000 --> 00:21:14,000
to find the total solution as
the sum of, v is simply the

314
00:21:14,000 --> 00:21:21,000
particular plus the homogenous
solution and then solve for

315
00:21:21,000 --> 00:21:27,000
constants based on the initial
conditions and the applied

316
00:21:27,000 --> 00:21:32,000
voltage.
So let's write down initial

317
00:21:32,000 --> 00:21:34,000
conditions.
Let's assume,

318
00:21:34,000 --> 00:21:37,000
for simplicity,
that my initial conditions are

319
00:21:37,000 --> 00:21:42,000
simply the voltage across the
capacitor is zero to begin and

320
00:21:42,000 --> 00:21:47,000
the current through my inductor
is also zero as I begin life.

321
00:21:47,000 --> 00:21:51,000
Now, this is what is called
"zero state".

322
00:21:51,000 --> 00:21:54,000
v and i are both zero,
and so the response of my

323
00:21:54,000 --> 00:21:58,000
circuit for some input is going
to be called ZSR.

324
00:21:58,000 --> 00:22:05,000
You've probably heard this term
in one of your recitations.

325
00:22:05,000 --> 00:22:10,000
So zero state response simply
says I start with my circuit at

326
00:22:10,000 --> 00:22:14,000
rest and looks at how it behaves
for some given input.

327
00:22:14,000 --> 00:22:18,000
That is a little term you may
end up using.

328
00:22:18,000 --> 00:22:22,000
My input next.
I am going to use the following

329
00:22:22,000 --> 00:22:25,000
input.
vI of t is going to be a step,

330
00:22:25,000 --> 00:22:31,000
is going to look like this.
My input is at t=0 v is going

331
00:22:31,000 --> 00:22:35,000
from zero to some voltage VI and
then stay at that voltage.

332
00:22:35,000 --> 00:22:37,000
It is going to be a step.
Kaboom.

333
00:22:37,000 --> 00:22:41,000
And you can see why I am going
with this set of variables,

334
00:22:41,000 --> 00:22:45,000
because I want make this
situation as close as possible

335
00:22:45,000 --> 00:22:48,000
to the funny behavior we
observed there.

336
00:22:48,000 --> 00:22:52,000
Remember we had a step,
and because of the step we had

337
00:22:52,000 --> 00:22:56,000
some behavior at that node?
So I will try to bring you as

338
00:22:56,000 --> 00:23:00,000
close to that.
In tomorrow's lecture,

339
00:23:00,000 --> 00:23:04,000
I am going to close the loop
around that and derive for you

340
00:23:04,000 --> 00:23:07,000
exactly the behavior we saw on
the scope.

341
00:23:07,000 --> 00:23:11,000
And to get there I am going to
be try to be as close as

342
00:23:11,000 --> 00:23:15,000
possible to the constants and
other parameters in the demo.

343
00:23:15,000 --> 00:23:19,000
So VI is a step and zero state.
Just in terms of notation,

344
00:23:19,000 --> 00:23:22,000
this kind of a step input
occurs pretty frequently.

345
00:23:22,000 --> 00:23:25,000
And we just have a special
notation for it.

346
00:23:25,000 --> 00:23:30,000
We simply call it VI is the
final value here.

347
00:23:30,000 --> 00:23:33,000
And we call it u(t).
So VIu(t), u(t) simply

348
00:23:33,000 --> 00:23:39,000
represents a step at time t=0,
steps from zero volts to VI.

349
00:23:39,000 --> 00:23:44,000
That is just a little more
notation that will come in handy

350
00:23:44,000 --> 00:23:47,000
at some point.
More math now.

351
00:23:47,000 --> 00:23:50,000
Three steps,
particular solution,

352
00:23:50,000 --> 00:23:54,000
homogenous solution,
total solution/constants.

353
00:23:54,000 --> 00:24:00,000
This is almost like a mantra
here, like a chorus.

354
00:24:00,000 --> 00:24:03,000
Homogenous solution we compute
using a four-step method.

355
00:24:03,000 --> 00:24:06,000
And four-step method for
homogenous solutions,

356
00:24:06,000 --> 00:24:09,000
it turns out that it happens to
be that way for all the

357
00:24:09,000 --> 00:24:12,000
equations we will see in our
course.

358
00:24:12,000 --> 00:24:16,000
The first step would be assume
a solution of the form Ae^st.

359
00:24:16,000 --> 00:24:19,000
Exactly as with RCs.
If you close your eyes and do

360
00:24:19,000 --> 00:24:23,000
exactly what you did for RCs you
will get to where you want to

361
00:24:23,000 --> 00:24:25,000
be.
You assume a solution of the

362
00:24:25,000 --> 00:24:27,000
form Ae^st.
Substitute that into your

363
00:24:27,000 --> 00:24:31,000
homogenous equation.
Obtain the characteristic

364
00:24:31,000 --> 00:24:34,000
equation.
Solve for the roots.

365
00:24:34,000 --> 00:24:37,000
And then write down your
homogenous solution.

366
00:24:37,000 --> 00:24:42,000
Same sort of steps again and
again and again until you get

367
00:24:42,000 --> 00:24:45,000
bored to tears.
Particular solution.

368
00:24:45,000 --> 00:24:49,000
For the particular solution,
I simply need to find a

369
00:24:49,000 --> 00:24:53,000
solution, any solution,
if not the most general one but

370
00:24:53,000 --> 00:24:57,000
any solution that satisfies the
particular equation which

371
00:24:57,000 --> 00:25:03,000
satisfies that equation.
LCd^2vP/dt^2+vP=VI.

372
00:25:03,000 --> 00:25:09,000
My input is a step and I am
going to look for the solution

373
00:25:09,000 --> 00:25:15,000
for time t greater than zero.
Notice that for time t less

374
00:25:15,000 --> 00:25:20,000
than or equal to zero,
v is going to be zero.

375
00:25:20,000 --> 00:25:26,000
So I am looking for a solution
greater than t=0.

376
00:25:26,000 --> 00:25:34,000
Here, if I substitute vP=VI,
that is a particular solution.

377
00:25:34,000 --> 00:25:39,000
Because if I substitute VI here
this goes to zero and then I get

378
00:25:39,000 --> 00:25:43,000
VI=VI, so this works.
I promised you this was going

379
00:25:43,000 --> 00:25:47,000
to be simple.
You cannot get any simpler than

380
00:25:47,000 --> 00:25:49,000
that.
I have done my first step.

381
00:25:49,000 --> 00:25:52,000
I found the particular
solution.

382
00:25:52,000 --> 00:25:57,000
And VI is a good enough
particular solution so I will

383
00:25:57,000 --> 00:26:03,000
use it, I will take it.
As my second step I am going to

384
00:26:03,000 --> 00:26:08,000
find vH or the solution to the
homogenous equation.

385
00:26:08,000 --> 00:26:15,000
And the homogenous equation is
simply that equation with drive

386
00:26:15,000 --> 00:26:18,000
set to zero.
What I get here is

387
00:26:18,000 --> 00:26:23,000
LCd^2vH/dt^2+vH=0.
That is my homogenous equation.

388
00:26:23,000 --> 00:26:28,000
I simply set the drive to be
zero.

389
00:26:28,000 --> 00:26:32,000
And to find the solution here,
I go through my four-step

390
00:26:32,000 --> 00:26:34,000
method.
Again, in 6.002 following the

391
00:26:34,000 --> 00:26:38,000
kind of Occam's principle,
we just show you the absolute

392
00:26:38,000 --> 00:26:41,000
minimum necessary to get to
where you want.

393
00:26:41,000 --> 00:26:45,000
The absolute minimum necessary
is it turns out that we can

394
00:26:45,000 --> 00:26:50,000
solve all our differential
equations that we use here by

395
00:26:50,000 --> 00:26:54,000
using the methods of homogenous
and particular solutions.

396
00:26:54,000 --> 00:26:58,000
And every homogenous solution
can be solved by a four-step

397
00:26:58,000 --> 00:27:04,000
method.
That is about as minimal as it

398
00:27:04,000 --> 00:27:09,000
can get.
So no extraneous stuff there.

399
00:27:09,000 --> 00:27:13,000
The four-step method,
four steps.

400
00:27:13,000 --> 00:27:21,000
The first step is assume a
solution of the form vH=Ae^st.

401
00:27:21,000 --> 00:27:28,000
What I have noticed is that
students starting out are

402
00:27:28,000 --> 00:27:35,000
usually scared of differential
equations.

403
00:27:35,000 --> 00:27:36,000
I know I was when I was a
student.

404
00:27:36,000 --> 00:27:40,000
And the trick with differential
equations is that it is all a

405
00:27:40,000 --> 00:27:42,000
matter of psych.
Just because you see some

406
00:27:42,000 --> 00:27:46,000
squigglies and squagglies and a
bunch of math and so on you say

407
00:27:46,000 --> 00:27:49,000
oh, that must be hard.
But differential equations are

408
00:27:49,000 --> 00:27:52,000
actually the simplest thing
there is because in a large

409
00:27:52,000 --> 00:27:55,000
majority of cases the way you
solve them is you assume you

410
00:27:55,000 --> 00:27:59,000
know the answer,
someone tells you the answer.

411
00:27:59,000 --> 00:28:02,000
And then all you are left to do
is shove the answer into the

412
00:28:02,000 --> 00:28:05,000
equation and find out the
constants that makes it the

413
00:28:05,000 --> 00:28:07,000
answer.
Just a matter of psych.

414
00:28:07,000 --> 00:28:09,000
Psych yourselves that this
stuff is easy,

415
00:28:09,000 --> 00:28:11,000
because I am telling you what
the solution is.

416
00:28:11,000 --> 00:28:14,000
All you have to do is
substitute and verify.

417
00:28:14,000 --> 00:28:17,000
If you think about differential
equations that way or a large

418
00:28:17,000 --> 00:28:20,000
majority of them,
it really is very simple if you

419
00:28:20,000 --> 00:28:22,000
can just get past the squigglies
here.

420
00:28:22,000 --> 00:28:26,000
Just get past the squigglies
and then just simply stick in

421
00:28:26,000 --> 00:28:31,000
some simple stuff and it works.
I mean it just cannot get any

422
00:28:31,000 --> 00:28:34,000
easier.
I cannot think of any other

423
00:28:34,000 --> 00:28:39,000
field where the way you find a
solution is assume you know the

424
00:28:39,000 --> 00:28:43,000
solution and stick it in.
It has never made any sense to

425
00:28:43,000 --> 00:28:48,000
me but that is how it is.
So we assume the solution to

426
00:28:48,000 --> 00:28:51,000
the form Ae^st,
you stick it in there,

427
00:28:51,000 --> 00:28:55,000
and you have to find out the A
and s that make it so.

428
00:28:55,000 --> 00:28:58,000
It cannot get any simpler than
that.

429
00:28:58,000 --> 00:29:02,000
Let's stick the sucker in here
and see what we can get.

430
00:29:02,000 --> 00:29:07,000
Substitute Ae^st here I get
LCA, and second derivative,

431
00:29:07,000 --> 00:29:12,000
so it's s^2 e^st.
And Ae^st on this one here.

432
00:29:12,000 --> 00:29:17,000
And that equals zero.
And then let me just solve for

433
00:29:17,000 --> 00:29:22,000
whatever I can find.
Assuming I don't take the

434
00:29:22,000 --> 00:29:26,000
trivial case A=0,
I cancel these guys out.

435
00:29:26,000 --> 00:29:32,000
And what I am left with is
simply LCs^2+1=0.

436
00:29:32,000 --> 00:29:36,000
In other words,
what I end up getting is B,

437
00:29:36,000 --> 00:29:38,000
s^2=-1/LC.
My first step was,

438
00:29:38,000 --> 00:29:43,000
I am giving you solutions,
stick them in there,

439
00:29:43,000 --> 00:29:48,000
assume a solution of this form.
Second step is get the

440
00:29:48,000 --> 00:29:53,000
characteristic equation.
And the way you get the

441
00:29:53,000 --> 00:29:59,000
characteristic equation is that
you simply stick this guy in

442
00:29:59,000 --> 00:30:04,000
there.
And what you end up getting is

443
00:30:04,000 --> 00:30:09,000
some equation in s^2.
Do you remember what you got

444
00:30:09,000 --> 00:30:13,000
for first order circuits?
What s was?

445
00:30:13,000 --> 00:30:17,000
What is s?
For first order circuits,

446
00:30:17,000 --> 00:30:21,000
what did you get as a
characteristic equation?

447
00:30:21,000 --> 00:30:24,000
s+1/RC=0.
The same thing.

448
00:30:24,000 --> 00:30:30,000
Just remember to blindly apply
the steps.

449
00:30:30,000 --> 00:30:33,000
It will lead you to the answer.
This is called the

450
00:30:33,000 --> 00:30:37,000
"characteristic equation".
This is incredibly important.

451
00:30:37,000 --> 00:30:41,000
You will see in about a couple
weeks from now that once you

452
00:30:41,000 --> 00:30:44,000
write the characteristic
equation down for a circuit,

453
00:30:44,000 --> 00:30:48,000
it tells you all there is to
know about the circuit.

454
00:30:48,000 --> 00:30:51,000
And often times you can stop
solving right here.

455
00:30:51,000 --> 00:30:54,000
To experienced circuit
designers this tells me

456
00:30:54,000 --> 00:30:59,000
everything there is to know.
This is really key.

457
00:30:59,000 --> 00:31:02,000
That's why it's called a
characteristic equation.

458
00:31:02,000 --> 00:31:05,000
I believe in problem number
three of the homework that will

459
00:31:05,000 --> 00:31:09,000
be coming out this week,
that is exactly what you are

460
00:31:09,000 --> 00:31:11,000
going to do.
I am going to give you a

461
00:31:11,000 --> 00:31:15,000
circuit, ask you to get to the
characteristic equation quickly

462
00:31:15,000 --> 00:31:18,000
and then from there intuit the
solution.

463
00:31:18,000 --> 00:31:21,000
Write the characteristic
equation and then just intuit

464
00:31:21,000 --> 00:31:25,000
solution, it's that simple.
So, step A, assume a solution

465
00:31:25,000 --> 00:31:27,000
of the form, step B,
write the characteristic

466
00:31:27,000 --> 00:31:33,000
equation down.
And let me just simplify that a

467
00:31:33,000 --> 00:31:37,000
little bit.
I go ahead and find my roots.

468
00:31:37,000 --> 00:31:42,000
And my roots here,
remember that j is the square

469
00:31:42,000 --> 00:31:47,000
root of minus one.
And so what I end up getting

470
00:31:47,000 --> 00:31:53,000
is, my two roots here are,
plus j square root of 1/LC and

471
00:31:53,000 --> 00:31:59,000
minus j square root of 1/LC.
Two roots.

472
00:31:59,000 --> 00:32:03,000
And just as a shorthand
notation, much like I had a

473
00:32:03,000 --> 00:32:08,000
shorthand notation for RC,
what was my shorthand notation

474
00:32:08,000 --> 00:32:09,000
for RC?
Tau.

475
00:32:09,000 --> 00:32:15,000
Just as tau was big in first
order, we have a corresponding

476
00:32:15,000 --> 00:32:20,000
thing that is big in second
order and that is omega nought.

477
00:32:20,000 --> 00:32:24,000
Omega nought is simply square
root 1/LC.

478
00:32:24,000 --> 00:32:28,000
Just as tau was RC,
omega nought is a shorthand

479
00:32:28,000 --> 00:32:33,000
here.
And so s is simply plus or

480
00:32:33,000 --> 00:32:40,000
minus j omega nought.
Notice that in this equation

481
00:32:40,000 --> 00:32:48,000
here, if you take the square
root of LC there that has units

482
00:32:48,000 --> 00:32:56,000
of time, so one divided by that
has units of frequency.

483
00:32:56,000 --> 00:33:04,000
Notice that this guy is a
frequency in radians.

484
00:33:04,000 --> 00:33:09,000
I end up getting my roots of
the homogenous equation,

485
00:33:09,000 --> 00:33:13,000
and that is my third step.
And as my fourth step,

486
00:33:13,000 --> 00:33:18,000
I simply write down the
homogenous solution as

487
00:33:18,000 --> 00:33:23,000
substituting s with its roots
and writing the most general

488
00:33:23,000 --> 00:33:29,000
possible form of the solution,
and that would be A1e^(j omega

489
00:33:29,000 --> 00:33:34,000
nought t)+A2e^(-j omega nought
t).

490
00:33:34,000 --> 00:33:36,000
Done.
Some constant times this

491
00:33:36,000 --> 00:33:38,000
solution plus some other
constant times,

492
00:33:38,000 --> 00:33:41,000
the other solution.
Plus zero omega nought.

493
00:33:41,000 --> 00:33:44,000
Remember it comes from here,
Ae^st.

494
00:33:44,000 --> 00:33:48,000
I assume the solution of this
form, so my solution in this

495
00:33:48,000 --> 00:33:52,000
most general case would be s
being j omega nought in one

496
00:33:52,000 --> 00:33:55,000
case, minus j omega nought in
the other case,

497
00:33:55,000 --> 00:34:00,000
and I sum the two to get the
most general solution.

498
00:34:05,000 --> 00:34:10,000
So blasting ahead.
I now have my homogenous

499
00:34:10,000 --> 00:34:14,000
solution.
And as my third step of

500
00:34:14,000 --> 00:34:21,000
solution to differential
equations I write down the total

501
00:34:21,000 --> 00:34:27,000
solution, v=vP+vH,
particular plus the homogenous

502
00:34:27,000 --> 00:34:32,000
solutions.
And v=VI, was my particular

503
00:34:32,000 --> 00:34:38,000
solution, +A1e^(j omega nought
t)+A2e^(-j omega nought t) is my

504
00:34:38,000 --> 00:34:41,000
complete solution.
The final step,

505
00:34:41,000 --> 00:34:46,000
write down the total solution
and find the constants from the

506
00:34:46,000 --> 00:34:51,000
initial conditions.
To find the constants from the

507
00:34:51,000 --> 00:34:54,000
initial conditions,
let's start with,

508
00:34:54,000 --> 00:34:59,000
the voltage is zero to begin
with.

509
00:34:59,000 --> 00:35:03,000
This equation governs the
characteristics of v,

510
00:35:03,000 --> 00:35:08,000
so I need to find the initial
conditions.

511
00:35:08,000 --> 00:35:12,000
First of all,
I know that know that v(0)=0.

512
00:35:12,000 --> 00:35:17,000
From there I substitute t=0.
And so this goes to one,

513
00:35:17,000 --> 00:35:21,000
this goes to one,
and I end up getting

514
00:35:21,000 --> 00:35:25,000
0=VI+A1+A2.
That is my first expression.

515
00:35:25,000 --> 00:35:31,000
And then I am also given that
i(0)=0.

516
00:35:31,000 --> 00:35:37,000
And so I can get that as well.
How do I get i?

517
00:35:37,000 --> 00:35:41,000
This is v.
I know that i=Cdv/dt,

518
00:35:41,000 --> 00:35:47,000
so I can get i by simply
multiplying by C and

519
00:35:47,000 --> 00:35:52,000
differentiating this with
respect to t.

520
00:35:52,000 --> 00:36:00,000
I get C, this guy vanishes so I
get d/dt of this.

521
00:36:00,000 --> 00:36:07,000
So it is CA1(j omega nought)
e^(j omega nought t)+CA2(-j

522
00:36:07,000 --> 00:36:12,000
omega nought)e^(-j omega nought
t).

523
00:36:12,000 --> 00:36:21,000
From here I am given that that
is zero, and so therefore this

524
00:36:21,000 --> 00:36:26,000
guy becomes a one,
this guy becomes a one,

525
00:36:26,000 --> 00:36:34,000
j omega nought,
j omega nought cancel out.

526
00:36:34,000 --> 00:36:43,000
What I end up getting is A1=A2.
From the second initial

527
00:36:43,000 --> 00:36:50,000
condition I get A1=A2.
From these two,

528
00:36:50,000 --> 00:36:58,000
if I substitute here for A2,
I get VI + 2A1 = 0,

529
00:36:58,000 --> 00:37:05,000
or A1=-VI/2.
That is also equal to A2.

530
00:37:05,000 --> 00:37:13,000
Therefore, my total solution
now can be written in terms of

531
00:37:13,000 --> 00:37:19,000
the actual values of the
constants I have obtained.

532
00:37:19,000 --> 00:37:24,000
I get VI-VI/2.
So A1 and A2 are equal.

533
00:37:24,000 --> 00:37:33,000
I just pull them outside.
I pull VI-2 outside and I stick

534
00:37:33,000 --> 00:37:38,000
these two guys in parenthesis
in.

535
00:37:38,000 --> 00:37:46,000
Again, I promised you no more
circuits from here on until the

536
00:37:46,000 --> 00:37:52,000
very last board or something
like that.

537
00:37:52,000 --> 00:37:57,000
It is all math,
so not much else happening

538
00:37:57,000 --> 00:38:01,000
there.
More math.

539
00:38:01,000 --> 00:38:06,000
If you would like,
I could skip all the way to the

540
00:38:06,000 --> 00:38:13,000
end and show you the answer.
But I just love to write

541
00:38:13,000 --> 00:38:19,000
equations on the board so let me
just go through that.

542
00:38:19,000 --> 00:38:25,000
I am going to simplify this a
little further here.

543
00:38:25,000 --> 00:38:31,000
And we should remember this
form by the Euler relation,

544
00:38:31,000 --> 00:38:38,000
ejx=cos x+j sin x.
And by the same token,

545
00:38:38,000 --> 00:38:47,000
(e^jx + e^-jx)/2=cos x.
You all should know this from

546
00:38:47,000 --> 00:38:54,000
the Euler relation.
So were are using this guy

547
00:38:54,000 --> 00:39:04,000
here, ej^x + e^-jx=2cos x.
And so this one is 2 cosine of

548
00:39:04,000 --> 00:39:08,000
omega nought t,
2 and 2 cancel out,

549
00:39:08,000 --> 00:39:16,000
and what I am left with is
v(t)=VI-VI cos( omega nought t).

550
00:39:16,000 --> 00:39:24,000
And the current is Cdv/dt,
which is simply CVI sin( omega

551
00:39:24,000 --> 00:39:30,000
nought t).
Just remember that omega nought

552
00:39:30,000 --> 00:39:36,000
is the square root of 1/LC.
We are done.

553
00:39:36,000 --> 00:39:44,000
In fact, I did not give that
answer the importance that was

554
00:39:44,000 --> 00:39:48,000
due so let me just draw.

555
00:39:57,000 --> 00:40:00,000
There.
That is better.

556
00:40:00,000 --> 00:40:02,000
Enough math.
In a nutshell,

557
00:40:02,000 --> 00:40:06,000
what did we do.
We wrote the node method,

558
00:40:06,000 --> 00:40:10,000
it's a very simple circuit,
to write down the equation

559
00:40:10,000 --> 00:40:15,000
governing that circuit.
And then we grunged through a

560
00:40:15,000 --> 00:40:18,000
bunch of math.
Not a whole lot here.

561
00:40:18,000 --> 00:40:23,000
It is pretty simple.
And ended up with a relation

562
00:40:23,000 --> 00:40:28,000
that says the voltage across the
capacitor for a step input,

563
00:40:28,000 --> 00:40:33,000
assuming zero state,
is a constant VI-VI cos omega

564
00:40:33,000 --> 00:40:36,000
t.
Notice that even though I have

565
00:40:36,000 --> 00:40:39,000
a step input,
the circuit dynamics are such

566
00:40:39,000 --> 00:40:42,000
that I get a cosine in there.
You can begin to see where

567
00:40:42,000 --> 00:40:44,000
these cosines are coming from
now.

568
00:40:44,000 --> 00:40:47,000
They come in here.
And if you recall the example I

569
00:40:47,000 --> 00:40:50,000
showed you earlier of the
inverter circuit,

570
00:40:50,000 --> 00:40:52,000
remember there was a cosine
that decayed,

571
00:40:52,000 --> 00:40:56,000
that was sort of losing energy
and kind of dying out?

572
00:40:56,000 --> 00:41:00,000
So you can see where the
cosines are coming from.

573
00:41:00,000 --> 00:41:11,000
And just to draw you a little
sketch here.

574
00:41:11,000 --> 00:41:26,000
Let me draw v and i for you and
let me plot omega t,

575
00:41:26,000 --> 00:41:35,000
pi/2, pi and so on.
Let me plot VI.

576
00:41:35,000 --> 00:41:41,000
When time t=0,
VI=0, cosine omega t is one,

577
00:41:41,000 --> 00:41:48,000
and so VI-VI=0.
That is simply a cosine that

578
00:41:48,000 --> 00:41:56,000
starts out at zero here,
and at pi I get cosine omega t

579
00:41:56,000 --> 00:42:02,000
is minus one,
so I get plus VI on the other

580
00:42:02,000 --> 00:42:08,000
side.
So I end up at +2VI.

581
00:42:08,000 --> 00:42:13,000
At this point the voltage is
here.

582
00:42:13,000 --> 00:42:19,000
And notice that this guy looks
like this.

583
00:42:19,000 --> 00:42:27,000
It is a cosine that is
translated up so that its mean

584
00:42:27,000 --> 00:42:37,000
value is not zero but VI.
It is just a translation up of

585
00:42:37,000 --> 00:42:42,000
a cosine.
Similarly, in this case for the

586
00:42:42,000 --> 00:42:48,000
current it is a sinusoidal
characteristic.

587
00:42:48,000 --> 00:42:56,000
And it looks something like
this where the peak is given by

588
00:42:56,000 --> 00:43:00,000
CVI, oh, I messed up.

589
00:43:08,000 --> 00:43:14,000
When I differentiated this is
missed the omega nought out

590
00:43:14,000 --> 00:43:15,000
there.

591
00:43:25,000 --> 00:43:30,000
What I would like to do now --
This is the form of the output

592
00:43:30,000 --> 00:43:33,000
for a step input.
What I would like to do next is

593
00:43:33,000 --> 00:43:36,000
show you a demo.
But before I show you a demo,

594
00:43:36,000 --> 00:43:40,000
I always found it strange that
I have a step input and then I

595
00:43:40,000 --> 00:43:44,000
have two little elements,
how can I get a sine coming out

596
00:43:44,000 --> 00:43:47,000
of the output?
I would like to get some

597
00:43:47,000 --> 00:43:50,000
intuition as to why things
behave the way they are.

598
00:43:50,000 --> 00:43:54,000
I could go and pray to find
out, but let me just give you

599
00:43:54,000 --> 00:43:58,000
some very basic insight as to
why this behaves the way it

600
00:43:58,000 --> 00:44:02,000
does.
Let me draw the circuit for you

601
00:44:02,000 --> 00:44:05,000
here.
And this is my inductor L and

602
00:44:05,000 --> 00:44:08,000
capacitance C.
Remember this is v.

603
00:44:08,000 --> 00:44:13,000
Let me just walk you through
what is happening there and get

604
00:44:13,000 --> 00:44:17,000
you to understand this.
Now, you have seen sines occur

605
00:44:17,000 --> 00:44:20,000
before.
If you go and write down the

606
00:44:20,000 --> 00:44:24,000
equation of motion of a
pendulum, you know,

607
00:44:24,000 --> 00:44:28,000
you have a pendulum,
you move it to one side,

608
00:44:28,000 --> 00:44:31,000
let go.
It is also governed by

609
00:44:31,000 --> 00:44:35,000
sinusoidal characteristics.
And you will find that the

610
00:44:35,000 --> 00:44:39,000
equation governing its motion is
very much of the same form,

611
00:44:39,000 --> 00:44:43,000
and you get the sinusoid where
you have energy that is sloshing

612
00:44:43,000 --> 00:44:47,000
back and forth between maximum
potential energy to maximum

613
00:44:47,000 --> 00:44:51,000
kinetic energy and zero
potential energy back to maximum

614
00:44:51,000 --> 00:44:53,000
potential energy,
zero kinetic.

615
00:44:53,000 --> 00:44:56,000
So it is energy sloshing back
and forth.

616
00:44:56,000 --> 00:45:01,000
The same way here.
Capacitors and inductors store

617
00:45:01,000 --> 00:45:04,000
energy.
Let's walk through and see what

618
00:45:04,000 --> 00:45:06,000
happens.
I start off with both of them

619
00:45:06,000 --> 00:45:09,000
having the stage zero,
zero current,

620
00:45:09,000 --> 00:45:11,000
zero voltage.
I apply a step here.

621
00:45:11,000 --> 00:45:14,000
Boom, the step comes
instanteously to VI.

622
00:45:14,000 --> 00:45:18,000
I notice that the capacitor
voltage cannot change instantly

623
00:45:18,000 --> 00:45:22,000
unless there is an infinite
pulse of a sort,

624
00:45:22,000 --> 00:45:24,000
so this guy cannot change
instantly.

625
00:45:24,000 --> 00:45:29,000
And so its voltage starts off
being zero.

626
00:45:29,000 --> 00:45:32,000
So the entire voltage here,
KVL must be true no matter

627
00:45:32,000 --> 00:45:34,000
what.
They are absolutely fundamental

628
00:45:34,000 --> 00:45:36,000
principles from Maxwell's
equations.

629
00:45:36,000 --> 00:45:38,000
KVL must hold,
which means that the entire

630
00:45:38,000 --> 00:45:40,000
voltage VI must appear across
the inductor.

631
00:45:40,000 --> 00:45:44,000
I put a big voltage across the
inductor and its current begins

632
00:45:44,000 --> 00:45:45,000
to build up.
There you go.

633
00:45:45,000 --> 00:45:49,000
A voltage across the inductor,
its current begins to build up.

634
00:45:49,000 --> 00:45:52,000
As its current begins to build
up that current must flow

635
00:45:52,000 --> 00:45:54,000
through the capacitor,
too.

636
00:45:54,000 --> 00:45:57,000
And as current flows through a
capacitor it is depositing

637
00:45:57,000 --> 00:46:02,000
charge into the capacitor.
As the capacitor begins to get

638
00:46:02,000 --> 00:46:07,000
charge deposited on it,
its voltage begins to rise.

639
00:46:07,000 --> 00:46:12,000
Let's see what happens here.
Its voltage keeps rising.

640
00:46:12,000 --> 00:46:16,000
At some point,
the voltage across the

641
00:46:16,000 --> 00:46:21,000
capacitor is equal to VI.
But then VI equals this VI

642
00:46:21,000 --> 00:46:24,000
here.
So when the two become VI,

643
00:46:24,000 --> 00:46:29,000
the inductor has zero volts
across it.

644
00:46:29,000 --> 00:46:32,000
So there is no longer a
potential difference that is

645
00:46:32,000 --> 00:46:35,000
increasing the current in that
direction.

646
00:46:35,000 --> 00:46:37,000
At that point,
at pi divided by 2,

647
00:46:37,000 --> 00:46:42,000
I have some current going into
the inductor so there is no

648
00:46:42,000 --> 00:46:46,000
longer a pressure that is
forcing more current through the

649
00:46:46,000 --> 00:46:49,000
inductor because this voltage
reaches VI.

650
00:46:49,000 --> 00:46:53,000
But remember capacitors like to
sit around holding voltages.

651
00:46:53,000 --> 00:46:57,000
Just remember that demo.
That rinky-dink capacitor sat

652
00:46:57,000 --> 00:47:01,000
there stubbornly holding its
voltage.

653
00:47:01,000 --> 00:47:03,000
And it had a huge spark towards
the end.

654
00:47:03,000 --> 00:47:05,000
It just sat there holding its
voltage.

655
00:47:05,000 --> 00:47:09,000
In the same manner,
inductors love to sit around

656
00:47:09,000 --> 00:47:12,000
holding a current.
They will do whatever they can

657
00:47:12,000 --> 00:47:14,000
to keep the current going
through them.

658
00:47:14,000 --> 00:47:17,000
It has got the current going
through.

659
00:47:17,000 --> 00:47:19,000
And few forces on earth can
change that.

660
00:47:19,000 --> 00:47:22,000
And so therefore,
even though the capacitor

661
00:47:22,000 --> 00:47:26,000
voltage is VI and the voltage
drop across the inductor is

662
00:47:26,000 --> 00:47:30,000
zero, it still keeps supplying a
current.

663
00:47:30,000 --> 00:47:32,000
It has got the current.
It's got inertia.

664
00:47:32,000 --> 00:47:34,000
It keeps going.
It is like a runaway train.

665
00:47:34,000 --> 00:47:37,000
You may not be pushing the
train from the back,

666
00:47:37,000 --> 00:47:41,000
but once it is running it has
got kinetic energy and is going

667
00:47:41,000 --> 00:47:43,000
to run no matter what for a
least some more time,

668
00:47:43,000 --> 00:47:46,000
even if you take away the force
on the train.

669
00:47:46,000 --> 00:47:49,000
So I have taken away the force
on the punching more current

670
00:47:49,000 --> 00:47:52,000
through, but it has kinetic
energy.

671
00:47:52,000 --> 00:47:55,000
It has current flowing through
it so it continues to supply a

672
00:47:55,000 --> 00:47:57,000
current.
Because it continues to supply

673
00:47:57,000 --> 00:48:02,000
the current the capacitor
voltage keeps increasing.

674
00:48:02,000 --> 00:48:05,000
This is a subtle insight which
is absolutely spectacular that

675
00:48:05,000 --> 00:48:09,000
with zero volts across it,
it still keeps pumping that

676
00:48:09,000 --> 00:48:11,000
current.
Capacitor voltage has gone up.

677
00:48:11,000 --> 00:48:14,000
And guess what?
The voltage on this side is

678
00:48:14,000 --> 00:48:17,000
higher now but this guy is still
pumping a current.

679
00:48:17,000 --> 00:48:20,000
Man, I have been born to do
this, you know,

680
00:48:20,000 --> 00:48:24,000
I shall pump a current.
However, because the voltage

681
00:48:24,000 --> 00:48:29,000
has now gone up here gradually
the current begins to diminish.

682
00:48:29,000 --> 00:48:33,000
So the capacitor is concerned.
You pump a current into me,

683
00:48:33,000 --> 00:48:36,000
my voltage goes up.
At some point,

684
00:48:36,000 --> 00:48:39,000
like a runaway train,
it comes to a halt.

685
00:48:39,000 --> 00:48:44,000
The current through the
capacitor drains and now goes to

686
00:48:44,000 --> 00:48:47,000
zero and the capacitor voltage
reaches 2VI.

687
00:48:47,000 --> 00:48:50,000
So this is at 2VI now and this
is at VI.

688
00:48:50,000 --> 00:48:53,000
Now the situation is not in
equilibrium.

689
00:48:53,000 --> 00:48:57,000
At this point there is zero
current through it,

690
00:48:57,000 --> 00:49:01,000
but guess what?
I have a VI pumping in this

691
00:49:01,000 --> 00:49:05,000
direction now.
I have the same VI punching in

692
00:49:05,000 --> 00:49:07,000
this direction.
So guess what?

693
00:49:07,000 --> 00:49:11,000
Its current must now build up
in this direction and its

694
00:49:11,000 --> 00:49:14,000
current begins to build up in
that direction.

695
00:49:14,000 --> 00:49:18,000
That begins to discharge the
capacitor and the capacitor then

696
00:49:18,000 --> 00:49:22,000
goes on to a negative,
or the current goes down to a

697
00:49:22,000 --> 00:49:26,000
maximum negative current,
and this process continues.

698
00:49:26,000 --> 00:49:30,000
What you are seeing here is
energy.

699
00:49:30,000 --> 00:49:33,000
It is sloshing back and forth
between the two,

700
00:49:33,000 --> 00:49:37,000
and that is kind of a key.
I will just quickly put up a

701
00:49:37,000 --> 00:49:40,000
demo that you can watch as you
are walking out.

702
00:49:40,000 --> 00:49:44,000
With a step input,
notice the green is the voltage

703
00:49:44,000 --> 00:49:48,000
across the capacitor and the
orange is the current through

704
00:49:48,000 --> 00:49:51,000
the capacitor.