1
00:00:00 --> 00:00:32


2
00:00:32 --> 00:00:33
OK.
Good morning.

3
00:00:33 --> 00:00:38
Let's get going.
As always, I will start with a

4
00:00:38 --> 00:00:41
review.
And today we embark on another

5
00:00:41 --> 00:00:46
major milestone in our analysis
of lumped circuits.

6
00:00:46 --> 00:00:51
And it is called the
"Sinusoidal Steady-state".

7
00:00:51 --> 00:00:57
Again, I believe this will be
the second and the last lecture

8
00:00:57 --> 00:01:02
for which I will be using view
graphs.

9
00:01:02 --> 00:01:05
And the idea here is that,
just like in the last lecture,

10
00:01:05 --> 00:01:09
there is a bunch of
mathematical grunge that needs

11
00:01:09 --> 00:01:12
to be gone through.
And I want to show you a

12
00:01:12 --> 00:01:16
sequence in a little chart today
that talks about the effort

13
00:01:16 --> 00:01:19
level in doing some problems
based on time domain

14
00:01:19 --> 00:01:23
differential equations,
as in the last lecture,

15
00:01:23 --> 00:01:27
something slightly different
today and something much better

16
00:01:27 --> 00:01:30
on Thursday.
And so, in some sense,

17
00:01:30 --> 00:01:34
Thursday's lecture and today's
lecture involve talking about

18
00:01:34 --> 00:01:38
the foundations of the behavior
of certain types of circuits.

19
00:01:38 --> 00:01:43
And it is good for you to have
this foundation as background,

20
00:01:43 --> 00:01:47
but when you actually go to
solve circuits you don't quite

21
00:01:47 --> 00:01:50
use these methods.
You use much easier techniques,

22
00:01:50 --> 00:01:53
which I will talk about next
Thursday.

23
00:01:53 --> 00:01:57
Let's start with a quick
review, and then we will go into

24
00:01:57 --> 00:02:02
sinusoidal steady state.
The last lecture we talked

25
00:02:02 --> 00:02:07
about this circuit and we did
the same two lectures ago on

26
00:02:07 --> 00:02:11
Tuesday.
We had one inverter driving

27
00:02:11 --> 00:02:15
another inverter.
And we said that the wire over

28
00:02:15 --> 00:02:20
ground had some inductance.
CGS is the capacitor of the

29
00:02:20 --> 00:02:26
gate and R is the resistance at
the drain of the first inverter.

30
00:02:26 --> 00:02:31
And if you look at this
circuit, that circuit formed an

31
00:02:31 --> 00:02:36
RLC pattern.
And what we did was we said

32
00:02:36 --> 00:02:41
let's drive this with a one to
zero transition at the input.

33
00:02:41 --> 00:02:47
And the one to zero transition
at the input would cause this

34
00:02:47 --> 00:02:52
transistor to switch off,
and this node would then go

35
00:02:52 --> 00:02:55
from a very low value to a high
value.

36
00:02:55 --> 00:03:01
So it as if a 5 volt step was
applied at this input.

37
00:03:01 --> 00:03:05
We also saw that using time
domain differential equations

38
00:03:05 --> 00:03:09
that by applying a step input
here the output looked like

39
00:03:09 --> 00:03:11
this.
The output would show some

40
00:03:11 --> 00:03:16
oscillatory behavior when we
have a capacitor and inductor.

41
00:03:16 --> 00:03:20
I also gave you some insight as
to why it oscillates like this.

42
00:03:20 --> 00:03:24
And you also heard in
recitation that the reason for

43
00:03:24 --> 00:03:30
this oscillation was because of
these two storage elements.

44
00:03:30 --> 00:03:34
Each of these storage elements
tries to hold onto its state

45
00:03:34 --> 00:03:35
variable.
For example,

46
00:03:35 --> 00:03:40
the capacitor tries to maintain
its voltage while the inductor

47
00:03:40 --> 00:03:45
tries to maintain its current.
And, much like a pendulum which

48
00:03:45 --> 00:03:48
oscillates back and forth,
it swaps potential energy

49
00:03:48 --> 00:03:54
versus kinetic energy down here
and swings back and forth.

50
00:03:54 --> 00:03:57
In the same way,
in an LC circuit like this,

51
00:03:57 --> 00:04:01
energy swaps back and forth
between a potential energy and a

52
00:04:01 --> 00:04:05
kinetic energy equivalent,
which swaps back and forth

53
00:04:05 --> 00:04:09
between energy stored in the
inductor and energy stored in

54
00:04:09 --> 00:04:12
the capacitor and sloshes back
and forth.

55
00:04:12 --> 00:04:16
And because of this resistor
the energy eventually dissipates

56
00:04:16 --> 00:04:21
and you end up getting a final
value which corresponds to the 5

57
00:04:21 --> 00:04:24
volts appearing here.
And why is that?

58
00:04:24 --> 00:04:28
That is because remember the
capacitor is a long-term open

59
00:04:28 --> 00:04:31
for DC.
It is a DC voltage.

60
00:04:31 --> 00:04:35
After a long time this
capacitor looks like an open

61
00:04:35 --> 00:04:39
circuit and the inductor looks
like a complete short circuit,

62
00:04:39 --> 00:04:43
an ideal inductor as a complete
short circuit for DC.

63
00:04:43 --> 00:04:48
And so therefore in the
long-term it is as if this guy

64
00:04:48 --> 00:04:52
is a short, this guy is an open,
so 5 volts simply appears here.

65
00:04:52 --> 00:04:55
And this is the transient
behavior.

66
00:04:55 --> 00:05:00
Then we just switch the first
transistor off.

67
00:05:00 --> 00:05:03
In the last lecture,
I left off with intuitive

68
00:05:03 --> 00:05:06
analysis.
Let me quickly cover that and

69
00:05:06 --> 00:05:09
redo the intuitive analysis for
you.

70
00:05:09 --> 00:05:14
I left it the last time by
having you think about whether

71
00:05:14 --> 00:05:19
the transient response would
begin by going down or begin by

72
00:05:19 --> 00:05:22
going up.
And I will cover that today.

73
00:05:22 --> 00:05:25
This was the circuit that we
analyzed.

74
00:05:25 --> 00:05:30
A VI input with a step and an
RLC out here.

75
00:05:30 --> 00:05:32
And we were plotting the
capacitor voltage.

76
00:05:32 --> 00:05:36
And intuitively we can plot
this in the following way.

77
00:05:36 --> 00:05:40
I have also marked for you the
section number in the course

78
00:05:40 --> 00:05:44
notes which has a discussion of
this intuitive analysis,

79
00:05:44 --> 00:05:47
(Section 13.8).
Let's do the easy stuff first.

80
00:05:47 --> 00:05:50
Notice that the capacitor wants
to hold its voltage.

81
00:05:50 --> 00:05:54
And so given that we don't have
any infinite impulse here,

82
00:05:54 --> 00:05:58
I am going to start out with
the capacitor voltage being

83
00:05:58 --> 00:06:02
where it is.
And the initial conditions are

84
00:06:02 --> 00:06:04
given to you.
You are given that the

85
00:06:04 --> 00:06:08
capacitor voltage starts out
being positive at v zero and the

86
00:06:08 --> 00:06:11
current starts out being
negative at time zero.

87
00:06:11 --> 00:06:15
So I am telling you that there
is a voltage v across the

88
00:06:15 --> 00:06:19
capacitor at time t=0 and there
is a current that is flowing.

89
00:06:19 --> 00:06:23
Since i is negative there is a
current initially that is

90
00:06:23 --> 00:06:27
flowing in the opposite
direction to this arrow here.

91
00:06:27 --> 00:06:31
The i zero is negative.
In light of that,

92
00:06:31 --> 00:06:35
I can start plotting my curve
here by intuition.

93
00:06:35 --> 00:06:40
I start by saying at time t=0 I
am told that the initial voltage

94
00:06:40 --> 00:06:44
of the capacitor is at zero.
This is about as simple as it

95
00:06:44 --> 00:06:46
gets.
Completely intuitive.

96
00:06:46 --> 00:06:51
I also know that after a long
time, can someone tell me after

97
00:06:51 --> 00:06:55
a long time what the voltage
will be at the end of the

98
00:06:55 --> 00:06:58
capacitor?
You should be able to get his

99
00:06:58 --> 00:07:01
by observation?
VI.

100
00:07:01 --> 00:07:05
And why is it VI?
It is vI because this is a DC

101
00:07:05 --> 00:07:08
value VI.
And after a long time this guy

102
00:07:08 --> 00:07:11
behaves like an open circuit to
DC.

103
00:07:11 --> 00:07:15
This guy behaves like a short
circuit to DC.

104
00:07:15 --> 00:07:20
So since this is an open
circuit, VI will appear here

105
00:07:20 --> 00:07:23
after a long time.
And so therefore,

106
00:07:23 --> 00:07:27
after a long time,
the capacitor voltage is going

107
00:07:27 --> 00:07:32
to be at VI.
And I just showed you that.

108
00:07:32 --> 00:07:34
There you go.
You already have the two

109
00:07:34 --> 00:07:38
endpoints of your curve
completely by observation,

110
00:07:38 --> 00:07:40
intuition.
No DEs, no nothing.

111
00:07:40 --> 00:07:45
Just by staring at it and
understanding the fundamentals

112
00:07:45 --> 00:07:48
of how simple primitive circuit
elements work.

113
00:07:48 --> 00:07:52
Absolutely simple stuff.
So you've nailed the two ends

114
00:07:52 --> 00:07:57
now and you cannot go wrong with
the stuff in the middle.

115
00:07:57 --> 00:08:01
Let's see.
As a next step what I do is I

116
00:08:01 --> 00:08:05
need to understand what the
dynamics of the circuit looks

117
00:08:05 --> 00:08:08
like here.
What I do is I develop the

118
00:08:08 --> 00:08:12
characteristic equation.
And initially you will write

119
00:08:12 --> 00:08:16
the differential equation and
then substitute e^st and get

120
00:08:16 --> 00:08:20
this characteristic equation.
But you could also remember it

121
00:08:20 --> 00:08:23
as a pattern.
For a series RLC circuit you

122
00:08:23 --> 00:08:27
always get an equation of this
form, always.

123
00:08:27 --> 00:08:30
If this were R,
L and C.

124
00:08:30 --> 00:08:33
And whether you are looking at
L up here or C up here,

125
00:08:33 --> 00:08:37
as long as you're looking at
the capacitor voltage,

126
00:08:37 --> 00:08:40
the capacitor voltage is going
to behave the same.

127
00:08:40 --> 00:08:44
And for this circuit the
characteristic equation remains

128
00:08:44 --> 00:08:47
the same as well for a series
RLC.

129
00:08:47 --> 00:08:50
It is exactly this.
And I write the standard

130
00:08:50 --> 00:08:54
canonic form as s squared plus
two alpha s + omega nought

131
00:08:54 --> 00:08:56
squared.
And omega nought is simply one

132
00:08:56 --> 00:09:01
by square root of LC and alpha
is simply R divided by L and I

133
00:09:01 --> 00:09:05
have two in the denominator as
well.

134
00:09:05 --> 00:09:09
And then I get omega d which is
my damped frequency given by

135
00:09:09 --> 00:09:12
omega nought squared minus alpha
squared.

136
00:09:12 --> 00:09:14
And Q is simply called the
quality factor.

137
00:09:14 --> 00:09:18
And we will learn about Q in a
lot more detail in about a

138
00:09:18 --> 00:09:22
couple of lectures from today.
That is given where omega

139
00:09:22 --> 00:09:26
nought divided by two alpha.
These parameters,

140
00:09:26 --> 00:09:29
alpha, omega nought,
Q and omega d pretty much

141
00:09:29 --> 00:09:33
characterize everything else
that I need to know about the

142
00:09:33 --> 00:09:36
circuit.
First of all,

143
00:09:36 --> 00:09:39
omega d is the frequency of
oscillation.

144
00:09:39 --> 00:09:43
And so since omega d is a
frequency of oscillation then I

145
00:09:43 --> 00:09:48
know that the time period of
oscillation is two pi by omega

146
00:09:48 --> 00:09:49
d.
Omega is in radians.

147
00:09:49 --> 00:09:54
Notice that for typical values
of circuits like this when R is

148
00:09:54 --> 00:09:58
pretty small,
alpha squared is going to be

149
00:09:58 --> 00:10:01
very small.
It's a common case for

150
00:10:01 --> 00:10:07
underdamped circuit that omega d
will more or less be equal to

151
00:10:07 --> 00:10:10
omega nought.
Commonly that is going to be

152
00:10:10 --> 00:10:14
the case.
This frequency is governed by

153
00:10:14 --> 00:10:16
LC.
And if R is large it is

154
00:10:16 --> 00:10:21
governed by this omega d here.
So I have the frequency of

155
00:10:21 --> 00:10:24
oscillation.
I also know that Q is related

156
00:10:24 --> 00:10:30
to the frequency of oscillation
divided by alpha.

157
00:10:30 --> 00:10:34
It is a ratio of the frequency
divided by how badly I am being

158
00:10:34 --> 00:10:37
damped.
So it is a fight between the

159
00:10:37 --> 00:10:41
frequency of oscillation and now
heavily I damp.

160
00:10:41 --> 00:10:45
And the ratio of that is an
indication of how many cycles I

161
00:10:45 --> 00:10:48
ring.
So Q tells me that the ringing

162
00:10:48 --> 00:10:50
stops approximately after Q
cycles.

163
00:10:50 --> 00:10:54
These four values,
omega d, Q, alpha and omega

164
00:10:54 --> 00:10:57
nought are telling me more and
more now.

165
00:10:57 --> 00:11:01
So I have got these two
factors.

166
00:11:01 --> 00:11:04
So I know now,
based on omega d and Q,

167
00:11:04 --> 00:11:08
that it is going to look
something like this.

168
00:11:08 --> 00:11:12
Some ringing here and then I
stop at this point.

169
00:11:12 --> 00:11:17
The last thing that is left to
do here for me for now is to

170
00:11:17 --> 00:11:22
figure out whether I start out
going down or I start out going

171
00:11:22 --> 00:11:25
up.
I start out going down or I

172
00:11:25 --> 00:11:30
start out going up?
I don't know that yet.

173
00:11:30 --> 00:11:34
And I stopped at this point in
the last lecture and let you

174
00:11:34 --> 00:11:38
think about how you can stare at
the circuit and intuitively

175
00:11:38 --> 00:11:41
figure out whether this goes
down or that goes up.

176
00:11:41 --> 00:11:44
So here is the insight.
Let's stare at this for a

177
00:11:44 --> 00:11:49
minute purely through intuition.
The capacitor has a voltage v

178
00:11:49 --> 00:11:52
across it, right?
And that is because I am

179
00:11:52 --> 00:11:55
telling you that it has an
initial voltage v.

180
00:11:55 --> 00:11:58
Now, I want to find out at
prime t equals zero plus,

181
00:11:58 --> 00:12:03
in which direction does a
capacitor voltage go?

182
00:12:03 --> 00:12:05
Increase or decrease?
What do I do?

183
00:12:05 --> 00:12:08
Let me take a look at the
inductor current.

184
00:12:08 --> 00:12:12
I am told that the inductor
current is negative which means

185
00:12:12 --> 00:12:16
I am told that the inductor
current is going in this

186
00:12:16 --> 00:12:19
direction initially.
The inductor current is pushing

187
00:12:19 --> 00:12:21
in this direction.
Now, remember,

188
00:12:21 --> 00:12:26
just as the capacitor is one
stubborn nut trying to hold its

189
00:12:26 --> 00:12:28
voltage, the inductor is as
stubborn.

190
00:12:28 --> 00:12:32
It is trying to hold its
current.

191
00:12:32 --> 00:12:34
It is trying to maintain its
current.

192
00:12:34 --> 00:12:37
And its initial current i(0) is
in this direction.

193
00:12:37 --> 00:12:40
Capacitor has a voltage here,
that is v(0),

194
00:12:40 --> 00:12:44
and the inductor is yanking the
current in that direction.

195
00:12:44 --> 00:12:48
So what should happen to the
capacitor voltage initially?

196
00:12:48 --> 00:12:51
If I am at v(0) and someone is
yanking current out,

197
00:12:51 --> 00:12:55
at least initially in that
direction, what should the

198
00:12:55 --> 00:13:00
initial dynamics of the
capacitor voltage look like?

199
00:13:00 --> 00:13:01
Pardon?
It should drop,

200
00:13:01 --> 00:13:07
which means that if the initial
current is being pulled in that

201
00:13:07 --> 00:13:12
direction the capacitor voltage
must droop to begin with.

202
00:13:12 --> 00:13:16
Completely through intuition.
No math here.

203
00:13:16 --> 00:13:18
This means that i(0) is
negative.

204
00:13:18 --> 00:13:23
It is as if i(0) is being
pulled out in this manner,

205
00:13:23 --> 00:13:27
so the capacitor voltage must
drop to begin life.

206
00:13:27 --> 00:13:33
Therefore, the dynamics look
somewhat like this.

207
00:13:33 --> 00:13:40
Notice that this is very
reminiscent of the ringing that

208
00:13:40 --> 00:13:47
we saw at the gate node of the
second inverter.

209
00:13:47 --> 00:13:54
Let's stop here in terms of
time domain analysis of RLC,

210
00:13:54 --> 00:14:02
and today let's take another
big step forward.

211
00:14:02 --> 00:14:05
Today marks another transition
in life here.

212
00:14:05 --> 00:14:09
This is actually a huge
transition so I want to just

213
00:14:09 --> 00:14:14
pause and take like ten seconds
of a breather just to clearly

214
00:14:14 --> 00:14:19
demarcate the fact that we have
a huge transition coming up.

215
00:14:19 --> 00:14:24
The key to this transition is
that I want to look at today the

216
00:14:24 --> 00:14:30
steady-state response of
networks to a sinusoidal drive.

217
00:14:30 --> 00:14:33
We are going to do two things
differently starting today on

218
00:14:33 --> 00:14:36
this new journey of ours.
In the past,

219
00:14:36 --> 00:14:39
we looked at time domain
behavior of circuits.

220
00:14:39 --> 00:14:42
For RLC, for example,
we looked at the transient

221
00:14:42 --> 00:14:44
response.
And what happened the instant I

222
00:14:44 --> 00:14:47
turn something on?
The transient response.

223
00:14:47 --> 00:14:50
Today we are going to look at a
steady-state response.

224
00:14:50 --> 00:14:54
Steady-state response says that
if I wait long enough,

225
00:14:54 --> 00:14:58
for whatever the circuit wants
to do in the beginning of life

226
00:14:58 --> 00:15:01
to die out.
If I wait long enough,

227
00:15:01 --> 00:15:04
how is the circuit going to
behave after a long time?

228
00:15:04 --> 00:15:07
I will tell you why that is
important in a second.

229
00:15:07 --> 00:15:09
I look at the steady-state
behavior.

230
00:15:09 --> 00:15:13
Second, I am going to look
today at sinusoidal drive.

231
00:15:13 --> 00:15:16
Two things that are different
from, say for example,

232
00:15:16 --> 00:15:18
what I covered in the past ten
minutes.

233
00:15:18 --> 00:15:22
In the past ten minutes I
covered two things which were

234
00:15:22 --> 00:15:24
different.
One is that I looked at the

235
00:15:24 --> 00:15:28
transient response and then
steady-state.

236
00:15:28 --> 00:15:31
And remember for a DC input,
for a DC voltage the

237
00:15:31 --> 00:15:34
steady-state was a DC voltage
across the capacitor,

238
00:15:34 --> 00:15:37
correct?
So the steady-state was pretty

239
00:15:37 --> 00:15:41
boring, it was steady-state DC.
But what we are going to do

240
00:15:41 --> 00:15:45
today is instead of having DC
inputs or step inputs which

241
00:15:45 --> 00:15:49
settle to a DC value after some
time, we are going to drive a

242
00:15:49 --> 00:15:52
circuit through the sinusoidal
input.

243
00:15:52 --> 00:15:54
So you may ask me,
Agarwal, are you nuts?

244
00:15:54 --> 00:15:58
Why do you want to drive
something with a sinusoidal

245
00:15:58 --> 00:16:02
input?
Why not just steps in DC and so

246
00:16:02 --> 00:16:04
on?
That was painful enough.

247
00:16:04 --> 00:16:08
Why sinusoidal?
Why not do triangular or why

248
00:16:08 --> 00:16:13
not do some other exponentially
decaying stuff or something

249
00:16:13 --> 00:16:16
really cool like a whacko music
signal?

250
00:16:16 --> 00:16:19
What is special about
sinusoidal stuff?

251
00:16:19 --> 00:16:23
The key thing to realize is
that, well, let me ask you a

252
00:16:23 --> 00:16:27
question first.
How many people here know about

253
00:16:27 --> 00:16:30
Fourier series?
Good.

254
00:16:30 --> 00:16:34
It looks like some of you have
taken the prerequisites.

255
00:16:34 --> 00:16:36
Good.
Need I say more as to why this

256
00:16:36 --> 00:16:40
is important?
Just that question should give

257
00:16:40 --> 00:16:41
you the answer,
right?

258
00:16:41 --> 00:16:44
You've learned about Fourier
series.

259
00:16:44 --> 00:16:47
And when you learned about
Fourier series you were

260
00:16:47 --> 00:16:51
wondering why on earth are we
learning about Fourier series?

261
00:16:51 --> 00:16:55
Who cares that you can
represent the periodic signals

262
00:16:55 --> 00:17:00
as a summation of a series of
sine waves?

263
00:17:00 --> 00:17:03
Why is that interesting?
Why are you telling me that I

264
00:17:03 --> 00:17:06
can take a square wave and
represent that as a summation of

265
00:17:06 --> 00:17:10
periodic square waves and
represent that as a summation of

266
00:17:10 --> 00:17:12
sines?
Who cares that I can take a

267
00:17:12 --> 00:17:16
sequence of pulses with a fixed
period and represent that as a

268
00:17:16 --> 00:17:19
sum of sines?
Who cares that I can take a

269
00:17:19 --> 00:17:22
triangular wave and represent
that as a sum of sines?

270
00:17:22 --> 00:17:26
I am not sure what answer your
math professors gave you when

271
00:17:26 --> 00:17:30
they taught you Fourier series.
But in math they are purists.

272
00:17:30 --> 00:17:32
They don't care about
applications.

273
00:17:32 --> 00:17:36
The answer could well have been
because it is aesthetically

274
00:17:36 --> 00:17:38
pleasing.
I mean isn't it cool that you

275
00:17:38 --> 00:17:42
can represent a sequence of
pulses as a sum of sines?

276
00:17:42 --> 00:17:44
That is good enough for
mathematicians.

277
00:17:44 --> 00:17:47
But I am an engineer.
If it I cannot see how it helps

278
00:17:47 --> 00:17:51
humanity in the short-term then
I probably don't care too much

279
00:17:51 --> 00:17:53
about it.
Let me give you some practical

280
00:17:53 --> 00:17:56
significance of this.
So it turns out that,

281
00:17:56 --> 00:17:59
since we know that we can
represent periodic signals with

282
00:17:59 --> 00:18:04
sums of sines.
What this means is that if I

283
00:18:04 --> 00:18:10
can figure out the behavior of
networks to a sinusoidal input,

284
00:18:10 --> 00:18:14
if I can understand how to
analyze a network for a

285
00:18:14 --> 00:18:20
sinusoidal input that means that
if the circuit is linear I can

286
00:18:20 --> 00:18:25
then compute the response of the
circuit to any periodic

287
00:18:25 --> 00:18:29
waveform.
Here is the argument.

288
00:18:29 --> 00:18:33
I can represent any periodic
waveform as a sum of sines.

289
00:18:33 --> 00:18:36
The Fourier series,
remember?

290
00:18:36 --> 00:18:40
If I just figure out the
response of my network for a

291
00:18:40 --> 00:18:44
sine wave, then if this is a
linear network,

292
00:18:44 --> 00:18:49
I can compute the response to
any sequence of scaled sum of

293
00:18:49 --> 00:18:50
sines.
A some sine,

294
00:18:50 --> 00:18:55
B sine two, omega whatever,
C sine something or the other.

295
00:18:55 --> 00:19:01
I can simply take the response
of the one sine.

296
00:19:01 --> 00:19:05
And from there I can go ahead,
and knowing the response of a

297
00:19:05 --> 00:19:09
sine wave I can compute the
response to a sum of sines.

298
00:19:09 --> 00:19:12
That is pretty cool.
Therefore, doing it for

299
00:19:12 --> 00:19:14
sinusoidal drives is really
important.

300
00:19:14 --> 00:19:18
Why steady-state now?
Hopefully, I have convinced you

301
00:19:18 --> 00:19:22
that looking at the response of
circuits to sinusoidal drive is

302
00:19:22 --> 00:19:26
important and interesting
because we can long ways from

303
00:19:26 --> 00:19:30
there.
What about steady-state?

304
00:19:30 --> 00:19:32
Well, it turns out that,
and I am going to show you,

305
00:19:32 --> 00:19:35
when you listen to music,
you have an amplifier and

306
00:19:35 --> 00:19:38
listen to music,
what you are observing by and

307
00:19:38 --> 00:19:41
large is the steady-state
behavior of the amplifier.

308
00:19:41 --> 00:19:44
You are listening to something
over many seconds or many hours.

309
00:19:44 --> 00:19:48
And the transients used for
most of our common circuits,

310
00:19:48 --> 00:19:50
the transients die out pretty
quickly.

311
00:19:50 --> 00:19:53
And so the transients are quite
complicated and they die out

312
00:19:53 --> 00:19:55
quickly.
We say we are engineers.

313
00:19:55 --> 00:19:59
Let's focus on what is
practically important.

314
00:19:59 --> 00:20:02
And we focus on the
steady-state behavior as a large

315
00:20:02 --> 00:20:05
part of our analysis and just
completely ignore the transient

316
00:20:05 --> 00:20:09
response when we care about the
response to sinusoidal input.

317
00:20:09 --> 00:20:12
The transient response will die
away, and I will show that

318
00:20:12 --> 00:20:15
mathematically to you in a few
seconds.

319
00:20:15 --> 00:20:18
And let's focus on the
steady-state because that what I

320
00:20:18 --> 00:20:20
am going to hear most of the
time.

321
00:20:20 --> 00:20:23
I am going to listen to an
average over long periods of

322
00:20:23 --> 00:20:25
time.
That's the motivation behind

323
00:20:25 --> 00:20:27
this.
And let me put this in

324
00:20:27 --> 00:20:31
perspective for you.
By now this should bring

325
00:20:31 --> 00:20:35
memories to your mind.
This is the playground that we

326
00:20:35 --> 00:20:37
are in.
This is the lumped circuit

327
00:20:37 --> 00:20:40
playground here.
Remember we came from the

328
00:20:40 --> 00:20:44
playground of nature to the
playground of EECS where we made

329
00:20:44 --> 00:20:48
the big leap from Maxwell's
equations to lumped circuits,

330
00:20:48 --> 00:20:50
that's lumped circuit
abstraction.

331
00:20:50 --> 00:20:54
And within there we spent a
large part of the last couple of

332
00:20:54 --> 00:20:58
months doing linear circuits.
We also analyzed nonlinear

333
00:20:58 --> 00:21:02
circuits.
Remember the amplifier circuit

334
00:21:02 --> 00:21:05
of the MOSFET large signal
analysis was nonlinear?

335
00:21:05 --> 00:21:07
Well, there is linear and
nonlinear.

336
00:21:07 --> 00:21:11
Within linear we also showed
that if you take a digital

337
00:21:11 --> 00:21:14
circuit, at least as we
understood them,

338
00:21:14 --> 00:21:18
and draw the subcircuit for a
given set of switch settings,

339
00:21:18 --> 00:21:22
if I set the switches in a
given way what I was left with

340
00:21:22 --> 00:21:26
was another linear circuit for a
given value of all the switch

341
00:21:26 --> 00:21:29
settings.
My small signal analysis was

342
00:21:29 --> 00:21:32
also linear.
If I focused on small signals I

343
00:21:32 --> 00:21:36
also had linear analysis.
Today what we are going to do

344
00:21:36 --> 00:21:38
is this.
We are going to articulate a

345
00:21:38 --> 00:21:40
different part of the
playground.

346
00:21:40 --> 00:21:42
This was a big linear
playground.

347
00:21:42 --> 00:21:44
We've done this.
We've done this.

348
00:21:44 --> 00:21:47
We are going to explore this
territory.

349
00:21:47 --> 00:21:51
This is that territory of the
playground in which we have

350
00:21:51 --> 00:21:53
sinusoidal inputs to circuits.
Furthermore,

351
00:21:53 --> 00:21:57
we are going to look at a
subcircuit of that region which

352
00:21:57 --> 00:22:02
is steady-state.
We will look at sinusoidal

353
00:22:02 --> 00:22:06
input and in the steady-state.
So that is going to be our

354
00:22:06 --> 00:22:11
focus for the next two or three
lectures just to give you a

355
00:22:11 --> 00:22:14
perspective of where we are.
To motivate this,

356
00:22:14 --> 00:22:18
what I would like to do is
consider your amplifier.

357
00:22:18 --> 00:22:22
This is our friend the
amplifier circuit,

358
00:22:22 --> 00:22:24
this part here.
And remember,

359
00:22:24 --> 00:22:28
even though this is an
amplifier, I am using a MOSFET

360
00:22:28 --> 00:22:30
here.
And a MOSFET,

361
00:22:30 --> 00:22:33
as you know,
has this gate capacitance CGS.

362
00:22:33 --> 00:22:36
I am explicitly drawing it out
for you here.

363
00:22:36 --> 00:22:40
And I am going to drive this
with a bias voltage plus some

364
00:22:40 --> 00:22:43
small signal vI,
the usual template for

365
00:22:43 --> 00:22:45
amplifiers.
And there is some Thevenin

366
00:22:45 --> 00:22:48
resistance attached to that
source.

367
00:22:48 --> 00:22:51
I am going to model my source
as a bias voltage,

368
00:22:51 --> 00:22:54
a small signal plus some source
resistance.

369
00:22:54 --> 00:22:58
And I want to apply a sine wave
here and I am going to look at

370
00:22:58 --> 00:23:02
what this looks like.
You may think,

371
00:23:02 --> 00:23:05
look, this is a linear
amplifier.

372
00:23:05 --> 00:23:10
And so if I apply a sine wave
here I should see some response

373
00:23:10 --> 00:23:15
here, and that should be the
same amplitude if I feed the

374
00:23:15 --> 00:23:18
same amplitude here over any
frequency.

375
00:23:18 --> 00:23:22
But let's see what happens.
Keep a look at,

376
00:23:22 --> 00:23:27
switch over to that view graph
while I show you a little

377
00:23:27 --> 00:23:32
demonstration here.
What you see here are three

378
00:23:32 --> 00:23:36
sine waves, a yellow sine wave
which is the input here,

379
00:23:36 --> 00:23:41
you see a green sine wave which
is the input vC at the gate of

380
00:23:41 --> 00:23:47
the MOSFET, and then you see the
blue which is the output here.

381
00:23:47 --> 00:23:51
For now simply focus on the
yellow and the blue.

382
00:23:51 --> 00:23:55
The yellow is the input and the
blue is the output.

383
00:23:55 --> 00:24:00
So I apply some input and I get
an output that looks more or

384
00:24:00 --> 00:24:05
less some linear function of
this input here.

385
00:24:05 --> 00:24:07
It is a small signal.
What I am going to do is I am

386
00:24:07 --> 00:24:10
going to change the frequency of
the input.

387
00:24:10 --> 00:24:13
Remember, I want to look at the
response of the circuit to a

388
00:24:13 --> 00:24:16
sinusoid.
I am feeding a sinusoid here.

389
00:24:16 --> 00:24:19
I look at the response.
I am going to change the

390
00:24:19 --> 00:24:21
frequency.
That is a big shift that we are

391
00:24:21 --> 00:24:25
making in that the curve that we
drew in the last lecture had to

392
00:24:25 --> 00:24:27
do with varying time.
Now I am going to focus on

393
00:24:27 --> 00:24:31
sinusoids and vary their
frequency.

394
00:24:31 --> 00:24:33
I am going to change the
frequency.

395
00:24:33 --> 00:24:37
Stare at the blue curve as I
increase the frequency and just

396
00:24:37 --> 00:24:41
think of what you might expect.
Based on the knowledge you have

397
00:24:41 --> 00:24:46
so far you would expect that
look, as I change the frequency,

398
00:24:46 --> 00:24:50
the frequency should change but
I should see the same amplitude.

399
00:24:50 --> 00:24:53
OK but take a look.
Let me increase the frequency

400
00:24:53 --> 00:24:56
of the input.
What do you see at the output?

401
00:24:56 --> 00:25:00
I am increasing the frequency.

402
00:25:00 --> 00:25:10


403
00:25:10 --> 00:25:12
What do you see happening
there?

404
00:25:12 --> 00:25:16
If you don't see anything
changing there you will need to

405
00:25:16 --> 00:25:19
see an optometrist.
What do we see here?

406
00:25:19 --> 00:25:24
As I changed the frequency,
as I increased the frequency

407
00:25:24 --> 00:25:28
what happened to the blue?
The blue kept decreasing in

408
00:25:28 --> 00:25:31
amplitude.
And you are saying whoa,

409
00:25:31 --> 00:25:35
what is happening here?
We don't have the tools to deal

410
00:25:35 --> 00:25:38
with this.
I expected that when I changed

411
00:25:38 --> 00:25:41
my frequency,
my frequency here should change

412
00:25:41 --> 00:25:44
of course, but why is the
amplitude changing?

413
00:25:44 --> 00:25:47
What is happening here?
That is weird.

414
00:25:47 --> 00:25:51
I noticed that this amplitude
became smaller because the

415
00:25:51 --> 00:25:54
amplitude of the green became
smaller.

416
00:25:54 --> 00:25:58
And remember the green was the
voltage across the capacitor.

417
00:25:58 --> 00:26:02
So this is your RC.
And here is my input.

418
00:26:02 --> 00:26:07
My input has the amplitude,
which I am holding more or less

419
00:26:07 --> 00:26:10
constant.
And notice that vC decreased in

420
00:26:10 --> 00:26:12
value as I increased my
frequency.

421
00:26:12 --> 00:26:16
Just hold that thought.
As I increased the frequency of

422
00:26:16 --> 00:26:20
my input the amplitude of the
output kept diminishing.

423
00:26:20 --> 00:26:24
In other words,
the gain of the system seemed

424
00:26:24 --> 00:26:27
to have decreased as I increased
by frequency.

425
00:26:27 --> 00:26:31
And today we will look at why
that is so and how we can

426
00:26:31 --> 00:26:37
analyze that.
The other thing that is not so

427
00:26:37 --> 00:26:42
obvious here is there is a phase
shift.

428
00:26:42 --> 00:26:50
What I am going to do is try to
see if we can observe the phase

429
00:26:50 --> 00:26:53
shift here.

430
00:26:53 --> 00:27:02


431
00:27:02 --> 00:27:05
Notice here.
What we have been used to is

432
00:27:05 --> 00:27:09
for the amplifier we get a
complete inversion at the

433
00:27:09 --> 00:27:11
output.
Inversion means a phase

434
00:27:11 --> 00:27:15
difference of 180 degrees for a
sine wave, right?

435
00:27:15 --> 00:27:19
This peak should have been
here, but notice that there is a

436
00:27:19 --> 00:27:22
shifting of the peak.
In other words,

437
00:27:22 --> 00:27:26
if the yellow was my input my
output should have had its

438
00:27:26 --> 00:27:31
minimum when the input had its
maximum.

439
00:27:31 --> 00:27:35
But notice there is a shifting
of the signal such that the

440
00:27:35 --> 00:27:39
output is a maximum,
not quite at the point where

441
00:27:39 --> 00:27:43
the input is a minimum but a
little bit after that.

442
00:27:43 --> 00:27:46
Also weird.
Not only has this little

443
00:27:46 --> 00:27:51
circuit here lost its gain
somehow, but also it seems to

444
00:27:51 --> 00:27:53
have shifted the signal in
phase.

445
00:27:53 --> 00:27:57
That is weird.
And today we will look at why

446
00:27:57 --> 00:28:02
that is so and try to understand
the frequency behavior of this

447
00:28:02 --> 00:28:09
little subcomponent here.
Notice that vC is exactly 180

448
00:28:09 --> 00:28:15
out of phase with vO.
So vO is faithfully an inverted

449
00:28:15 --> 00:28:23
amplified form of the input vC.
However, vC itself should have

450
00:28:23 --> 00:28:30
been the same as vI but it looks
quite different.

451
00:28:30 --> 00:28:32
So let's understand why that is
so.

452
00:28:32 --> 00:28:36
The subcircuit to model is the
subcircuit comprising the

453
00:28:36 --> 00:28:38
source, resistor and the
capacitor.

454
00:28:38 --> 00:28:41
And I am just showing that to
you here.

455
00:28:41 --> 00:28:44
I have a vI,
a resistor and capacitor.

456
00:28:44 --> 00:28:47
And I am going to understand
how this behaves.

457
00:28:47 --> 00:28:51
And it is a first order
circuit, single capacitor.

458
00:28:51 --> 00:28:53
My input is a vI cosine of
omega t.

459
00:28:53 --> 00:28:58
vI is a real number for t
greater than zero.

460
00:28:58 --> 00:29:02
And I am telling you that the
capacitor voltage starts out

461
00:29:02 --> 00:29:05
being zero.
And my vI is a sinusoid.

462
00:29:05 --> 00:29:08
It's not a step this time.
It's a sinusoid.

463
00:29:08 --> 00:29:13
So vI is a sinusoid and I want
to find out what vC looks like.

464
00:29:13 --> 00:29:17
The behavior here tells me,
I will give you the answer,

465
00:29:17 --> 00:29:21
that when I feed a sinusoidal
input as the frequency

466
00:29:21 --> 00:29:25
increases, vC should be
decreasing somehow and also be

467
00:29:25 --> 00:29:28
shifting in phase.
Let's do the derivation for

468
00:29:28 --> 00:29:34
that and see what happens.
To give you some insight as to

469
00:29:34 --> 00:29:39
how to go about analyzing this
let me draw a little graph as to

470
00:29:39 --> 00:29:44
the effort level of doing this.
To determine vC of t on the

471
00:29:44 --> 00:29:48
y-axis here is our effort.
How hard do we have to work to

472
00:29:48 --> 00:29:52
solve this circuit for a
sinusoidal input?

473
00:29:52 --> 00:29:56
And on this graph,
down here is easy and up here

474
00:29:56 --> 00:30:00
is pure agony.
Sheer agony up here.

475
00:30:00 --> 00:30:03
So it's the scale of effort
level ranging from easy to

476
00:30:03 --> 00:30:06
complete agony.
And this is the time axis.

477
00:30:06 --> 00:30:10
And the time axis starts out at
11 o'clock, the early part of

478
00:30:10 --> 00:30:12
today's lecture,
and ends at roughly 12,

479
00:30:12 --> 00:30:16
that is today's lecture and
this is next lecture.

480
00:30:16 --> 00:30:19
What I am going to show you
today is a method that uses a

481
00:30:19 --> 00:30:23
usual differential equation
approach, and this is going to

482
00:30:23 --> 00:30:26
be pure agony.
If you thought last Thursday

483
00:30:26 --> 00:30:30
was agony watch today.
This is going to be sheer,

484
00:30:30 --> 00:30:33
sheer, sheer hell.
So I am going to grunge through

485
00:30:33 --> 00:30:37
that, and I think I will sort of
give up halfway because it's

486
00:30:37 --> 00:30:39
just too painful even for me
here.

487
00:30:39 --> 00:30:42
Then what I am going to do is
at the end of this lecture I am

488
00:30:42 --> 00:30:45
going to show you an approach
that I give a cutesy name.

489
00:30:45 --> 00:30:47
I call it the "sneaky
approach".

490
00:30:47 --> 00:30:51
We are going to sneak something
in there it is going to be a lot

491
00:30:51 --> 00:30:53
easier.
And then in the next lecture I

492
00:30:53 --> 00:30:57
am going to show you an even
sneakier approach that is just

493
00:30:57 --> 00:31:01
going to be absolute bliss.
So let's start here.

494
00:31:01 --> 00:31:04
Indulge me as I go through the
agony part.

495
00:31:04 --> 00:31:08
I am going to blast through it
because that is not of how we

496
00:31:08 --> 00:31:12
are going to do things,
but you just need to know that

497
00:31:12 --> 00:31:15
that is agony.
Let's do a usual differential

498
00:31:15 --> 00:31:17
equation approach.
Steps one, two,

499
00:31:17 --> 00:31:20
three and four.
Set up differential equation,

500
00:31:20 --> 00:31:24
find the particular solution,
find the homogenous solution,

501
00:31:24 --> 00:31:27
add up the two and solve for
the unknowns.

502
00:31:27 --> 00:31:31
It's a mantra.
The four-step manta.

503
00:31:31 --> 00:31:34
Let's do it.
Step one, write the DE.

504
00:31:34 --> 00:31:37
That's easy.
We have done this before the RC

505
00:31:37 --> 00:31:40
circuit.
It's RC dvc/dt+vc=vI.

506
00:31:40 --> 00:31:45
This is no different from what
you got from what you got from

507
00:31:45 --> 00:31:49
your RC circuit with a step
input, just that my input is VI

508
00:31:49 --> 00:31:55
cosine of omega t in this case.
It is not just a DC voltage VI.

509
00:31:55 --> 00:31:58
Stare at that.
Enjoy it while the going is

510
00:31:58 --> 00:32:02
easy.
It's like traversing rapids,

511
00:32:02 --> 00:32:06
and before you come to a class
five, you have calm and raging

512
00:32:06 --> 00:32:09
waters there,
you kind of sit there saying

513
00:32:09 --> 00:32:14
oh, the scenery around here
looks really good and so on.

514
00:32:14 --> 00:32:18
All you are doing is stalling
before you have dive down to

515
00:32:18 --> 00:32:21
class five.
We will get to class five

516
00:32:21 --> 00:32:24
rapids in a few seconds here,
so just enjoy this.

517
00:32:24 --> 00:32:27
RC dvc/dt+vC=vI.
You've seen this before.

518
00:32:27 --> 00:32:31
Nothing fancy.
Good old stuff.

519
00:32:31 --> 00:32:34
VI cosine of omega t.
You want to hold onto your

520
00:32:34 --> 00:32:36
seatbelts?
OK.

521
00:32:36 --> 00:32:40
Let's find the particular
solution to the cosine input.

522
00:32:40 --> 00:32:44
Let's use our standard method.
What I will do is just so,

523
00:32:44 --> 00:32:48
there is going to be so much
crapola up there,

524
00:32:48 --> 00:32:53
so that I draw your attention
to vP, which is what we are

525
00:32:53 --> 00:32:56
trying to get,
I am just going to put a box

526
00:32:56 --> 00:33:01
around vP in red.
If you see like all sorts of

527
00:33:01 --> 00:33:04
garbage appear,
just look for the red square.

528
00:33:04 --> 00:33:06
That is what we are trying to
get at.

529
00:33:06 --> 00:33:09
That's the equation.
Let's try.

530
00:33:09 --> 00:33:12
First try, A worked before.
A constant value A worked

531
00:33:12 --> 00:33:16
before for DC inputs.
Let's try that again.

532
00:33:16 --> 00:33:18
If it worked then it may work
now.

533
00:33:18 --> 00:33:22
If I use A, a constant value,
and I substitute it here,

534
00:33:22 --> 00:33:26
I get dA/dt goes to zero,
vP is A, but on the right-hand

535
00:33:26 --> 00:33:32
side I have VI cosine omega t.
So clearly A doesn't work.

536
00:33:32 --> 00:33:34
Sorry.
I struck out.

537
00:33:34 --> 00:33:39
Well, cosine omega t here,
let's try A cosine omega T as

538
00:33:39 --> 00:33:43
my particular solution.
Things are getting a little

539
00:33:43 --> 00:33:46
harder now, a little more
painful.

540
00:33:46 --> 00:33:49
So substitute A cosine omega t
here.

541
00:33:49 --> 00:33:55
So I do get A cosine omega T
for vP, but out here I get RCA

542
00:33:55 --> 00:34:00
sine omega t times omega times
minus one.

543
00:34:00 --> 00:34:05
So I have a sine and a cosine,
and I have a cosine on the

544
00:34:05 --> 00:34:09
right-hand side.
Sorry, it doesn't work.

545
00:34:09 --> 00:34:15
Nope, doesn't work either.
Well, let's try A cosine omega

546
00:34:15 --> 00:34:19
t plus phi.
We are now embarking into the

547
00:34:19 --> 00:34:22
rapids here.
You can begin feeling the

548
00:34:22 --> 00:34:26
pressure.
Just to refresh your memories

549
00:34:26 --> 00:34:32
of sines and cosines.
A is the amplitude of the

550
00:34:32 --> 00:34:34
cosine.
Omega is the frequency.

551
00:34:34 --> 00:34:38
Phi is the phase.
Remember the signal I showed

552
00:34:38 --> 00:34:41
you earlier?
If something happens to the

553
00:34:41 --> 00:34:45
amplitude of the sine,
something happens to the phase.

554
00:34:45 --> 00:34:50
A cosine omega t plus phi.
Let me plug it in here and go

555
00:34:50 --> 00:34:53
by standard practice.
Here is what I get.

556
00:34:53 --> 00:34:56
I plug in A cosine omega t to
this equation,

557
00:34:56 --> 00:35:01
and this is what I get.
I differentiate it.

558
00:35:01 --> 00:35:05
I get omega out minus sine,
sine of negative d plus phi,

559
00:35:05 --> 00:35:09
A cosine omega t plus phi
equals VI cosine omega t.

560
00:35:09 --> 00:35:12
That might work.
Now we get into the class six

561
00:35:12 --> 00:35:16
part of the class five.
All class fives have a little

562
00:35:16 --> 00:35:20
bit of class six rapids.
Remember, the rapids go up on

563
00:35:20 --> 00:35:23
an exponential scale so it like
earthquakes.

564
00:35:23 --> 00:35:27
What I do now is expand out
sine omega t plus phi,

565
00:35:27 --> 00:35:31
blah, blah, blah,
it goes on and on.

566
00:35:31 --> 00:35:35
I could go on and on,
but this is even tiring me.

567
00:35:35 --> 00:35:41
This can be made to work,
but I am not sure I want to put

568
00:35:41 --> 00:35:45
all of us through this trig
nightmare here.

569
00:35:45 --> 00:35:50
If I am really,
really nasty I could give this

570
00:35:50 --> 00:35:54
to you as a homework assignment
or something,

571
00:35:54 --> 00:36:00
but I am not that nasty so you
won't see that.

572
00:36:00 --> 00:36:03
But if I go down this path it
will get me to the answer,

573
00:36:03 --> 00:36:06
but I would have to soon
negotiate class six,

574
00:36:06 --> 00:36:09
class seven rapids to get to
where I want.

575
00:36:09 --> 00:36:12
So let me punt on it,
let me start from scratch.

576
00:36:12 --> 00:36:15
I am at step two,
let me start from scratch.

577
00:36:15 --> 00:36:18
Instead what I would like to do
is let's get sneaky here.

578
00:36:18 --> 00:36:21
Rather than negotiating the
class five rapids,

579
00:36:21 --> 00:36:25
what we can say is ah-ha,
we can take our canoes and jump

580
00:36:25 --> 00:36:30
onto shore and run down and then
get back onto the river.

581
00:36:30 --> 00:36:32
Let's do that.
That is called the sneaky

582
00:36:32 --> 00:36:34
approach.
So that all our friends who are

583
00:36:34 --> 00:36:37
behind us think we are
negotiating the rapids,

584
00:36:37 --> 00:36:41
but what we are going to do is
get sneaky and take the shore

585
00:36:41 --> 00:36:42
path.
Let's get sneaky,

586
00:36:42 --> 00:36:45
just walk down the shore and
see what is there.

587
00:36:45 --> 00:36:47
I want to do something
completely weird here.

588
00:36:47 --> 00:36:51
I want to look at solving this
equation through the shore

589
00:36:51 --> 00:36:53
method.
S stands for shore or S stands

590
00:36:53 --> 00:36:57
for sneaky, whatever you want.
What I am going to do is rather

591
00:36:57 --> 00:37:01
than trying to solve for VI
cosine omega t.

592
00:37:01 --> 00:37:04
I am going to say let's try a
different input all together.

593
00:37:04 --> 00:37:06
And you will understand why in
a second.

594
00:37:06 --> 00:37:09
It's like I am the captain of
my canoe and I tell my

595
00:37:09 --> 00:37:12
teammates, hey,
let's not negotiate the rapids,

596
00:37:12 --> 00:37:15
let's go and explore the shore.
Maybe down the shore we can

597
00:37:15 --> 00:37:19
find a path that gets us across
to the other side more easily.

598
00:37:19 --> 00:37:22
So here is me and my colleagues
carrying our canoe and getting

599
00:37:22 --> 00:37:24
onto shore and taking a sneaky
path.

600
00:37:24 --> 00:37:26
This is not what I set out to
solve.

601
00:37:26 --> 00:37:30
I don't know where this will
lead me.

602
00:37:30 --> 00:37:33
But let's see where the shore
path leads us.

603
00:37:33 --> 00:37:36
I want to try solving this
equation Vie^st.

604
00:37:36 --> 00:37:40
S stands for shore or sneaky or
whatever you want.

605
00:37:40 --> 00:37:45
I don't know where I am going,
but let's see where this leads

606
00:37:45 --> 00:37:46
us.
Let's explore.

607
00:37:46 --> 00:37:50
Make believe you are Columbus
or something.

608
00:37:50 --> 00:37:53
I don't know.
Let's use the usual techniques

609
00:37:53 --> 00:37:57
and see how this works out.
Let's try a particular

610
00:37:57 --> 00:38:01
solution, Vpe^st.
My input is Vie^st.

611
00:38:01 --> 00:38:06
I am trying to solve the
circuit for a different input.

612
00:38:06 --> 00:38:11
And let me try solution Vpe^st
and see if that works out

613
00:38:11 --> 00:38:15
nicely.
I substitute Vpe^st into my

614
00:38:15 --> 00:38:18
equation here,
so RCVpe^st blah blah blah.

615
00:38:18 --> 00:38:23
What I get here is Vie^st,
Vpe^st stays the same,

616
00:38:23 --> 00:38:28
but here vP comes out,
s comes out and e^st stays the

617
00:38:28 --> 00:38:31
same.
That is nice property of

618
00:38:31 --> 00:38:34
exponentials.
This is what I get.

619
00:38:34 --> 00:38:37
A really cool property of
exponentials is that when I

620
00:38:37 --> 00:38:41
differentiate it I get the
exponential back.

621
00:38:41 --> 00:38:45
Unlike a cosine I got a sine,
and for a sine I got a cosine.

622
00:38:45 --> 00:38:49
Exponentials are very plain and
simple, are straightforward.

623
00:38:49 --> 00:38:53
What you see is what you get.
You differentiate it.

624
00:38:53 --> 00:38:56
You get the same thing,
you get scaling vP,

625
00:38:56 --> 00:39:00
S and so on,
and some scaling here.

626
00:39:00 --> 00:39:05
You get S scaling here,
but the basic form stays the

627
00:39:05 --> 00:39:07
same.
This is quite nice.

628
00:39:07 --> 00:39:12
I have e^st in all three
places, so I can cancel those

629
00:39:12 --> 00:39:17
out and I get this expression.
And I get this.

630
00:39:17 --> 00:39:21
Wow.
So if I go down the shore I get

631
00:39:21 --> 00:39:25
some place fast.
I don't know where I am yet,

632
00:39:25 --> 00:39:30
but whatever I did,
it was easy.

633
00:39:30 --> 00:39:33
I am just exploring this path,
down the shore path.

634
00:39:33 --> 00:39:36
I am making progress.
I don't know where I have

635
00:39:36 --> 00:39:39
gotten yet.
We will see where we got to in

636
00:39:39 --> 00:39:43
a second, but I got some place
quickly, fast.

637
00:39:43 --> 00:39:47
I was able to solve for this
input Vie^st and get this

638
00:39:47 --> 00:39:50
solution very quickly.
So what happened here?

639
00:39:50 --> 00:39:54
I assumed the solution of the
form Vpe^st, substituted it

640
00:39:54 --> 00:39:58
back, and found that if vP were
equal to Vi/(1+sRC) then Vpe^st

641
00:39:58 --> 00:40:03
is a solution.
What I have done here is that

642
00:40:03 --> 00:40:09
Vi/(1+sRC) is a particular
solution to this equation for

643
00:40:09 --> 00:40:14
the input Vie^st.
I put a box around it because

644
00:40:14 --> 00:40:17
this is important.
This was easy.

645
00:40:17 --> 00:40:22
We went down the shore,
and said let's try this other

646
00:40:22 --> 00:40:26
input.
We made rapid progress on shore

647
00:40:26 --> 00:40:31
and I got some place.
I don't know where I am yet.

648
00:40:31 --> 00:40:34
I got this.
Let me pause here and let me

649
00:40:34 --> 00:40:37
give you the final answer.
I am going to show you over the

650
00:40:37 --> 00:40:40
next five minutes that this is
our answer.

651
00:40:40 --> 00:40:42
You are staring at the answer
already.

652
00:40:42 --> 00:40:45
I am a party,
I have taken a shore path and

653
00:40:45 --> 00:40:48
we have gotten some place.
We see the river there,

654
00:40:48 --> 00:40:51
so it turns out we are exactly
where we want to be,

655
00:40:51 --> 00:40:54
just after the rapids.
All I have to do now is get my

656
00:40:54 --> 00:40:58
colleagues into the river with
myself in the canoe and we are

657
00:40:58 --> 00:41:01
done.
You don't know that yet.

658
00:41:01 --> 00:41:04
My colleagues and I are sitting
on the shore looking at the

659
00:41:04 --> 00:41:05
river.
We've gotten some place.

660
00:41:05 --> 00:41:08
There are no rapids there.
We have gotten some place.

661
00:41:08 --> 00:41:11
We don't quite know is this
just after the rapids or not.

662
00:41:11 --> 00:41:14
We don't know yet,
but I got there very quickly.

663
00:41:14 --> 00:41:17
And I will tell you right now,
that is the place we wanted to

664
00:41:17 --> 00:41:19
go to.
The next five view graphs I am

665
00:41:19 --> 00:41:21
going to blast through.
There is going to be more pain

666
00:41:21 --> 00:41:24
and agony to show you why that
is the case.

667
00:41:24 --> 00:41:27
It's me thinking I am Columbus
and proving to my colleagues

668
00:41:27 --> 00:41:30
that this is where we want to
be.

669
00:41:30 --> 00:41:33
And pulling out my sextant,
and the compasses and so on

670
00:41:33 --> 00:41:37
that cartographers and people
use to prove to my colleagues

671
00:41:37 --> 00:41:41
that this is where I want to be.
This is the answer.

672
00:41:41 --> 00:41:45
The next five view graphs will
be demonstrating that this is

673
00:41:45 --> 00:41:48
indeed the answer,
or close enough to the answer

674
00:41:48 --> 00:41:51
that we will be satisfied.
Isn't this spectacular?

675
00:41:51 --> 00:41:56
I am going to show you in about
five minutes that this gives us

676
00:41:56 --> 00:42:00
all the information we need to
know to compute the sinusoidal

677
00:42:00 --> 00:42:05
steady-state response to this
differential equation.

678
00:42:05 --> 00:42:09
Let me write that down here
just so you know.

679
00:42:09 --> 00:42:21


680
00:42:21 --> 00:42:24
Just so you remember.
I am going to put a marker on

681
00:42:24 --> 00:42:28
the shore that says this is
where we are right now.

682
00:42:28 --> 00:42:31
Now let me prove to you.
As I just said,

683
00:42:31 --> 00:42:39
vPS is Vi, it's this stuff here
multiplied by e^st is the

684
00:42:39 --> 00:42:45
solution to Vie^st.
This guy here is a solution for

685
00:42:45 --> 00:42:52
Vie^st and vP is simply the
coefficient that multiplies

686
00:42:52 --> 00:42:56
e^st.
Similarly, if I substitute S

687
00:42:56 --> 00:43:02
equals j omega.
I told you five view graphs of

688
00:43:02 --> 00:43:07
more hell, but I am just going
to prove to you that this is it.

689
00:43:07 --> 00:43:10
I am substituting S equals j
omega.

690
00:43:10 --> 00:43:15
This is Columbus giving a big
speech at the end convincing his

691
00:43:15 --> 00:43:18
colleagues that we are where we
want to be.

692
00:43:18 --> 00:43:22
I substitute j omega for S and
this is what I get.

693
00:43:22 --> 00:43:27
This is a solution for e to the
st, and so this is a solution

694
00:43:27 --> 00:43:33
for e to the j omega t.
And let me mark this for you as

695
00:43:33 --> 00:43:37
something to remember.
Vi/(1+j omega RC).

696
00:43:37 --> 00:43:43
In terms of that,
I am substituting j omega for

697
00:43:43 --> 00:43:46
S.
And this is a complex number.

698
00:43:46 --> 00:43:52
It is a complex amplitude.
It is a complex number because

699
00:43:52 --> 00:43:58
of j here, and it multiplies e
to the j omega t.

700
00:43:58 --> 00:44:03
Just keep this in mind.
So that was easy.

701
00:44:03 --> 00:44:08
The steps were easy.
I am still proving to you that

702
00:44:08 --> 00:44:13
this is where we want to be.
Now let me draw the connection

703
00:44:13 --> 00:44:17
back to vP.
And the first fact was that

704
00:44:17 --> 00:44:21
finding the response to Vie^(j
omega t) was easy.

705
00:44:21 --> 00:44:25
We know that.
Second was the following

706
00:44:25 --> 00:44:31
observation that Vi cosine omega
t is simply the real part of

707
00:44:31 --> 00:44:37
this number here.
So Vi cosine omega t is simply

708
00:44:37 --> 00:44:42
the real part of Vie^(j omega t)
from the Euler relation.

709
00:44:42 --> 00:44:48
So cosine omega t is simply the
real part of this guy.

710
00:44:48 --> 00:44:51
Light bulbs beginning to go
off?

711
00:44:51 --> 00:44:58
The first fast was that finding
the response to Vie^(j omega t)

712
00:44:58 --> 00:45:02
was easy.
And the response was this,

713
00:45:02 --> 00:45:05
right?
Times e to the j omega t.

714
00:45:05 --> 00:45:07
That was easy.
All right.

715
00:45:07 --> 00:45:12
And the second part is that the
real part of this is Vi cosine

716
00:45:12 --> 00:45:17
omega t was our input.
Draw the connection between two

717
00:45:17 --> 00:45:21
steps.
Finding the response to Vie^(j

718
00:45:21 --> 00:45:25
omega t) was easy.
The real part of that was the

719
00:45:25 --> 00:45:31
input we cared about.
Are light bulbs going off?

720
00:45:31 --> 00:45:35
Let me draw you a little
picture here to show you what is

721
00:45:35 --> 00:45:37
happening.
Response to vI is vP.

722
00:45:37 --> 00:45:41
It's the particular response we
are looking for.

723
00:45:41 --> 00:45:46
Remember the red square?
But we threw in a sneaky input

724
00:45:46 --> 00:45:49
vIS and we formed the response
vPS to that.

725
00:45:49 --> 00:45:52
This step was easy.
This step was hard.

726
00:45:52 --> 00:45:55
vI to vP was hard,
trig nightmare,

727
00:45:55 --> 00:45:59
remember?
But vIS to vPS was easy.

728
00:45:59 --> 00:46:05
It was a simple Vpe^st thing.
We also know that the real part

729
00:46:05 --> 00:46:10
of vIS is vI.
The real part of this is simply

730
00:46:10 --> 00:46:14
vI.
If you have a real circuit,

731
00:46:14 --> 00:46:20
if you have a real linear
circuit, for a linear circuit,

732
00:46:20 --> 00:46:27
if the real part of this gives
me vI then the real part of the

733
00:46:27 --> 00:46:33
solution should give me vP.
So computing vPS was easy.

734
00:46:33 --> 00:46:37
If I take the real part of
this, I take the corresponding

735
00:46:37 --> 00:46:40
real part of this.
This is sort of an inverse

736
00:46:40 --> 00:46:43
superposition argument.
Superposition,

737
00:46:43 --> 00:46:47
I said, take the response for
A, take the response for B,

738
00:46:47 --> 00:46:51
add them up and you get the
response for A plus B.

739
00:46:51 --> 00:46:55
Here what we are saying is that
get the response to A plus B,

740
00:46:55 --> 00:46:59
or to A plus jB and the real
part of the input will produce

741
00:46:59 --> 00:47:04
the response given by the real
part of the output.

742
00:47:04 --> 00:47:07
So this is an inverse
superposition argument.

743
00:47:07 --> 00:47:10
If it is a linear circuit,
then if vI is the real part of

744
00:47:10 --> 00:47:15
this sneaky input then I find
the response to the sneaky input

745
00:47:15 --> 00:47:17
and take its real part I should
get vP.

746
00:47:17 --> 00:47:21
Here I am, Columbus,
staring down at the entrance to

747
00:47:21 --> 00:47:24
this part of the river.
I just proved to my colleagues

748
00:47:24 --> 00:47:30
that all we have to do is take
the real part of what we have.

749
00:47:30 --> 00:47:34
We can just jump right back
into the river and get back to

750
00:47:34 --> 00:47:36
vP.
And what I am going to do next

751
00:47:36 --> 00:47:40
is just grind through the math
and just show you that.

752
00:47:40 --> 00:47:44
I will just blast through it.
It is not important,

753
00:47:44 --> 00:47:48
but you have it in your notes.
I am telling you that vP is

754
00:47:48 --> 00:47:51
simply the real part of the
sneaky output.

755
00:47:51 --> 00:47:54
And I take the real part of vP
e to the j omega t.

756
00:47:54 --> 00:48:00
And I take the real part.
And just a bunch of math here.

757
00:48:00 --> 00:48:04
I am just taking the real part
and doing a bunch of complex

758
00:48:04 --> 00:48:06
math.
Remember vP was given by this

759
00:48:06 --> 00:48:09
quantity here.
And I take the real part and I

760
00:48:09 --> 00:48:13
end up with vP is simply this
quantity multiplied by cosine

761
00:48:13 --> 00:48:16
omega t plus phi,
where phi is given by is given

762
00:48:16 --> 00:48:20
by tan inverse of omega RC,
and this is the coefficient

763
00:48:20 --> 00:48:24
multiplying the cosine.
So by taking the sneaky path

764
00:48:24 --> 00:48:27
and then taking the real part of
that output answer,

765
00:48:27 --> 00:48:33
I was able to very quickly get
to where I wanted to be.

766
00:48:33 --> 00:48:35
So from here to here it is only
math.

767
00:48:35 --> 00:48:38
Recall, that vP,
the thing in the red was what

768
00:48:38 --> 00:48:41
we set out to find out,
which was the particular

769
00:48:41 --> 00:48:43
response to VI cosine of omega
t.

770
00:48:43 --> 00:48:46
And remember that two grunge is
all of this stuff.

771
00:48:46 --> 00:48:50
I am going to blast through two
or three more view graphs that

772
00:48:50 --> 00:48:54
just give you more insight and
more math, nothing particular.

773
00:48:54 --> 00:48:57
And remember to solve the
equation we have to find a

774
00:48:57 --> 00:49:00
homogenous solution,
too.

775
00:49:00 --> 00:49:04
But recall that the homogenous
solution for an RC circuit is of

776
00:49:04 --> 00:49:07
the form Ae^-t/RC.
This means that as time becomes

777
00:49:07 --> 00:49:10
very large this part goes to
zero.

778
00:49:10 --> 00:49:13
As time becomes large in the
steady state,

779
00:49:13 --> 00:49:15
remember I care about the
steady state?

780
00:49:15 --> 00:49:18
This goes to zero.
I don't care about the

781
00:49:18 --> 00:49:21
homogenous solution.
Isn't that fantastic?

782
00:49:21 --> 00:49:25
Most the circuits we will deal
with, except for purely

783
00:49:25 --> 00:49:30
oscillatory ones,
the homogenous part dies away.

784
00:49:30 --> 00:49:33
You have something like e to
the -t whatever.

785
00:49:33 --> 00:49:35
It just dies away.
It's gone.

786
00:49:35 --> 00:49:38
So the total solution has vH
going away.

787
00:49:38 --> 00:49:40
And what I end up with is just
vP.

788
00:49:40 --> 00:49:44
My total solution in the steady
state is simply vP.

789
00:49:44 --> 00:49:48
And A is given by this that we
just calculated.

790
00:49:48 --> 00:49:52
I just have a bunch more
insight that I talk about that

791
00:49:52 --> 00:49:55
you can look through in your
notes.

792
00:49:55 --> 00:50:00
And I just want to show you a
very quick summary.

793
00:50:00 --> 00:50:04
In summary, what we have is we
computed vP.

794
00:50:04 --> 00:50:08
It was a complex coefficient.
And all these steps,

795
00:50:08 --> 00:50:12
2 grunge, 3 and 4 were a waste
of time.

796
00:50:12 --> 00:50:18
And what I showed you was that
for the input VI the coefficient

797
00:50:18 --> 00:50:23
vP was complex.
And I can take the ratio and

798
00:50:23 --> 00:50:26
represent it in this manner as
well.

799
00:50:26 --> 00:50:29
And from vP,
I can then compute the

800
00:50:29 --> 00:50:35
multiplier for the cosine as
follows.

801
00:50:35 --> 00:50:38
I divide by vP here.
Remember the cosine was

802
00:50:38 --> 00:50:41
multiplied by,
in the mathematical step that I

803
00:50:41 --> 00:50:44
did, VI divided one plus,
this stuff here,

804
00:50:44 --> 00:50:48
so I could get the magnitude
and phase of the transfer

805
00:50:48 --> 00:50:51
function of this circuit in the
following manner.

806
00:50:51 --> 00:50:55
And to wrap up very quickly,
I am going to cover this again

807
00:50:55 --> 00:51:00
the next time and show you a
magnitude plot.

808
00:51:00 --> 00:51:02
Notice here that if I plot
Vp/Vi.

809
00:51:02 --> 00:51:06
Remember this was Vp here.
That's the answer.

810
00:51:06 --> 00:51:12
The magnitude looks like this.
On a log scale Vp/Vi for small

811
00:51:12 --> 00:51:17
frequencies omega is at one,
but as omega increases Vp/Vi

812
00:51:17 --> 00:51:20
keeps decreasing.
That is the output.

813
00:51:20 --> 00:51:24
Remember Vp was the amplitude
of the output?

814
00:51:24 --> 00:51:30
That keeps decreasing.
And this is the reason why.

815
00:51:30 --> 00:51:34
As I increase the frequency,
the amplitude of my output

816
00:51:34 --> 00:51:39
cosine kept decreasing.
I could also plot the phase for

817
00:51:39 --> 00:51:41
you.
And the phase,

818
00:51:41 --> 00:51:46
in the same manner as omega
increased, my phase also kept

819
00:51:46 --> 00:51:50
shifting from zero initially to
pi/2 finally.

820
00:51:50 --> 00:51:55
Let me stop here and start with
this the next time and revisit

821
00:51:55 --> 00:51:57
this.
Unfortunately,

822
00:51:57 --> 00:52:02
I won't have time for the demo.
I will show it to you next

823
00:52:02 --> 52:05
time.