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OK.
Good morning.

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Let's get going.
As always, I will start with a

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review.
And today we embark on another

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major milestone in our analysis
of lumped circuits.

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And it is called the
"Sinusoidal Steady-state".

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Again, I believe this will be
the second and the last lecture

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for which I will be using view
graphs.

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And the idea here is that,
just like in the last lecture,

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there is a bunch of
mathematical grunge that needs

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to be gone through.
And I want to show you a

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sequence in a little chart today
that talks about the effort

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level in doing some problems
based on time domain

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differential equations,
as in the last lecture,

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something slightly different
today and something much better

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on Thursday.
And so, in some sense,

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Thursday's lecture and today's
lecture involve talking about

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the foundations of the behavior
of certain types of circuits.

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And it is good for you to have
this foundation as background,

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but when you actually go to
solve circuits you don't quite

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use these methods.
You use much easier techniques,

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which I will talk about next
Thursday.

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Let's start with a quick
review, and then we will go into

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sinusoidal steady state.
The last lecture we talked

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about this circuit and we did
the same two lectures ago on

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Tuesday.
We had one inverter driving

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another inverter.
And we said that the wire over

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ground had some inductance.
CGS is the capacitor of the

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gate and R is the resistance at
the drain of the first inverter.

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And if you look at this
circuit, that circuit formed an

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RLC pattern.
And what we did was we said

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let's drive this with a one to
zero transition at the input.

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And the one to zero transition
at the input would cause this

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transistor to switch off,
and this node would then go

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from a very low value to a high
value.

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So it as if a 5 volt step was
applied at this input.

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We also saw that using time
domain differential equations

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that by applying a step input
here the output looked like

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this.
The output would show some

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oscillatory behavior when we
have a capacitor and inductor.

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I also gave you some insight as
to why it oscillates like this.

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And you also heard in
recitation that the reason for

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this oscillation was because of
these two storage elements.

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Each of these storage elements
tries to hold onto its state

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variable.
For example,

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the capacitor tries to maintain
its voltage while the inductor

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tries to maintain its current.
And, much like a pendulum which

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oscillates back and forth,
it swaps potential energy

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versus kinetic energy down here
and swings back and forth.

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In the same way,
in an LC circuit like this,

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energy swaps back and forth
between a potential energy and a

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kinetic energy equivalent,
which swaps back and forth

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between energy stored in the
inductor and energy stored in

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the capacitor and sloshes back
and forth.

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And because of this resistor
the energy eventually dissipates

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and you end up getting a final
value which corresponds to the 5

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volts appearing here.
And why is that?

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That is because remember the
capacitor is a long-term open

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for DC.
It is a DC voltage.

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After a long time this
capacitor looks like an open

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circuit and the inductor looks
like a complete short circuit,

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an ideal inductor as a complete
short circuit for DC.

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And so therefore in the
long-term it is as if this guy

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is a short, this guy is an open,
so 5 volts simply appears here.

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And this is the transient
behavior.

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Then we just switch the first
transistor off.

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In the last lecture,
I left off with intuitive

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analysis.
Let me quickly cover that and

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redo the intuitive analysis for
you.

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I left it the last time by
having you think about whether

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the transient response would
begin by going down or begin by

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going up.
And I will cover that today.

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This was the circuit that we
analyzed.

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A VI input with a step and an
RLC out here.

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And we were plotting the
capacitor voltage.

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And intuitively we can plot
this in the following way.

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I have also marked for you the
section number in the course

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notes which has a discussion of
this intuitive analysis,

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(Section 13.8).
Let's do the easy stuff first.

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Notice that the capacitor wants
to hold its voltage.

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And so given that we don't have
any infinite impulse here,

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I am going to start out with
the capacitor voltage being

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where it is.
And the initial conditions are

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given to you.
You are given that the

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capacitor voltage starts out
being positive at v zero and the

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current starts out being
negative at time zero.

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So I am telling you that there
is a voltage v across the

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capacitor at time t=0 and there
is a current that is flowing.

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Since i is negative there is a
current initially that is

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flowing in the opposite
direction to this arrow here.

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The i zero is negative.
In light of that,

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I can start plotting my curve
here by intuition.

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I start by saying at time t=0 I
am told that the initial voltage

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of the capacitor is at zero.
This is about as simple as it

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gets.
Completely intuitive.

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I also know that after a long
time, can someone tell me after

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a long time what the voltage
will be at the end of the

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capacitor?
You should be able to get his

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by observation?
VI.

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And why is it VI?
It is vI because this is a DC

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value VI.
And after a long time this guy

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behaves like an open circuit to
DC.

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This guy behaves like a short
circuit to DC.

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So since this is an open
circuit, VI will appear here

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after a long time.
And so therefore,

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after a long time,
the capacitor voltage is going

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to be at VI.
And I just showed you that.

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There you go.
You already have the two

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endpoints of your curve
completely by observation,

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intuition.
No DEs, no nothing.

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Just by staring at it and
understanding the fundamentals

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of how simple primitive circuit
elements work.

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Absolutely simple stuff.
So you've nailed the two ends

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now and you cannot go wrong with
the stuff in the middle.

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Let's see.
As a next step what I do is I

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need to understand what the
dynamics of the circuit looks

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like here.
What I do is I develop the

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characteristic equation.
And initially you will write

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the differential equation and
then substitute e^st and get

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this characteristic equation.
But you could also remember it

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as a pattern.
For a series RLC circuit you

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always get an equation of this
form, always.

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If this were R,
L and C.

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And whether you are looking at
L up here or C up here,

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as long as you're looking at
the capacitor voltage,

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the capacitor voltage is going
to behave the same.

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And for this circuit the
characteristic equation remains

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the same as well for a series
RLC.

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It is exactly this.
And I write the standard

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canonic form as s squared plus
two alpha s + omega nought

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squared.
And omega nought is simply one

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by square root of LC and alpha
is simply R divided by L and I

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have two in the denominator as
well.

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And then I get omega d which is
my damped frequency given by

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omega nought squared minus alpha
squared.

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And Q is simply called the
quality factor.

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And we will learn about Q in a
lot more detail in about a

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couple of lectures from today.
That is given where omega

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nought divided by two alpha.
These parameters,

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alpha, omega nought,
Q and omega d pretty much

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characterize everything else
that I need to know about the

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circuit.
First of all,

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omega d is the frequency of
oscillation.

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And so since omega d is a
frequency of oscillation then I

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know that the time period of
oscillation is two pi by omega

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d.
Omega is in radians.

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Notice that for typical values
of circuits like this when R is

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pretty small,
alpha squared is going to be

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very small.
It's a common case for

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underdamped circuit that omega d
will more or less be equal to

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omega nought.
Commonly that is going to be

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the case.
This frequency is governed by

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LC.
And if R is large it is

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governed by this omega d here.
So I have the frequency of

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oscillation.
I also know that Q is related

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to the frequency of oscillation
divided by alpha.

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It is a ratio of the frequency
divided by how badly I am being

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damped.
So it is a fight between the

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frequency of oscillation and now
heavily I damp.

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And the ratio of that is an
indication of how many cycles I

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ring.
So Q tells me that the ringing

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stops approximately after Q
cycles.

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These four values,
omega d, Q, alpha and omega

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nought are telling me more and
more now.

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So I have got these two
factors.

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So I know now,
based on omega d and Q,

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that it is going to look
something like this.

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Some ringing here and then I
stop at this point.

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The last thing that is left to
do here for me for now is to

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figure out whether I start out
going down or I start out going

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up.
I start out going down or I

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start out going up?
I don't know that yet.

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And I stopped at this point in
the last lecture and let you

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think about how you can stare at
the circuit and intuitively

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figure out whether this goes
down or that goes up.

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So here is the insight.
Let's stare at this for a

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minute purely through intuition.
The capacitor has a voltage v

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across it, right?
And that is because I am

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telling you that it has an
initial voltage v.

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Now, I want to find out at
prime t equals zero plus,

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in which direction does a
capacitor voltage go?

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Increase or decrease?
What do I do?

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Let me take a look at the
inductor current.

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I am told that the inductor
current is negative which means

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I am told that the inductor
current is going in this

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direction initially.
The inductor current is pushing

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in this direction.
Now, remember,

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just as the capacitor is one
stubborn nut trying to hold its

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voltage, the inductor is as
stubborn.

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It is trying to hold its
current.

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It is trying to maintain its
current.

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And its initial current i(0) is
in this direction.

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Capacitor has a voltage here,
that is v(0),

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and the inductor is yanking the
current in that direction.

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So what should happen to the
capacitor voltage initially?

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If I am at v(0) and someone is
yanking current out,

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at least initially in that
direction, what should the

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initial dynamics of the
capacitor voltage look like?

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Pardon?
It should drop,

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which means that if the initial
current is being pulled in that

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direction the capacitor voltage
must droop to begin with.

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Completely through intuition.
No math here.

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This means that i(0) is
negative.

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It is as if i(0) is being
pulled out in this manner,

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so the capacitor voltage must
drop to begin life.

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Therefore, the dynamics look
somewhat like this.

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Notice that this is very
reminiscent of the ringing that

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we saw at the gate node of the
second inverter.

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00:13:47,000 --> 00:13:54,000
Let's stop here in terms of
time domain analysis of RLC,

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and today let's take another
big step forward.

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Today marks another transition
in life here.

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This is actually a huge
transition so I want to just

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pause and take like ten seconds
of a breather just to clearly

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demarcate the fact that we have
a huge transition coming up.

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The key to this transition is
that I want to look at today the

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00:14:24,000 --> 00:14:30,000
steady-state response of
networks to a sinusoidal drive.

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We are going to do two things
differently starting today on

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this new journey of ours.
In the past,

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we looked at time domain
behavior of circuits.

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For RLC, for example,
we looked at the transient

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response.
And what happened the instant I

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turn something on?
The transient response.

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Today we are going to look at a
steady-state response.

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Steady-state response says that
if I wait long enough,

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for whatever the circuit wants
to do in the beginning of life

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to die out.
If I wait long enough,

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how is the circuit going to
behave after a long time?

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00:15:04,000 --> 00:15:07,000
I will tell you why that is
important in a second.

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I look at the steady-state
behavior.

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Second, I am going to look
today at sinusoidal drive.

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Two things that are different
from, say for example,

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what I covered in the past ten
minutes.

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00:15:18,000 --> 00:15:22,000
In the past ten minutes I
covered two things which were

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different.
One is that I looked at the

234
00:15:24,000 --> 00:15:28,000
transient response and then
steady-state.

235
00:15:28,000 --> 00:15:31,000
And remember for a DC input,
for a DC voltage the

236
00:15:31,000 --> 00:15:34,000
steady-state was a DC voltage
across the capacitor,

237
00:15:34,000 --> 00:15:37,000
correct?
So the steady-state was pretty

238
00:15:37,000 --> 00:15:41,000
boring, it was steady-state DC.
But what we are going to do

239
00:15:41,000 --> 00:15:45,000
today is instead of having DC
inputs or step inputs which

240
00:15:45,000 --> 00:15:49,000
settle to a DC value after some
time, we are going to drive a

241
00:15:49,000 --> 00:15:52,000
circuit through the sinusoidal
input.

242
00:15:52,000 --> 00:15:54,000
So you may ask me,
Agarwal, are you nuts?

243
00:15:54,000 --> 00:15:58,000
Why do you want to drive
something with a sinusoidal

244
00:15:58,000 --> 00:16:02,000
input?
Why not just steps in DC and so

245
00:16:02,000 --> 00:16:04,000
on?
That was painful enough.

246
00:16:04,000 --> 00:16:08,000
Why sinusoidal?
Why not do triangular or why

247
00:16:08,000 --> 00:16:13,000
not do some other exponentially
decaying stuff or something

248
00:16:13,000 --> 00:16:16,000
really cool like a whacko music
signal?

249
00:16:16,000 --> 00:16:19,000
What is special about
sinusoidal stuff?

250
00:16:19,000 --> 00:16:23,000
The key thing to realize is
that, well, let me ask you a

251
00:16:23,000 --> 00:16:27,000
question first.
How many people here know about

252
00:16:27,000 --> 00:16:30,000
Fourier series?
Good.

253
00:16:30,000 --> 00:16:34,000
It looks like some of you have
taken the prerequisites.

254
00:16:34,000 --> 00:16:36,000
Good.
Need I say more as to why this

255
00:16:36,000 --> 00:16:40,000
is important?
Just that question should give

256
00:16:40,000 --> 00:16:41,000
you the answer,
right?

257
00:16:41,000 --> 00:16:44,000
You've learned about Fourier
series.

258
00:16:44,000 --> 00:16:47,000
And when you learned about
Fourier series you were

259
00:16:47,000 --> 00:16:51,000
wondering why on earth are we
learning about Fourier series?

260
00:16:51,000 --> 00:16:55,000
Who cares that you can
represent the periodic signals

261
00:16:55,000 --> 00:17:00,000
as a summation of a series of
sine waves?

262
00:17:00,000 --> 00:17:03,000
Why is that interesting?
Why are you telling me that I

263
00:17:03,000 --> 00:17:06,000
can take a square wave and
represent that as a summation of

264
00:17:06,000 --> 00:17:10,000
periodic square waves and
represent that as a summation of

265
00:17:10,000 --> 00:17:12,000
sines?
Who cares that I can take a

266
00:17:12,000 --> 00:17:16,000
sequence of pulses with a fixed
period and represent that as a

267
00:17:16,000 --> 00:17:19,000
sum of sines?
Who cares that I can take a

268
00:17:19,000 --> 00:17:22,000
triangular wave and represent
that as a sum of sines?

269
00:17:22,000 --> 00:17:26,000
I am not sure what answer your
math professors gave you when

270
00:17:26,000 --> 00:17:30,000
they taught you Fourier series.
But in math they are purists.

271
00:17:30,000 --> 00:17:32,000
They don't care about
applications.

272
00:17:32,000 --> 00:17:36,000
The answer could well have been
because it is aesthetically

273
00:17:36,000 --> 00:17:38,000
pleasing.
I mean isn't it cool that you

274
00:17:38,000 --> 00:17:42,000
can represent a sequence of
pulses as a sum of sines?

275
00:17:42,000 --> 00:17:44,000
That is good enough for
mathematicians.

276
00:17:44,000 --> 00:17:47,000
But I am an engineer.
If it I cannot see how it helps

277
00:17:47,000 --> 00:17:51,000
humanity in the short-term then
I probably don't care too much

278
00:17:51,000 --> 00:17:53,000
about it.
Let me give you some practical

279
00:17:53,000 --> 00:17:56,000
significance of this.
So it turns out that,

280
00:17:56,000 --> 00:17:59,000
since we know that we can
represent periodic signals with

281
00:17:59,000 --> 00:18:04,000
sums of sines.
What this means is that if I

282
00:18:04,000 --> 00:18:10,000
can figure out the behavior of
networks to a sinusoidal input,

283
00:18:10,000 --> 00:18:14,000
if I can understand how to
analyze a network for a

284
00:18:14,000 --> 00:18:20,000
sinusoidal input that means that
if the circuit is linear I can

285
00:18:20,000 --> 00:18:25,000
then compute the response of the
circuit to any periodic

286
00:18:25,000 --> 00:18:29,000
waveform.
Here is the argument.

287
00:18:29,000 --> 00:18:33,000
I can represent any periodic
waveform as a sum of sines.

288
00:18:33,000 --> 00:18:36,000
The Fourier series,
remember?

289
00:18:36,000 --> 00:18:40,000
If I just figure out the
response of my network for a

290
00:18:40,000 --> 00:18:44,000
sine wave, then if this is a
linear network,

291
00:18:44,000 --> 00:18:49,000
I can compute the response to
any sequence of scaled sum of

292
00:18:49,000 --> 00:18:50,000
sines.
A some sine,

293
00:18:50,000 --> 00:18:55,000
B sine two, omega whatever,
C sine something or the other.

294
00:18:55,000 --> 00:19:01,000
I can simply take the response
of the one sine.

295
00:19:01,000 --> 00:19:05,000
And from there I can go ahead,
and knowing the response of a

296
00:19:05,000 --> 00:19:09,000
sine wave I can compute the
response to a sum of sines.

297
00:19:09,000 --> 00:19:12,000
That is pretty cool.
Therefore, doing it for

298
00:19:12,000 --> 00:19:14,000
sinusoidal drives is really
important.

299
00:19:14,000 --> 00:19:18,000
Why steady-state now?
Hopefully, I have convinced you

300
00:19:18,000 --> 00:19:22,000
that looking at the response of
circuits to sinusoidal drive is

301
00:19:22,000 --> 00:19:26,000
important and interesting
because we can long ways from

302
00:19:26,000 --> 00:19:30,000
there.
What about steady-state?

303
00:19:30,000 --> 00:19:32,000
Well, it turns out that,
and I am going to show you,

304
00:19:32,000 --> 00:19:35,000
when you listen to music,
you have an amplifier and

305
00:19:35,000 --> 00:19:38,000
listen to music,
what you are observing by and

306
00:19:38,000 --> 00:19:41,000
large is the steady-state
behavior of the amplifier.

307
00:19:41,000 --> 00:19:44,000
You are listening to something
over many seconds or many hours.

308
00:19:44,000 --> 00:19:48,000
And the transients used for
most of our common circuits,

309
00:19:48,000 --> 00:19:50,000
the transients die out pretty
quickly.

310
00:19:50,000 --> 00:19:53,000
And so the transients are quite
complicated and they die out

311
00:19:53,000 --> 00:19:55,000
quickly.
We say we are engineers.

312
00:19:55,000 --> 00:19:59,000
Let's focus on what is
practically important.

313
00:19:59,000 --> 00:20:02,000
And we focus on the
steady-state behavior as a large

314
00:20:02,000 --> 00:20:05,000
part of our analysis and just
completely ignore the transient

315
00:20:05,000 --> 00:20:09,000
response when we care about the
response to sinusoidal input.

316
00:20:09,000 --> 00:20:12,000
The transient response will die
away, and I will show that

317
00:20:12,000 --> 00:20:15,000
mathematically to you in a few
seconds.

318
00:20:15,000 --> 00:20:18,000
And let's focus on the
steady-state because that what I

319
00:20:18,000 --> 00:20:20,000
am going to hear most of the
time.

320
00:20:20,000 --> 00:20:23,000
I am going to listen to an
average over long periods of

321
00:20:23,000 --> 00:20:25,000
time.
That's the motivation behind

322
00:20:25,000 --> 00:20:27,000
this.
And let me put this in

323
00:20:27,000 --> 00:20:31,000
perspective for you.
By now this should bring

324
00:20:31,000 --> 00:20:35,000
memories to your mind.
This is the playground that we

325
00:20:35,000 --> 00:20:37,000
are in.
This is the lumped circuit

326
00:20:37,000 --> 00:20:40,000
playground here.
Remember we came from the

327
00:20:40,000 --> 00:20:44,000
playground of nature to the
playground of EECS where we made

328
00:20:44,000 --> 00:20:48,000
the big leap from Maxwell's
equations to lumped circuits,

329
00:20:48,000 --> 00:20:50,000
that's lumped circuit
abstraction.

330
00:20:50,000 --> 00:20:54,000
And within there we spent a
large part of the last couple of

331
00:20:54,000 --> 00:20:58,000
months doing linear circuits.
We also analyzed nonlinear

332
00:20:58,000 --> 00:21:02,000
circuits.
Remember the amplifier circuit

333
00:21:02,000 --> 00:21:05,000
of the MOSFET large signal
analysis was nonlinear?

334
00:21:05,000 --> 00:21:07,000
Well, there is linear and
nonlinear.

335
00:21:07,000 --> 00:21:11,000
Within linear we also showed
that if you take a digital

336
00:21:11,000 --> 00:21:14,000
circuit, at least as we
understood them,

337
00:21:14,000 --> 00:21:18,000
and draw the subcircuit for a
given set of switch settings,

338
00:21:18,000 --> 00:21:22,000
if I set the switches in a
given way what I was left with

339
00:21:22,000 --> 00:21:26,000
was another linear circuit for a
given value of all the switch

340
00:21:26,000 --> 00:21:29,000
settings.
My small signal analysis was

341
00:21:29,000 --> 00:21:32,000
also linear.
If I focused on small signals I

342
00:21:32,000 --> 00:21:36,000
also had linear analysis.
Today what we are going to do

343
00:21:36,000 --> 00:21:38,000
is this.
We are going to articulate a

344
00:21:38,000 --> 00:21:40,000
different part of the
playground.

345
00:21:40,000 --> 00:21:42,000
This was a big linear
playground.

346
00:21:42,000 --> 00:21:44,000
We've done this.
We've done this.

347
00:21:44,000 --> 00:21:47,000
We are going to explore this
territory.

348
00:21:47,000 --> 00:21:51,000
This is that territory of the
playground in which we have

349
00:21:51,000 --> 00:21:53,000
sinusoidal inputs to circuits.
Furthermore,

350
00:21:53,000 --> 00:21:57,000
we are going to look at a
subcircuit of that region which

351
00:21:57,000 --> 00:22:02,000
is steady-state.
We will look at sinusoidal

352
00:22:02,000 --> 00:22:06,000
input and in the steady-state.
So that is going to be our

353
00:22:06,000 --> 00:22:11,000
focus for the next two or three
lectures just to give you a

354
00:22:11,000 --> 00:22:14,000
perspective of where we are.
To motivate this,

355
00:22:14,000 --> 00:22:18,000
what I would like to do is
consider your amplifier.

356
00:22:18,000 --> 00:22:22,000
This is our friend the
amplifier circuit,

357
00:22:22,000 --> 00:22:24,000
this part here.
And remember,

358
00:22:24,000 --> 00:22:28,000
even though this is an
amplifier, I am using a MOSFET

359
00:22:28,000 --> 00:22:30,000
here.
And a MOSFET,

360
00:22:30,000 --> 00:22:33,000
as you know,
has this gate capacitance CGS.

361
00:22:33,000 --> 00:22:36,000
I am explicitly drawing it out
for you here.

362
00:22:36,000 --> 00:22:40,000
And I am going to drive this
with a bias voltage plus some

363
00:22:40,000 --> 00:22:43,000
small signal vI,
the usual template for

364
00:22:43,000 --> 00:22:45,000
amplifiers.
And there is some Thevenin

365
00:22:45,000 --> 00:22:48,000
resistance attached to that
source.

366
00:22:48,000 --> 00:22:51,000
I am going to model my source
as a bias voltage,

367
00:22:51,000 --> 00:22:54,000
a small signal plus some source
resistance.

368
00:22:54,000 --> 00:22:58,000
And I want to apply a sine wave
here and I am going to look at

369
00:22:58,000 --> 00:23:02,000
what this looks like.
You may think,

370
00:23:02,000 --> 00:23:05,000
look, this is a linear
amplifier.

371
00:23:05,000 --> 00:23:10,000
And so if I apply a sine wave
here I should see some response

372
00:23:10,000 --> 00:23:15,000
here, and that should be the
same amplitude if I feed the

373
00:23:15,000 --> 00:23:18,000
same amplitude here over any
frequency.

374
00:23:18,000 --> 00:23:22,000
But let's see what happens.
Keep a look at,

375
00:23:22,000 --> 00:23:27,000
switch over to that view graph
while I show you a little

376
00:23:27,000 --> 00:23:32,000
demonstration here.
What you see here are three

377
00:23:32,000 --> 00:23:36,000
sine waves, a yellow sine wave
which is the input here,

378
00:23:36,000 --> 00:23:41,000
you see a green sine wave which
is the input vC at the gate of

379
00:23:41,000 --> 00:23:47,000
the MOSFET, and then you see the
blue which is the output here.

380
00:23:47,000 --> 00:23:51,000
For now simply focus on the
yellow and the blue.

381
00:23:51,000 --> 00:23:55,000
The yellow is the input and the
blue is the output.

382
00:23:55,000 --> 00:24:00,000
So I apply some input and I get
an output that looks more or

383
00:24:00,000 --> 00:24:05,000
less some linear function of
this input here.

384
00:24:05,000 --> 00:24:07,000
It is a small signal.
What I am going to do is I am

385
00:24:07,000 --> 00:24:10,000
going to change the frequency of
the input.

386
00:24:10,000 --> 00:24:13,000
Remember, I want to look at the
response of the circuit to a

387
00:24:13,000 --> 00:24:16,000
sinusoid.
I am feeding a sinusoid here.

388
00:24:16,000 --> 00:24:19,000
I look at the response.
I am going to change the

389
00:24:19,000 --> 00:24:21,000
frequency.
That is a big shift that we are

390
00:24:21,000 --> 00:24:25,000
making in that the curve that we
drew in the last lecture had to

391
00:24:25,000 --> 00:24:27,000
do with varying time.
Now I am going to focus on

392
00:24:27,000 --> 00:24:31,000
sinusoids and vary their
frequency.

393
00:24:31,000 --> 00:24:33,000
I am going to change the
frequency.

394
00:24:33,000 --> 00:24:37,000
Stare at the blue curve as I
increase the frequency and just

395
00:24:37,000 --> 00:24:41,000
think of what you might expect.
Based on the knowledge you have

396
00:24:41,000 --> 00:24:46,000
so far you would expect that
look, as I change the frequency,

397
00:24:46,000 --> 00:24:50,000
the frequency should change but
I should see the same amplitude.

398
00:24:50,000 --> 00:24:53,000
OK but take a look.
Let me increase the frequency

399
00:24:53,000 --> 00:24:56,000
of the input.
What do you see at the output?

400
00:24:56,000 --> 00:25:00,000
I am increasing the frequency.

401
00:25:10,000 --> 00:25:12,000
What do you see happening
there?

402
00:25:12,000 --> 00:25:16,000
If you don't see anything
changing there you will need to

403
00:25:16,000 --> 00:25:19,000
see an optometrist.
What do we see here?

404
00:25:19,000 --> 00:25:24,000
As I changed the frequency,
as I increased the frequency

405
00:25:24,000 --> 00:25:28,000
what happened to the blue?
The blue kept decreasing in

406
00:25:28,000 --> 00:25:31,000
amplitude.
And you are saying whoa,

407
00:25:31,000 --> 00:25:35,000
what is happening here?
We don't have the tools to deal

408
00:25:35,000 --> 00:25:38,000
with this.
I expected that when I changed

409
00:25:38,000 --> 00:25:41,000
my frequency,
my frequency here should change

410
00:25:41,000 --> 00:25:44,000
of course, but why is the
amplitude changing?

411
00:25:44,000 --> 00:25:47,000
What is happening here?
That is weird.

412
00:25:47,000 --> 00:25:51,000
I noticed that this amplitude
became smaller because the

413
00:25:51,000 --> 00:25:54,000
amplitude of the green became
smaller.

414
00:25:54,000 --> 00:25:58,000
And remember the green was the
voltage across the capacitor.

415
00:25:58,000 --> 00:26:02,000
So this is your RC.
And here is my input.

416
00:26:02,000 --> 00:26:07,000
My input has the amplitude,
which I am holding more or less

417
00:26:07,000 --> 00:26:10,000
constant.
And notice that vC decreased in

418
00:26:10,000 --> 00:26:12,000
value as I increased my
frequency.

419
00:26:12,000 --> 00:26:16,000
Just hold that thought.
As I increased the frequency of

420
00:26:16,000 --> 00:26:20,000
my input the amplitude of the
output kept diminishing.

421
00:26:20,000 --> 00:26:24,000
In other words,
the gain of the system seemed

422
00:26:24,000 --> 00:26:27,000
to have decreased as I increased
by frequency.

423
00:26:27,000 --> 00:26:31,000
And today we will look at why
that is so and how we can

424
00:26:31,000 --> 00:26:37,000
analyze that.
The other thing that is not so

425
00:26:37,000 --> 00:26:42,000
obvious here is there is a phase
shift.

426
00:26:42,000 --> 00:26:50,000
What I am going to do is try to
see if we can observe the phase

427
00:26:50,000 --> 00:26:53,000
shift here.

428
00:27:02,000 --> 00:27:05,000
Notice here.
What we have been used to is

429
00:27:05,000 --> 00:27:09,000
for the amplifier we get a
complete inversion at the

430
00:27:09,000 --> 00:27:11,000
output.
Inversion means a phase

431
00:27:11,000 --> 00:27:15,000
difference of 180 degrees for a
sine wave, right?

432
00:27:15,000 --> 00:27:19,000
This peak should have been
here, but notice that there is a

433
00:27:19,000 --> 00:27:22,000
shifting of the peak.
In other words,

434
00:27:22,000 --> 00:27:26,000
if the yellow was my input my
output should have had its

435
00:27:26,000 --> 00:27:31,000
minimum when the input had its
maximum.

436
00:27:31,000 --> 00:27:35,000
But notice there is a shifting
of the signal such that the

437
00:27:35,000 --> 00:27:39,000
output is a maximum,
not quite at the point where

438
00:27:39,000 --> 00:27:43,000
the input is a minimum but a
little bit after that.

439
00:27:43,000 --> 00:27:46,000
Also weird.
Not only has this little

440
00:27:46,000 --> 00:27:51,000
circuit here lost its gain
somehow, but also it seems to

441
00:27:51,000 --> 00:27:53,000
have shifted the signal in
phase.

442
00:27:53,000 --> 00:27:57,000
That is weird.
And today we will look at why

443
00:27:57,000 --> 00:28:02,000
that is so and try to understand
the frequency behavior of this

444
00:28:02,000 --> 00:28:09,000
little subcomponent here.
Notice that vC is exactly 180

445
00:28:09,000 --> 00:28:15,000
out of phase with vO.
So vO is faithfully an inverted

446
00:28:15,000 --> 00:28:23,000
amplified form of the input vC.
However, vC itself should have

447
00:28:23,000 --> 00:28:30,000
been the same as vI but it looks
quite different.

448
00:28:30,000 --> 00:28:32,000
So let's understand why that is
so.

449
00:28:32,000 --> 00:28:36,000
The subcircuit to model is the
subcircuit comprising the

450
00:28:36,000 --> 00:28:38,000
source, resistor and the
capacitor.

451
00:28:38,000 --> 00:28:41,000
And I am just showing that to
you here.

452
00:28:41,000 --> 00:28:44,000
I have a vI,
a resistor and capacitor.

453
00:28:44,000 --> 00:28:47,000
And I am going to understand
how this behaves.

454
00:28:47,000 --> 00:28:51,000
And it is a first order
circuit, single capacitor.

455
00:28:51,000 --> 00:28:53,000
My input is a vI cosine of
omega t.

456
00:28:53,000 --> 00:28:58,000
vI is a real number for t
greater than zero.

457
00:28:58,000 --> 00:29:02,000
And I am telling you that the
capacitor voltage starts out

458
00:29:02,000 --> 00:29:05,000
being zero.
And my vI is a sinusoid.

459
00:29:05,000 --> 00:29:08,000
It's not a step this time.
It's a sinusoid.

460
00:29:08,000 --> 00:29:13,000
So vI is a sinusoid and I want
to find out what vC looks like.

461
00:29:13,000 --> 00:29:17,000
The behavior here tells me,
I will give you the answer,

462
00:29:17,000 --> 00:29:21,000
that when I feed a sinusoidal
input as the frequency

463
00:29:21,000 --> 00:29:25,000
increases, vC should be
decreasing somehow and also be

464
00:29:25,000 --> 00:29:28,000
shifting in phase.
Let's do the derivation for

465
00:29:28,000 --> 00:29:34,000
that and see what happens.
To give you some insight as to

466
00:29:34,000 --> 00:29:39,000
how to go about analyzing this
let me draw a little graph as to

467
00:29:39,000 --> 00:29:44,000
the effort level of doing this.
To determine vC of t on the

468
00:29:44,000 --> 00:29:48,000
y-axis here is our effort.
How hard do we have to work to

469
00:29:48,000 --> 00:29:52,000
solve this circuit for a
sinusoidal input?

470
00:29:52,000 --> 00:29:56,000
And on this graph,
down here is easy and up here

471
00:29:56,000 --> 00:30:00,000
is pure agony.
Sheer agony up here.

472
00:30:00,000 --> 00:30:03,000
So it's the scale of effort
level ranging from easy to

473
00:30:03,000 --> 00:30:06,000
complete agony.
And this is the time axis.

474
00:30:06,000 --> 00:30:10,000
And the time axis starts out at
11 o'clock, the early part of

475
00:30:10,000 --> 00:30:12,000
today's lecture,
and ends at roughly 12,

476
00:30:12,000 --> 00:30:16,000
that is today's lecture and
this is next lecture.

477
00:30:16,000 --> 00:30:19,000
What I am going to show you
today is a method that uses a

478
00:30:19,000 --> 00:30:23,000
usual differential equation
approach, and this is going to

479
00:30:23,000 --> 00:30:26,000
be pure agony.
If you thought last Thursday

480
00:30:26,000 --> 00:30:30,000
was agony watch today.
This is going to be sheer,

481
00:30:30,000 --> 00:30:33,000
sheer, sheer hell.
So I am going to grunge through

482
00:30:33,000 --> 00:30:37,000
that, and I think I will sort of
give up halfway because it's

483
00:30:37,000 --> 00:30:39,000
just too painful even for me
here.

484
00:30:39,000 --> 00:30:42,000
Then what I am going to do is
at the end of this lecture I am

485
00:30:42,000 --> 00:30:45,000
going to show you an approach
that I give a cutesy name.

486
00:30:45,000 --> 00:30:47,000
I call it the "sneaky
approach".

487
00:30:47,000 --> 00:30:51,000
We are going to sneak something
in there it is going to be a lot

488
00:30:51,000 --> 00:30:53,000
easier.
And then in the next lecture I

489
00:30:53,000 --> 00:30:57,000
am going to show you an even
sneakier approach that is just

490
00:30:57,000 --> 00:31:01,000
going to be absolute bliss.
So let's start here.

491
00:31:01,000 --> 00:31:04,000
Indulge me as I go through the
agony part.

492
00:31:04,000 --> 00:31:08,000
I am going to blast through it
because that is not of how we

493
00:31:08,000 --> 00:31:12,000
are going to do things,
but you just need to know that

494
00:31:12,000 --> 00:31:15,000
that is agony.
Let's do a usual differential

495
00:31:15,000 --> 00:31:17,000
equation approach.
Steps one, two,

496
00:31:17,000 --> 00:31:20,000
three and four.
Set up differential equation,

497
00:31:20,000 --> 00:31:24,000
find the particular solution,
find the homogenous solution,

498
00:31:24,000 --> 00:31:27,000
add up the two and solve for
the unknowns.

499
00:31:27,000 --> 00:31:31,000
It's a mantra.
The four-step manta.

500
00:31:31,000 --> 00:31:34,000
Let's do it.
Step one, write the DE.

501
00:31:34,000 --> 00:31:37,000
That's easy.
We have done this before the RC

502
00:31:37,000 --> 00:31:40,000
circuit.
It's RC dvc/dt+vc=vI.

503
00:31:40,000 --> 00:31:45,000
This is no different from what
you got from what you got from

504
00:31:45,000 --> 00:31:49,000
your RC circuit with a step
input, just that my input is VI

505
00:31:49,000 --> 00:31:55,000
cosine of omega t in this case.
It is not just a DC voltage VI.

506
00:31:55,000 --> 00:31:58,000
Stare at that.
Enjoy it while the going is

507
00:31:58,000 --> 00:32:02,000
easy.
It's like traversing rapids,

508
00:32:02,000 --> 00:32:06,000
and before you come to a class
five, you have calm and raging

509
00:32:06,000 --> 00:32:09,000
waters there,
you kind of sit there saying

510
00:32:09,000 --> 00:32:14,000
oh, the scenery around here
looks really good and so on.

511
00:32:14,000 --> 00:32:18,000
All you are doing is stalling
before you have dive down to

512
00:32:18,000 --> 00:32:21,000
class five.
We will get to class five

513
00:32:21,000 --> 00:32:24,000
rapids in a few seconds here,
so just enjoy this.

514
00:32:24,000 --> 00:32:27,000
RC dvc/dt+vC=vI.
You've seen this before.

515
00:32:27,000 --> 00:32:31,000
Nothing fancy.
Good old stuff.

516
00:32:31,000 --> 00:32:34,000
VI cosine of omega t.
You want to hold onto your

517
00:32:34,000 --> 00:32:36,000
seatbelts?
OK.

518
00:32:36,000 --> 00:32:40,000
Let's find the particular
solution to the cosine input.

519
00:32:40,000 --> 00:32:44,000
Let's use our standard method.
What I will do is just so,

520
00:32:44,000 --> 00:32:48,000
there is going to be so much
crapola up there,

521
00:32:48,000 --> 00:32:53,000
so that I draw your attention
to vP, which is what we are

522
00:32:53,000 --> 00:32:56,000
trying to get,
I am just going to put a box

523
00:32:56,000 --> 00:33:01,000
around vP in red.
If you see like all sorts of

524
00:33:01,000 --> 00:33:04,000
garbage appear,
just look for the red square.

525
00:33:04,000 --> 00:33:06,000
That is what we are trying to
get at.

526
00:33:06,000 --> 00:33:09,000
That's the equation.
Let's try.

527
00:33:09,000 --> 00:33:12,000
First try, A worked before.
A constant value A worked

528
00:33:12,000 --> 00:33:16,000
before for DC inputs.
Let's try that again.

529
00:33:16,000 --> 00:33:18,000
If it worked then it may work
now.

530
00:33:18,000 --> 00:33:22,000
If I use A, a constant value,
and I substitute it here,

531
00:33:22,000 --> 00:33:26,000
I get dA/dt goes to zero,
vP is A, but on the right-hand

532
00:33:26,000 --> 00:33:32,000
side I have VI cosine omega t.
So clearly A doesn't work.

533
00:33:32,000 --> 00:33:34,000
Sorry.
I struck out.

534
00:33:34,000 --> 00:33:39,000
Well, cosine omega t here,
let's try A cosine omega T as

535
00:33:39,000 --> 00:33:43,000
my particular solution.
Things are getting a little

536
00:33:43,000 --> 00:33:46,000
harder now, a little more
painful.

537
00:33:46,000 --> 00:33:49,000
So substitute A cosine omega t
here.

538
00:33:49,000 --> 00:33:55,000
So I do get A cosine omega T
for vP, but out here I get RCA

539
00:33:55,000 --> 00:34:00,000
sine omega t times omega times
minus one.

540
00:34:00,000 --> 00:34:05,000
So I have a sine and a cosine,
and I have a cosine on the

541
00:34:05,000 --> 00:34:09,000
right-hand side.
Sorry, it doesn't work.

542
00:34:09,000 --> 00:34:15,000
Nope, doesn't work either.
Well, let's try A cosine omega

543
00:34:15,000 --> 00:34:19,000
t plus phi.
We are now embarking into the

544
00:34:19,000 --> 00:34:22,000
rapids here.
You can begin feeling the

545
00:34:22,000 --> 00:34:26,000
pressure.
Just to refresh your memories

546
00:34:26,000 --> 00:34:32,000
of sines and cosines.
A is the amplitude of the

547
00:34:32,000 --> 00:34:34,000
cosine.
Omega is the frequency.

548
00:34:34,000 --> 00:34:38,000
Phi is the phase.
Remember the signal I showed

549
00:34:38,000 --> 00:34:41,000
you earlier?
If something happens to the

550
00:34:41,000 --> 00:34:45,000
amplitude of the sine,
something happens to the phase.

551
00:34:45,000 --> 00:34:50,000
A cosine omega t plus phi.
Let me plug it in here and go

552
00:34:50,000 --> 00:34:53,000
by standard practice.
Here is what I get.

553
00:34:53,000 --> 00:34:56,000
I plug in A cosine omega t to
this equation,

554
00:34:56,000 --> 00:35:01,000
and this is what I get.
I differentiate it.

555
00:35:01,000 --> 00:35:05,000
I get omega out minus sine,
sine of negative d plus phi,

556
00:35:05,000 --> 00:35:09,000
A cosine omega t plus phi
equals VI cosine omega t.

557
00:35:09,000 --> 00:35:12,000
That might work.
Now we get into the class six

558
00:35:12,000 --> 00:35:16,000
part of the class five.
All class fives have a little

559
00:35:16,000 --> 00:35:20,000
bit of class six rapids.
Remember, the rapids go up on

560
00:35:20,000 --> 00:35:23,000
an exponential scale so it like
earthquakes.

561
00:35:23,000 --> 00:35:27,000
What I do now is expand out
sine omega t plus phi,

562
00:35:27,000 --> 00:35:31,000
blah, blah, blah,
it goes on and on.

563
00:35:31,000 --> 00:35:35,000
I could go on and on,
but this is even tiring me.

564
00:35:35,000 --> 00:35:41,000
This can be made to work,
but I am not sure I want to put

565
00:35:41,000 --> 00:35:45,000
all of us through this trig
nightmare here.

566
00:35:45,000 --> 00:35:50,000
If I am really,
really nasty I could give this

567
00:35:50,000 --> 00:35:54,000
to you as a homework assignment
or something,

568
00:35:54,000 --> 00:36:00,000
but I am not that nasty so you
won't see that.

569
00:36:00,000 --> 00:36:03,000
But if I go down this path it
will get me to the answer,

570
00:36:03,000 --> 00:36:06,000
but I would have to soon
negotiate class six,

571
00:36:06,000 --> 00:36:09,000
class seven rapids to get to
where I want.

572
00:36:09,000 --> 00:36:12,000
So let me punt on it,
let me start from scratch.

573
00:36:12,000 --> 00:36:15,000
I am at step two,
let me start from scratch.

574
00:36:15,000 --> 00:36:18,000
Instead what I would like to do
is let's get sneaky here.

575
00:36:18,000 --> 00:36:21,000
Rather than negotiating the
class five rapids,

576
00:36:21,000 --> 00:36:25,000
what we can say is ah-ha,
we can take our canoes and jump

577
00:36:25,000 --> 00:36:30,000
onto shore and run down and then
get back onto the river.

578
00:36:30,000 --> 00:36:32,000
Let's do that.
That is called the sneaky

579
00:36:32,000 --> 00:36:34,000
approach.
So that all our friends who are

580
00:36:34,000 --> 00:36:37,000
behind us think we are
negotiating the rapids,

581
00:36:37,000 --> 00:36:41,000
but what we are going to do is
get sneaky and take the shore

582
00:36:41,000 --> 00:36:42,000
path.
Let's get sneaky,

583
00:36:42,000 --> 00:36:45,000
just walk down the shore and
see what is there.

584
00:36:45,000 --> 00:36:47,000
I want to do something
completely weird here.

585
00:36:47,000 --> 00:36:51,000
I want to look at solving this
equation through the shore

586
00:36:51,000 --> 00:36:53,000
method.
S stands for shore or S stands

587
00:36:53,000 --> 00:36:57,000
for sneaky, whatever you want.
What I am going to do is rather

588
00:36:57,000 --> 00:37:01,000
than trying to solve for VI
cosine omega t.

589
00:37:01,000 --> 00:37:04,000
I am going to say let's try a
different input all together.

590
00:37:04,000 --> 00:37:06,000
And you will understand why in
a second.

591
00:37:06,000 --> 00:37:09,000
It's like I am the captain of
my canoe and I tell my

592
00:37:09,000 --> 00:37:12,000
teammates, hey,
let's not negotiate the rapids,

593
00:37:12,000 --> 00:37:15,000
let's go and explore the shore.
Maybe down the shore we can

594
00:37:15,000 --> 00:37:19,000
find a path that gets us across
to the other side more easily.

595
00:37:19,000 --> 00:37:22,000
So here is me and my colleagues
carrying our canoe and getting

596
00:37:22,000 --> 00:37:24,000
onto shore and taking a sneaky
path.

597
00:37:24,000 --> 00:37:26,000
This is not what I set out to
solve.

598
00:37:26,000 --> 00:37:30,000
I don't know where this will
lead me.

599
00:37:30,000 --> 00:37:33,000
But let's see where the shore
path leads us.

600
00:37:33,000 --> 00:37:36,000
I want to try solving this
equation Vie^st.

601
00:37:36,000 --> 00:37:40,000
S stands for shore or sneaky or
whatever you want.

602
00:37:40,000 --> 00:37:45,000
I don't know where I am going,
but let's see where this leads

603
00:37:45,000 --> 00:37:46,000
us.
Let's explore.

604
00:37:46,000 --> 00:37:50,000
Make believe you are Columbus
or something.

605
00:37:50,000 --> 00:37:53,000
I don't know.
Let's use the usual techniques

606
00:37:53,000 --> 00:37:57,000
and see how this works out.
Let's try a particular

607
00:37:57,000 --> 00:38:01,000
solution, Vpe^st.
My input is Vie^st.

608
00:38:01,000 --> 00:38:06,000
I am trying to solve the
circuit for a different input.

609
00:38:06,000 --> 00:38:11,000
And let me try solution Vpe^st
and see if that works out

610
00:38:11,000 --> 00:38:15,000
nicely.
I substitute Vpe^st into my

611
00:38:15,000 --> 00:38:18,000
equation here,
so RCVpe^st blah blah blah.

612
00:38:18,000 --> 00:38:23,000
What I get here is Vie^st,
Vpe^st stays the same,

613
00:38:23,000 --> 00:38:28,000
but here vP comes out,
s comes out and e^st stays the

614
00:38:28,000 --> 00:38:31,000
same.
That is nice property of

615
00:38:31,000 --> 00:38:34,000
exponentials.
This is what I get.

616
00:38:34,000 --> 00:38:37,000
A really cool property of
exponentials is that when I

617
00:38:37,000 --> 00:38:41,000
differentiate it I get the
exponential back.

618
00:38:41,000 --> 00:38:45,000
Unlike a cosine I got a sine,
and for a sine I got a cosine.

619
00:38:45,000 --> 00:38:49,000
Exponentials are very plain and
simple, are straightforward.

620
00:38:49,000 --> 00:38:53,000
What you see is what you get.
You differentiate it.

621
00:38:53,000 --> 00:38:56,000
You get the same thing,
you get scaling vP,

622
00:38:56,000 --> 00:39:00,000
S and so on,
and some scaling here.

623
00:39:00,000 --> 00:39:05,000
You get S scaling here,
but the basic form stays the

624
00:39:05,000 --> 00:39:07,000
same.
This is quite nice.

625
00:39:07,000 --> 00:39:12,000
I have e^st in all three
places, so I can cancel those

626
00:39:12,000 --> 00:39:17,000
out and I get this expression.
And I get this.

627
00:39:17,000 --> 00:39:21,000
Wow.
So if I go down the shore I get

628
00:39:21,000 --> 00:39:25,000
some place fast.
I don't know where I am yet,

629
00:39:25,000 --> 00:39:30,000
but whatever I did,
it was easy.

630
00:39:30,000 --> 00:39:33,000
I am just exploring this path,
down the shore path.

631
00:39:33,000 --> 00:39:36,000
I am making progress.
I don't know where I have

632
00:39:36,000 --> 00:39:39,000
gotten yet.
We will see where we got to in

633
00:39:39,000 --> 00:39:43,000
a second, but I got some place
quickly, fast.

634
00:39:43,000 --> 00:39:47,000
I was able to solve for this
input Vie^st and get this

635
00:39:47,000 --> 00:39:50,000
solution very quickly.
So what happened here?

636
00:39:50,000 --> 00:39:54,000
I assumed the solution of the
form Vpe^st, substituted it

637
00:39:54,000 --> 00:39:58,000
back, and found that if vP were
equal to Vi/(1+sRC) then Vpe^st

638
00:39:58,000 --> 00:40:03,000
is a solution.
What I have done here is that

639
00:40:03,000 --> 00:40:09,000
Vi/(1+sRC) is a particular
solution to this equation for

640
00:40:09,000 --> 00:40:14,000
the input Vie^st.
I put a box around it because

641
00:40:14,000 --> 00:40:17,000
this is important.
This was easy.

642
00:40:17,000 --> 00:40:22,000
We went down the shore,
and said let's try this other

643
00:40:22,000 --> 00:40:26,000
input.
We made rapid progress on shore

644
00:40:26,000 --> 00:40:31,000
and I got some place.
I don't know where I am yet.

645
00:40:31,000 --> 00:40:34,000
I got this.
Let me pause here and let me

646
00:40:34,000 --> 00:40:37,000
give you the final answer.
I am going to show you over the

647
00:40:37,000 --> 00:40:40,000
next five minutes that this is
our answer.

648
00:40:40,000 --> 00:40:42,000
You are staring at the answer
already.

649
00:40:42,000 --> 00:40:45,000
I am a party,
I have taken a shore path and

650
00:40:45,000 --> 00:40:48,000
we have gotten some place.
We see the river there,

651
00:40:48,000 --> 00:40:51,000
so it turns out we are exactly
where we want to be,

652
00:40:51,000 --> 00:40:54,000
just after the rapids.
All I have to do now is get my

653
00:40:54,000 --> 00:40:58,000
colleagues into the river with
myself in the canoe and we are

654
00:40:58,000 --> 00:41:01,000
done.
You don't know that yet.

655
00:41:01,000 --> 00:41:04,000
My colleagues and I are sitting
on the shore looking at the

656
00:41:04,000 --> 00:41:05,000
river.
We've gotten some place.

657
00:41:05,000 --> 00:41:08,000
There are no rapids there.
We have gotten some place.

658
00:41:08,000 --> 00:41:11,000
We don't quite know is this
just after the rapids or not.

659
00:41:11,000 --> 00:41:14,000
We don't know yet,
but I got there very quickly.

660
00:41:14,000 --> 00:41:17,000
And I will tell you right now,
that is the place we wanted to

661
00:41:17,000 --> 00:41:19,000
go to.
The next five view graphs I am

662
00:41:19,000 --> 00:41:21,000
going to blast through.
There is going to be more pain

663
00:41:21,000 --> 00:41:24,000
and agony to show you why that
is the case.

664
00:41:24,000 --> 00:41:27,000
It's me thinking I am Columbus
and proving to my colleagues

665
00:41:27,000 --> 00:41:30,000
that this is where we want to
be.

666
00:41:30,000 --> 00:41:33,000
And pulling out my sextant,
and the compasses and so on

667
00:41:33,000 --> 00:41:37,000
that cartographers and people
use to prove to my colleagues

668
00:41:37,000 --> 00:41:41,000
that this is where I want to be.
This is the answer.

669
00:41:41,000 --> 00:41:45,000
The next five view graphs will
be demonstrating that this is

670
00:41:45,000 --> 00:41:48,000
indeed the answer,
or close enough to the answer

671
00:41:48,000 --> 00:41:51,000
that we will be satisfied.
Isn't this spectacular?

672
00:41:51,000 --> 00:41:56,000
I am going to show you in about
five minutes that this gives us

673
00:41:56,000 --> 00:42:00,000
all the information we need to
know to compute the sinusoidal

674
00:42:00,000 --> 00:42:05,000
steady-state response to this
differential equation.

675
00:42:05,000 --> 00:42:09,000
Let me write that down here
just so you know.

676
00:42:21,000 --> 00:42:24,000
Just so you remember.
I am going to put a marker on

677
00:42:24,000 --> 00:42:28,000
the shore that says this is
where we are right now.

678
00:42:28,000 --> 00:42:31,000
Now let me prove to you.
As I just said,

679
00:42:31,000 --> 00:42:39,000
vPS is Vi, it's this stuff here
multiplied by e^st is the

680
00:42:39,000 --> 00:42:45,000
solution to Vie^st.
This guy here is a solution for

681
00:42:45,000 --> 00:42:52,000
Vie^st and vP is simply the
coefficient that multiplies

682
00:42:52,000 --> 00:42:56,000
e^st.
Similarly, if I substitute S

683
00:42:56,000 --> 00:43:02,000
equals j omega.
I told you five view graphs of

684
00:43:02,000 --> 00:43:07,000
more hell, but I am just going
to prove to you that this is it.

685
00:43:07,000 --> 00:43:10,000
I am substituting S equals j
omega.

686
00:43:10,000 --> 00:43:15,000
This is Columbus giving a big
speech at the end convincing his

687
00:43:15,000 --> 00:43:18,000
colleagues that we are where we
want to be.

688
00:43:18,000 --> 00:43:22,000
I substitute j omega for S and
this is what I get.

689
00:43:22,000 --> 00:43:27,000
This is a solution for e to the
st, and so this is a solution

690
00:43:27,000 --> 00:43:33,000
for e to the j omega t.
And let me mark this for you as

691
00:43:33,000 --> 00:43:37,000
something to remember.
Vi/(1+j omega RC).

692
00:43:37,000 --> 00:43:43,000
In terms of that,
I am substituting j omega for

693
00:43:43,000 --> 00:43:46,000
S.
And this is a complex number.

694
00:43:46,000 --> 00:43:52,000
It is a complex amplitude.
It is a complex number because

695
00:43:52,000 --> 00:43:58,000
of j here, and it multiplies e
to the j omega t.

696
00:43:58,000 --> 00:44:03,000
Just keep this in mind.
So that was easy.

697
00:44:03,000 --> 00:44:08,000
The steps were easy.
I am still proving to you that

698
00:44:08,000 --> 00:44:13,000
this is where we want to be.
Now let me draw the connection

699
00:44:13,000 --> 00:44:17,000
back to vP.
And the first fact was that

700
00:44:17,000 --> 00:44:21,000
finding the response to Vie^(j
omega t) was easy.

701
00:44:21,000 --> 00:44:25,000
We know that.
Second was the following

702
00:44:25,000 --> 00:44:31,000
observation that Vi cosine omega
t is simply the real part of

703
00:44:31,000 --> 00:44:37,000
this number here.
So Vi cosine omega t is simply

704
00:44:37,000 --> 00:44:42,000
the real part of Vie^(j omega t)
from the Euler relation.

705
00:44:42,000 --> 00:44:48,000
So cosine omega t is simply the
real part of this guy.

706
00:44:48,000 --> 00:44:51,000
Light bulbs beginning to go
off?

707
00:44:51,000 --> 00:44:58,000
The first fast was that finding
the response to Vie^(j omega t)

708
00:44:58,000 --> 00:45:02,000
was easy.
And the response was this,

709
00:45:02,000 --> 00:45:05,000
right?
Times e to the j omega t.

710
00:45:05,000 --> 00:45:07,000
That was easy.
All right.

711
00:45:07,000 --> 00:45:12,000
And the second part is that the
real part of this is Vi cosine

712
00:45:12,000 --> 00:45:17,000
omega t was our input.
Draw the connection between two

713
00:45:17,000 --> 00:45:21,000
steps.
Finding the response to Vie^(j

714
00:45:21,000 --> 00:45:25,000
omega t) was easy.
The real part of that was the

715
00:45:25,000 --> 00:45:31,000
input we cared about.
Are light bulbs going off?

716
00:45:31,000 --> 00:45:35,000
Let me draw you a little
picture here to show you what is

717
00:45:35,000 --> 00:45:37,000
happening.
Response to vI is vP.

718
00:45:37,000 --> 00:45:41,000
It's the particular response we
are looking for.

719
00:45:41,000 --> 00:45:46,000
Remember the red square?
But we threw in a sneaky input

720
00:45:46,000 --> 00:45:49,000
vIS and we formed the response
vPS to that.

721
00:45:49,000 --> 00:45:52,000
This step was easy.
This step was hard.

722
00:45:52,000 --> 00:45:55,000
vI to vP was hard,
trig nightmare,

723
00:45:55,000 --> 00:45:59,000
remember?
But vIS to vPS was easy.

724
00:45:59,000 --> 00:46:05,000
It was a simple Vpe^st thing.
We also know that the real part

725
00:46:05,000 --> 00:46:10,000
of vIS is vI.
The real part of this is simply

726
00:46:10,000 --> 00:46:14,000
vI.
If you have a real circuit,

727
00:46:14,000 --> 00:46:20,000
if you have a real linear
circuit, for a linear circuit,

728
00:46:20,000 --> 00:46:27,000
if the real part of this gives
me vI then the real part of the

729
00:46:27,000 --> 00:46:33,000
solution should give me vP.
So computing vPS was easy.

730
00:46:33,000 --> 00:46:37,000
If I take the real part of
this, I take the corresponding

731
00:46:37,000 --> 00:46:40,000
real part of this.
This is sort of an inverse

732
00:46:40,000 --> 00:46:43,000
superposition argument.
Superposition,

733
00:46:43,000 --> 00:46:47,000
I said, take the response for
A, take the response for B,

734
00:46:47,000 --> 00:46:51,000
add them up and you get the
response for A plus B.

735
00:46:51,000 --> 00:46:55,000
Here what we are saying is that
get the response to A plus B,

736
00:46:55,000 --> 00:46:59,000
or to A plus jB and the real
part of the input will produce

737
00:46:59,000 --> 00:47:04,000
the response given by the real
part of the output.

738
00:47:04,000 --> 00:47:07,000
So this is an inverse
superposition argument.

739
00:47:07,000 --> 00:47:10,000
If it is a linear circuit,
then if vI is the real part of

740
00:47:10,000 --> 00:47:15,000
this sneaky input then I find
the response to the sneaky input

741
00:47:15,000 --> 00:47:17,000
and take its real part I should
get vP.

742
00:47:17,000 --> 00:47:21,000
Here I am, Columbus,
staring down at the entrance to

743
00:47:21,000 --> 00:47:24,000
this part of the river.
I just proved to my colleagues

744
00:47:24,000 --> 00:47:30,000
that all we have to do is take
the real part of what we have.

745
00:47:30,000 --> 00:47:34,000
We can just jump right back
into the river and get back to

746
00:47:34,000 --> 00:47:36,000
vP.
And what I am going to do next

747
00:47:36,000 --> 00:47:40,000
is just grind through the math
and just show you that.

748
00:47:40,000 --> 00:47:44,000
I will just blast through it.
It is not important,

749
00:47:44,000 --> 00:47:48,000
but you have it in your notes.
I am telling you that vP is

750
00:47:48,000 --> 00:47:51,000
simply the real part of the
sneaky output.

751
00:47:51,000 --> 00:47:54,000
And I take the real part of vP
e to the j omega t.

752
00:47:54,000 --> 00:48:00,000
And I take the real part.
And just a bunch of math here.

753
00:48:00,000 --> 00:48:04,000
I am just taking the real part
and doing a bunch of complex

754
00:48:04,000 --> 00:48:06,000
math.
Remember vP was given by this

755
00:48:06,000 --> 00:48:09,000
quantity here.
And I take the real part and I

756
00:48:09,000 --> 00:48:13,000
end up with vP is simply this
quantity multiplied by cosine

757
00:48:13,000 --> 00:48:16,000
omega t plus phi,
where phi is given by is given

758
00:48:16,000 --> 00:48:20,000
by tan inverse of omega RC,
and this is the coefficient

759
00:48:20,000 --> 00:48:24,000
multiplying the cosine.
So by taking the sneaky path

760
00:48:24,000 --> 00:48:27,000
and then taking the real part of
that output answer,

761
00:48:27,000 --> 00:48:33,000
I was able to very quickly get
to where I wanted to be.

762
00:48:33,000 --> 00:48:35,000
So from here to here it is only
math.

763
00:48:35,000 --> 00:48:38,000
Recall, that vP,
the thing in the red was what

764
00:48:38,000 --> 00:48:41,000
we set out to find out,
which was the particular

765
00:48:41,000 --> 00:48:43,000
response to VI cosine of omega
t.

766
00:48:43,000 --> 00:48:46,000
And remember that two grunge is
all of this stuff.

767
00:48:46,000 --> 00:48:50,000
I am going to blast through two
or three more view graphs that

768
00:48:50,000 --> 00:48:54,000
just give you more insight and
more math, nothing particular.

769
00:48:54,000 --> 00:48:57,000
And remember to solve the
equation we have to find a

770
00:48:57,000 --> 00:49:00,000
homogenous solution,
too.

771
00:49:00,000 --> 00:49:04,000
But recall that the homogenous
solution for an RC circuit is of

772
00:49:04,000 --> 00:49:07,000
the form Ae^-t/RC.
This means that as time becomes

773
00:49:07,000 --> 00:49:10,000
very large this part goes to
zero.

774
00:49:10,000 --> 00:49:13,000
As time becomes large in the
steady state,

775
00:49:13,000 --> 00:49:15,000
remember I care about the
steady state?

776
00:49:15,000 --> 00:49:18,000
This goes to zero.
I don't care about the

777
00:49:18,000 --> 00:49:21,000
homogenous solution.
Isn't that fantastic?

778
00:49:21,000 --> 00:49:25,000
Most the circuits we will deal
with, except for purely

779
00:49:25,000 --> 00:49:30,000
oscillatory ones,
the homogenous part dies away.

780
00:49:30,000 --> 00:49:33,000
You have something like e to
the -t whatever.

781
00:49:33,000 --> 00:49:35,000
It just dies away.
It's gone.

782
00:49:35,000 --> 00:49:38,000
So the total solution has vH
going away.

783
00:49:38,000 --> 00:49:40,000
And what I end up with is just
vP.

784
00:49:40,000 --> 00:49:44,000
My total solution in the steady
state is simply vP.

785
00:49:44,000 --> 00:49:48,000
And A is given by this that we
just calculated.

786
00:49:48,000 --> 00:49:52,000
I just have a bunch more
insight that I talk about that

787
00:49:52,000 --> 00:49:55,000
you can look through in your
notes.

788
00:49:55,000 --> 00:50:00,000
And I just want to show you a
very quick summary.

789
00:50:00,000 --> 00:50:04,000
In summary, what we have is we
computed vP.

790
00:50:04,000 --> 00:50:08,000
It was a complex coefficient.
And all these steps,

791
00:50:08,000 --> 00:50:12,000
2 grunge, 3 and 4 were a waste
of time.

792
00:50:12,000 --> 00:50:18,000
And what I showed you was that
for the input VI the coefficient

793
00:50:18,000 --> 00:50:23,000
vP was complex.
And I can take the ratio and

794
00:50:23,000 --> 00:50:26,000
represent it in this manner as
well.

795
00:50:26,000 --> 00:50:29,000
And from vP,
I can then compute the

796
00:50:29,000 --> 00:50:35,000
multiplier for the cosine as
follows.

797
00:50:35,000 --> 00:50:38,000
I divide by vP here.
Remember the cosine was

798
00:50:38,000 --> 00:50:41,000
multiplied by,
in the mathematical step that I

799
00:50:41,000 --> 00:50:44,000
did, VI divided one plus,
this stuff here,

800
00:50:44,000 --> 00:50:48,000
so I could get the magnitude
and phase of the transfer

801
00:50:48,000 --> 00:50:51,000
function of this circuit in the
following manner.

802
00:50:51,000 --> 00:50:55,000
And to wrap up very quickly,
I am going to cover this again

803
00:50:55,000 --> 00:51:00,000
the next time and show you a
magnitude plot.

804
00:51:00,000 --> 00:51:02,000
Notice here that if I plot
Vp/Vi.

805
00:51:02,000 --> 00:51:06,000
Remember this was Vp here.
That's the answer.

806
00:51:06,000 --> 00:51:12,000
The magnitude looks like this.
On a log scale Vp/Vi for small

807
00:51:12,000 --> 00:51:17,000
frequencies omega is at one,
but as omega increases Vp/Vi

808
00:51:17,000 --> 00:51:20,000
keeps decreasing.
That is the output.

809
00:51:20,000 --> 00:51:24,000
Remember Vp was the amplitude
of the output?

810
00:51:24,000 --> 00:51:30,000
That keeps decreasing.
And this is the reason why.

811
00:51:30,000 --> 00:51:34,000
As I increase the frequency,
the amplitude of my output

812
00:51:34,000 --> 00:51:39,000
cosine kept decreasing.
I could also plot the phase for

813
00:51:39,000 --> 00:51:41,000
you.
And the phase,

814
00:51:41,000 --> 00:51:46,000
in the same manner as omega
increased, my phase also kept

815
00:51:46,000 --> 00:51:50,000
shifting from zero initially to
pi/2 finally.

816
00:51:50,000 --> 00:51:55,000
Let me stop here and start with
this the next time and revisit

817
00:51:55,000 --> 00:51:57,000
this.
Unfortunately,

818
00:51:57,000 --> 00:52:02,000
I won't have time for the demo.
I will show it to you next

819
00:52:02,000 --> 00:52:05,000
time.