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Before I begin today,
I thought I would take the

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first five minutes and show you
some fun stuff I have been

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hacking on for the past three
years.

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This has to do with 6.002 and
circuits and all that stuff,

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but this is completely
optional, this is for fun,

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this is to go build your
intuition, this is to check your

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answers, whatever you want.
This is not a required part of

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the course.
Just for fun.

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There is this URL out here that
I put down here.

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I have been hacking on this
system for the past three years,

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and for the first time this
year and very tentatively and

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gingerly introducing it to
students.

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The idea here is that it is a,
that is kind of defocused.

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Any chance of focusing that a
little bit better?

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The idea of this is that it is
a Web-based interactive

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simulation package that I have
pulled together.

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And what you can do is you can
pull up a bunch of circuits.

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Notice that the URL is up here.
It is

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euryale.lcs.mit.edu/websim.
And there is the pointer to it.

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So you have a bunch of fun
things you can play with.

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And we have gone through all of
these things in lecture.

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Let's pick the MOSFET
amplifier.

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You come to this page.
This is something you have seen

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in class.
And let's play with this little

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circuit.
And you see the mouse?

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Good.
You can set up a bunch of

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parameters.
You can set up the MOSFET

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parameters VT and K.
You can set up the value of R

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for your resistor,
you can establish a bias

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voltage, and you can have an
input voltage vIN.

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So you can apply a bunch of
input voltages.

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You can apply a zero input,
unit in pulse,

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unit step, sine wave,
square waves.

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Or this was the part that took
me the longest to get right.

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You can also input a bunch of
music.

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And so far I just have two
clips, so you are going to get

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bored listening to them.
Good.

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So you can also input music.
And what you can do is you can

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watch the waveforms,
you can listen to the output

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and do a bunch of fun stuff.
One experiment I would love for

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you guys to try out.
Again, remember,

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this is completely optional.
Just for fun.

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You can apply some input.
Step input, for example,

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to an RLC circuit and spend 30
seconds thinking about what

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should the output look like.
I divine that the output should

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look like this and then do this
and see if what you thought was

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correct.
And it's fun to kind of play

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around with it.
Let me start with,

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just as an example,
let's say I input classical

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music.
And let us say I would like to

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listen to the output here that
is the voltage at the drain

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terminal of the MOSFET.
For listening it sets up a

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default timeframe to listen to,
so you go ahead and do it.

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This shows you the time domain
waveform of a clip of the music

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and then you can listen to it.
Lot's of distortion,

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right?
As you can see,

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there is a bunch of distortion.
And that is as you expect

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because the peak-to-peak voltage
is 1 volt, the bias is 2.5,

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and so this is clipping at the
lower end, plus the MOSFET is

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nonlinear.
You can play around with a

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bunch of things and you can have
a lot of fun.

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And the reason I created this
is that MIT is putting a bunch

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of its courses on the Web.
And one of the hottest things

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about courses like this is the
lab component.

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If you are beaming a course to,
say, a Third World country or

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something, how do you get people
to set up the massive lab

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infrastructure?
I know you hate your

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oscilloscopes,
I know you hate your wires,

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I know you hate the clips,
but the fact is you have them.

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I know a lot of places those
are way too expensive to pull

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together, which is why I have
been creating this Web-based

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kind of interactive laboratory
so that people can learn this

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stuff over the Web.
Let's go do another example

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very quickly.
Let's say you learned about,

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well, let's do RC circuits.
Here is the parallel RC

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circuit.
And you can set up capacitor

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values, resistor values,
you can set up input.

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Here, let me look at the time
domain waveform for the voltage

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across the capacitor.
And this time around let me

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play a unit step.
And let's see what the output

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is going to look like.
You can think in your minds

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what should the output look
like, and then you can go and

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plot it.
There you go.

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That's what the output looks
like.

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So you can play around with it
and have fun.

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That's all the good news.
The bad news is that so far I

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just have one Pentium III
machine behind us.

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It is a Linux box,
so don't all of you try it at

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once.
However, what I have also done,

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and that took me another six
months of hacking in the small

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amount of time professors have
to hack on stuff,

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I've hacked an incredibly
elaborate cashing system so that

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once anyone in class tries out
some combination of parameters

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it goes and squirrels away all
the outputs.

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If anybody else types in the
same sets of parameters it will

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just get all the output and play
it back to you.

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So if enough of you play with
over time, we may end up cashing

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all the important waveforms and
music clips and all of that

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stuff.
I have allocated a few

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gigabytes of storage,
so I am hoping that it may

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work.
Go forth.

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Play with it.
And this is completely my

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fault, so if there are any bugs
or anything simply email them to

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me.
This is the first time this is

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coming alive so bear with it.
Now let me switch back to the

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scheduled presentation for
today.

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All right, hope and pray that
this works.

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Yes.
Good.

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I am going to do today's
lecture using view graphs.

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And the reason I am going to do
that and not do my usual

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blackboard presentation which I
way, way, way prefer to a view

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graph presentation.
The only reason I am going to

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do this for today,
and maybe one more lecture,

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is that there is just a huge
amount of math grunge in this

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lecture.
What I want to do is kind of

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blast through that,
but you will have it all in the

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notes that you have,
so that you don't waste time in

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class as you watch me stumbling
over twiddles and tildes and all

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that stuff.
The key thing here is that the

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insight is actually very simple.
The beginning and the end are

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connected very tightly and very
simple.

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There is a bunch of math grunge
in the middle that we are going

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to work through and,
again, follows a complete old

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established pattern.
So, in that sense,

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there is really nothing
dramatically new in there.

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Let me spend the next five
minutes reviewing for you how we

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got here, what have we covered
so far and set up the

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presentation.
The first ten view graphs I am

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going to blast through and just
tell you where we are in terms

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of LC and RLC circuits.
I began by showing you this

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little demo, two inverters,
one driving.

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I can model the inductance here
with a little inductor,

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the capacitor of the gate here.
And recall that when I wanted

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to speed this up by introducing
a 50 ohm smaller resistance,

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I got some really strange
behavior.

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Just to remind you,
for Tuesday's lecture it would

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help if you quickly reviewed the
appendix on complex algebra in

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the course notes.
Remember all the real and

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imaginary j and omega stuff?
It would be good to very

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quickly skim through that.
It is a couple of pages.

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Remember this demo?
And the relevant circuit that

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is of interest to us is this one
here.

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It is the resistor,
there is the inductor and there

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is a capacitor.
This is Page 3.

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I am just going to blast
through the first ten view

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graphs.
It is all old stuff.

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Then we observed the following
output.

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We applied this input at VA and
we got this output,

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a very slowly rising waveform
because of the RC transient.

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And because of that you saw a
delay.

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Notice that this delay was
because of the slowly rising

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transient.
This waveform took some time to

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hit the threshold of the
neighboring transistor.

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So we say ah-ha,
let's try to speed this sucker

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up by reducing the resistance in
the collector of the first

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inverter.
And so I had this input.

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Now, to my surprise,
instead of seeing a nice little

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much higher and much faster
transitioning circuit,

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well, I did see a much faster
transitioning circuit but I got

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all this strange behavior on the
output that I was interested in.

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And because of that,
if these excursions were low

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enough, I could actually trigger
the output and get a whole bunch

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of false ones here because of
these negative excursions which

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should not really be there.
That was kind of strange.

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In the last lecture we said
let's take this one step at a

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time.
Let's not jump into an RLC

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circuit.
Let's go step by step.

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Let's start with an LC,
understand the behavior.

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We started off with an LC
circuit of this sort,

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and using the node equation we
showed that this was the

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equation that governed the
behavior of the circuit.

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And then we said that for a
step input and for zero initial

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conditions, that is the zero
state response,

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let's find out what the output,
the voltage across the

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capacitor looks like.
And so we obtained the total

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solution to be this.
And there was a sinusoidal term

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in there.
And the omega nought which was

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one by square root of LC.
And this was the circuit.

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And so for this step input
notice that the output looked

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like this.
So far an input step I had an

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output that went like this.
Notice that it is indeed

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possible for the output voltage
to actually go above the input

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value VI.
This is kind of non-intuitive

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but this can happen.
So this waveform jumps up and

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down.
But the steady state value,

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on average if you will,
is VI.

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On the other hand,
it does have sinusoidal

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excursions and this kind of goes
on because there is nothing to

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dissipate the energy inside that
circuit.

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By the way, the fact that the
capacitor voltage shoots above

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the input voltage is actually a
very important property.

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We won't dwell on it in 6.002,
but just squirrel that away in

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your brain somewhere.
I promise you that some time in

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your life you will have to
create a little design somewhere

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that will need a higher voltage
than your DC input.

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And you can use this primitive
fact to actually produce a DC

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voltage higher than you are
given, and then use that

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somehow.
In fact, there is a whole

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research area of what are called
DC to DC converters,

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voltage converters.
Let's say you have 1.5 volt

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battery, a AA battery,
but let's say a circuit needs

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1.8 volts.
The Pentium IIIs,

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for example,
needed 1.8 volts.

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And the strong arm is another
chip that required 1.8 volts a

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few years ago,
but the AA cell was 1.5 volts.

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How do get 1.8 from 1.5?
Well, you have to step it up

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somehow.
And this basic principle where

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the voltage can jump up above
the input is actually used,

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of course with additional
circuitry, to kind of get higher

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voltages.
It is a really key point that

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you can squirrel away.
This was pretty much where we

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got to in the last lecture.
This starts off the material

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for today.
What we are going to do is take

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that same circuit,
but instead we are going to put

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in this little resistor here.
This is what we set out to

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analyze.
And for details you can read

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the course notes Section 13.6.
The green curve here was the

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behavior of the LC circuit.
And what we are going to show

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today is that the moment we
introduce R this sinusoid here

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gets damp.
It kind of loses energy.

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And I am going to show you that
the behavior is going to look

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like this.
By introducing R this guy

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doesn't keep oscillating
forever.

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Rather it begins to oscillate
and then kind of loses energy

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and kind of gets tired and
settles down at VI.

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And remember the demo.
This is exactly what you saw in

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the demo.
You had a step input and you

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had this funny behavior.
And for the RLC that is exactly

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what it was.
So today's lecture will close

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the loop on what you saw in the
demo and the weird behavior,

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00:14:43,000 --> 00:14:48,000
and I am going to show you the
mathematics foundations for that

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00:14:48,000 --> 00:14:50,000
today.
Let's go ahead and analyze the

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00:14:50,000 --> 00:14:53,000
RLC circuit.
I purposely created the entire

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00:14:53,000 --> 00:14:57,000
presentation to follow as
closely as possible both the

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00:14:57,000 --> 00:15:02,000
discussion of the RC networks
and the LC networks so that the

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00:15:02,000 --> 00:15:06,000
math is all the same.
Exactly the same steps in the

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00:15:06,000 --> 00:15:10,000
mathematics are in the
exposition of the analysis.

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00:15:10,000 --> 00:15:13,000
What's different are the
results because the circuit is

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00:15:13,000 --> 00:15:15,000
different.
So don't get bogged down or

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00:15:15,000 --> 00:15:19,000
whatever in the mathematics.
Just remember it is the same

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00:15:19,000 --> 00:15:22,000
set of steps that you are going
to be applying.

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00:15:22,000 --> 00:15:26,000
We start by writing down the
element rules for our elements.

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00:15:26,000 --> 00:15:30,000
Nothing new here.
For the inductor V is Ldi/dt.

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00:15:30,000 --> 00:15:33,000
The integral form which is
simply 1/L integral vLdt=i.

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00:15:33,000 --> 00:15:37,000
We saw this the last time.
And for the capacitor,

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00:15:37,000 --> 00:15:41,000
the current through the
capacitor is simply Cdv/dt.

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00:15:41,000 --> 00:15:45,000
Those are the two element rules
for the capacitor and inductor.

248
00:15:45,000 --> 00:15:49,000
The element rule for the
resistor, of course,

249
00:15:49,000 --> 00:15:50,000
is V=iR.
You know that.

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00:15:50,000 --> 00:15:53,000
And for the voltage source we
know that, too,

251
00:15:53,000 --> 00:15:57,000
the voltage is a constant.
Just follow the same

252
00:15:57,000 --> 00:16:02,000
established pattern.
By the way, just so you are

253
00:16:02,000 --> 00:16:08,000
aware, I have booby trapped the
presentation a little bit to

254
00:16:08,000 --> 00:16:14,000
prevent you from falling asleep.
You see the dash lines here?

255
00:16:14,000 --> 00:16:19,000
Whenever you see a dash line,
that stuff needs to be copied

256
00:16:19,000 --> 00:16:22,000
down.
Don't trip over that.

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00:16:22,000 --> 00:16:27,000
Don't say I didn't warn you.
We start by using the usual

258
00:16:27,000 --> 00:16:31,000
node method.
And I have two nodes in this

259
00:16:31,000 --> 00:16:33,000
case.
Unlike the LC circuits,

260
00:16:33,000 --> 00:16:37,000
I have two unknown nodes.
One is this node here with the

261
00:16:37,000 --> 00:16:42,000
node voltage vA and the second
node is the node with voltage

262
00:16:42,000 --> 00:16:44,000
vT.
Let me start with vA and write

263
00:16:44,000 --> 00:16:47,000
the node equation for that.
It is simply 1/L,

264
00:16:47,000 --> 00:16:51,000
the node equation for this is
the current going in this

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00:16:51,000 --> 00:16:55,000
direction with is vI-vA integral
and that equals the current

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00:16:55,000 --> 00:17:00,000
going this way which is vA-v/R,
node equation.

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00:17:00,000 --> 00:17:03,000
I then write the node equation
for the node v,

268
00:17:03,000 --> 00:17:06,000
for this node here,
and that is simply

269
00:17:06,000 --> 00:17:09,000
(vA-v)/R=Cdvdt.
And that is what I have here,

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00:17:09,000 --> 00:17:13,000
two node equations.
Let me summarize the results

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00:17:13,000 --> 00:17:17,000
for you and then show you a view
graph where I grind through the

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00:17:17,000 --> 00:17:22,000
math as to how I got the result.
Here is the result I am going

273
00:17:22,000 --> 00:17:24,000
to get.
If I take these two node

274
00:17:24,000 --> 00:17:28,000
equations and I massage some of
the mathematics,

275
00:17:28,000 --> 00:17:34,000
I am going to get this result.
And I will show you that in a

276
00:17:34,000 --> 00:17:37,000
second.
By grinding through some math

277
00:17:37,000 --> 00:17:42,000
and solving these two equations
and expressing this in terms of

278
00:17:42,000 --> 00:17:46,000
v, I get a second order
differential equation,

279
00:17:46,000 --> 00:17:48,000
d^2v blah, blah,
blah.

280
00:17:48,000 --> 00:17:52,000
Notice that this is different
from the LC in this term.

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00:17:52,000 --> 00:17:57,000
Every step of the way you can
check to see if I am lying or I

282
00:17:57,000 --> 00:18:01,000
am correct.
I will indulge you,

283
00:18:01,000 --> 00:18:04,000
indulge myself rather with a
little story here.

284
00:18:04,000 --> 00:18:06,000
Richard Fineman was a known
smart guy.

285
00:18:06,000 --> 00:18:10,000
And one of the reasons that he
was that was in the middle of

286
00:18:10,000 --> 00:18:14,000
talks he was known to get up and
ask some of the darndest,

287
00:18:14,000 --> 00:18:17,000
hardest questions and say
ah-ha, you have a bug in this

288
00:18:17,000 --> 00:18:21,000
proof here or a bug in this
equation that is not right.

289
00:18:21,000 --> 00:18:23,000
And usually he would be
correct.

290
00:18:23,000 --> 00:18:27,000
So his trick in doing this and
which is one reason how he

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00:18:27,000 --> 00:18:31,000
became a known smart guy.
What he would do is,

292
00:18:31,000 --> 00:18:34,000
as the speaker went on talking
he would kind of follow along

293
00:18:34,000 --> 00:18:36,000
and think of a simple initial
primitive case.

294
00:18:36,000 --> 00:18:38,000
In this case,
I have an RLC circuit.

295
00:18:38,000 --> 00:18:40,000
So think of a simpler case of
this.

296
00:18:40,000 --> 00:18:43,000
A simpler case of this is R=0.
Whenever you set R to be zero,

297
00:18:43,000 --> 00:18:46,000
you should get exactly what we
got in the last lecture,

298
00:18:46,000 --> 00:18:48,000
correct?
That is what Fineman would do.

299
00:18:48,000 --> 00:18:50,000
He would boil this down to a
simpler case,

300
00:18:50,000 --> 00:18:52,000
make some assumptions and just
follow along.

301
00:18:52,000 --> 00:18:55,000
And whenever he found a
discrepancy between the math

302
00:18:55,000 --> 00:18:57,000
here and his simple case he
would say oh,

303
00:18:57,000 --> 00:19:02,000
there is a bug there.
If you want you can catch me

304
00:19:02,000 --> 00:19:04,000
that way.
Here, what Fineman would do is

305
00:19:04,000 --> 00:19:08,000
replace R being zero,
and notice then this equation

306
00:19:08,000 --> 00:19:12,000
here is exactly what we got the
last time with R being zero.

307
00:19:12,000 --> 00:19:14,000
Just remember that Fineman
trick.

308
00:19:14,000 --> 00:19:18,000
This is the equation we get,
the second-order differential

309
00:19:18,000 --> 00:19:21,000
equation with an R term in
there.

310
00:19:21,000 --> 00:19:25,000
And let me just grind through
the math and show you how I got

311
00:19:25,000 --> 00:19:28,000
this from this.
So the two node equations

312
00:19:28,000 --> 00:19:32,000
again.
And what I do is I start by

313
00:19:32,000 --> 00:19:38,000
taking these two equations and
differentiating this with

314
00:19:38,000 --> 00:19:42,000
respect to t and this is what I
get.

315
00:19:42,000 --> 00:19:48,000
And, at the same time,
I have replaced (vA-v)/R here

316
00:19:48,000 --> 00:19:52,000
by this term.
I replace this with this and

317
00:19:52,000 --> 00:19:57,000
differentiate.
Then I simply divide the whole

318
00:19:57,000 --> 00:20:02,000
thing by C.
Then I take this expression

319
00:20:02,000 --> 00:20:07,000
here and write down vA is equal
to this stuff here.

320
00:20:07,000 --> 00:20:12,000
Next I am going to substitute
this back for vA and eliminate

321
00:20:12,000 --> 00:20:14,000
vA.
So I take this vA,

322
00:20:14,000 --> 00:20:19,000
stick the sucker in here,
and thereby eliminate vA and

323
00:20:19,000 --> 00:20:23,000
get this.
And then I simplify it and here

324
00:20:23,000 --> 00:20:26,000
is what I get.
That is what I get.

325
00:20:26,000 --> 00:20:33,000
I just grind through the two
equations and get that result.

326
00:20:33,000 --> 00:20:36,000
So like a stuck record I will
repeat our mantra here,

327
00:20:36,000 --> 00:20:40,000
which is here is how we solve
the equations that we run across

328
00:20:40,000 --> 00:20:43,000
in this course,
the same three steps.

329
00:20:43,000 --> 00:20:47,000
Find the particular solution.
Find the homogenous solution.

330
00:20:47,000 --> 00:20:51,000
Find the total solution and
then find the constants using

331
00:20:51,000 --> 00:20:53,000
the initial conditions.
Same steps.

332
00:20:53,000 --> 00:20:56,000
You could recite this in your
sleep.

333
00:20:56,000 --> 00:21:00,000
And the homogenous solution is
obtained using a further four

334
00:21:00,000 --> 00:21:04,000
steps.
Let's just go through and apply

335
00:21:04,000 --> 00:21:07,000
this method to our equation and
get the results.

336
00:21:07,000 --> 00:21:11,000
vP is a particular solution and
vH is the homogenous solution.

337
00:21:11,000 --> 00:21:13,000
With a particular solution,
oh.

338
00:21:13,000 --> 00:21:17,000
Before I go on to do that,
let me set up my inputs and my

339
00:21:17,000 --> 00:21:21,000
state variables.
My input is going to be a step.

340
00:21:21,000 --> 00:21:25,000
Remember, I am trying to take
you to the point where the demo

341
00:21:25,000 --> 00:21:27,000
left off.
The demo had a step input,

342
00:21:27,000 --> 00:21:32,000
so I am going to use a step
input rising to vI.

343
00:21:32,000 --> 00:21:36,000
And I am going to with the
initial conditions being all

344
00:21:36,000 --> 00:21:39,000
zeros.
So the capacitor voltage is

345
00:21:39,000 --> 00:21:43,000
zero, inductor current,
another state variable is also

346
00:21:43,000 --> 00:21:48,000
zero, and therefore this is also
fondly called the ZSR or the

347
00:21:48,000 --> 00:21:53,000
zero state response because
there is only an input but zero

348
00:21:53,000 --> 00:21:55,000
state.
Again, remember the dashed

349
00:21:55,000 --> 00:22:00,000
lines here.
Don't say I didn't warn you.

350
00:22:00,000 --> 00:22:02,000
Let's start with a particular
solution.

351
00:22:02,000 --> 00:22:05,000
This is as simple as it gets.
I simply write down the

352
00:22:05,000 --> 00:22:07,000
particular equation and stick my
specific input.

353
00:22:07,000 --> 00:22:11,000
And remember the solution to
the particular equation is any

354
00:22:11,000 --> 00:22:13,000
old solution,
it doesn't have to be a general

355
00:22:13,000 --> 00:22:16,000
solution, any old solution that
satisfies it.

356
00:22:16,000 --> 00:22:18,000
And I am going to find a simple
solution here.

357
00:22:18,000 --> 00:22:20,000
And V particular is a constant
VI.

358
00:22:20,000 --> 00:22:22,000
It works.
Because remember this has been

359
00:22:22,000 --> 00:22:25,000
working all along.
And I am going to keep pushing

360
00:22:25,000 --> 00:22:30,000
this and see if this works until
the end of the course.

361
00:22:30,000 --> 00:22:31,000
Guess what?
It will.

362
00:22:31,000 --> 00:22:33,000
So this is a solution.
I'm done.

363
00:22:33,000 --> 00:22:36,000
That is my particular solution.
Simple.

364
00:22:36,000 --> 00:22:39,000
Second, I go and do my
homogenous solution.

365
00:22:39,000 --> 00:22:43,000
And the homogenous equation,
remember, is the same old

366
00:22:43,000 --> 00:22:47,000
differential equation with the
drive set to zero.

367
00:22:47,000 --> 00:22:51,000
Remember that sometimes this
equation with the drive set to

368
00:22:51,000 --> 00:22:55,000
zero is the entire equation you
have to deal with in situations

369
00:22:55,000 --> 00:23:00,000
where you have zero input,
for example.

370
00:23:00,000 --> 00:23:03,000
Or in other situations in which
you have an impulse at the

371
00:23:03,000 --> 00:23:06,000
input.
And the impulse simply sets up

372
00:23:06,000 --> 00:23:09,000
the initial conditions like a
charge in the capacitor or

373
00:23:09,000 --> 00:23:12,000
something like that.
So we are going to blast

374
00:23:12,000 --> 00:23:16,000
through this four-step method.
The method simply says that

375
00:23:16,000 --> 00:23:20,000
four steps, I am going to assume
a solution of the form Ae^st.

376
00:23:20,000 --> 00:23:23,000
And if you think you've seen
that before, yes,

377
00:23:23,000 --> 00:23:26,000
you have seen it many times
before.

378
00:23:26,000 --> 00:23:30,000
And you will see it again,
again and again.

379
00:23:30,000 --> 00:23:34,000
And we need to find A and s.
We want to form the

380
00:23:34,000 --> 00:23:39,000
characteristic equation,
find the roots of the equation

381
00:23:39,000 --> 00:23:44,000
and then write down the general
solution to the homogenous

382
00:23:44,000 --> 00:23:47,000
equation as this.
Same old same old.

383
00:23:47,000 --> 00:23:50,000
Let me just walk through the
steps here.

384
00:23:50,000 --> 00:23:54,000
Step A, assume a solution to
the form Ae^st.

385
00:23:54,000 --> 00:23:59,000
And so I substitute Ae^st as my
tentative solution to the

386
00:23:59,000 --> 00:24:04,000
equation.
Again, let me remind you that

387
00:24:04,000 --> 00:24:08,000
the differential equations that
we solve here are really easy

388
00:24:08,000 --> 00:24:12,000
because the way you solve them
is you begin by assuming you

389
00:24:12,000 --> 00:24:17,000
know the solution and stick it
in and find out what makes it

390
00:24:17,000 --> 00:24:19,000
work.
I am going to stick Ae^st into

391
00:24:19,000 --> 00:24:23,000
this differential equation,
and A comes out here.

392
00:24:23,000 --> 00:24:27,000
Differentiate this d squared,
I get s squared down here,

393
00:24:27,000 --> 00:24:31,000
A s here and this simply gets
stuck down here with the 1/LC

394
00:24:31,000 --> 00:24:35,000
coefficient.
The next step I begin

395
00:24:35,000 --> 00:24:39,000
eliminating what I can,
so I eliminate the A's,

396
00:24:39,000 --> 00:24:44,000
then eliminate the e^st's,
and I end up with this equation

397
00:24:44,000 --> 00:24:47,000
here.
I end up with this equation.

398
00:24:47,000 --> 00:24:50,000
This is my characteristic
equation.

399
00:24:50,000 --> 00:24:54,000
It is an equation in s.
Do people remember the

400
00:24:54,000 --> 00:25:00,000
characteristic equation we got
for the LC circuit?

401
00:25:00,000 --> 00:25:03,000
Remember the Fineman trick?
That's right,

402
00:25:03,000 --> 00:25:04,000
LC.
S^2+1/LC=0.

403
00:25:04,000 --> 00:25:08,000
This thing wasn't there.
All you do is simply follow the

404
00:25:08,000 --> 00:25:10,000
R.
Just follow the R.

405
00:25:10,000 --> 00:25:14,000
Just imagine this is a dollar
sign and kind of follow it.

406
00:25:14,000 --> 00:25:19,000
And you will see what the
differences are between the LC

407
00:25:19,000 --> 00:25:22,000
and the RLC.
So this is the characteristic

408
00:25:22,000 --> 00:25:24,000
equation.
What I am going to do,

409
00:25:24,000 --> 00:25:28,000
iss much as I wrote the
characteristic equation for the

410
00:25:28,000 --> 00:25:35,000
LC circuit, by substituting
omega nought squared for 1/LC.

411
00:25:35,000 --> 00:25:39,000
Let me do the same thing here
but introduce something for R

412
00:25:39,000 --> 00:25:42,000
and L as well.
What I will do is let me give

413
00:25:42,000 --> 00:25:46,000
you this canonic form.
The very first second-order

414
00:25:46,000 --> 00:25:50,000
equation I learned about when I
was a kid was this one,

415
00:25:50,000 --> 00:25:53,000
S^2+2AS+B^2 or something like
that.

416
00:25:53,000 --> 00:25:57,000
Let me write it in that form
where I get 2 alpha s plus omega

417
00:25:57,000 --> 00:26:02,000
nought squared.
Again, remember the alpha comes

418
00:26:02,000 --> 00:26:07,000
about because of R.
So omega nought squared is 1/LC

419
00:26:07,000 --> 00:26:11,000
and alpha is RL/2.
Omega nought squared is 1/LC

420
00:26:11,000 --> 00:26:17,000
and R/L is equal to two alpha.
I am just writing this in a

421
00:26:17,000 --> 00:26:22,000
simpler form so that from now on
going forward I am just going to

422
00:26:22,000 --> 00:26:27,000
deal with alphas and omega
noughts.

423
00:26:27,000 --> 00:26:29,000
Once I get to this
characteristic equation,

424
00:26:29,000 --> 00:26:32,000
after that I can give you one
generic way of solving it.

425
00:26:32,000 --> 00:26:35,000
And depending on the kind of
circuit you have,

426
00:26:35,000 --> 00:26:37,000
a series RLC,
which is what we have,

427
00:26:37,000 --> 00:26:40,000
or a parallel RLC we will
simply get different

428
00:26:40,000 --> 00:26:44,000
coefficients for the alpha term.
This is going to stay the same

429
00:26:44,000 --> 00:26:47,000
but this term will look
different, alpha is going to

430
00:26:47,000 --> 00:26:50,000
look different.
There is a real pattern here.

431
00:26:50,000 --> 00:26:53,000
And what I am doing is simply
focusing on what is important,

432
00:26:53,000 --> 00:26:56,000
what the differences are
between the pattern.

433
00:26:56,000 --> 00:27:01,000
You learned the LC situation
and the RLC situation.

434
00:27:01,000 --> 00:27:04,000
Given this I can now write
down, I am just simply replacing

435
00:27:04,000 --> 00:27:08,000
this as my characteristic
equation in dealing with alphas

436
00:27:08,000 --> 00:27:10,000
and omegas.
I will give you a physical

437
00:27:10,000 --> 00:27:12,000
significance of alpha in a
little bit.

438
00:27:12,000 --> 00:27:16,000
Do you remember the physical
significance of omega nought?

439
00:27:16,000 --> 00:27:18,000
That was the oscillation
frequency.

440
00:27:18,000 --> 00:27:20,000
In other words,
given an inductor and

441
00:27:20,000 --> 00:27:24,000
capacitor, you put some charge
on the capacitor and you watch

442
00:27:24,000 --> 00:27:27,000
it, it will oscillate.
And its oscillation frequency

443
00:27:27,000 --> 00:27:31,000
will be one by a square root of
LC.

444
00:27:31,000 --> 00:27:35,000
The magnitude of the initial
conditions will determine how

445
00:27:35,000 --> 00:27:40,000
high are the oscillations or
what the phase is in terms of

446
00:27:40,000 --> 00:27:43,000
when it starts,
but the frequency is going to

447
00:27:43,000 --> 00:27:46,000
be the same.
Step three, to solve the

448
00:27:46,000 --> 00:27:50,000
homogenous equation,
is find the roots of the

449
00:27:50,000 --> 00:27:53,000
equation, s1 and s2,
and here are my roots.

450
00:27:53,000 --> 00:27:56,000
Good old roots for a
second-order,

451
00:27:56,000 --> 00:28:02,000
little s squared equation here.
Finally, given that I have the

452
00:28:02,000 --> 00:28:06,000
roots, I can write down the
general homogenous solution.

453
00:28:06,000 --> 00:28:09,000
So general solution is simply
A1e^s1t, A2e^s2t.

454
00:28:09,000 --> 00:28:12,000
That's it.
That's the solution.

455
00:28:12,000 --> 00:28:16,000
This looks big and corny,
but we are going to make some

456
00:28:16,000 --> 00:28:20,000
simplifications as we go along
and show that it ends up boiling

457
00:28:20,000 --> 00:28:25,000
down to something cos omega t.
The math is kind of involved

458
00:28:25,000 --> 00:28:30,000
but we get down to something
very simple, a cosine.

459
00:28:30,000 --> 00:28:34,000
Hold this general solution.
From that, as a step three of

460
00:28:34,000 --> 00:28:38,000
the differential equation
solution, I write the total

461
00:28:38,000 --> 00:28:42,000
solution down.
And my total solution is the

462
00:28:42,000 --> 00:28:47,000
sum of the particular and the
homogenous, so therefore I get

463
00:28:47,000 --> 00:28:49,000
this.
VI was my particular and this

464
00:28:49,000 --> 00:28:52,000
term here is my homogenous
solution.

465
00:28:52,000 --> 00:28:57,000
Now, if I wasn't doing circuits
and simply trying to solve this

466
00:28:57,000 --> 00:29:02,000
mathematically here is what I
would do.

467
00:29:02,000 --> 00:29:06,000
I would find the unknown from
the initial conditions,

468
00:29:06,000 --> 00:29:09,000
so I know that v(0)=0.
And so therefore if I

469
00:29:09,000 --> 00:29:12,000
substitute zero for V(0) I get
this.

470
00:29:12,000 --> 00:29:15,000
If I substitute zero here,
t is 0, t is 0,

471
00:29:15,000 --> 00:29:19,000
so I simply get V1+A1+A2.
And let me just blast through

472
00:29:19,000 --> 00:29:22,000
because I am going to redo this
differently.

473
00:29:22,000 --> 00:29:25,000
i=Cdv/dt.
And so that's what I get.

474
00:29:25,000 --> 00:29:30,000
I substitute zero and this is
what I would get.

475
00:29:30,000 --> 00:29:32,000
I hurried through this.
Don't worry.

476
00:29:32,000 --> 00:29:35,000
I'm going to do it again.
If you just do it

477
00:29:35,000 --> 00:29:37,000
mathematically,
you can solve this equation

478
00:29:37,000 --> 00:29:42,000
here and these two simultaneous
equations in a1 and a2 and get

479
00:29:42,000 --> 00:29:44,000
the coefficients and you are
done.

480
00:29:44,000 --> 00:29:48,000
But it doesn't give us a whole
lot of insight into the behavior

481
00:29:48,000 --> 00:29:51,000
of these terms here.
What I am going to do for now

482
00:29:51,000 --> 00:29:55,000
is kind of ignore that.
Ignore I did that and instead

483
00:29:55,000 --> 00:30:00,000
try to go down a path that is a
little bit more intuitive.

484
00:30:00,000 --> 00:30:05,000
Let's stare at this expression
we got for the total solution.

485
00:30:05,000 --> 00:30:09,000
That is the expression we got.
All I did is,

486
00:30:09,000 --> 00:30:13,000
I had alpha in there,
I simply pulled out the alpha

487
00:30:13,000 --> 00:30:17,000
outside.
So this is my total solution,

488
00:30:17,000 --> 00:30:21,000
V1-A1e^(-alpha t) something
else and something else.

489
00:30:21,000 --> 00:30:26,000
Three cases to consider
depending on the relative values

490
00:30:26,000 --> 00:30:32,000
of alpha and omega nought.
If alpha is greater than omega

491
00:30:32,000 --> 00:30:35,000
nought then I get a real
quantity here.

492
00:30:35,000 --> 00:30:40,000
The square root of a positive
number, I get a real number,

493
00:30:40,000 --> 00:30:45,000
and that number will add up to
the minus alpha and I am going

494
00:30:45,000 --> 00:30:49,000
to get a solution that will look
like, oh, I'm sorry.

495
00:30:49,000 --> 00:30:53,000
Let me just do it a little
differently.

496
00:30:53,000 --> 00:30:55,000
There are three situations
here.

497
00:30:55,000 --> 00:31:00,000
One is alpha greater than omega
nought.

498
00:31:00,000 --> 00:31:04,000
Alpha equal to omega nought.
Alpha less than omega nought.

499
00:31:04,000 --> 00:31:06,000
Alpha is greater,
alpha is less,

500
00:31:06,000 --> 00:31:10,000
alpha is equal to this term
inside the square root sign.

501
00:31:10,000 --> 00:31:14,000
For reasons you will understand
shortly, we call this

502
00:31:14,000 --> 00:31:18,000
"overdamped" case,
the "underdamped" case and the

503
00:31:18,000 --> 00:31:22,000
"critically damped" case.
When alpha is greater than

504
00:31:22,000 --> 00:31:26,000
omega nought this term gives me
a real number,

505
00:31:26,000 --> 00:31:30,000
and I get something as simple
as this.

506
00:31:30,000 --> 00:31:33,000
Remember, for the series RLC
circuit, alpha was R/2L.

507
00:31:33,000 --> 00:31:36,000
So if R is big,
in other words,

508
00:31:36,000 --> 00:31:40,000
if in my RLC circuit R is huge
then I am going to get this

509
00:31:40,000 --> 00:31:43,000
situation.
My output voltage on the

510
00:31:43,000 --> 00:31:47,000
capacitor is going to look like
this, the sum of two

511
00:31:47,000 --> 00:31:50,000
exponentials.
And if I were to plot it very

512
00:31:50,000 --> 00:31:52,000
quickly for you,
for a VI step,

513
00:31:52,000 --> 00:31:56,000
V would look like this.
So v would simply look like

514
00:31:56,000 --> 00:32:02,000
this because it is the sum of a
couple of exponentials.

515
00:32:02,000 --> 00:32:04,000
All right.
Now, alpha is positive here.

516
00:32:04,000 --> 00:32:08,000
Remember alpha1 and alpha2 are
both positive.

517
00:32:08,000 --> 00:32:11,000
These two added up,
because of this constant VI,

518
00:32:11,000 --> 00:32:15,000
give rise to something that
increases in the following

519
00:32:15,000 --> 00:32:18,000
manner.
Let's look at the situation

520
00:32:18,000 --> 00:32:22,000
where alpha is less than omega
nought, where the term inside

521
00:32:22,000 --> 00:32:24,000
the square root sign is
negative.

522
00:32:24,000 --> 00:32:29,000
What I can do is pull the
negative sign out and express it

523
00:32:29,000 --> 00:32:32,000
this way.
What I am going to do is since

524
00:32:32,000 --> 00:32:36,000
alpha is less than omega nought,
I am going to reverse these two

525
00:32:36,000 --> 00:32:40,000
and pull out square root of
minus one to the outside.

526
00:32:40,000 --> 00:32:43,000
This is what I get.
I am just playing around with

527
00:32:43,000 --> 00:32:47,000
this so that whatever is under
the square root sign ends up

528
00:32:47,000 --> 00:32:49,000
giving me a positive real
number.

529
00:32:49,000 --> 00:32:52,000
So I pull the j outside and
this is what I get.

530
00:32:52,000 --> 00:32:55,000
Now, let me blast through a
bunch of math and end up with

531
00:32:55,000 --> 00:32:57,000
something very,
very simple for this

532
00:32:57,000 --> 00:33:02,000
underdamped case.
Let me define a few other

533
00:33:02,000 --> 00:33:05,000
terms.
I am going to call omega nought

534
00:33:05,000 --> 00:33:09,000
minus alpha squared the square
root of that.

535
00:33:09,000 --> 00:33:14,000
I am going to call it omega d.
And here is what I get.

536
00:33:14,000 --> 00:33:18,000
So I have defined three things
for you now, alpha,

537
00:33:18,000 --> 00:33:23,000
omega nought and omega d.
And I get this equation in

538
00:33:23,000 --> 00:33:28,000
terms of alpha and omega d.
And then, remember from your

539
00:33:28,000 --> 00:33:34,000
good-old Euler relationship?
e to the j omega d is simply

540
00:33:34,000 --> 00:33:37,000
cosine plus a j sine.
I am just going to blast

541
00:33:37,000 --> 00:33:40,000
through a bunch of math rather
quickly.

542
00:33:40,000 --> 00:33:44,000
Once I replace this in terms of
a cosine and sine,

543
00:33:44,000 --> 00:33:48,000
cosine and a j sine and then
collect all the coefficients

544
00:33:48,000 --> 00:33:52,000
together, I get an equation of
the form VI plus some constant e

545
00:33:52,000 --> 00:33:56,000
to the minus alpha t,
cosine, the sum of the constant

546
00:33:56,000 --> 00:34:00,000
e to the minus alpha t,
sine.

547
00:34:00,000 --> 00:34:02,000
Remember the sines and cosines
are coming out,

548
00:34:02,000 --> 00:34:06,000
but because of my R I am
getting this funny alpha here,

549
00:34:06,000 --> 00:34:09,000
e to the minus alpha here.
So I am getting sums of sine

550
00:34:09,000 --> 00:34:11,000
and cosine.
And K1 and K2 are some

551
00:34:11,000 --> 00:34:15,000
constants which I will need to
determine for my initial

552
00:34:15,000 --> 00:34:17,000
conditions.
I am going to continue on with

553
00:34:17,000 --> 00:34:21,000
this and keep on simplifying it
because, as I promised you,

554
00:34:21,000 --> 00:34:24,000
I want to get to something that
is just a cosine.

555
00:34:24,000 --> 00:34:27,000
I want to go down this path.
I am not going to cover this

556
00:34:27,000 --> 00:34:31,000
case, the critically damped
case.

557
00:34:31,000 --> 00:34:34,000
And I will touch upon it later
but not dwell on it.

558
00:34:34,000 --> 00:34:39,000
Let me continue down the path
of the underdamped case,

559
00:34:39,000 --> 00:34:43,000
and this is what we have.
Continuing with the math,

560
00:34:43,000 --> 00:34:47,000
let's start with the initial
conditions, v nought equals

561
00:34:47,000 --> 00:34:50,000
zero, and that gives me K1 is
simply -VI.

562
00:34:50,000 --> 00:34:54,000
So at v(0)=0 t is zero,
so this terms goes away,

563
00:34:54,000 --> 00:34:58,000
the cosine becomes a 1,
e^(alpha t) goes away,

564
00:34:58,000 --> 00:35:04,000
and K1=-VI.
Then I know that i(0) and i is

565
00:35:04,000 --> 00:35:08,000
simply Cdv/dt.
And I get this nasty

566
00:35:08,000 --> 00:35:13,000
expression.
I substitute t=0 and I get

567
00:35:13,000 --> 00:35:20,000
something that looks like this.
I know what K1 is,

568
00:35:20,000 --> 00:35:27,000
and so therefore K2 is simply
-V1alpha divided by omega

569
00:35:27,000 --> 00:35:31,000
nought.
I have taken this expression

570
00:35:31,000 --> 00:35:34,000
where the unknowns K1 and K2 are
to be found.

571
00:35:34,000 --> 00:35:39,000
I set the initial conditions
down at t=0 and I get K1 and K2

572
00:35:39,000 --> 00:35:42,000
as follows, which gives me the
following solution.

573
00:35:42,000 --> 00:35:46,000
This is the solution I get
where I do not have any unknowns

574
00:35:46,000 --> 00:35:49,000
anymore.
Remember that omega d and alpha

575
00:35:49,000 --> 00:35:52,000
are directly related to circuit
parameters.

576
00:35:52,000 --> 00:35:56,000
Alpha was R/2L and omega d was
square root of alpha squared

577
00:35:56,000 --> 00:35:59,000
minus omega nought squared.
** omega d = sqrt(alpha^2 -

578
00:35:59,000 --> 00:36:04,000
omega_0^2) **
And omega nought squared was 1

579
00:36:04,000 --> 00:36:07,000
by square root of LC.
So I know it all now.

580
00:36:07,000 --> 00:36:12,000
I still have sines and cosines
here, so I am going to simplify

581
00:36:12,000 --> 00:36:17,000
this a little further.
Oh, before I go on to do that,

582
00:36:17,000 --> 00:36:21,000
let's do the Fineman trick
again and notice if I am still

583
00:36:21,000 --> 00:36:25,000
true to the LC circuit I did the
last time.

584
00:36:25,000 --> 00:36:30,000
Remember when R goes to zero
alpha goes to zero.

585
00:36:30,000 --> 00:36:33,000
Because alpha is R divided by
2L.

586
00:36:33,000 --> 00:36:38,000
If alpha was zero what happens?
If alpha was zero,

587
00:36:38,000 --> 00:36:43,000
this guy goes to one,
this whole term goes to zero

588
00:36:43,000 --> 00:36:48,000
and omega dt now ends up
becoming omega nought,

589
00:36:48,000 --> 00:36:53,000
and I get this term here.
I get VI-VIcosine(omega t),

590
00:36:53,000 --> 00:37:00,000
which is exactly what I
expected in my equation.

591
00:37:00,000 --> 00:37:07,000
This is the same as the LC case
that I got.

592
00:37:07,000 --> 00:37:17,000
Let's go back to this situation
and simply if further.

593
00:37:17,000 --> 00:37:25,000
If you look at Appendix B.7 in
your course notes,

594
00:37:25,000 --> 00:37:34,000
Appendix B.7 is a quick
tutorial on trig.

595
00:37:34,000 --> 00:37:37,000
And in that trig tutorial you
will see that,

596
00:37:37,000 --> 00:37:41,000
and you have probably seen this
before, too, multiple times,

597
00:37:41,000 --> 00:37:43,000
the scaled sum of sines are
also sines.

598
00:37:43,000 --> 00:37:47,000
This is an incredibly cool fact
of sinusoids.

599
00:37:47,000 --> 00:37:51,000
If you take two sinusoids of
the same frequency and you scale

600
00:37:51,000 --> 00:37:55,000
them up in any which way and add
them up you also end up with a

601
00:37:55,000 --> 00:37:58,000
sinusoid.
It is hard to believe but it is

602
00:37:58,000 --> 00:38:02,000
true.
It is an incredible property of

603
00:38:02,000 --> 00:38:04,000
sinusoids.
Take any two sinusoids,

604
00:38:04,000 --> 00:38:07,000
scale them in any way you like,
same frequency,

605
00:38:07,000 --> 00:38:10,000
add them up,
you will get a sinusoid.

606
00:38:10,000 --> 00:38:13,000
What that is saying is that,
look, here is a sinusoid,

607
00:38:13,000 --> 00:38:17,000
here is a sinusoidal function,
and I am scaling them up in

608
00:38:17,000 --> 00:38:21,000
some manner.
So I should be able to add them

609
00:38:21,000 --> 00:38:24,000
up and be able to express that
as single sine.

610
00:38:24,000 --> 00:38:27,000
And to be sure you can,
look at the Appendix,

611
00:38:27,000 --> 00:38:31,000
and there is an expression for
a1 sinX plus a2 cosX is equal to

612
00:38:31,000 --> 00:38:35,000
a cosine of blah,
blah, blah.

613
00:38:35,000 --> 00:38:38,000
This is what you get.
No magic here.

614
00:38:38,000 --> 00:38:42,000
Just math.
From here I directly get this.

615
00:38:42,000 --> 00:38:47,000
And look at what I have.
It is absolutely unbelievable.

616
00:38:47,000 --> 00:38:51,000
v(t) is simply VI,
there is a constant here,

617
00:38:51,000 --> 00:38:58,000
this an e to the minus alpha
term and there is a cosine.

618
00:38:58,000 --> 00:39:02,000
Again, to pull the Fineman
trick, if this alpha were to go

619
00:39:02,000 --> 00:39:06,000
to zero here then you would end
up with the expression you had

620
00:39:06,000 --> 00:39:10,000
for the LC situation.
Let's stare at this a little

621
00:39:10,000 --> 00:39:12,000
while longer.
There is a constant plus a

622
00:39:12,000 --> 00:39:16,000
minus, a cosine term,
so there is a sinusoid at the

623
00:39:16,000 --> 00:39:21,000
output, and there is an e to the
minus alpha which ends up giving

624
00:39:21,000 --> 00:39:23,000
you the decay you have seen
before.

625
00:39:23,000 --> 00:39:25,000
In other words,
to a step input,

626
00:39:25,000 --> 00:39:30,000
the LC circuit would give you a
sinusoid.

627
00:39:30,000 --> 00:39:34,000
That is what the LC circuit
would do if alpha was zero.

628
00:39:34,000 --> 00:39:39,000
But because of this alpha term
here, e to the minus alpha t,

629
00:39:39,000 --> 00:39:43,000
that gives rise to a damping
effect, so this causes this

630
00:39:43,000 --> 00:39:48,000
thing to become smaller and
smaller as time goes by until

631
00:39:48,000 --> 00:39:51,000
this term goes to zero at t
equals infinity.

632
00:39:51,000 --> 00:39:56,000
This guy damps down and so
therefore you end up getting the

633
00:39:56,000 --> 00:40:02,000
curve that you saw like this.
Twenty minutes of juggling math

634
00:40:02,000 --> 00:40:06,000
solving a second-order
differential equation,

635
00:40:06,000 --> 00:40:11,000
but what ends up is the same
sinusoid but it is damped in the

636
00:40:11,000 --> 00:40:17,000
following manner such that the
frequency, rather the amplitude

637
00:40:17,000 --> 00:40:22,000
keeps decaying until it starts
off at zero and then settles

638
00:40:22,000 --> 00:40:25,000
down at vI.
This is exactly what you saw in

639
00:40:25,000 --> 00:40:30,000
the demo that we showed you
earlier.

640
00:40:30,000 --> 00:40:34,000
The critically damped case,
I am not going to do it here.

641
00:40:34,000 --> 00:40:37,000
I am going to point you to the
following insight.

642
00:40:37,000 --> 00:40:40,000
The underdamped case looked
like this.

643
00:40:40,000 --> 00:40:43,000
It was a sinusoid that kind of
decayed out.

644
00:40:43,000 --> 00:40:47,000
That is the underdamped case.
And then I showed you the

645
00:40:47,000 --> 00:40:51,000
overdamped case.
The overdamped case looked like

646
00:40:51,000 --> 00:40:53,000
this.
And, as you might expect,

647
00:40:53,000 --> 00:40:58,000
the critically damped case is
kind of in the middle and looks

648
00:40:58,000 --> 00:41:02,000
like this.
So the overdamped case would

649
00:41:02,000 --> 00:41:04,000
look like this,
underdamped like this,

650
00:41:04,000 --> 00:41:09,000
and the critically damped case
kind of goes up and kind of

651
00:41:09,000 --> 00:41:14,000
settles down almost immediately.
This is when alpha equals omega

652
00:41:14,000 --> 00:41:16,000
nought.
I won't do that case here,

653
00:41:16,000 --> 00:41:19,000
but I will simply point you to
Section 13.2.3.

654
00:41:19,000 --> 00:41:24,000
Just to tie things together,
recall this demo here that we

655
00:41:24,000 --> 00:41:28,000
showed you in class yesterday.
This is exactly the kind of

656
00:41:28,000 --> 00:41:34,000
form of the sinusoid you saw
because of that input step.

657
00:41:34,000 --> 00:41:39,000
If you want to see a complete
analysis of inverter pairs and

658
00:41:39,000 --> 00:41:43,000
look at the delays and so on
because of that,

659
00:41:43,000 --> 00:41:46,000
you can look at Page 170 and
example 898.

660
00:41:46,000 --> 00:41:51,000
In the next five or six
minutes, what I would like to do

661
00:41:51,000 --> 00:41:56,000
is stare at the RLC circuit.
And much like I showed you some

662
00:41:56,000 --> 00:42:01,000
intuitive methods to get the RC
response, what we are going to

663
00:42:01,000 --> 00:42:06,000
do is do the same thing for the
RLC.

664
00:42:06,000 --> 00:42:09,000
In the RLC situation,
much like the RC situation,

665
00:42:09,000 --> 00:42:13,000
experts don't go around writing
15 pages of differential

666
00:42:13,000 --> 00:42:18,000
equations and solving them each
time they see an RLC circuit.

667
00:42:18,000 --> 00:42:21,000
They stare at it and boom,
the response pops out,

668
00:42:21,000 --> 00:42:25,000
the sketch pops out.
This one is going to be another

669
00:42:25,000 --> 00:42:30,000
one like our Bend it Like
Beckham series here.

670
00:42:30,000 --> 00:42:34,000
And this one is in honor of
Leslie Kolodziejski.

671
00:42:34,000 --> 00:42:38,000
And I call it "Konquer it like
Kolodziejski".

672
00:42:38,000 --> 00:42:42,000
Again, as I said,
experts don't go around solving

673
00:42:42,000 --> 00:42:47,000
long differential equations and
spending ten pages of notes

674
00:42:47,000 --> 00:42:51,000
trying to get a sinusoid.
They look at a circuit and

675
00:42:51,000 --> 00:42:55,000
sketch response.
I am going to show you how to

676
00:42:55,000 --> 00:42:59,000
do that, too.
And what you can do is,

677
00:42:59,000 --> 00:43:02,000
to practice,
go to Websim and try out

678
00:43:02,000 --> 00:43:06,000
various combinations of inputs
and initial conditions and

679
00:43:06,000 --> 00:43:10,000
sketch it, time yourself,
give yourself 30 seconds or a

680
00:43:10,000 --> 00:43:13,000
minute if you like,
and sketch it and check it

681
00:43:13,000 --> 00:43:18,000
against the Websim response.
If it doesn't match either you

682
00:43:18,000 --> 00:43:20,000
are wrong or there is a bug in
Websim.

683
00:43:20,000 --> 00:43:24,000
What I am going to do is,
the response to the critically

684
00:43:24,000 --> 00:43:30,000
damped and underdamped case was
very easy to sketch out.

685
00:43:30,000 --> 00:43:33,000
You started with an initial
condition, you settled at VI and

686
00:43:33,000 --> 00:43:36,000
just kind of drew it like that.
The interesting case is the

687
00:43:36,000 --> 00:43:39,000
underdamped case,
and that is what I am going to

688
00:43:39,000 --> 00:43:41,000
dwell on.
Before we go on and I show you

689
00:43:41,000 --> 00:43:45,000
the intuitive method,
as a first step I would like to

690
00:43:45,000 --> 00:43:47,000
build some intuition.
Let's stare at this response

691
00:43:47,000 --> 00:43:50,000
here and try to understand what
is going on.

692
00:43:50,000 --> 00:43:52,000
This is the response that we
saw.

693
00:43:52,000 --> 00:43:55,000
And this fact that you see an
oscillation happening is also

694
00:43:55,000 --> 00:43:58,000
called "ringing".
You say that your circuit is

695
00:43:58,000 --> 00:44:00,000
ringing.
All right.

696
00:44:00,000 --> 00:44:04,000
You see some interesting facts.
You see that frequency of the

697
00:44:04,000 --> 00:44:08,000
ringing is given by omega d.
This cosine omega d,

698
00:44:08,000 --> 00:44:10,000
so that is the frequency omega
d.

699
00:44:10,000 --> 00:44:13,000
So the time is 2 pi divided by
omega d.

700
00:44:13,000 --> 00:44:17,000
The oscillation frequency is
omega d, but omega d is simply

701
00:44:17,000 --> 00:44:20,000
omega nought squared minus alpha
squared.

702
00:44:20,000 --> 00:44:24,000
Once you have a big value of R
alpha becomes very small and

703
00:44:24,000 --> 00:44:30,000
omega d is very commonly equal
to, very close to omega nought.

704
00:44:30,000 --> 00:44:34,000
So omega d and omega nought
very commonly are very close

705
00:44:34,000 --> 00:44:37,000
together.
And when that happens this

706
00:44:37,000 --> 00:44:40,000
frequency is directly omega
nought.

707
00:44:40,000 --> 00:44:43,000
Alpha governs how quickly your
sinusoid decays.

708
00:44:43,000 --> 00:44:48,000
e to the alpha t here is the
envelope that governs how

709
00:44:48,000 --> 00:44:52,000
quickly my sinusoid decays.
And notice that each of these

710
00:44:52,000 --> 00:44:56,000
terms, alpha and omega nought,
comes directly from my

711
00:44:56,000 --> 00:45:01,000
characteristic equation.
Which means that once you get

712
00:45:01,000 --> 00:45:05,000
your characteristic equation you
really don't have to do much

713
00:45:05,000 --> 00:45:07,000
else.
And up until now you still have

714
00:45:07,000 --> 00:45:10,000
to write the differential
equation to get the

715
00:45:10,000 --> 00:45:14,000
characteristic equation,
so you still have to do some

716
00:45:14,000 --> 00:45:17,000
differential equation stuff,
but in two lectures I am going

717
00:45:17,000 --> 00:45:20,000
to show you a way that you can
even write down the

718
00:45:20,000 --> 00:45:23,000
characteristic equation by
inspection.

719
00:45:23,000 --> 00:45:26,000
Look at your circuit and boom,
in 15 seconds or less write

720
00:45:26,000 --> 00:45:30,000
down the characteristic
equation.

721
00:45:30,000 --> 00:45:34,000
It is absolutely unbelievable.
What are the other factors that

722
00:45:34,000 --> 00:45:37,000
are interesting here?
Of course I need to find out

723
00:45:37,000 --> 00:45:39,000
initial values.
I start off at zero.

724
00:45:39,000 --> 00:45:43,000
This is my capacitor voltage.
If I don't have an infinite

725
00:45:43,000 --> 00:45:48,000
spike or an impulse my capacitor
voltage tries to stay where it

726
00:45:48,000 --> 00:45:51,000
is and starts off at zero.
And the final value is given by

727
00:45:51,000 --> 00:45:56,000
VI, the capacitor is a long-term
open so therefore VI appears

728
00:45:56,000 --> 00:45:59,000
across the capacitor.
In the long-term my final value

729
00:45:59,000 --> 00:46:04,000
is going to be VI.
There is one other interesting

730
00:46:04,000 --> 00:46:09,000
parameter, which I will simply
define today but dwell on about

731
00:46:09,000 --> 00:46:12,000
a week from today,
and that is called the Q.

732
00:46:12,000 --> 00:46:17,000
Some of you may have heard the
term oh, that's a high Q

733
00:46:17,000 --> 00:46:20,000
circuit.
Q is an indication of how ringy

734
00:46:20,000 --> 00:46:23,000
the circuit is.
And Q is defined as omega

735
00:46:23,000 --> 00:46:27,000
nought by 2 alpha.
It is called the "quality

736
00:46:27,000 --> 00:46:30,000
factor".
And it turns out that Q is

737
00:46:30,000 --> 00:46:33,000
approximately the number of
cycles of ringing.

738
00:46:33,000 --> 00:46:37,000
So if you have a high Q you
ring for a long time and if you

739
00:46:37,000 --> 00:46:39,000
have a low Q you ring for a very
short time.

740
00:46:39,000 --> 00:46:43,000
That is called the quality
factor defined by omega nought

741
00:46:43,000 --> 00:46:44,000
by 2 alpha.
Notice that Q,

742
00:46:44,000 --> 00:46:46,000
omega nought,
alpha, omega d,

743
00:46:46,000 --> 00:46:50,000
all of these come from the
terms in the characteristic

744
00:46:50,000 --> 00:46:52,000
equation.
We will spend more time on Q

745
00:46:52,000 --> 00:46:54,000
later.
With this insight here is how I

746
00:46:54,000 --> 00:46:58,000
can go about very quickly
sketching out the form of the

747
00:46:58,000 --> 00:47:01,000
response.
Here is my circuit.

748
00:47:01,000 --> 00:47:05,000
I want to sketch the form of
the response for a step input at

749
00:47:05,000 --> 00:47:08,000
vI.
Zero to vI step input here,

750
00:47:08,000 --> 00:47:11,000
I want to find out what happens
at this point.

751
00:47:11,000 --> 00:47:16,000
This is described to you in a
lot more detail in Section 13.8

752
00:47:16,000 --> 00:47:19,000
in your course notes.
Let's go through the steps.

753
00:47:19,000 --> 00:47:23,000
Let's do the really simple
situation first.

754
00:47:23,000 --> 00:47:27,000
Let's also assume for fun that
you are given that v(0) starts

755
00:47:27,000 --> 00:47:33,000
out being some positive value.
Some v(0) which is a positive

756
00:47:33,000 --> 00:47:35,000
number.
And, to make it harder on

757
00:47:35,000 --> 00:47:39,000
ourselves, let's say i(0) starts
out being some negative number.

758
00:47:39,000 --> 00:47:42,000
So i(0) is some negative
current.

759
00:47:42,000 --> 00:47:46,000
The first thing I know is v(0),
the capacitor voltage starts

760
00:47:46,000 --> 00:47:48,000
out here, which can change
suddenly.

761
00:47:48,000 --> 00:47:52,000
And I also know that in the
long-term this is an open

762
00:47:52,000 --> 00:47:55,000
circuit.
So that this voltage vI will

763
00:47:55,000 --> 00:48:00,000
appear directly across the
capacitor in the long-term.

764
00:48:00,000 --> 00:48:03,000
So I get starting out at v(0),
ending at vI,

765
00:48:03,000 --> 00:48:07,000
I am also half the way there.
I know the initial and ending

766
00:48:07,000 --> 00:48:10,000
point of the curve.
And then I know that somewhere

767
00:48:10,000 --> 00:48:13,000
in here there must be some funny
gyrations here,

768
00:48:13,000 --> 00:48:17,000
because remember I am dealing
with the underdamped case.

769
00:48:17,000 --> 00:48:21,000
And you can determine that from
alpha and omega nought.

770
00:48:21,000 --> 00:48:25,000
If alpha is less than omega
nought, you know that you are in

771
00:48:25,000 --> 00:48:30,000
the underdamped case and this is
what you get.

772
00:48:30,000 --> 00:48:33,000
Let's compute and write the
characteristic equation down.

773
00:48:33,000 --> 00:48:37,000
A week from today you will
write it by inspection,

774
00:48:37,000 --> 00:48:40,000
but for now you will do it by
writing down a differential

775
00:48:40,000 --> 00:48:43,000
equation.
And from the characteristic

776
00:48:43,000 --> 00:48:46,000
equation you will get omega d,
you will get alpha,

777
00:48:46,000 --> 00:48:49,000
omega nought and Q.
So omega d gives you the

778
00:48:49,000 --> 00:48:53,000
frequency of oscillations.
My frequency of oscillation is

779
00:48:53,000 --> 00:48:55,000
now known.
From Q I know how long it

780
00:48:55,000 --> 00:49:00,000
rings, because I know it rings
for about Q cycles.

781
00:49:00,000 --> 00:49:02,000
I know that ringing stops
approximately here.

782
00:49:02,000 --> 00:49:06,000
And then I know that between
that the start and end point my

783
00:49:06,000 --> 00:49:10,000
curve kind of looks like this,
something like this.

784
00:49:10,000 --> 00:49:12,000
Right there we are 95% of the
way there.

785
00:49:12,000 --> 00:49:16,000
The only question is I do not
know if it goes like this or it

786
00:49:16,000 --> 00:49:19,000
goes like this.
I am not quite sure yet if it

787
00:49:19,000 --> 00:49:22,000
starts off going high or starts
off going low.

788
00:49:22,000 --> 00:49:25,000
Not quite clear.
I also do not know what the

789
00:49:25,000 --> 00:49:30,000
maximum amplitude is.
It turns out this is rather

790
00:49:30,000 --> 00:49:34,000
complicated to determine so we
won't deal with that.

791
00:49:34,000 --> 00:49:37,000
Just simply so you can draw a
rough sketch.

792
00:49:37,000 --> 00:49:40,000
The questions is which way does
it start?

793
00:49:40,000 --> 00:49:43,000
I could leave it for you to
think about.

794
00:49:43,000 --> 00:49:47,000
Yeah, let me do that.
It is given on this page so

795
00:49:47,000 --> 00:49:50,000
don't look at it.
Think about it,

796
00:49:50,000 --> 00:49:55,000
and think about how you can
determine whether it goes up or

797
00:49:55,000 --> 00:49:57,000
down.
It turns out that in this case

798
00:49:57,000 --> 00:50:02,000
it is going to down and then
ring.

799
00:50:02,000 --> 00:50:06,000
See if you can figure it out
for yourselves and then we will

800
00:50:06,000 --> 00:50:09,000
talk about it next week.