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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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and I am going to show you the
mathematics foundations for that

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today.
Let's go ahead and analyze the

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RLC circuit.
I purposely created the entire

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presentation to follow as
closely as possible both the

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discussion of the RC networks
and the LC networks so that the

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math is all the same.
Exactly the same steps in the

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mathematics are in the
exposition of the analysis.

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What's different are the
results because the circuit is

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different.
So don't get bogged down or

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whatever in the mathematics.
Just remember it is the same

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set of steps that you are going
to be applying.

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We start by writing down the
element rules for our elements.

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Nothing new here.
For the inductor V is Ldi/dt.

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The integral form which is
simply 1/L integral vLdt=i.

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We saw this the last time.
And for the capacitor,

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the current through the
capacitor is simply Cdv/dt.

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Those are the two element rules
for the capacitor and inductor.

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The element rule for the
resistor, of course,

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is V=iR.
You know that.

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And for the voltage source we
know that, too,

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the voltage is a constant.
Just follow the same

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established pattern.
By the way, just so you are

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aware, I have booby trapped the
presentation a little bit to

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prevent you from falling asleep.
You see the dash lines here?

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Whenever you see a dash line,
that stuff needs to be copied

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down.
Don't trip over that.

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Don't say I didn't warn you.
We start by using the usual

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node method.
And I have two nodes in this

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

262
00:16:33 --> 00:16:37
I have two unknown nodes.
One is this node here with the

263
00:16:37 --> 00:16:42
node voltage vA and the second
node is the node with voltage

264
00:16:42 --> 00:16:44
vT.
Let me start with vA and write

265
00:16:44 --> 00:16:47
the node equation for that.
It is simply 1/L,

266
00:16:47 --> 00:16:51
the node equation for this is
the current going in this

267
00:16:51 --> 00:16:55
direction with is vI-vA integral
and that equals the current

268
00:16:55 --> 00:17:00
going this way which is vA-v/R,
node equation.

269
00:17:00 --> 00:17:03
I then write the node equation
for the node v,

270
00:17:03 --> 00:17:06
for this node here,
and that is simply

271
00:17:06 --> 00:17:09
(vA-v)/R=Cdvdt.
And that is what I have here,

272
00:17:09 --> 00:17:13
two node equations.
Let me summarize the results

273
00:17:13 --> 00:17:17
for you and then show you a view
graph where I grind through the

274
00:17:17 --> 00:17:22
math as to how I got the result.
Here is the result I am going

275
00:17:22 --> 00:17:24
to get.
If I take these two node

276
00:17:24 --> 00:17:28
equations and I massage some of
the mathematics,

277
00:17:28 --> 00:17:34
I am going to get this result.
And I will show you that in a

278
00:17:34 --> 00:17:37
second.
By grinding through some math

279
00:17:37 --> 00:17:42
and solving these two equations
and expressing this in terms of

280
00:17:42 --> 00:17:46
v, I get a second order
differential equation,

281
00:17:46 --> 00:17:48
d^2v blah, blah,
blah.

282
00:17:48 --> 00:17:52
Notice that this is different
from the LC in this term.

283
00:17:52 --> 00:17:57
Every step of the way you can
check to see if I am lying or I

284
00:17:57 --> 00:18:01
am correct.
I will indulge you,

285
00:18:01 --> 00:18:04
indulge myself rather with a
little story here.

286
00:18:04 --> 00:18:06
Richard Fineman was a known
smart guy.

287
00:18:06 --> 00:18:10
And one of the reasons that he
was that was in the middle of

288
00:18:10 --> 00:18:14
talks he was known to get up and
ask some of the darndest,

289
00:18:14 --> 00:18:17
hardest questions and say
ah-ha, you have a bug in this

290
00:18:17 --> 00:18:21
proof here or a bug in this
equation that is not right.

291
00:18:21 --> 00:18:23
And usually he would be
correct.

292
00:18:23 --> 00:18:27
So his trick in doing this and
which is one reason how he

293
00:18:27 --> 00:18:31
became a known smart guy.
What he would do is,

294
00:18:31 --> 00:18:34
as the speaker went on talking
he would kind of follow along

295
00:18:34 --> 00:18:36
and think of a simple initial
primitive case.

296
00:18:36 --> 00:18:38
In this case,
I have an RLC circuit.

297
00:18:38 --> 00:18:40
So think of a simpler case of
this.

298
00:18:40 --> 00:18:43
A simpler case of this is R=0.
Whenever you set R to be zero,

299
00:18:43 --> 00:18:46
you should get exactly what we
got in the last lecture,

300
00:18:46 --> 00:18:48
correct?
That is what Fineman would do.

301
00:18:48 --> 00:18:50
He would boil this down to a
simpler case,

302
00:18:50 --> 00:18:52
make some assumptions and just
follow along.

303
00:18:52 --> 00:18:55
And whenever he found a
discrepancy between the math

304
00:18:55 --> 00:18:57
here and his simple case he
would say oh,

305
00:18:57 --> 00:19:02
there is a bug there.
If you want you can catch me

306
00:19:02 --> 00:19:04
that way.
Here, what Fineman would do is

307
00:19:04 --> 00:19:08
replace R being zero,
and notice then this equation

308
00:19:08 --> 00:19:12
here is exactly what we got the
last time with R being zero.

309
00:19:12 --> 00:19:14
Just remember that Fineman
trick.

310
00:19:14 --> 00:19:18
This is the equation we get,
the second-order differential

311
00:19:18 --> 00:19:21
equation with an R term in
there.

312
00:19:21 --> 00:19:25
And let me just grind through
the math and show you how I got

313
00:19:25 --> 00:19:28
this from this.
So the two node equations

314
00:19:28 --> 00:19:32
again.
And what I do is I start by

315
00:19:32 --> 00:19:38
taking these two equations and
differentiating this with

316
00:19:38 --> 00:19:42
respect to t and this is what I
get.

317
00:19:42 --> 00:19:48
And, at the same time,
I have replaced (vA-v)/R here

318
00:19:48 --> 00:19:52
by this term.
I replace this with this and

319
00:19:52 --> 00:19:57
differentiate.
Then I simply divide the whole

320
00:19:57 --> 00:20:02
thing by C.
Then I take this expression

321
00:20:02 --> 00:20:07
here and write down vA is equal
to this stuff here.

322
00:20:07 --> 00:20:12
Next I am going to substitute
this back for vA and eliminate

323
00:20:12 --> 00:20:14
vA.
So I take this vA,

324
00:20:14 --> 00:20:19
stick the sucker in here,
and thereby eliminate vA and

325
00:20:19 --> 00:20:23
get this.
And then I simplify it and here

326
00:20:23 --> 00:20:26
is what I get.
That is what I get.

327
00:20:26 --> 00:20:33
I just grind through the two
equations and get that result.

328
00:20:33 --> 00:20:36
So like a stuck record I will
repeat our mantra here,

329
00:20:36 --> 00:20:40
which is here is how we solve
the equations that we run across

330
00:20:40 --> 00:20:43
in this course,
the same three steps.

331
00:20:43 --> 00:20:47
Find the particular solution.
Find the homogenous solution.

332
00:20:47 --> 00:20:51
Find the total solution and
then find the constants using

333
00:20:51 --> 00:20:53
the initial conditions.
Same steps.

334
00:20:53 --> 00:20:56
You could recite this in your
sleep.

335
00:20:56 --> 00:21:00
And the homogenous solution is
obtained using a further four

336
00:21:00 --> 00:21:04
steps.
Let's just go through and apply

337
00:21:04 --> 00:21:07
this method to our equation and
get the results.

338
00:21:07 --> 00:21:11
vP is a particular solution and
vH is the homogenous solution.

339
00:21:11 --> 00:21:13
With a particular solution,
oh.

340
00:21:13 --> 00:21:17
Before I go on to do that,
let me set up my inputs and my

341
00:21:17 --> 00:21:21
state variables.
My input is going to be a step.

342
00:21:21 --> 00:21:25
Remember, I am trying to take
you to the point where the demo

343
00:21:25 --> 00:21:27
left off.
The demo had a step input,

344
00:21:27 --> 00:21:32
so I am going to use a step
input rising to vI.

345
00:21:32 --> 00:21:36
And I am going to with the
initial conditions being all

346
00:21:36 --> 00:21:39
zeros.
So the capacitor voltage is

347
00:21:39 --> 00:21:43
zero, inductor current,
another state variable is also

348
00:21:43 --> 00:21:48
zero, and therefore this is also
fondly called the ZSR or the

349
00:21:48 --> 00:21:53
zero state response because
there is only an input but zero

350
00:21:53 --> 00:21:55
state.
Again, remember the dashed

351
00:21:55 --> 00:22:00
lines here.
Don't say I didn't warn you.

352
00:22:00 --> 00:22:02
Let's start with a particular
solution.

353
00:22:02 --> 00:22:05
This is as simple as it gets.
I simply write down the

354
00:22:05 --> 00:22:07
particular equation and stick my
specific input.

355
00:22:07 --> 00:22:11
And remember the solution to
the particular equation is any

356
00:22:11 --> 00:22:13
old solution,
it doesn't have to be a general

357
00:22:13 --> 00:22:16
solution, any old solution that
satisfies it.

358
00:22:16 --> 00:22:18
And I am going to find a simple
solution here.

359
00:22:18 --> 00:22:20
And V particular is a constant
VI.

360
00:22:20 --> 00:22:22
It works.
Because remember this has been

361
00:22:22 --> 00:22:25
working all along.
And I am going to keep pushing

362
00:22:25 --> 00:22:30
this and see if this works until
the end of the course.

363
00:22:30 --> 00:22:31
Guess what?
It will.

364
00:22:31 --> 00:22:33
So this is a solution.
I'm done.

365
00:22:33 --> 00:22:36
That is my particular solution.
Simple.

366
00:22:36 --> 00:22:39
Second, I go and do my
homogenous solution.

367
00:22:39 --> 00:22:43
And the homogenous equation,
remember, is the same old

368
00:22:43 --> 00:22:47
differential equation with the
drive set to zero.

369
00:22:47 --> 00:22:51
Remember that sometimes this
equation with the drive set to

370
00:22:51 --> 00:22:55
zero is the entire equation you
have to deal with in situations

371
00:22:55 --> 00:23:00
where you have zero input,
for example.

372
00:23:00 --> 00:23:03
Or in other situations in which
you have an impulse at the

373
00:23:03 --> 00:23:06
input.
And the impulse simply sets up

374
00:23:06 --> 00:23:09
the initial conditions like a
charge in the capacitor or

375
00:23:09 --> 00:23:12
something like that.
So we are going to blast

376
00:23:12 --> 00:23:16
through this four-step method.
The method simply says that

377
00:23:16 --> 00:23:20
four steps, I am going to assume
a solution of the form Ae^st.

378
00:23:20 --> 00:23:23
And if you think you've seen
that before, yes,

379
00:23:23 --> 00:23:26
you have seen it many times
before.

380
00:23:26 --> 00:23:30
And you will see it again,
again and again.

381
00:23:30 --> 00:23:34
And we need to find A and s.
We want to form the

382
00:23:34 --> 00:23:39
characteristic equation,
find the roots of the equation

383
00:23:39 --> 00:23:44
and then write down the general
solution to the homogenous

384
00:23:44 --> 00:23:47
equation as this.
Same old same old.

385
00:23:47 --> 00:23:50
Let me just walk through the
steps here.

386
00:23:50 --> 00:23:54
Step A, assume a solution to
the form Ae^st.

387
00:23:54 --> 00:23:59
And so I substitute Ae^st as my
tentative solution to the

388
00:23:59 --> 00:24:04
equation.
Again, let me remind you that

389
00:24:04 --> 00:24:08
the differential equations that
we solve here are really easy

390
00:24:08 --> 00:24:12
because the way you solve them
is you begin by assuming you

391
00:24:12 --> 00:24:17
know the solution and stick it
in and find out what makes it

392
00:24:17 --> 00:24:19
work.
I am going to stick Ae^st into

393
00:24:19 --> 00:24:23
this differential equation,
and A comes out here.

394
00:24:23 --> 00:24:27
Differentiate this d squared,
I get s squared down here,

395
00:24:27 --> 00:24:31
A s here and this simply gets
stuck down here with the 1/LC

396
00:24:31 --> 00:24:35
coefficient.
The next step I begin

397
00:24:35 --> 00:24:39
eliminating what I can,
so I eliminate the A's,

398
00:24:39 --> 00:24:44
then eliminate the e^st's,
and I end up with this equation

399
00:24:44 --> 00:24:47
here.
I end up with this equation.

400
00:24:47 --> 00:24:50
This is my characteristic
equation.

401
00:24:50 --> 00:24:54
It is an equation in s.
Do people remember the

402
00:24:54 --> 00:25:00
characteristic equation we got
for the LC circuit?

403
00:25:00 --> 00:25:03
Remember the Fineman trick?
That's right,

404
00:25:03 --> 00:25:04
LC.
S^2+1/LC=0.

405
00:25:04 --> 00:25:08
This thing wasn't there.
All you do is simply follow the

406
00:25:08 --> 00:25:10
R.
Just follow the R.

407
00:25:10 --> 00:25:14
Just imagine this is a dollar
sign and kind of follow it.

408
00:25:14 --> 00:25:19
And you will see what the
differences are between the LC

409
00:25:19 --> 00:25:22
and the RLC.
So this is the characteristic

410
00:25:22 --> 00:25:24
equation.
What I am going to do,

411
00:25:24 --> 00:25:28
iss much as I wrote the
characteristic equation for the

412
00:25:28 --> 00:25:35
LC circuit, by substituting
omega nought squared for 1/LC.

413
00:25:35 --> 00:25:39
Let me do the same thing here
but introduce something for R

414
00:25:39 --> 00:25:42
and L as well.
What I will do is let me give

415
00:25:42 --> 00:25:46
you this canonic form.
The very first second-order

416
00:25:46 --> 00:25:50
equation I learned about when I
was a kid was this one,

417
00:25:50 --> 00:25:53
S^2+2AS+B^2 or something like
that.

418
00:25:53 --> 00:25:57
Let me write it in that form
where I get 2 alpha s plus omega

419
00:25:57 --> 00:26:02
nought squared.
Again, remember the alpha comes

420
00:26:02 --> 00:26:07
about because of R.
So omega nought squared is 1/LC

421
00:26:07 --> 00:26:11
and alpha is RL/2.
Omega nought squared is 1/LC

422
00:26:11 --> 00:26:17
and R/L is equal to two alpha.
I am just writing this in a

423
00:26:17 --> 00:26:22
simpler form so that from now on
going forward I am just going to

424
00:26:22 --> 00:26:27
deal with alphas and omega
noughts.

425
00:26:27 --> 00:26:29
Once I get to this
characteristic equation,

426
00:26:29 --> 00:26:32
after that I can give you one
generic way of solving it.

427
00:26:32 --> 00:26:35
And depending on the kind of
circuit you have,

428
00:26:35 --> 00:26:37
a series RLC,
which is what we have,

429
00:26:37 --> 00:26:40
or a parallel RLC we will
simply get different

430
00:26:40 --> 00:26:44
coefficients for the alpha term.
This is going to stay the same

431
00:26:44 --> 00:26:47
but this term will look
different, alpha is going to

432
00:26:47 --> 00:26:50
look different.
There is a real pattern here.

433
00:26:50 --> 00:26:53
And what I am doing is simply
focusing on what is important,

434
00:26:53 --> 00:26:56
what the differences are
between the pattern.

435
00:26:56 --> 00:27:01
You learned the LC situation
and the RLC situation.

436
00:27:01 --> 00:27:04
Given this I can now write
down, I am just simply replacing

437
00:27:04 --> 00:27:08
this as my characteristic
equation in dealing with alphas

438
00:27:08 --> 00:27:10
and omegas.
I will give you a physical

439
00:27:10 --> 00:27:12
significance of alpha in a
little bit.

440
00:27:12 --> 00:27:16
Do you remember the physical
significance of omega nought?

441
00:27:16 --> 00:27:18
That was the oscillation
frequency.

442
00:27:18 --> 00:27:20
In other words,
given an inductor and

443
00:27:20 --> 00:27:24
capacitor, you put some charge
on the capacitor and you watch

444
00:27:24 --> 00:27:27
it, it will oscillate.
And its oscillation frequency

445
00:27:27 --> 00:27:31
will be one by a square root of
LC.

446
00:27:31 --> 00:27:35
The magnitude of the initial
conditions will determine how

447
00:27:35 --> 00:27:40
high are the oscillations or
what the phase is in terms of

448
00:27:40 --> 00:27:43
when it starts,
but the frequency is going to

449
00:27:43 --> 00:27:46
be the same.
Step three, to solve the

450
00:27:46 --> 00:27:50
homogenous equation,
is find the roots of the

451
00:27:50 --> 00:27:53
equation, s1 and s2,
and here are my roots.

452
00:27:53 --> 00:27:56
Good old roots for a
second-order,

453
00:27:56 --> 00:28:02
little s squared equation here.
Finally, given that I have the

454
00:28:02 --> 00:28:06
roots, I can write down the
general homogenous solution.

455
00:28:06 --> 00:28:09
So general solution is simply
A1e^s1t, A2e^s2t.

456
00:28:09 --> 00:28:12
That's it.
That's the solution.

457
00:28:12 --> 00:28:16
This looks big and corny,
but we are going to make some

458
00:28:16 --> 00:28:20
simplifications as we go along
and show that it ends up boiling

459
00:28:20 --> 00:28:25
down to something cos omega t.
The math is kind of involved

460
00:28:25 --> 00:28:30
but we get down to something
very simple, a cosine.

461
00:28:30 --> 00:28:34
Hold this general solution.
From that, as a step three of

462
00:28:34 --> 00:28:38
the differential equation
solution, I write the total

463
00:28:38 --> 00:28:42
solution down.
And my total solution is the

464
00:28:42 --> 00:28:47
sum of the particular and the
homogenous, so therefore I get

465
00:28:47 --> 00:28:49
this.
VI was my particular and this

466
00:28:49 --> 00:28:52
term here is my homogenous
solution.

467
00:28:52 --> 00:28:57
Now, if I wasn't doing circuits
and simply trying to solve this

468
00:28:57 --> 00:29:02
mathematically here is what I
would do.

469
00:29:02 --> 00:29:06
I would find the unknown from
the initial conditions,

470
00:29:06 --> 00:29:09
so I know that v(0)=0.
And so therefore if I

471
00:29:09 --> 00:29:12
substitute zero for V(0) I get
this.

472
00:29:12 --> 00:29:15
If I substitute zero here,
t is 0, t is 0,

473
00:29:15 --> 00:29:19
so I simply get V1+A1+A2.
And let me just blast through

474
00:29:19 --> 00:29:22
because I am going to redo this
differently.

475
00:29:22 --> 00:29:25
i=Cdv/dt.
And so that's what I get.

476
00:29:25 --> 00:29:30
I substitute zero and this is
what I would get.

477
00:29:30 --> 00:29:32
I hurried through this.
Don't worry.

478
00:29:32 --> 00:29:35
I'm going to do it again.
If you just do it

479
00:29:35 --> 00:29:37
mathematically,
you can solve this equation

480
00:29:37 --> 00:29:42
here and these two simultaneous
equations in a1 and a2 and get

481
00:29:42 --> 00:29:44
the coefficients and you are
done.

482
00:29:44 --> 00:29:48
But it doesn't give us a whole
lot of insight into the behavior

483
00:29:48 --> 00:29:51
of these terms here.
What I am going to do for now

484
00:29:51 --> 00:29:55
is kind of ignore that.
Ignore I did that and instead

485
00:29:55 --> 00:30:00
try to go down a path that is a
little bit more intuitive.

486
00:30:00 --> 00:30:05
Let's stare at this expression
we got for the total solution.

487
00:30:05 --> 00:30:09
That is the expression we got.
All I did is,

488
00:30:09 --> 00:30:13
I had alpha in there,
I simply pulled out the alpha

489
00:30:13 --> 00:30:17
outside.
So this is my total solution,

490
00:30:17 --> 00:30:21
V1-A1e^(-alpha t) something
else and something else.

491
00:30:21 --> 00:30:26
Three cases to consider
depending on the relative values

492
00:30:26 --> 00:30:32
of alpha and omega nought.
If alpha is greater than omega

493
00:30:32 --> 00:30:35
nought then I get a real
quantity here.

494
00:30:35 --> 00:30:40
The square root of a positive
number, I get a real number,

495
00:30:40 --> 00:30:45
and that number will add up to
the minus alpha and I am going

496
00:30:45 --> 00:30:49
to get a solution that will look
like, oh, I'm sorry.

497
00:30:49 --> 00:30:53
Let me just do it a little
differently.

498
00:30:53 --> 00:30:55
There are three situations
here.

499
00:30:55 --> 00:31:00
One is alpha greater than omega
nought.

500
00:31:00 --> 00:31:04
Alpha equal to omega nought.
Alpha less than omega nought.

501
00:31:04 --> 00:31:06
Alpha is greater,
alpha is less,

502
00:31:06 --> 00:31:10
alpha is equal to this term
inside the square root sign.

503
00:31:10 --> 00:31:14
For reasons you will understand
shortly, we call this

504
00:31:14 --> 00:31:18
"overdamped" case,
the "underdamped" case and the

505
00:31:18 --> 00:31:22
"critically damped" case.
When alpha is greater than

506
00:31:22 --> 00:31:26
omega nought this term gives me
a real number,

507
00:31:26 --> 00:31:30
and I get something as simple
as this.

508
00:31:30 --> 00:31:33
Remember, for the series RLC
circuit, alpha was R/2L.

509
00:31:33 --> 00:31:36
So if R is big,
in other words,

510
00:31:36 --> 00:31:40
if in my RLC circuit R is huge
then I am going to get this

511
00:31:40 --> 00:31:43
situation.
My output voltage on the

512
00:31:43 --> 00:31:47
capacitor is going to look like
this, the sum of two

513
00:31:47 --> 00:31:50
exponentials.
And if I were to plot it very

514
00:31:50 --> 00:31:52
quickly for you,
for a VI step,

515
00:31:52 --> 00:31:56
V would look like this.
So v would simply look like

516
00:31:56 --> 00:32:02
this because it is the sum of a
couple of exponentials.

517
00:32:02 --> 00:32:04
All right.
Now, alpha is positive here.

518
00:32:04 --> 00:32:08
Remember alpha1 and alpha2 are
both positive.

519
00:32:08 --> 00:32:11
These two added up,
because of this constant VI,

520
00:32:11 --> 00:32:15
give rise to something that
increases in the following

521
00:32:15 --> 00:32:18
manner.
Let's look at the situation

522
00:32:18 --> 00:32:22
where alpha is less than omega
nought, where the term inside

523
00:32:22 --> 00:32:24
the square root sign is
negative.

524
00:32:24 --> 00:32:29
What I can do is pull the
negative sign out and express it

525
00:32:29 --> 00:32:32
this way.
What I am going to do is since

526
00:32:32 --> 00:32:36
alpha is less than omega nought,
I am going to reverse these two

527
00:32:36 --> 00:32:40
and pull out square root of
minus one to the outside.

528
00:32:40 --> 00:32:43
This is what I get.
I am just playing around with

529
00:32:43 --> 00:32:47
this so that whatever is under
the square root sign ends up

530
00:32:47 --> 00:32:49
giving me a positive real
number.

531
00:32:49 --> 00:32:52
So I pull the j outside and
this is what I get.

532
00:32:52 --> 00:32:55
Now, let me blast through a
bunch of math and end up with

533
00:32:55 --> 00:32:57
something very,
very simple for this

534
00:32:57 --> 00:33:02
underdamped case.
Let me define a few other

535
00:33:02 --> 00:33:05
terms.
I am going to call omega nought

536
00:33:05 --> 00:33:09
minus alpha squared the square
root of that.

537
00:33:09 --> 00:33:14
I am going to call it omega d.
And here is what I get.

538
00:33:14 --> 00:33:18
So I have defined three things
for you now, alpha,

539
00:33:18 --> 00:33:23
omega nought and omega d.
And I get this equation in

540
00:33:23 --> 00:33:28
terms of alpha and omega d.
And then, remember from your

541
00:33:28 --> 00:33:34
good-old Euler relationship?
e to the j omega d is simply

542
00:33:34 --> 00:33:37
cosine plus a j sine.
I am just going to blast

543
00:33:37 --> 00:33:40
through a bunch of math rather
quickly.

544
00:33:40 --> 00:33:44
Once I replace this in terms of
a cosine and sine,

545
00:33:44 --> 00:33:48
cosine and a j sine and then
collect all the coefficients

546
00:33:48 --> 00:33:52
together, I get an equation of
the form VI plus some constant e

547
00:33:52 --> 00:33:56
to the minus alpha t,
cosine, the sum of the constant

548
00:33:56 --> 00:34:00
e to the minus alpha t,
sine.

549
00:34:00 --> 00:34:02
Remember the sines and cosines
are coming out,

550
00:34:02 --> 00:34:06
but because of my R I am
getting this funny alpha here,

551
00:34:06 --> 00:34:09
e to the minus alpha here.
So I am getting sums of sine

552
00:34:09 --> 00:34:11
and cosine.
And K1 and K2 are some

553
00:34:11 --> 00:34:15
constants which I will need to
determine for my initial

554
00:34:15 --> 00:34:17
conditions.
I am going to continue on with

555
00:34:17 --> 00:34:21
this and keep on simplifying it
because, as I promised you,

556
00:34:21 --> 00:34:24
I want to get to something that
is just a cosine.

557
00:34:24 --> 00:34:27
I want to go down this path.
I am not going to cover this

558
00:34:27 --> 00:34:31
case, the critically damped
case.

559
00:34:31 --> 00:34:34
And I will touch upon it later
but not dwell on it.

560
00:34:34 --> 00:34:39
Let me continue down the path
of the underdamped case,

561
00:34:39 --> 00:34:43
and this is what we have.
Continuing with the math,

562
00:34:43 --> 00:34:47
let's start with the initial
conditions, v nought equals

563
00:34:47 --> 00:34:50
zero, and that gives me K1 is
simply -VI.

564
00:34:50 --> 00:34:54
So at v(0)=0 t is zero,
so this terms goes away,

565
00:34:54 --> 00:34:58
the cosine becomes a 1,
e^(alpha t) goes away,

566
00:34:58 --> 00:35:04
and K1=-VI.
Then I know that i(0) and i is

567
00:35:04 --> 00:35:08
simply Cdv/dt.
And I get this nasty

568
00:35:08 --> 00:35:13
expression.
I substitute t=0 and I get

569
00:35:13 --> 00:35:20
something that looks like this.
I know what K1 is,

570
00:35:20 --> 00:35:27
and so therefore K2 is simply
-V1alpha divided by omega

571
00:35:27 --> 00:35:31
nought.
I have taken this expression

572
00:35:31 --> 00:35:34
where the unknowns K1 and K2 are
to be found.

573
00:35:34 --> 00:35:39
I set the initial conditions
down at t=0 and I get K1 and K2

574
00:35:39 --> 00:35:42
as follows, which gives me the
following solution.

575
00:35:42 --> 00:35:46
This is the solution I get
where I do not have any unknowns

576
00:35:46 --> 00:35:49
anymore.
Remember that omega d and alpha

577
00:35:49 --> 00:35:52
are directly related to circuit
parameters.

578
00:35:52 --> 00:35:56
Alpha was R/2L and omega d was
square root of alpha squared

579
00:35:56 --> 00:35:59
minus omega nought squared.
** omega d = sqrt(alpha^2 -

580
00:35:59 --> 00:36:04
omega_0^2) **
And omega nought squared was 1

581
00:36:04 --> 00:36:07
by square root of LC.
So I know it all now.

582
00:36:07 --> 00:36:12
I still have sines and cosines
here, so I am going to simplify

583
00:36:12 --> 00:36:17
this a little further.
Oh, before I go on to do that,

584
00:36:17 --> 00:36:21
let's do the Fineman trick
again and notice if I am still

585
00:36:21 --> 00:36:25
true to the LC circuit I did the
last time.

586
00:36:25 --> 00:36:30
Remember when R goes to zero
alpha goes to zero.

587
00:36:30 --> 00:36:33
Because alpha is R divided by
2L.

588
00:36:33 --> 00:36:38
If alpha was zero what happens?
If alpha was zero,

589
00:36:38 --> 00:36:43
this guy goes to one,
this whole term goes to zero

590
00:36:43 --> 00:36:48
and omega dt now ends up
becoming omega nought,

591
00:36:48 --> 00:36:53
and I get this term here.
I get VI-VIcosine(omega t),

592
00:36:53 --> 00:37:00
which is exactly what I
expected in my equation.

593
00:37:00 --> 00:37:07
This is the same as the LC case
that I got.

594
00:37:07 --> 00:37:17
Let's go back to this situation
and simply if further.

595
00:37:17 --> 00:37:25
If you look at Appendix B.7 in
your course notes,

596
00:37:25 --> 00:37:34
Appendix B.7 is a quick
tutorial on trig.

597
00:37:34 --> 00:37:37
And in that trig tutorial you
will see that,

598
00:37:37 --> 00:37:41
and you have probably seen this
before, too, multiple times,

599
00:37:41 --> 00:37:43
the scaled sum of sines are
also sines.

600
00:37:43 --> 00:37:47
This is an incredibly cool fact
of sinusoids.

601
00:37:47 --> 00:37:51
If you take two sinusoids of
the same frequency and you scale

602
00:37:51 --> 00:37:55
them up in any which way and add
them up you also end up with a

603
00:37:55 --> 00:37:58
sinusoid.
It is hard to believe but it is

604
00:37:58 --> 00:38:02
true.
It is an incredible property of

605
00:38:02 --> 00:38:04
sinusoids.
Take any two sinusoids,

606
00:38:04 --> 00:38:07
scale them in any way you like,
same frequency,

607
00:38:07 --> 00:38:10
add them up,
you will get a sinusoid.

608
00:38:10 --> 00:38:13
What that is saying is that,
look, here is a sinusoid,

609
00:38:13 --> 00:38:17
here is a sinusoidal function,
and I am scaling them up in

610
00:38:17 --> 00:38:21
some manner.
So I should be able to add them

611
00:38:21 --> 00:38:24
up and be able to express that
as single sine.

612
00:38:24 --> 00:38:27
And to be sure you can,
look at the Appendix,

613
00:38:27 --> 00:38:31
and there is an expression for
a1 sinX plus a2 cosX is equal to

614
00:38:31 --> 00:38:35
a cosine of blah,
blah, blah.

615
00:38:35 --> 00:38:38
This is what you get.
No magic here.

616
00:38:38 --> 00:38:42
Just math.
From here I directly get this.

617
00:38:42 --> 00:38:47
And look at what I have.
It is absolutely unbelievable.

618
00:38:47 --> 00:38:51
v(t) is simply VI,
there is a constant here,

619
00:38:51 --> 00:38:58
this an e to the minus alpha
term and there is a cosine.

620
00:38:58 --> 00:39:02
Again, to pull the Fineman
trick, if this alpha were to go

621
00:39:02 --> 00:39:06
to zero here then you would end
up with the expression you had

622
00:39:06 --> 00:39:10
for the LC situation.
Let's stare at this a little

623
00:39:10 --> 00:39:12
while longer.
There is a constant plus a

624
00:39:12 --> 00:39:16
minus, a cosine term,
so there is a sinusoid at the

625
00:39:16 --> 00:39:21
output, and there is an e to the
minus alpha which ends up giving

626
00:39:21 --> 00:39:23
you the decay you have seen
before.

627
00:39:23 --> 00:39:25
In other words,
to a step input,

628
00:39:25 --> 00:39:30
the LC circuit would give you a
sinusoid.

629
00:39:30 --> 00:39:34
That is what the LC circuit
would do if alpha was zero.

630
00:39:34 --> 00:39:39
But because of this alpha term
here, e to the minus alpha t,

631
00:39:39 --> 00:39:43
that gives rise to a damping
effect, so this causes this

632
00:39:43 --> 00:39:48
thing to become smaller and
smaller as time goes by until

633
00:39:48 --> 00:39:51
this term goes to zero at t
equals infinity.

634
00:39:51 --> 00:39:56
This guy damps down and so
therefore you end up getting the

635
00:39:56 --> 00:40:02
curve that you saw like this.
Twenty minutes of juggling math

636
00:40:02 --> 00:40:06
solving a second-order
differential equation,

637
00:40:06 --> 00:40:11
but what ends up is the same
sinusoid but it is damped in the

638
00:40:11 --> 00:40:17
following manner such that the
frequency, rather the amplitude

639
00:40:17 --> 00:40:22
keeps decaying until it starts
off at zero and then settles

640
00:40:22 --> 00:40:25
down at vI.
This is exactly what you saw in

641
00:40:25 --> 00:40:30
the demo that we showed you
earlier.

642
00:40:30 --> 00:40:34
The critically damped case,
I am not going to do it here.

643
00:40:34 --> 00:40:37
I am going to point you to the
following insight.

644
00:40:37 --> 00:40:40
The underdamped case looked
like this.

645
00:40:40 --> 00:40:43
It was a sinusoid that kind of
decayed out.

646
00:40:43 --> 00:40:47
That is the underdamped case.
And then I showed you the

647
00:40:47 --> 00:40:51
overdamped case.
The overdamped case looked like

648
00:40:51 --> 00:40:53
this.
And, as you might expect,

649
00:40:53 --> 00:40:58
the critically damped case is
kind of in the middle and looks

650
00:40:58 --> 00:41:02
like this.
So the overdamped case would

651
00:41:02 --> 00:41:04
look like this,
underdamped like this,

652
00:41:04 --> 00:41:09
and the critically damped case
kind of goes up and kind of

653
00:41:09 --> 00:41:14
settles down almost immediately.
This is when alpha equals omega

654
00:41:14 --> 00:41:16
nought.
I won't do that case here,

655
00:41:16 --> 00:41:19
but I will simply point you to
Section 13.2.3.

656
00:41:19 --> 00:41:24
Just to tie things together,
recall this demo here that we

657
00:41:24 --> 00:41:28
showed you in class yesterday.
This is exactly the kind of

658
00:41:28 --> 00:41:34
form of the sinusoid you saw
because of that input step.

659
00:41:34 --> 00:41:39
If you want to see a complete
analysis of inverter pairs and

660
00:41:39 --> 00:41:43
look at the delays and so on
because of that,

661
00:41:43 --> 00:41:46
you can look at Page 170 and
example 898.

662
00:41:46 --> 00:41:51
In the next five or six
minutes, what I would like to do

663
00:41:51 --> 00:41:56
is stare at the RLC circuit.
And much like I showed you some

664
00:41:56 --> 00:42:01
intuitive methods to get the RC
response, what we are going to

665
00:42:01 --> 00:42:06
do is do the same thing for the
RLC.

666
00:42:06 --> 00:42:09
In the RLC situation,
much like the RC situation,

667
00:42:09 --> 00:42:13
experts don't go around writing
15 pages of differential

668
00:42:13 --> 00:42:18
equations and solving them each
time they see an RLC circuit.

669
00:42:18 --> 00:42:21
They stare at it and boom,
the response pops out,

670
00:42:21 --> 00:42:25
the sketch pops out.
This one is going to be another

671
00:42:25 --> 00:42:30
one like our Bend it Like
Beckham series here.

672
00:42:30 --> 00:42:34
And this one is in honor of
Leslie Kolodziejski.

673
00:42:34 --> 00:42:38
And I call it "Konquer it like
Kolodziejski".

674
00:42:38 --> 00:42:42
Again, as I said,
experts don't go around solving

675
00:42:42 --> 00:42:47
long differential equations and
spending ten pages of notes

676
00:42:47 --> 00:42:51
trying to get a sinusoid.
They look at a circuit and

677
00:42:51 --> 00:42:55
sketch response.
I am going to show you how to

678
00:42:55 --> 00:42:59
do that, too.
And what you can do is,

679
00:42:59 --> 00:43:02
to practice,
go to Websim and try out

680
00:43:02 --> 00:43:06
various combinations of inputs
and initial conditions and

681
00:43:06 --> 00:43:10
sketch it, time yourself,
give yourself 30 seconds or a

682
00:43:10 --> 00:43:13
minute if you like,
and sketch it and check it

683
00:43:13 --> 00:43:18
against the Websim response.
If it doesn't match either you

684
00:43:18 --> 00:43:20
are wrong or there is a bug in
Websim.

685
00:43:20 --> 00:43:24
What I am going to do is,
the response to the critically

686
00:43:24 --> 00:43:30
damped and underdamped case was
very easy to sketch out.

687
00:43:30 --> 00:43:33
You started with an initial
condition, you settled at VI and

688
00:43:33 --> 00:43:36
just kind of drew it like that.
The interesting case is the

689
00:43:36 --> 00:43:39
underdamped case,
and that is what I am going to

690
00:43:39 --> 00:43:41
dwell on.
Before we go on and I show you

691
00:43:41 --> 00:43:45
the intuitive method,
as a first step I would like to

692
00:43:45 --> 00:43:47
build some intuition.
Let's stare at this response

693
00:43:47 --> 00:43:50
here and try to understand what
is going on.

694
00:43:50 --> 00:43:52
This is the response that we
saw.

695
00:43:52 --> 00:43:55
And this fact that you see an
oscillation happening is also

696
00:43:55 --> 00:43:58
called "ringing".
You say that your circuit is

697
00:43:58 --> 00:44:00
ringing.
All right.

698
00:44:00 --> 00:44:04
You see some interesting facts.
You see that frequency of the

699
00:44:04 --> 00:44:08
ringing is given by omega d.
This cosine omega d,

700
00:44:08 --> 00:44:10
so that is the frequency omega
d.

701
00:44:10 --> 00:44:13
So the time is 2 pi divided by
omega d.

702
00:44:13 --> 00:44:17
The oscillation frequency is
omega d, but omega d is simply

703
00:44:17 --> 00:44:20
omega nought squared minus alpha
squared.

704
00:44:20 --> 00:44:24
Once you have a big value of R
alpha becomes very small and

705
00:44:24 --> 00:44:30
omega d is very commonly equal
to, very close to omega nought.

706
00:44:30 --> 00:44:34
So omega d and omega nought
very commonly are very close

707
00:44:34 --> 00:44:37
together.
And when that happens this

708
00:44:37 --> 00:44:40
frequency is directly omega
nought.

709
00:44:40 --> 00:44:43
Alpha governs how quickly your
sinusoid decays.

710
00:44:43 --> 00:44:48
e to the alpha t here is the
envelope that governs how

711
00:44:48 --> 00:44:52
quickly my sinusoid decays.
And notice that each of these

712
00:44:52 --> 00:44:56
terms, alpha and omega nought,
comes directly from my

713
00:44:56 --> 00:45:01
characteristic equation.
Which means that once you get

714
00:45:01 --> 00:45:05
your characteristic equation you
really don't have to do much

715
00:45:05 --> 00:45:07
else.
And up until now you still have

716
00:45:07 --> 00:45:10
to write the differential
equation to get the

717
00:45:10 --> 00:45:14
characteristic equation,
so you still have to do some

718
00:45:14 --> 00:45:17
differential equation stuff,
but in two lectures I am going

719
00:45:17 --> 00:45:20
to show you a way that you can
even write down the

720
00:45:20 --> 00:45:23
characteristic equation by
inspection.

721
00:45:23 --> 00:45:26
Look at your circuit and boom,
in 15 seconds or less write

722
00:45:26 --> 00:45:30
down the characteristic
equation.

723
00:45:30 --> 00:45:34
It is absolutely unbelievable.
What are the other factors that

724
00:45:34 --> 00:45:37
are interesting here?
Of course I need to find out

725
00:45:37 --> 00:45:39
initial values.
I start off at zero.

726
00:45:39 --> 00:45:43
This is my capacitor voltage.
If I don't have an infinite

727
00:45:43 --> 00:45:48
spike or an impulse my capacitor
voltage tries to stay where it

728
00:45:48 --> 00:45:51
is and starts off at zero.
And the final value is given by

729
00:45:51 --> 00:45:56
VI, the capacitor is a long-term
open so therefore VI appears

730
00:45:56 --> 00:45:59
across the capacitor.
In the long-term my final value

731
00:45:59 --> 00:46:04
is going to be VI.
There is one other interesting

732
00:46:04 --> 00:46:09
parameter, which I will simply
define today but dwell on about

733
00:46:09 --> 00:46:12
a week from today,
and that is called the Q.

734
00:46:12 --> 00:46:17
Some of you may have heard the
term oh, that's a high Q

735
00:46:17 --> 00:46:20
circuit.
Q is an indication of how ringy

736
00:46:20 --> 00:46:23
the circuit is.
And Q is defined as omega

737
00:46:23 --> 00:46:27
nought by 2 alpha.
It is called the "quality

738
00:46:27 --> 00:46:30
factor".
And it turns out that Q is

739
00:46:30 --> 00:46:33
approximately the number of
cycles of ringing.

740
00:46:33 --> 00:46:37
So if you have a high Q you
ring for a long time and if you

741
00:46:37 --> 00:46:39
have a low Q you ring for a very
short time.

742
00:46:39 --> 00:46:43
That is called the quality
factor defined by omega nought

743
00:46:43 --> 00:46:44
by 2 alpha.
Notice that Q,

744
00:46:44 --> 00:46:46
omega nought,
alpha, omega d,

745
00:46:46 --> 00:46:50
all of these come from the
terms in the characteristic

746
00:46:50 --> 00:46:52
equation.
We will spend more time on Q

747
00:46:52 --> 00:46:54
later.
With this insight here is how I

748
00:46:54 --> 00:46:58
can go about very quickly
sketching out the form of the

749
00:46:58 --> 00:47:01
response.
Here is my circuit.

750
00:47:01 --> 00:47:05
I want to sketch the form of
the response for a step input at

751
00:47:05 --> 00:47:08
vI.
Zero to vI step input here,

752
00:47:08 --> 00:47:11
I want to find out what happens
at this point.

753
00:47:11 --> 00:47:16
This is described to you in a
lot more detail in Section 13.8

754
00:47:16 --> 00:47:19
in your course notes.
Let's go through the steps.

755
00:47:19 --> 00:47:23
Let's do the really simple
situation first.

756
00:47:23 --> 00:47:27
Let's also assume for fun that
you are given that v(0) starts

757
00:47:27 --> 00:47:33
out being some positive value.
Some v(0) which is a positive

758
00:47:33 --> 00:47:35
number.
And, to make it harder on

759
00:47:35 --> 00:47:39
ourselves, let's say i(0) starts
out being some negative number.

760
00:47:39 --> 00:47:42
So i(0) is some negative
current.

761
00:47:42 --> 00:47:46
The first thing I know is v(0),
the capacitor voltage starts

762
00:47:46 --> 00:47:48
out here, which can change
suddenly.

763
00:47:48 --> 00:47:52
And I also know that in the
long-term this is an open

764
00:47:52 --> 00:47:55
circuit.
So that this voltage vI will

765
00:47:55 --> 00:48:00
appear directly across the
capacitor in the long-term.

766
00:48:00 --> 00:48:03
So I get starting out at v(0),
ending at vI,

767
00:48:03 --> 00:48:07
I am also half the way there.
I know the initial and ending

768
00:48:07 --> 00:48:10
point of the curve.
And then I know that somewhere

769
00:48:10 --> 00:48:13
in here there must be some funny
gyrations here,

770
00:48:13 --> 00:48:17
because remember I am dealing
with the underdamped case.

771
00:48:17 --> 00:48:21
And you can determine that from
alpha and omega nought.

772
00:48:21 --> 00:48:25
If alpha is less than omega
nought, you know that you are in

773
00:48:25 --> 00:48:30
the underdamped case and this is
what you get.

774
00:48:30 --> 00:48:33
Let's compute and write the
characteristic equation down.

775
00:48:33 --> 00:48:37
A week from today you will
write it by inspection,

776
00:48:37 --> 00:48:40
but for now you will do it by
writing down a differential

777
00:48:40 --> 00:48:43
equation.
And from the characteristic

778
00:48:43 --> 00:48:46
equation you will get omega d,
you will get alpha,

779
00:48:46 --> 00:48:49
omega nought and Q.
So omega d gives you the

780
00:48:49 --> 00:48:53
frequency of oscillations.
My frequency of oscillation is

781
00:48:53 --> 00:48:55
now known.
From Q I know how long it

782
00:48:55 --> 00:49:00
rings, because I know it rings
for about Q cycles.

783
00:49:00 --> 00:49:02
I know that ringing stops
approximately here.

784
00:49:02 --> 00:49:06
And then I know that between
that the start and end point my

785
00:49:06 --> 00:49:10
curve kind of looks like this,
something like this.

786
00:49:10 --> 00:49:12
Right there we are 95% of the
way there.

787
00:49:12 --> 00:49:16
The only question is I do not
know if it goes like this or it

788
00:49:16 --> 00:49:19
goes like this.
I am not quite sure yet if it

789
00:49:19 --> 00:49:22
starts off going high or starts
off going low.

790
00:49:22 --> 00:49:25
Not quite clear.
I also do not know what the

791
00:49:25 --> 00:49:30
maximum amplitude is.
It turns out this is rather

792
00:49:30 --> 00:49:34
complicated to determine so we
won't deal with that.

793
00:49:34 --> 00:49:37
Just simply so you can draw a
rough sketch.

794
00:49:37 --> 00:49:40
The questions is which way does
it start?

795
00:49:40 --> 00:49:43
I could leave it for you to
think about.

796
00:49:43 --> 00:49:47
Yeah, let me do that.
It is given on this page so

797
00:49:47 --> 00:49:50
don't look at it.
Think about it,

798
00:49:50 --> 00:49:55
and think about how you can
determine whether it goes up or

799
00:49:55 --> 00:49:57
down.
It turns out that in this case

800
00:49:57 --> 00:50:02
it is going to down and then
ring.

801
00:50:02 --> 00:50:06
See if you can figure it out
for yourselves and then we will

802
00:50:06 --> 50:09
talk about it next week.