OK. Good morning.
Let's get going. As always, I will start with a
review. And today we embark on another
major milestone in our analysis of lumped circuits.
And it is called the "Sinusoidal Steady-state".
Again, I believe this will be the second and the last lecture
for which I will be using view graphs.
And the idea here is that, just like in the last lecture,
there is a bunch of mathematical grunge that needs
to be gone through. And I want to show you a
sequence in a little chart today that talks about the effort
level in doing some problems based on time domain
differential equations, as in the last lecture,
something slightly different today and something much better
on Thursday. And so, in some sense,
Thursday's lecture and today's lecture involve talking about
the foundations of the behavior of certain types of circuits.
And it is good for you to have this foundation as background,
but when you actually go to solve circuits you don't quite
use these methods. You use much easier techniques,
which I will talk about next Thursday.
Let's start with a quick review, and then we will go into
sinusoidal steady state. The last lecture we talked
about this circuit and we did the same two lectures ago on
Tuesday. We had one inverter driving
another inverter. And we said that the wire over
ground had some inductance. CGS is the capacitor of the
gate and R is the resistance at the drain of the first inverter.
And if you look at this circuit, that circuit formed an
RLC pattern. And what we did was we said
let's drive this with a one to zero transition at the input.
And the one to zero transition at the input would cause this
transistor to switch off, and this node would then go
from a very low value to a high value.
So it as if a 5 volt step was applied at this input.
We also saw that using time domain differential equations
that by applying a step input here the output looked like
this. The output would show some
oscillatory behavior when we have a capacitor and inductor.
I also gave you some insight as to why it oscillates like this.
And you also heard in recitation that the reason for
this oscillation was because of these two storage elements.
Each of these storage elements tries to hold onto its state
variable. For example,
the capacitor tries to maintain its voltage while the inductor
tries to maintain its current. And, much like a pendulum which
oscillates back and forth, it swaps potential energy
versus kinetic energy down here and swings back and forth.
In the same way, in an LC circuit like this,
energy swaps back and forth between a potential energy and a
kinetic energy equivalent, which swaps back and forth
between energy stored in the inductor and energy stored in
the capacitor and sloshes back and forth.
And because of this resistor the energy eventually dissipates
and you end up getting a final value which corresponds to the 5
volts appearing here. And why is that?
That is because remember the capacitor is a long-term open
for DC. It is a DC voltage.
After a long time this capacitor looks like an open
circuit and the inductor looks like a complete short circuit,
an ideal inductor as a complete short circuit for DC.
And so therefore in the long-term it is as if this guy
is a short, this guy is an open, so 5 volts simply appears here.
And this is the transient behavior.
Then we just switch the first transistor off.
In the last lecture, I left off with intuitive
analysis. Let me quickly cover that and
redo the intuitive analysis for you.
I left it the last time by having you think about whether
the transient response would begin by going down or begin by
going up. And I will cover that today.
This was the circuit that we analyzed.
A VI input with a step and an RLC out here.
And we were plotting the capacitor voltage.
And intuitively we can plot this in the following way.
I have also marked for you the section number in the course
notes which has a discussion of this intuitive analysis,
(Section 13.8). Let's do the easy stuff first.
Notice that the capacitor wants to hold its voltage.
And so given that we don't have any infinite impulse here,
I am going to start out with the capacitor voltage being
where it is. And the initial conditions are
given to you. You are given that the
capacitor voltage starts out being positive at v zero and the
current starts out being negative at time zero.
So I am telling you that there is a voltage v across the
capacitor at time t=0 and there is a current that is flowing.
Since i is negative there is a current initially that is
flowing in the opposite direction to this arrow here.
The i zero is negative. In light of that,
I can start plotting my curve here by intuition.
I start by saying at time t=0 I am told that the initial voltage
of the capacitor is at zero. This is about as simple as it
gets. Completely intuitive.
I also know that after a long time, can someone tell me after
a long time what the voltage will be at the end of the
capacitor? You should be able to get his
by observation? VI.
And why is it VI? It is vI because this is a DC
value VI. And after a long time this guy
behaves like an open circuit to DC.
This guy behaves like a short circuit to DC.
So since this is an open circuit, VI will appear here
after a long time. And so therefore,
after a long time, the capacitor voltage is going
to be at VI. And I just showed you that.
There you go. You already have the two
endpoints of your curve completely by observation,
intuition. No DEs, no nothing.
Just by staring at it and understanding the fundamentals
of how simple primitive circuit elements work.
Absolutely simple stuff. So you've nailed the two ends
now and you cannot go wrong with the stuff in the middle.
Let's see. As a next step what I do is I
need to understand what the dynamics of the circuit looks
like here. What I do is I develop the
characteristic equation. And initially you will write
the differential equation and then substitute e^st and get
this characteristic equation. But you could also remember it
as a pattern. For a series RLC circuit you
always get an equation of this form, always.
If this were R, L and C.
And whether you are looking at L up here or C up here,
as long as you're looking at the capacitor voltage,
the capacitor voltage is going to behave the same.
And for this circuit the characteristic equation remains
the same as well for a series RLC.
It is exactly this. And I write the standard
canonic form as s squared plus two alpha s + omega nought
squared. And omega nought is simply one
by square root of LC and alpha is simply R divided by L and I
have two in the denominator as well.
And then I get omega d which is my damped frequency given by
omega nought squared minus alpha squared.
And Q is simply called the quality factor.
And we will learn about Q in a lot more detail in about a
couple of lectures from today. That is given where omega
nought divided by two alpha. These parameters,
alpha, omega nought, Q and omega d pretty much
characterize everything else that I need to know about the
circuit. First of all,
omega d is the frequency of oscillation.
And so since omega d is a frequency of oscillation then I
know that the time period of oscillation is two pi by omega
d. Omega is in radians.
Notice that for typical values of circuits like this when R is
pretty small, alpha squared is going to be
very small. It's a common case for
underdamped circuit that omega d will more or less be equal to
omega nought. Commonly that is going to be
the case. This frequency is governed by
LC. And if R is large it is
governed by this omega d here. So I have the frequency of
oscillation. I also know that Q is related
to the frequency of oscillation divided by alpha.
It is a ratio of the frequency divided by how badly I am being
damped. So it is a fight between the
frequency of oscillation and now heavily I damp.
And the ratio of that is an indication of how many cycles I
ring. So Q tells me that the ringing
stops approximately after Q cycles.
These four values, omega d, Q, alpha and omega
nought are telling me more and more now.
So I have got these two factors.
So I know now, based on omega d and Q,
that it is going to look something like this.
Some ringing here and then I stop at this point.
The last thing that is left to do here for me for now is to
figure out whether I start out going down or I start out going
up. I start out going down or I
start out going up? I don't know that yet.
And I stopped at this point in the last lecture and let you
think about how you can stare at the circuit and intuitively
figure out whether this goes down or that goes up.
So here is the insight. Let's stare at this for a
minute purely through intuition. The capacitor has a voltage v
across it, right? And that is because I am
telling you that it has an initial voltage v.
Now, I want to find out at prime t equals zero plus,
in which direction does a capacitor voltage go?
Increase or decrease? What do I do?
Let me take a look at the inductor current.
I am told that the inductor current is negative which means
I am told that the inductor current is going in this
direction initially. The inductor current is pushing
in this direction. Now, remember,
just as the capacitor is one stubborn nut trying to hold its
voltage, the inductor is as stubborn.
It is trying to hold its current.
It is trying to maintain its current.
And its initial current i(0) is in this direction.
Capacitor has a voltage here, that is v(0),
and the inductor is yanking the current in that direction.
So what should happen to the capacitor voltage initially?
If I am at v(0) and someone is yanking current out,
at least initially in that direction, what should the
initial dynamics of the capacitor voltage look like?
Pardon? It should drop,
which means that if the initial current is being pulled in that
direction the capacitor voltage must droop to begin with.
Completely through intuition. No math here.
This means that i(0) is negative.
It is as if i(0) is being pulled out in this manner,
so the capacitor voltage must drop to begin life.
Therefore, the dynamics look somewhat like this.
Notice that this is very reminiscent of the ringing that
we saw at the gate node of the second inverter.
Let's stop here in terms of time domain analysis of RLC,
and today let's take another big step forward.
Today marks another transition in life here.
This is actually a huge transition so I want to just
pause and take like ten seconds of a breather just to clearly
demarcate the fact that we have a huge transition coming up.
The key to this transition is that I want to look at today the
steady-state response of networks to a sinusoidal drive.
We are going to do two things differently starting today on
this new journey of ours. In the past,
we looked at time domain behavior of circuits.
For RLC, for example, we looked at the transient
response. And what happened the instant I
turn something on? The transient response.
Today we are going to look at a steady-state response.
Steady-state response says that if I wait long enough,
for whatever the circuit wants to do in the beginning of life
to die out. If I wait long enough,
how is the circuit going to behave after a long time?
I will tell you why that is important in a second.
I look at the steady-state behavior.
Second, I am going to look today at sinusoidal drive.
Two things that are different from, say for example,
what I covered in the past ten minutes.
In the past ten minutes I covered two things which were
different. One is that I looked at the
transient response and then steady-state.
And remember for a DC input, for a DC voltage the
steady-state was a DC voltage across the capacitor,
correct? So the steady-state was pretty
boring, it was steady-state DC. But what we are going to do
today is instead of having DC inputs or step inputs which
settle to a DC value after some time, we are going to drive a
circuit through the sinusoidal input.
So you may ask me, Agarwal, are you nuts?
Why do you want to drive something with a sinusoidal
input? Why not just steps in DC and so
on? That was painful enough.
Why sinusoidal? Why not do triangular or why
not do some other exponentially decaying stuff or something
really cool like a whacko music signal?
What is special about sinusoidal stuff?
The key thing to realize is that, well, let me ask you a
question first. How many people here know about
Fourier series? Good.
It looks like some of you have taken the prerequisites.
Good. Need I say more as to why this
is important? Just that question should give
you the answer, right?
You've learned about Fourier series.
And when you learned about Fourier series you were
wondering why on earth are we learning about Fourier series?
Who cares that you can represent the periodic signals
as a summation of a series of sine waves?
Why is that interesting? Why are you telling me that I
can take a square wave and represent that as a summation of
periodic square waves and represent that as a summation of
sines? Who cares that I can take a
sequence of pulses with a fixed period and represent that as a
sum of sines? Who cares that I can take a
triangular wave and represent that as a sum of sines?
I am not sure what answer your math professors gave you when
they taught you Fourier series. But in math they are purists.
They don't care about applications.
The answer could well have been because it is aesthetically
pleasing. I mean isn't it cool that you
can represent a sequence of pulses as a sum of sines?
That is good enough for mathematicians.
But I am an engineer. If it I cannot see how it helps
humanity in the short-term then I probably don't care too much
about it. Let me give you some practical
significance of this. So it turns out that,
since we know that we can represent periodic signals with
sums of sines. What this means is that if I
can figure out the behavior of networks to a sinusoidal input,
if I can understand how to analyze a network for a
sinusoidal input that means that if the circuit is linear I can
then compute the response of the circuit to any periodic
waveform. Here is the argument.
I can represent any periodic waveform as a sum of sines.
The Fourier series, remember?
If I just figure out the response of my network for a
sine wave, then if this is a linear network,
I can compute the response to any sequence of scaled sum of
sines. A some sine,
B sine two, omega whatever, C sine something or the other.
I can simply take the response of the one sine.
And from there I can go ahead, and knowing the response of a
sine wave I can compute the response to a sum of sines.
That is pretty cool. Therefore, doing it for
sinusoidal drives is really important.
Why steady-state now? Hopefully, I have convinced you
that looking at the response of circuits to sinusoidal drive is
important and interesting because we can long ways from
there. What about steady-state?
Well, it turns out that, and I am going to show you,
when you listen to music, you have an amplifier and
listen to music, what you are observing by and
large is the steady-state behavior of the amplifier.
You are listening to something over many seconds or many hours.
And the transients used for most of our common circuits,
the transients die out pretty quickly.
And so the transients are quite complicated and they die out
quickly. We say we are engineers.
Let's focus on what is practically important.
And we focus on the steady-state behavior as a large
part of our analysis and just completely ignore the transient
response when we care about the response to sinusoidal input.
The transient response will die away, and I will show that
mathematically to you in a few seconds.
And let's focus on the steady-state because that what I
am going to hear most of the time.
I am going to listen to an average over long periods of
time. That's the motivation behind
this. And let me put this in
perspective for you. By now this should bring
memories to your mind. This is the playground that we
are in. This is the lumped circuit
playground here. Remember we came from the
playground of nature to the playground of EECS where we made
the big leap from Maxwell's equations to lumped circuits,
that's lumped circuit abstraction.
And within there we spent a large part of the last couple of
months doing linear circuits. We also analyzed nonlinear
circuits. Remember the amplifier circuit
of the MOSFET large signal analysis was nonlinear?
Well, there is linear and nonlinear.
Within linear we also showed that if you take a digital
circuit, at least as we understood them,
and draw the subcircuit for a given set of switch settings,
if I set the switches in a given way what I was left with
was another linear circuit for a given value of all the switch
settings. My small signal analysis was
also linear. If I focused on small signals I
also had linear analysis. Today what we are going to do
is this. We are going to articulate a
different part of the playground.
This was a big linear playground.
We've done this. We've done this.
We are going to explore this territory.
This is that territory of the playground in which we have
sinusoidal inputs to circuits. Furthermore,
we are going to look at a subcircuit of that region which
is steady-state. We will look at sinusoidal
input and in the steady-state. So that is going to be our
focus for the next two or three lectures just to give you a
perspective of where we are. To motivate this,
what I would like to do is consider your amplifier.
This is our friend the amplifier circuit,
this part here. And remember,
even though this is an amplifier, I am using a MOSFET
here. And a MOSFET,
as you know, has this gate capacitance CGS.
I am explicitly drawing it out for you here.
And I am going to drive this with a bias voltage plus some
small signal vI, the usual template for
amplifiers. And there is some Thevenin
resistance attached to that source.
I am going to model my source as a bias voltage,
a small signal plus some source resistance.
And I want to apply a sine wave here and I am going to look at
what this looks like. You may think,
look, this is a linear amplifier.
And so if I apply a sine wave here I should see some response
here, and that should be the same amplitude if I feed the
same amplitude here over any frequency.
But let's see what happens. Keep a look at,
switch over to that view graph while I show you a little
demonstration here. What you see here are three
sine waves, a yellow sine wave which is the input here,
you see a green sine wave which is the input vC at the gate of
the MOSFET, and then you see the blue which is the output here.
For now simply focus on the yellow and the blue.
The yellow is the input and the blue is the output.
So I apply some input and I get an output that looks more or
less some linear function of this input here.
It is a small signal. What I am going to do is I am
going to change the frequency of the input.
Remember, I want to look at the response of the circuit to a
sinusoid. I am feeding a sinusoid here.
I look at the response. I am going to change the
frequency. That is a big shift that we are
making in that the curve that we drew in the last lecture had to
do with varying time. Now I am going to focus on
sinusoids and vary their frequency.
I am going to change the frequency.
Stare at the blue curve as I increase the frequency and just
think of what you might expect. Based on the knowledge you have
so far you would expect that look, as I change the frequency,
the frequency should change but I should see the same amplitude.
OK but take a look. Let me increase the frequency
of the input. What do you see at the output?
I am increasing the frequency.
What do you see happening there?
If you don't see anything changing there you will need to
see an optometrist. What do we see here?
As I changed the frequency, as I increased the frequency
what happened to the blue? The blue kept decreasing in
amplitude. And you are saying whoa,
what is happening here? We don't have the tools to deal
with this. I expected that when I changed
my frequency, my frequency here should change
of course, but why is the amplitude changing?
What is happening here? That is weird.
I noticed that this amplitude became smaller because the
amplitude of the green became smaller.
And remember the green was the voltage across the capacitor.
So this is your RC. And here is my input.
My input has the amplitude, which I am holding more or less
constant. And notice that vC decreased in
value as I increased my frequency.
Just hold that thought. As I increased the frequency of
my input the amplitude of the output kept diminishing.
In other words, the gain of the system seemed
to have decreased as I increased by frequency.
And today we will look at why that is so and how we can
analyze that. The other thing that is not so
obvious here is there is a phase shift.
What I am going to do is try to see if we can observe the phase
shift here.
Notice here. What we have been used to is
for the amplifier we get a complete inversion at the
output. Inversion means a phase
difference of 180 degrees for a sine wave, right?
This peak should have been here, but notice that there is a
shifting of the peak. In other words,
if the yellow was my input my output should have had its
minimum when the input had its maximum.
But notice there is a shifting of the signal such that the
output is a maximum, not quite at the point where
the input is a minimum but a little bit after that.
Also weird. Not only has this little
circuit here lost its gain somehow, but also it seems to
have shifted the signal in phase.
That is weird. And today we will look at why
that is so and try to understand the frequency behavior of this
little subcomponent here. Notice that vC is exactly 180
out of phase with vO. So vO is faithfully an inverted
amplified form of the input vC. However, vC itself should have
been the same as vI but it looks quite different.
So let's understand why that is so.
The subcircuit to model is the subcircuit comprising the
source, resistor and the capacitor.
And I am just showing that to you here.
I have a vI, a resistor and capacitor.
And I am going to understand how this behaves.
And it is a first order circuit, single capacitor.
My input is a vI cosine of omega t.
vI is a real number for t greater than zero.
And I am telling you that the capacitor voltage starts out
being zero. And my vI is a sinusoid.
It's not a step this time. It's a sinusoid.
So vI is a sinusoid and I want to find out what vC looks like.
The behavior here tells me, I will give you the answer,
that when I feed a sinusoidal input as the frequency
increases, vC should be decreasing somehow and also be
shifting in phase. Let's do the derivation for
that and see what happens. To give you some insight as to
how to go about analyzing this let me draw a little graph as to
the effort level of doing this. To determine vC of t on the
y-axis here is our effort. How hard do we have to work to
solve this circuit for a sinusoidal input?
And on this graph, down here is easy and up here
is pure agony. Sheer agony up here.
So it's the scale of effort level ranging from easy to
complete agony. And this is the time axis.
And the time axis starts out at 11 o'clock, the early part of
today's lecture, and ends at roughly 12,
that is today's lecture and this is next lecture.
What I am going to show you today is a method that uses a
usual differential equation approach, and this is going to
be pure agony. If you thought last Thursday
was agony watch today. This is going to be sheer,
sheer, sheer hell. So I am going to grunge through
that, and I think I will sort of give up halfway because it's
just too painful even for me here.
Then what I am going to do is at the end of this lecture I am
going to show you an approach that I give a cutesy name.
I call it the "sneaky approach".
We are going to sneak something in there it is going to be a lot
easier. And then in the next lecture I
am going to show you an even sneakier approach that is just
going to be absolute bliss. So let's start here.
Indulge me as I go through the agony part.
I am going to blast through it because that is not of how we
are going to do things, but you just need to know that
that is agony. Let's do a usual differential
equation approach. Steps one, two,
three and four. Set up differential equation,
find the particular solution, find the homogenous solution,
add up the two and solve for the unknowns.
It's a mantra. The four-step manta.
Let's do it. Step one, write the DE.
That's easy. We have done this before the RC
circuit. It's RC dvc/dt+vc=vI.
This is no different from what you got from what you got from
your RC circuit with a step input, just that my input is VI
cosine of omega t in this case. It is not just a DC voltage VI.
Stare at that. Enjoy it while the going is
easy. It's like traversing rapids,
and before you come to a class five, you have calm and raging
waters there, you kind of sit there saying
oh, the scenery around here looks really good and so on.
All you are doing is stalling before you have dive down to
class five. We will get to class five
rapids in a few seconds here, so just enjoy this.
RC dvc/dt+vC=vI. You've seen this before.
Nothing fancy. Good old stuff.
VI cosine of omega t. You want to hold onto your
seatbelts? OK.
Let's find the particular solution to the cosine input.
Let's use our standard method. What I will do is just so,
there is going to be so much crapola up there,
so that I draw your attention to vP, which is what we are
trying to get, I am just going to put a box
around vP in red. If you see like all sorts of
garbage appear, just look for the red square.
That is what we are trying to get at.
That's the equation. Let's try.
First try, A worked before. A constant value A worked
before for DC inputs. Let's try that again.
If it worked then it may work now.
If I use A, a constant value, and I substitute it here,
I get dA/dt goes to zero, vP is A, but on the right-hand
side I have VI cosine omega t. So clearly A doesn't work.
Sorry. I struck out.
Well, cosine omega t here, let's try A cosine omega T as
my particular solution. Things are getting a little
harder now, a little more painful.
So substitute A cosine omega t here.
So I do get A cosine omega T for vP, but out here I get RCA
sine omega t times omega times minus one.
So I have a sine and a cosine, and I have a cosine on the
right-hand side. Sorry, it doesn't work.
Nope, doesn't work either. Well, let's try A cosine omega
t plus phi. We are now embarking into the
rapids here. You can begin feeling the
pressure. Just to refresh your memories
of sines and cosines. A is the amplitude of the
cosine. Omega is the frequency.
Phi is the phase. Remember the signal I showed
you earlier? If something happens to the
amplitude of the sine, something happens to the phase.
A cosine omega t plus phi. Let me plug it in here and go
by standard practice. Here is what I get.
I plug in A cosine omega t to this equation,
and this is what I get. I differentiate it.
I get omega out minus sine, sine of negative d plus phi,
A cosine omega t plus phi equals VI cosine omega t.
That might work. Now we get into the class six
part of the class five. All class fives have a little
bit of class six rapids. Remember, the rapids go up on
an exponential scale so it like earthquakes.
What I do now is expand out sine omega t plus phi,
blah, blah, blah, it goes on and on.
I could go on and on, but this is even tiring me.
This can be made to work, but I am not sure I want to put
all of us through this trig nightmare here.
If I am really, really nasty I could give this
to you as a homework assignment or something,
but I am not that nasty so you won't see that.
But if I go down this path it will get me to the answer,
but I would have to soon negotiate class six,
class seven rapids to get to where I want.
So let me punt on it, let me start from scratch.
I am at step two, let me start from scratch.
Instead what I would like to do is let's get sneaky here.
Rather than negotiating the class five rapids,
what we can say is ah-ha, we can take our canoes and jump
onto shore and run down and then get back onto the river.
Let's do that. That is called the sneaky
approach. So that all our friends who are
behind us think we are negotiating the rapids,
but what we are going to do is get sneaky and take the shore
path. Let's get sneaky,
just walk down the shore and see what is there.
I want to do something completely weird here.
I want to look at solving this equation through the shore
method. S stands for shore or S stands
for sneaky, whatever you want. What I am going to do is rather
than trying to solve for VI cosine omega t.
I am going to say let's try a different input all together.
And you will understand why in a second.
It's like I am the captain of my canoe and I tell my
teammates, hey, let's not negotiate the rapids,
let's go and explore the shore. Maybe down the shore we can
find a path that gets us across to the other side more easily.
So here is me and my colleagues carrying our canoe and getting
onto shore and taking a sneaky path.
This is not what I set out to solve.
I don't know where this will lead me.
But let's see where the shore path leads us.
I want to try solving this equation Vie^st.
S stands for shore or sneaky or whatever you want.
I don't know where I am going, but let's see where this leads
us. Let's explore.
Make believe you are Columbus or something.
I don't know. Let's use the usual techniques
and see how this works out. Let's try a particular
solution, Vpe^st. My input is Vie^st.
I am trying to solve the circuit for a different input.
And let me try solution Vpe^st and see if that works out
nicely. I substitute Vpe^st into my
equation here, so RCVpe^st blah blah blah.
What I get here is Vie^st, Vpe^st stays the same,
but here vP comes out, s comes out and e^st stays the
same. That is nice property of
exponentials. This is what I get.
A really cool property of exponentials is that when I
differentiate it I get the exponential back.
Unlike a cosine I got a sine, and for a sine I got a cosine.
Exponentials are very plain and simple, are straightforward.
What you see is what you get. You differentiate it.
You get the same thing, you get scaling vP,
S and so on, and some scaling here.
You get S scaling here, but the basic form stays the
same. This is quite nice.
I have e^st in all three places, so I can cancel those
out and I get this expression. And I get this.
Wow. So if I go down the shore I get
some place fast. I don't know where I am yet,
but whatever I did, it was easy.
I am just exploring this path, down the shore path.
I am making progress. I don't know where I have
gotten yet. We will see where we got to in
a second, but I got some place quickly, fast.
I was able to solve for this input Vie^st and get this
solution very quickly. So what happened here?
I assumed the solution of the form Vpe^st, substituted it
back, and found that if vP were equal to Vi/(1+sRC) then Vpe^st
is a solution. What I have done here is that
Vi/(1+sRC) is a particular solution to this equation for
the input Vie^st. I put a box around it because
this is important. This was easy.
We went down the shore, and said let's try this other
input. We made rapid progress on shore
and I got some place. I don't know where I am yet.
I got this. Let me pause here and let me
give you the final answer. I am going to show you over the
next five minutes that this is our answer.
You are staring at the answer already.
I am a party, I have taken a shore path and
we have gotten some place. We see the river there,
so it turns out we are exactly where we want to be,
just after the rapids. All I have to do now is get my
colleagues into the river with myself in the canoe and we are
done. You don't know that yet.
My colleagues and I are sitting on the shore looking at the
river. We've gotten some place.
There are no rapids there. We have gotten some place.
We don't quite know is this just after the rapids or not.
We don't know yet, but I got there very quickly.
And I will tell you right now, that is the place we wanted to
go to. The next five view graphs I am
going to blast through. There is going to be more pain
and agony to show you why that is the case.
It's me thinking I am Columbus and proving to my colleagues
that this is where we want to be.
And pulling out my sextant, and the compasses and so on
that cartographers and people use to prove to my colleagues
that this is where I want to be. This is the answer.
The next five view graphs will be demonstrating that this is
indeed the answer, or close enough to the answer
that we will be satisfied. Isn't this spectacular?
I am going to show you in about five minutes that this gives us
all the information we need to know to compute the sinusoidal
steady-state response to this differential equation.
Let me write that down here just so you know.
Just so you remember. I am going to put a marker on
the shore that says this is where we are right now.
Now let me prove to you. As I just said,
vPS is Vi, it's this stuff here multiplied by e^st is the
solution to Vie^st. This guy here is a solution for
Vie^st and vP is simply the coefficient that multiplies
e^st. Similarly, if I substitute S
equals j omega. I told you five view graphs of
more hell, but I am just going to prove to you that this is it.
I am substituting S equals j omega.
This is Columbus giving a big speech at the end convincing his
colleagues that we are where we want to be.
I substitute j omega for S and this is what I get.
This is a solution for e to the st, and so this is a solution
for e to the j omega t. And let me mark this for you as
something to remember. Vi/(1+j omega RC).
In terms of that, I am substituting j omega for
S. And this is a complex number.
It is a complex amplitude. It is a complex number because
of j here, and it multiplies e to the j omega t.
Just keep this in mind. So that was easy.
The steps were easy. I am still proving to you that
this is where we want to be. Now let me draw the connection
back to vP. And the first fact was that
finding the response to Vie^(j omega t) was easy.
We know that. Second was the following
observation that Vi cosine omega t is simply the real part of
this number here. So Vi cosine omega t is simply
the real part of Vie^(j omega t) from the Euler relation.
So cosine omega t is simply the real part of this guy.
Light bulbs beginning to go off?
The first fast was that finding the response to Vie^(j omega t)
was easy. And the response was this,
right? Times e to the j omega t.
That was easy. All right.
And the second part is that the real part of this is Vi cosine
omega t was our input. Draw the connection between two
steps. Finding the response to Vie^(j
omega t) was easy. The real part of that was the
input we cared about. Are light bulbs going off?
Let me draw you a little picture here to show you what is
happening. Response to vI is vP.
It's the particular response we are looking for.
Remember the red square? But we threw in a sneaky input
vIS and we formed the response vPS to that.
This step was easy. This step was hard.
vI to vP was hard, trig nightmare,
remember? But vIS to vPS was easy.
It was a simple Vpe^st thing. We also know that the real part
of vIS is vI. The real part of this is simply
vI. If you have a real circuit,
if you have a real linear circuit, for a linear circuit,
if the real part of this gives me vI then the real part of the
solution should give me vP. So computing vPS was easy.
If I take the real part of this, I take the corresponding
real part of this. This is sort of an inverse
superposition argument. Superposition,
I said, take the response for A, take the response for B,
add them up and you get the response for A plus B.
Here what we are saying is that get the response to A plus B,
or to A plus jB and the real part of the input will produce
the response given by the real part of the output.
So this is an inverse superposition argument.
If it is a linear circuit, then if vI is the real part of
this sneaky input then I find the response to the sneaky input
and take its real part I should get vP.
Here I am, Columbus, staring down at the entrance to
this part of the river. I just proved to my colleagues
that all we have to do is take the real part of what we have.
We can just jump right back into the river and get back to
vP. And what I am going to do next
is just grind through the math and just show you that.
I will just blast through it. It is not important,
but you have it in your notes. I am telling you that vP is
simply the real part of the sneaky output.
And I take the real part of vP e to the j omega t.
And I take the real part. And just a bunch of math here.
I am just taking the real part and doing a bunch of complex
math. Remember vP was given by this
quantity here. And I take the real part and I
end up with vP is simply this quantity multiplied by cosine
omega t plus phi, where phi is given by is given
by tan inverse of omega RC, and this is the coefficient
multiplying the cosine. So by taking the sneaky path
and then taking the real part of that output answer,
I was able to very quickly get to where I wanted to be.
So from here to here it is only math.
Recall, that vP, the thing in the red was what
we set out to find out, which was the particular
response to VI cosine of omega t.
And remember that two grunge is all of this stuff.
I am going to blast through two or three more view graphs that
just give you more insight and more math, nothing particular.
And remember to solve the equation we have to find a
homogenous solution, too.
But recall that the homogenous solution for an RC circuit is of
the form Ae^-t/RC. This means that as time becomes
very large this part goes to zero.
As time becomes large in the steady state,
remember I care about the steady state?
This goes to zero. I don't care about the
homogenous solution. Isn't that fantastic?
Most the circuits we will deal with, except for purely
oscillatory ones, the homogenous part dies away.
You have something like e to the -t whatever.
It just dies away. It's gone.
So the total solution has vH going away.
And what I end up with is just vP.
My total solution in the steady state is simply vP.
And A is given by this that we just calculated.
I just have a bunch more insight that I talk about that
you can look through in your notes.
And I just want to show you a very quick summary.
In summary, what we have is we computed vP.
It was a complex coefficient. And all these steps,
2 grunge, 3 and 4 were a waste of time.
And what I showed you was that for the input VI the coefficient
vP was complex. And I can take the ratio and
represent it in this manner as well.
And from vP, I can then compute the
multiplier for the cosine as follows.
I divide by vP here. Remember the cosine was
multiplied by, in the mathematical step that I
did, VI divided one plus, this stuff here,
so I could get the magnitude and phase of the transfer
function of this circuit in the following manner.
And to wrap up very quickly, I am going to cover this again
the next time and show you a magnitude plot.
Notice here that if I plot Vp/Vi.
Remember this was Vp here. That's the answer.
The magnitude looks like this. On a log scale Vp/Vi for small
frequencies omega is at one, but as omega increases Vp/Vi
keeps decreasing. That is the output.
Remember Vp was the amplitude of the output?
That keeps decreasing. And this is the reason why.
As I increase the frequency, the amplitude of my output
cosine kept decreasing. I could also plot the phase for
you. And the phase,
in the same manner as omega increased, my phase also kept
shifting from zero initially to pi/2 finally.
Let me stop here and start with this the next time and revisit
this. Unfortunately,
I won't have time for the demo. I will show it to you next
time.