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PROFESSOR STRANG: OK, so I feel
I should have brought pizza
10
00:00:25 --> 00:00:26
instead of math problems.
11
00:00:26 --> 00:00:34
I mean, but this is our
final, final, hour.
12
00:00:34 --> 00:00:37
Open for any questions,
any discussion about
13
00:00:37 --> 00:00:42
Fourier topics.
14
00:00:42 --> 00:00:45
Looking toward tomorrow
evening's exam.
15
00:00:45 --> 00:00:51
And, well, hoping you'll find
the whole course useful at
16
00:00:51 --> 00:00:54
many times in the future.
17
00:00:54 --> 00:01:00
So, somebody emailed me that
the posted solution, so I just
18
00:01:00 --> 00:01:07
posted a quick solution to the
un-collected homework ten, and
19
00:01:07 --> 00:01:10
so I haven't even looked to
see what did it say
20
00:01:10 --> 00:01:11
for this question.
21
00:01:11 --> 00:01:12
But what is the answer?
22
00:01:12 --> 00:01:16
If I convolve the delta
function with the delta
23
00:01:16 --> 00:01:18
function, what do I get?
24
00:01:18 --> 00:01:20
Delta, right.
25
00:01:20 --> 00:01:24
If I convolve anything with
delta I get delta because I'm
26
00:01:24 --> 00:01:26
over in the other space.
27
00:01:26 --> 00:01:28
I'm multiplying by one.
28
00:01:28 --> 00:01:33
Yeah, so if I use the
convolution rule to go over to
29
00:01:33 --> 00:01:37
the other space, where this
becomes just one times one I'm
30
00:01:37 --> 00:01:46
getting one, and transform
back and I've got delta.
31
00:01:46 --> 00:01:49
Well that question won't
be on tomorrow's exam.
32
00:01:49 --> 00:01:51
But I thought we could
deal with that fast.
33
00:01:51 --> 00:01:56
Now I'm open to questions
about any topic.
34
00:01:56 --> 00:01:56
Homework or not.
35
00:01:56 --> 00:01:57
Yeah, thanks.
36
00:01:57 --> 00:02:01
AUDIENCE: [INAUDIBLE]
37
00:02:01 --> 00:02:02
PROFESSOR STRANG: I could try.
38
00:02:02 --> 00:02:04
Did I make up the exam?
39
00:02:04 --> 00:02:05
Probably.
40
00:02:05 --> 00:02:06
Yeah, I'll recognize it.
41
00:02:06 --> 00:02:14
OK, alright, I'll read it out.
42
00:02:14 --> 00:02:25
OK, so this an exam from
2005, not that far back.
43
00:02:25 --> 00:02:31
OK, so it has some questions.
44
00:02:31 --> 00:02:33
Shall I tell you what the whole
question is but it looks
45
00:02:33 --> 00:02:36
like this Part D is sort of
separate from the others.
46
00:02:36 --> 00:02:41
But why not, this gives us some
questions to think about.
47
00:02:41 --> 00:02:44
So if I just read
out the questions.
48
00:02:44 --> 00:02:50
One was, find the coefficients
of, the function is e^(-x).
49
00:02:52 --> 00:02:55
That would naturally occur to
me as a function that you
50
00:02:55 --> 00:02:58
could, so from zero to 2pi.
51
00:03:00 --> 00:03:02
And then periodic.
52
00:03:02 --> 00:03:04
Always good to graph it.
53
00:03:04 --> 00:03:06
So will e to the
minus x look like?
54
00:03:06 --> 00:03:12
It'll come down to, much
steeper than that,
55
00:03:12 --> 00:03:14
but that would do it.
56
00:03:14 --> 00:03:14
So that's e^(-x).
57
00:03:15 --> 00:03:18
Here it's one, and
here it's e^(-2pi).
58
00:03:18 --> 00:03:20
59
00:03:20 --> 00:03:23
And then repeat,
repeat, repeat.
60
00:03:23 --> 00:03:27
So find its Fourier
coefficients.
61
00:03:27 --> 00:03:30
I'll just say what the question
is and you can ask me.
62
00:03:30 --> 00:03:32
And then what's the decay rate.
63
00:03:32 --> 00:03:34
Oh, maybe you can tell
me the decay rate.
64
00:03:34 --> 00:03:37
Even before you tell
me the coefficient.
65
00:03:37 --> 00:03:42
What would the decay rate be
if I expressed this as a
66
00:03:42 --> 00:03:49
combination of Fourier
terms, harmonics.
67
00:03:49 --> 00:03:52
How quickly will those
coefficients drop off?
68
00:03:52 --> 00:03:55
So 1/k, only.
69
00:03:55 --> 00:04:02
That's right, 1/k, because this
function has a jump at zero.
70
00:04:02 --> 00:04:07
Well, at zero and again at 2pi
at every one of those points.
71
00:04:07 --> 00:04:11
It has a jump, I guess, of
about this distance, between
72
00:04:11 --> 00:04:12
one and that number.
73
00:04:12 --> 00:04:16
Which would be the jump that we
would see of course every time.
74
00:04:16 --> 00:04:19
Yeah, so the coefficients
would decay like 1/k.
75
00:04:20 --> 00:04:25
Then it asked to compute
the sum of those
76
00:04:25 --> 00:04:27
coefficients squared.
77
00:04:27 --> 00:04:32
OK, so how can we find the sum
of those coefficients squared
78
00:04:32 --> 00:04:37
even before I know
what they are?
79
00:04:37 --> 00:04:40
Well, you know the
answer, right?
80
00:04:40 --> 00:04:43
How am I going to
do this problem?
81
00:04:43 --> 00:04:47
Yeah, so I'm looking for
the sum of the squares
82
00:04:47 --> 00:04:51
of coefficients and then
there's this magic trick.
83
00:04:51 --> 00:04:56
What is it that helps me to
find that sum of squares?
84
00:04:56 --> 00:05:00
How do I go about that?
85
00:05:00 --> 00:05:05
I connected to the integral
of the function square.
86
00:05:05 --> 00:05:08
It can go zero to 2pi of
the function square.
87
00:05:08 --> 00:05:12
And you remember the reason,
and I've left a little space
88
00:05:12 --> 00:05:13
for the fudge factor.
89
00:05:13 --> 00:05:17
2pi, or 1/2pi, we can
figure out what it is by
90
00:05:17 --> 00:05:19
taking a simple case.
91
00:05:19 --> 00:05:22
Yeah, why don't we take a
simple case to figure out
92
00:05:22 --> 00:05:25
should there be a 2pi
or a 1/2pi or what?
93
00:05:25 --> 00:05:28
What's a simple case here?
94
00:05:28 --> 00:05:29
Suppose I take f(x)=1.
95
00:05:31 --> 00:05:36
Suppose I take the function
f(x)=1, then the right side.
96
00:05:36 --> 00:05:42
So if f(x) is one, the right
side would be 2pi, I guess.
97
00:05:42 --> 00:05:45
The integral of one, and the
left side would be what?
98
00:05:45 --> 00:05:50
What would be the Fourier
coefficients of the
99
00:05:50 --> 00:05:53
constant function?
100
00:05:53 --> 00:05:56
Well just c_0, the only one
we would have would be
101
00:05:56 --> 00:06:00
c_0, the constant term.
102
00:06:00 --> 00:06:00
The c_0*e^0.
103
00:06:02 --> 00:06:03
And it would be one.
104
00:06:03 --> 00:06:08
So I'd have a one there, when
k is zero, and otherwise
105
00:06:08 --> 00:06:09
I wouldn't have anything.
106
00:06:09 --> 00:06:12
So this would be one squared,
this would be a one.
107
00:06:12 --> 00:06:14
So I do need to divide by 2pi.
108
00:06:16 --> 00:06:19
Do you agree with that?
109
00:06:19 --> 00:06:26
To make it correct I'd better
have a 1/2pi to get that.
110
00:06:26 --> 00:06:27
To be right.
111
00:06:27 --> 00:06:33
OK, and then this gives me the
answer then, I just do this
112
00:06:33 --> 00:06:37
integral, e^(-2x),
then f is e^(-x).
113
00:06:38 --> 00:06:43
So f squared would be e^(-2x),
which I can certainly integrate
114
00:06:43 --> 00:06:45
and get some expression.
115
00:06:45 --> 00:06:46
Yeah, OK.
116
00:06:46 --> 00:06:49
So that's a
straightforward part.
117
00:06:49 --> 00:06:53
And now no to the Part D,
which you asked about
118
00:06:53 --> 00:06:56
which, is oh wow.
119
00:06:56 --> 00:07:05
OK, so now I'm asking about an
equation u, du/dx, plus u(x)
120
00:07:05 --> 00:07:08
equals a train of deltas.
121
00:07:08 --> 00:07:12
So delta(x) made periodic.
122
00:07:12 --> 00:07:14
Can I write it that way?
123
00:07:14 --> 00:07:16
So this is the periodic case.
124
00:07:16 --> 00:07:20
So I'm looking for
a periodic answer.
125
00:07:20 --> 00:07:26
So I'm in the Fourier series
world when I take, I'll be
126
00:07:26 --> 00:07:30
looking for coefficients,
not the integral transform.
127
00:07:30 --> 00:07:33
OK.
128
00:07:33 --> 00:07:36
So alright, so I'm
making it clear.
129
00:07:36 --> 00:07:40
So how do I proceed now,
with such a problem?
130
00:07:40 --> 00:07:44
So I've mentioned this
morning that a problem
131
00:07:44 --> 00:07:45
like this could appear.
132
00:07:45 --> 00:07:48
So I'm happy to discuss it.
133
00:07:48 --> 00:07:50
What do I do?
134
00:07:50 --> 00:07:51
Transform.
135
00:07:51 --> 00:07:52
Absolutely.
136
00:07:52 --> 00:07:54
Take Fourier transform.
137
00:07:54 --> 00:07:58
OK, so what happens when I
take Fourier transforms of u?
138
00:07:58 --> 00:08:01
Well, I guess so we're doing
periodic, so we're going to
139
00:08:01 --> 00:08:03
go to coefficients, right?
140
00:08:03 --> 00:08:10
I'll call them, shall I call,
shall I think of u(x), shall I
141
00:08:10 --> 00:08:12
just call its coefficient c?
142
00:08:12 --> 00:08:14
It's the only function and
I have around so I might
143
00:08:14 --> 00:08:17
as well use c for it.
144
00:08:17 --> 00:08:22
So here's the function, and
Fourier takes this to the
145
00:08:22 --> 00:08:25
set of coefficients, OK.
146
00:08:25 --> 00:08:30
And the point is that we can
separate - I mean this is
147
00:08:30 --> 00:08:33
the whole reason why
Fourier's so great.
148
00:08:33 --> 00:08:37
That we can do each
frequency separately.
149
00:08:37 --> 00:08:38
So let's see then.
150
00:08:38 --> 00:08:44
So this has coefficient c_k,
and what are the coefficients,
151
00:08:44 --> 00:08:50
what's the k'th coefficient, so
this is all the e^(ikx) term.
152
00:08:50 --> 00:09:00
What's the k'th coefficient of
the derivative? ikc_k, right?
153
00:09:00 --> 00:09:04
And what's the k'th coefficient
of the delta function?
154
00:09:04 --> 00:09:06
One, right?
155
00:09:06 --> 00:09:07
Is that right or
is there 1/2pi?
156
00:09:07 --> 00:09:09
157
00:09:09 --> 00:09:10
Is it 1/2pi?
158
00:09:10 --> 00:09:11
OK, 1/2pi.
159
00:09:13 --> 00:09:14
Yeah, I guess that's right.
160
00:09:14 --> 00:09:21
Because the coefficient, I
would have to take the integral
161
00:09:21 --> 00:09:24
times, yeah, the integral with
the delta function will give me
162
00:09:24 --> 00:09:26
the one, and then
there's a 1/2pi.
163
00:09:27 --> 00:09:28
Right, OK.
164
00:09:28 --> 00:09:33
So now I've got the answer.
165
00:09:33 --> 00:09:37
Now I have this
equation at each k.
166
00:09:37 --> 00:09:40
Following each eigenvector,
following each
167
00:09:40 --> 00:09:42
separate frequency.
168
00:09:42 --> 00:09:44
And it's easy to do.
169
00:09:44 --> 00:09:45
It's just 1/2pi.
170
00:09:46 --> 00:09:53
And there's one
plus ik, I think.
171
00:09:53 --> 00:09:57
OK, and now those
are the c_k's.
172
00:09:57 --> 00:10:00
So I think maybe this is it.
173
00:10:00 --> 00:10:05
Sum from minus infinity,
I know what the c_k is.
174
00:10:05 --> 00:10:10
2pi(1+ik), times e^(ikx).
175
00:10:10 --> 00:10:15
176
00:10:15 --> 00:10:18
That's my u(x), I think.
177
00:10:18 --> 00:10:19
Had to be, right?
178
00:10:19 --> 00:10:26
So I just did the sort
of automatic steps.
179
00:10:26 --> 00:10:28
When I made up this exam, I
don't know whether I was
180
00:10:28 --> 00:10:44
thinking I could find out can I
actually do that sum to produce
181
00:10:44 --> 00:10:47
some function u(x) that
I am familiar with.
182
00:10:47 --> 00:10:52
I'm not sure.
183
00:10:52 --> 00:10:56
Anybody reading the question,
so the question says solve
184
00:10:56 --> 00:10:58
this differential equation.
185
00:10:58 --> 00:11:02
Doesn't say what it
means by solve.
186
00:11:02 --> 00:11:03
Anyway, I've done it.
187
00:11:03 --> 00:11:05
I guess.
188
00:11:05 --> 00:11:13
Shall we agree that I
passed on this question?
189
00:11:13 --> 00:11:18
If we knew some tricky way
to do this addition, but I
190
00:11:18 --> 00:11:19
don't immediately see it.
191
00:11:19 --> 00:11:21
So that's good.
192
00:11:21 --> 00:11:23
OK, you're welcome.
193
00:11:23 --> 00:11:24
Thanks.
194
00:11:24 --> 00:11:27
OK, so that was a very
good question to ask.
195
00:11:27 --> 00:11:34
Has it got, it's got the method
of how do you deal with
196
00:11:34 --> 00:11:37
a differential equation.
197
00:11:37 --> 00:11:41
By the way, if this could be
a difference equation, too.
198
00:11:41 --> 00:11:42
That could be u(x+h)-u(x-h).
199
00:11:42 --> 00:11:46
200
00:11:46 --> 00:11:51
Shall I just emphasize that
that wouldn't have been any
201
00:11:51 --> 00:11:59
harder? u(x+h)-u(x-h)/2h,
let me take, for example,
202
00:11:59 --> 00:12:03
plus u equals delta.
203
00:12:03 --> 00:12:09
This is like I've taken a
center difference there,
204
00:12:09 --> 00:12:10
instead of a derivative.
205
00:12:10 --> 00:12:12
How would you write
a solution to that?
206
00:12:12 --> 00:12:18
Same method.
207
00:12:18 --> 00:12:19
So what would the method be?
208
00:12:19 --> 00:12:21
I'd take transforms.
209
00:12:21 --> 00:12:25
So here this would have
coefficient c_k, this would
210
00:12:25 --> 00:12:26
have coefficients 1/2pi.
211
00:12:28 --> 00:12:30
And what would be the
coefficients of these guys?
212
00:12:30 --> 00:12:31
Well, there's a 1/2h.
213
00:12:33 --> 00:12:35
So what are the Fourier
coefficients of u(x+h)?
214
00:12:35 --> 00:12:38
215
00:12:38 --> 00:12:42
Just to see that all these
things, I've got constant
216
00:12:42 --> 00:12:44
coefficients here.
217
00:12:44 --> 00:12:46
Why shouldn't Fourier work?
218
00:12:46 --> 00:12:53
So if it's u(x+h), so there's
an e to the something
219
00:12:53 --> 00:12:54
factor, right?
220
00:12:54 --> 00:12:58
When I've shifted a function
then in the transform I
221
00:12:58 --> 00:13:00
see an e to the i thing.
222
00:13:00 --> 00:13:01
So let's just see.
223
00:13:01 --> 00:13:14
If u(x), so if u(x) is the sum
of c_k*e^(ikx), then u(x+h)
224
00:13:14 --> 00:13:21
would be the sum of
c_k*e^ik(x+h), right?
225
00:13:21 --> 00:13:21
No.
226
00:13:21 --> 00:13:23
Nothing surprising there.
227
00:13:23 --> 00:13:25
So now I'm seeing the factor.
228
00:13:25 --> 00:13:29
There's c_k, it's multiplied
by the factor e^(ikh).
229
00:13:29 --> 00:13:34
230
00:13:34 --> 00:13:43
So this is the same c_k, but
an e^(ikh), from that one.
231
00:13:43 --> 00:13:46
And this would be the minus,
and this would shift
232
00:13:46 --> 00:13:47
the other way.
233
00:13:47 --> 00:13:50
So it would be the same c_k,
and probably e^(-ikh).
234
00:13:52 --> 00:13:54
If I did that right.
235
00:13:54 --> 00:13:55
Yeah.
236
00:13:55 --> 00:13:59
Anyway, once again we have
still linear problem.
237
00:13:59 --> 00:14:04
This is c_k times
something equals 1/2pi.
238
00:14:04 --> 00:14:08
239
00:14:08 --> 00:14:11
By the way, this thing is
what I would call the
240
00:14:11 --> 00:14:12
transfer function.
241
00:14:12 --> 00:14:14
You see that language?
242
00:14:14 --> 00:14:16
We haven't used that
language much.
243
00:14:16 --> 00:14:21
Or at all, it's sort of
more of a systems theory,
244
00:14:21 --> 00:14:23
control theory word.
245
00:14:23 --> 00:14:29
But hey, we're always doing
the same correct thing.
246
00:14:29 --> 00:14:35
That's the thing that transfers
the input to the output.
247
00:14:35 --> 00:14:38
So maybe the transfer
function is one over.
248
00:14:38 --> 00:14:41
Maybe I include the division
in the transfer function.
249
00:14:41 --> 00:14:48
So finally I get c_k is this
1/2pi guy, divided by whatever
250
00:14:48 --> 00:14:53
that was, 1/2h times that
number, minus 1/2h
251
00:14:53 --> 00:14:55
of that plus one.
252
00:14:55 --> 00:14:57
Whatever is multiplying
c_k there.
253
00:14:57 --> 00:14:59
It's a transfer function.
254
00:14:59 --> 00:14:59
Yeah.
255
00:14:59 --> 00:15:04
So there's another word which
we could have, and should
256
00:15:04 --> 00:15:06
have, used before today.
257
00:15:06 --> 00:15:09
But here it is.
258
00:15:09 --> 00:15:12
OK, so that's some thoughts
about that question.
259
00:15:12 --> 00:15:14
Any other direction to go?
260
00:15:14 --> 00:15:15
AUDIENCE: [INAUDIBLE]
261
00:15:15 --> 00:15:17
PROFESSOR STRANG:
Yes. go ahead.
262
00:15:17 --> 00:15:18
Yeah.
263
00:15:18 --> 00:15:22
AUDIENCE: [INAUDIBLE]
264
00:15:22 --> 00:15:24
PROFESSOR STRANG:
Lab problem 12, OK.
265
00:15:24 --> 00:15:28
4.5, Fourier
integrals, number 12.
266
00:15:28 --> 00:15:30
Oh yeah, maybe that was also.
267
00:15:30 --> 00:15:33
A question was raised
about was it right on
268
00:15:33 --> 00:15:42
the posted solution.
269
00:15:42 --> 00:15:45
So maybe not.
270
00:15:45 --> 00:15:46
Let's see what we want to do.
271
00:15:46 --> 00:15:50
Yeah I just thought that was
pretty - I could remember the
272
00:15:50 --> 00:15:51
day I thought of this equation.
273
00:15:51 --> 00:15:55
Integral of u minus derivative
of u equal delta(x).
274
00:15:55 --> 00:16:02
I thought that's kind of
a cool looking equation.
275
00:16:02 --> 00:16:05
But I don't know that I've
ever solved it, before.
276
00:16:05 --> 00:16:15
So integral of u, actually
yeah, I look cheerful, but so
277
00:16:15 --> 00:16:18
I'm finishing the fourth
edition of the linear
278
00:16:18 --> 00:16:19
algebra book.
279
00:16:19 --> 00:16:21
Introduction to Linear Algebra.
280
00:16:21 --> 00:16:23
So I love writing.
281
00:16:23 --> 00:16:27
I've done all the writing and
now comes the only horrible
282
00:16:27 --> 00:16:30
part, solving all those
stupid exercises.
283
00:16:30 --> 00:16:36
So that, you may think of me in
sunny Singapore or somewhere,
284
00:16:36 --> 00:16:40
but what I'll be doing
is not on the beach.
285
00:16:40 --> 00:16:47
It's in an office somewhere,
doing Problem 1.1.1, and
286
00:16:47 --> 00:16:50
two, three, it's not life.
287
00:16:50 --> 00:16:55
I mean, it's.
288
00:16:55 --> 00:16:57
But, OK.
289
00:16:57 --> 00:16:59
Anyway, that looks like
a pretty good equation.
290
00:16:59 --> 00:17:03
And certainly I'm going
to transform it.
291
00:17:03 --> 00:17:06
So I can see a little
question coming up.
292
00:17:06 --> 00:17:09
So am in the Fourier
integral world, yeah.
293
00:17:09 --> 00:17:15
So I should you use this
notation. u hat of k, right?
294
00:17:15 --> 00:17:22
Over ik, is that the right
thing when I've integrated?
295
00:17:22 --> 00:17:23
Yeah.
296
00:17:23 --> 00:17:24
That's made it smoother.
297
00:17:24 --> 00:17:28
That's dragging the
function down.
298
00:17:28 --> 00:17:32
And I guess the one point
that's a little tricky
299
00:17:32 --> 00:17:38
there is always, are
you dividing by zero.
300
00:17:38 --> 00:17:41
Because zero is one
of the frequencies.
301
00:17:41 --> 00:17:46
And so I sort of need u hat
of zero to be zero for this.
302
00:17:46 --> 00:17:50
So it's looking slightly
dangerous at k=0, but otherwise
303
00:17:50 --> 00:17:55
it's certainly improving things
by speeding up the decay rate.
304
00:17:55 --> 00:17:59
And then this would be minus ik
u hat of k, as we just said.
305
00:17:59 --> 00:18:04
And this would be the one,
maybe it's a one in the Fourier
306
00:18:04 --> 00:18:07
integral world, where the 2pi
went in a different point.
307
00:18:07 --> 00:18:11
So I think that's, yeah.
308
00:18:11 --> 00:18:14
So I would now just, the
usual thing was the
309
00:18:14 --> 00:18:15
transfer function.
310
00:18:15 --> 00:18:21
1/(ik)-ik, so that's
my transfer function.
311
00:18:21 --> 00:18:23
That's multiplying u hat.
312
00:18:23 --> 00:18:27
Maybe I can simplify that
by putting it all over
313
00:18:27 --> 00:18:30
ik, one minus ik.
314
00:18:30 --> 00:18:35
Is it plus k?
315
00:18:35 --> 00:18:41
Does that look good, or does
that look bad? k squared?
316
00:18:41 --> 00:18:44
Does that look better?
317
00:18:44 --> 00:18:53
So I'm wondering, so over here
is k squared over ik, is that
318
00:18:53 --> 00:18:54
the same as the minus ik?
319
00:18:54 --> 00:18:57
320
00:18:57 --> 00:18:58
Yeah.
321
00:18:58 --> 00:19:01
Yeah, bring it up and the
minus i squared is one
322
00:19:01 --> 00:19:02
and I've got k squared.
323
00:19:02 --> 00:19:04
Yeah, so this looks good.
324
00:19:04 --> 00:19:08
So that's then dividing
by it gives me because,
325
00:19:08 --> 00:19:10
I'm dividing the one.
326
00:19:10 --> 00:19:14
So u hat k is one over this.
327
00:19:14 --> 00:19:17
So it's the transfer function.
328
00:19:17 --> 00:19:19
One plus k squared.
329
00:19:19 --> 00:19:25
OK.
330
00:19:25 --> 00:19:28
I don't know whether I
had in mind to go any
331
00:19:28 --> 00:19:29
further than this.
332
00:19:29 --> 00:19:32
Did you get this far, or what?
333
00:19:32 --> 00:19:34
AUDIENCE: [INAUDIBLE]
334
00:19:34 --> 00:19:36
PROFESSOR STRANG: It said that?
335
00:19:36 --> 00:19:40
Oh jeez.
336
00:19:40 --> 00:19:43
I bitterly regret
saying these things.
337
00:19:43 --> 00:19:45
Yeah, OK.
338
00:19:45 --> 00:19:46
I don't know u(x).
339
00:19:46 --> 00:19:48
Did anybody have any luck?
340
00:19:48 --> 00:19:50
You did.
341
00:19:50 --> 00:19:53
Did I say it with a derivative?
342
00:19:53 --> 00:19:56
Oh one, over one plus
k squared, oh yeah.
343
00:19:56 --> 00:20:01
One over one plus k squared,
we know what to do.
344
00:20:01 --> 00:20:06
And then i k will
take its derivative.
345
00:20:06 --> 00:20:09
So what do we do if it's one
over one plus k squared?
346
00:20:09 --> 00:20:12
That's the one where, that
was the guy which was
347
00:20:12 --> 00:20:18
the e^(-x), right?
348
00:20:18 --> 00:20:20
Oh, it's just one side.
349
00:20:20 --> 00:20:22
No, two sides, isn't it?
350
00:20:22 --> 00:20:23
Two sides, yeah.
351
00:20:23 --> 00:20:25
Yeah, we just got a corner.
352
00:20:25 --> 00:20:26
Yeah, OK.
353
00:20:26 --> 00:20:28
That's looking good.
354
00:20:28 --> 00:20:35
And now maybe I need to divide
by two. a is one here but
355
00:20:35 --> 00:20:37
there is a division by two.
356
00:20:37 --> 00:20:39
Yeah, that looks good.
357
00:20:39 --> 00:20:44
OK, and now I take -
This function, its
358
00:20:44 --> 00:20:47
transform has that.
359
00:20:47 --> 00:20:49
And now if I want to multiply
by ik I have to take
360
00:20:49 --> 00:20:50
the derivative.
361
00:20:50 --> 00:20:54
So I believe with your help
here, that the derivative
362
00:20:54 --> 00:20:58
of that is the same.
363
00:20:58 --> 00:21:03
Can I just take the derivative
while we're looking at it?
364
00:21:03 --> 00:21:08
I mean e^x is the most - you
know, that was created to
365
00:21:08 --> 00:21:10
take its derivative, right?
366
00:21:10 --> 00:21:12
So its derivative is itself.
367
00:21:12 --> 00:21:15
The derivative of
this is a minus.
368
00:21:15 --> 00:21:22
So I think that maybe that's
the answer. e^x over two,
369
00:21:22 --> 00:21:25
coming up to 1/2, I guess.
370
00:21:25 --> 00:21:29
And then - oh, well the
picture won't look, because
371
00:21:29 --> 00:21:30
of that minus sign.
372
00:21:30 --> 00:21:31
Yeah.
373
00:21:31 --> 00:21:31
Huh.
374
00:21:31 --> 00:21:39
Good, because that minus sign
it's minus e^(-x) over two.
375
00:21:39 --> 00:21:41
So it starts at minus 1/2.
376
00:21:41 --> 00:21:43
So there's a jump of one.
377
00:21:43 --> 00:21:45
Or drop of one.
378
00:21:45 --> 00:21:51
Is that right, there should
be a drop of one in u hat?
379
00:21:51 --> 00:21:53
Probably.
380
00:21:53 --> 00:21:54
Yeah, yeah.
381
00:21:54 --> 00:21:58
The integral will be smooth,
at zero where the delta
382
00:21:58 --> 00:22:00
function's hitting us.
383
00:22:00 --> 00:22:02
I have a minus sign
and a derivative.
384
00:22:02 --> 00:22:06
But yeah, doesn't
that look good?
385
00:22:06 --> 00:22:11
If the derivative has a delta,
the function has a jump.
386
00:22:11 --> 00:22:15
And with that minus sign,
and with one delta the jump
387
00:22:15 --> 00:22:17
should be a drop of one.
388
00:22:17 --> 00:22:17
Which is that.
389
00:22:17 --> 00:22:18
Yeah.
390
00:22:18 --> 00:22:21
And the integral minus
the derivative should
391
00:22:21 --> 00:22:23
be otherwise zero.
392
00:22:23 --> 00:22:29
So again this is not
unrelated to exam questions.
393
00:22:29 --> 00:22:33
I did all this stuff, and
I really should check,
394
00:22:33 --> 00:22:34
did I get it right?
395
00:22:34 --> 00:22:37
Does that solve the
differential equation?
396
00:22:37 --> 00:22:39
I mean, you could say of
course it does if you took
397
00:22:39 --> 00:22:40
all these steps right.
398
00:22:40 --> 00:22:42
But it's wise to check.
399
00:22:42 --> 00:22:44
Plug it in.
400
00:22:44 --> 00:22:46
OK, it has the drop of one.
401
00:22:46 --> 00:22:49
So it deals correctly with
the delta because, the
402
00:22:49 --> 00:22:52
derivative has a delta.
403
00:22:52 --> 00:22:55
And things match.
404
00:22:55 --> 00:22:58
And now what else do
I have to check?
405
00:22:58 --> 00:23:00
I have to check out all
the other points, right?
406
00:23:00 --> 00:23:03
I've just checked that
yep, this answer is
407
00:23:03 --> 00:23:05
good at the jump.
408
00:23:05 --> 00:23:06
The tricky point.
409
00:23:06 --> 00:23:09
But now what about all
the other points?
410
00:23:09 --> 00:23:14
So where the delta is zero, I
should just check that the
411
00:23:14 --> 00:23:18
integral of u, so what
is the integral of u?
412
00:23:18 --> 00:23:24
Let me just check for the
points left of the origin.
413
00:23:24 --> 00:23:29
So I'm just going to
look at this part.
414
00:23:29 --> 00:23:31
Yeah, x negative.
415
00:23:31 --> 00:23:36
So its integral, what's the
integral of e^x over two?
416
00:23:36 --> 00:23:42
I guess e^x over two, right?
417
00:23:42 --> 00:23:43
Minus?
418
00:23:43 --> 00:23:45
And what's the
derivative of e^x?
419
00:23:45 --> 00:23:49
Oh, the derivative is also
e^x over two and I get
420
00:23:49 --> 00:23:51
zero, or x negative.
421
00:23:51 --> 00:23:53
As I want to.
422
00:23:53 --> 00:23:57
Because the delta is
zero in that left side.
423
00:23:57 --> 00:23:58
And similarly on
the right side.
424
00:23:58 --> 00:23:59
It should be OK.
425
00:23:59 --> 00:24:03
And then the key point was
that the jump was right.
426
00:24:03 --> 00:24:04
Yeah, so thank you.
427
00:24:04 --> 00:24:06
That's good.
428
00:24:06 --> 00:24:06
Yes please.
429
00:24:06 --> 00:24:09
AUDIENCE: [INAUDIBLE]
430
00:24:09 --> 00:24:11
PROFESSOR STRANG: What, sorry?
431
00:24:11 --> 00:24:14
Why did I take the derivative?
432
00:24:14 --> 00:24:18
What will be the
derivative of this?
433
00:24:18 --> 00:24:19
Yes, there is.
434
00:24:19 --> 00:24:20
That's right.
435
00:24:20 --> 00:24:22
What would be the
derivative of this?
436
00:24:22 --> 00:24:23
Huh, yeah, good question.
437
00:24:23 --> 00:24:27
What's the derivative
of this answer.
438
00:24:27 --> 00:24:33
So it's that, on the left. and
then there's a delta function,
439
00:24:33 --> 00:24:36
a minus delta, right,
because it's dropped down.
440
00:24:36 --> 00:24:39
And then the derivative
of that is e^(-x).
441
00:24:40 --> 00:24:43
Yeah, so if I graph
the derivative, cool.
442
00:24:43 --> 00:24:50
I graph the derivative, it
looks like the function.
443
00:24:50 --> 00:24:56
It has a spike
going down there.
444
00:24:56 --> 00:25:00
At the origin, and then the
derivative here is positive.
445
00:25:00 --> 00:25:01
Right?
446
00:25:01 --> 00:25:03
This function's coming up.
447
00:25:03 --> 00:25:05
It's e(-x) over two.
448
00:25:05 --> 00:25:06
It's positive, it's there.
449
00:25:06 --> 00:25:08
Yeah.
450
00:25:08 --> 00:25:11
Oh wow, that's a nice graph.
451
00:25:11 --> 00:25:17
So that's the derivative,
this is the graph of du/dx.
452
00:25:17 --> 00:25:19
Yeah, thanks.
453
00:25:19 --> 00:25:23
And similarly I could graph the
integral, the integral - well,
454
00:25:23 --> 00:25:26
would you want to
see the integral?
455
00:25:26 --> 00:25:28
Probably not.
456
00:25:28 --> 00:25:30
Maybe, yeah.
457
00:25:30 --> 00:25:34
The integral would probably
look pretty much just like that
458
00:25:34 --> 00:25:37
but without the delta, yeah.
459
00:25:37 --> 00:25:40
Cool, isn't that nice?
460
00:25:40 --> 00:25:44
It's artistic.
461
00:25:44 --> 00:25:46
That's the derivative, and
the integral is the same
462
00:25:46 --> 00:25:48
thing without the delta.
463
00:25:48 --> 00:25:52
And when you subtract,
you get the delta.
464
00:25:52 --> 00:25:57
OK, you sure you guys don't
want to come to Singapore
465
00:25:57 --> 00:26:01
and help with these
problem solutions?
466
00:26:01 --> 00:26:03
There's plenty for everybody.
467
00:26:03 --> 00:26:04
OK.
468
00:26:04 --> 00:26:07
Alright, so another question.
469
00:26:07 --> 00:26:10
Yeah.
470
00:26:10 --> 00:26:10
2005.
471
00:26:10 --> 00:26:11
AUDIENCE: [INAUDIBLE]
472
00:26:11 --> 00:26:14
PROFESSOR STRANG: OK, alright.
473
00:26:14 --> 00:26:14
Yeah, which one?
474
00:26:14 --> 00:26:19
AUDIENCE: [INAUDIBLE]
475
00:26:19 --> 00:26:22
PROFESSOR STRANG: OK, I'll just
read out the whole question.
476
00:26:22 --> 00:26:26
Yeah.
477
00:26:26 --> 00:26:32
So this is 2005, Problem 3.
478
00:26:32 --> 00:26:37
OK it's a half hat function.
479
00:26:37 --> 00:26:41
This is Fourier integral,
because it's on the whole line.
480
00:26:41 --> 00:26:46
And it's half a hat for
between zero and one.
481
00:26:46 --> 00:26:47
The function is coming down.
482
00:26:47 --> 00:26:53
And then otherwise it's zero.
483
00:26:53 --> 00:26:55
So its graph, done.
484
00:26:55 --> 00:26:56
Graph its derivative.
485
00:26:56 --> 00:27:01
OK, let's graph the
derivative of that function.
486
00:27:01 --> 00:27:04
Derivative is certainly
zero along there.
487
00:27:04 --> 00:27:07
Then the derivative
is minus one here.
488
00:27:07 --> 00:27:09
And then the
derivative is zero.
489
00:27:09 --> 00:27:16
So the derivative looks to
me like it's that, OK.
490
00:27:16 --> 00:27:21
So yeah, that function
- is that right?
491
00:27:21 --> 00:27:21
No.
492
00:27:21 --> 00:27:24
It's not right.
493
00:27:24 --> 00:27:25
There's a delta.
494
00:27:25 --> 00:27:28
That was the tricky
part of the problem.
495
00:27:28 --> 00:27:33
Right, there's a delta in
the derivative, right here
496
00:27:33 --> 00:27:35
because of that jump.
497
00:27:35 --> 00:27:37
There's a delta.
498
00:27:37 --> 00:27:37
OK.
499
00:27:37 --> 00:27:39
Good.
500
00:27:39 --> 00:27:42
Now, is that better?
501
00:27:42 --> 00:27:43
That's good.
502
00:27:43 --> 00:27:44
OK.
503
00:27:44 --> 00:27:47
What's its transform?
504
00:27:47 --> 00:27:52
The transform of
this derivative.
505
00:27:52 --> 00:27:54
What's the Fourier transform?
506
00:27:54 --> 00:27:56
Well, I guess we got two parts.
507
00:27:56 --> 00:27:58
So this is my function u'(x).
508
00:27:59 --> 00:28:00
The derivative.
509
00:28:00 --> 00:28:04
Now, what's u' transform?
510
00:28:04 --> 00:28:07
So it should be a
function of k, now.
511
00:28:07 --> 00:28:09
Alright, what do you think?
512
00:28:09 --> 00:28:12
So it's the sum of two things.
513
00:28:12 --> 00:28:17
That delta, which Fourier
transforms to one.
514
00:28:17 --> 00:28:25
And this guy down below between
zero and one, which was like
515
00:28:25 --> 00:28:30
the one we did today, we've got
the integral from zero to one.
516
00:28:30 --> 00:28:33
Now it happens to be a
minus one e^(-ikx)dx.
517
00:28:33 --> 00:28:37
518
00:28:37 --> 00:28:44
And that, of course, we can do.
519
00:28:44 --> 00:28:47
OK Oh, there's solutions here.
520
00:28:47 --> 00:28:49
AUDIENCE: [INAUDIBLE]
521
00:28:49 --> 00:28:55
PROFESSOR STRANG: Yeah.
522
00:28:55 --> 00:28:58
So OK, so I think
we're doing alright.
523
00:28:58 --> 00:29:03
This expression is familiar.
524
00:29:03 --> 00:29:04
It's (1-e^(-ik))/ik.
525
00:29:04 --> 00:29:09
526
00:29:09 --> 00:29:12
Is that right?
527
00:29:12 --> 00:29:13
We have a minus.
528
00:29:13 --> 00:29:18
Is the minus, yeah, I think
the minus looks good.
529
00:29:18 --> 00:29:23
And yeah, at least, that's
here and it's probably got
530
00:29:23 --> 00:29:26
the minus signs correct.
531
00:29:26 --> 00:29:27
So why was it?
532
00:29:27 --> 00:29:30
I plugged in x=1, and that's
why I got an e^(-ik).
533
00:29:32 --> 00:29:34
And then I plugged in
x=0 and that's where
534
00:29:34 --> 00:29:36
that one came from.
535
00:29:36 --> 00:29:42
OK, so but your question
was about - that was it?
536
00:29:42 --> 00:29:43
Oh, I see.
537
00:29:43 --> 00:29:45
OK.
538
00:29:45 --> 00:29:49
Right, and then the next part
asked about the transform
539
00:29:49 --> 00:29:51
of the original guy.
540
00:29:51 --> 00:29:53
The transform of
the original guy.
541
00:29:53 --> 00:30:00
Now, what would be the
transform of the original guy?
542
00:30:00 --> 00:30:11
The original u.
543
00:30:11 --> 00:30:17
Right, OK, yeah.
544
00:30:17 --> 00:30:21
The reason I'm sort of
stuttering is that if I'm going
545
00:30:21 --> 00:30:25
to integrate - I mean, what
are you going to tell me?
546
00:30:25 --> 00:30:27
What's a quick way to
find the transform of
547
00:30:27 --> 00:30:30
the original u now?
548
00:30:30 --> 00:30:31
Divide by ik.
549
00:30:31 --> 00:30:32
Divide by ik.
550
00:30:33 --> 00:30:36
And then I'm all ready
to do that except I'm
551
00:30:36 --> 00:30:37
worried that k=0.
552
00:30:39 --> 00:30:41
But it should come
out alright, now.
553
00:30:41 --> 00:30:44
So I have to hope that
this thing comes out
554
00:30:44 --> 00:30:48
to be zero at k=0.
555
00:30:48 --> 00:30:51
556
00:30:51 --> 00:30:54
Can you see that it does?
557
00:30:54 --> 00:30:57
What is, yeah?
558
00:30:57 --> 00:30:59
This is a good point.
559
00:30:59 --> 00:31:03
What does u hat at
zero represent?
560
00:31:03 --> 00:31:05
Have you thought about that?
561
00:31:05 --> 00:31:08
If I look at the Fourier
transform, which I've got here.
562
00:31:08 --> 00:31:10
So let me write what
I've got here.
563
00:31:10 --> 00:31:17
I've got here that Fourier
transform of this function.
564
00:31:17 --> 00:31:20
And if I take the Fourier
transform of any function, and
565
00:31:20 --> 00:31:22
I look at zero frequency.
566
00:31:22 --> 00:31:24
What am I seeing?
567
00:31:24 --> 00:31:25
The average, right.
568
00:31:25 --> 00:31:27
I'm saying the average.
569
00:31:27 --> 00:31:32
What's the average value
of that function?
570
00:31:32 --> 00:31:33
We never thought about that.
571
00:31:33 --> 00:31:35
But that's fun.
572
00:31:35 --> 00:31:37
What's the average of this guy?
573
00:31:37 --> 00:31:40
If I integrate over
the whole line.
574
00:31:40 --> 00:31:42
Do I get zero?
575
00:31:42 --> 00:31:43
Yes.
576
00:31:43 --> 00:31:48
Because I get the integral of
the delta part gives me a one.
577
00:31:48 --> 00:31:51
But the integral of this box
part gives me a minus one.
578
00:31:51 --> 00:31:56
So the integral, so this does
equal zero at k=0, and I can
579
00:31:56 --> 00:32:05
safely do that division and
get, so now u hat of k is this
580
00:32:05 --> 00:32:07
expression divided by ik.
581
00:32:07 --> 00:32:11
So it's one minus this thing.
582
00:32:11 --> 00:32:12
I'm just copying.
583
00:32:12 --> 00:32:16
Minus ik over ik, all
that divided by ik.
584
00:32:16 --> 00:32:20
And I can, of course,
maneuver that some more
585
00:32:20 --> 00:32:23
to make it look better.
586
00:32:23 --> 00:32:27
OK, these are good questions,
because it's giving me a few
587
00:32:27 --> 00:32:31
functions to do the
standard rules.
588
00:32:31 --> 00:32:39
Shifting, integrating,
differentiating.
589
00:32:39 --> 00:32:40
Oh, here's a Christmas present.
590
00:32:40 --> 00:32:42
What does that mean?
591
00:32:42 --> 00:32:47
Is the convolution of this
function with itself
592
00:32:47 --> 00:32:52
the whole hat?
593
00:32:52 --> 00:32:53
What is that, a
Christmas present?
594
00:32:53 --> 00:32:57
Is the convolution,
the convolution of
595
00:32:57 --> 00:33:00
that with itself.
596
00:33:00 --> 00:33:05
So first point, we don't want
to compute a convolution.
597
00:33:05 --> 00:33:07
You may have noticed, you have
not computed convolutions
598
00:33:07 --> 00:33:09
or some things like this.
599
00:33:09 --> 00:33:13
It's not a lot of fun.
600
00:33:13 --> 00:33:15
Because you have that
integral to do.
601
00:33:15 --> 00:33:19
And you'd have to separate out
the parts where it's this and
602
00:33:19 --> 00:33:22
the part where it's this and
the parts where it's that.
603
00:33:22 --> 00:33:23
It's not nice.
604
00:33:23 --> 00:33:27
Much better to multiply
in the transform domain.
605
00:33:27 --> 00:33:28
Much better.
606
00:33:28 --> 00:33:34
OK, so if I multiply in the
transform domain, I don't
607
00:33:34 --> 00:33:40
think I get the Fourier
transform of the hat.
608
00:33:40 --> 00:33:43
Of the whole hat.
609
00:33:43 --> 00:33:47
But can you give me a
convincing argument for why -
610
00:33:47 --> 00:33:51
this is here I'm in the
transform domain and I'm going
611
00:33:51 --> 00:33:59
to convolve with itself, so I'm
just going to square it.
612
00:33:59 --> 00:34:05
With the decay rate, what's
the decay rate as it is now?
613
00:34:05 --> 00:34:09
The decay rate is, yes.
614
00:34:09 --> 00:34:11
The decay rate is a good key.
615
00:34:11 --> 00:34:14
So what is the decay
rate right now?
616
00:34:14 --> 00:34:15
Here's a function with a jump.
617
00:34:15 --> 00:34:18
So I know that even though has
a slightly messy looking
618
00:34:18 --> 00:34:21
transform, I know
the decay rate.
619
00:34:21 --> 00:34:24
It's what, with a jump is 1/k.
620
00:34:25 --> 00:34:28
OK, so this must be, and I
sort of do see a k down
621
00:34:28 --> 00:34:30
here, so that's sensible.
622
00:34:30 --> 00:34:35
OK, now, if I convolve, then
I multiply in this domain.
623
00:34:35 --> 00:34:38
So that would give me the
k squared, as you said.
624
00:34:38 --> 00:34:43
So could that be the
transform of the whole hat?
625
00:34:43 --> 00:34:50
Does the whole hat have
a - ooh, could be.
626
00:34:50 --> 00:34:53
I'm pretty sure
the answer's no.
627
00:34:53 --> 00:34:54
Yeah, no way.
628
00:34:54 --> 00:34:59
Yeah, I think that
that's the best answer.
629
00:34:59 --> 00:35:03
I've lost the reason.
630
00:35:03 --> 00:35:06
Because I'm getting the
k squared to k rate.
631
00:35:06 --> 00:35:10
If I convolve something with
itself, that would give me a k
632
00:35:10 --> 00:35:14
squared, and the decay rate for
the transform of the
633
00:35:14 --> 00:35:16
hat is a k squared.
634
00:35:16 --> 00:35:18
But they wouldn't
be the same, yeah.
635
00:35:18 --> 00:35:21
OK, yeah, I won't.
636
00:35:21 --> 00:35:22
Yes, please.
637
00:35:22 --> 00:35:30
AUDIENCE: [INAUDIBLE]
638
00:35:30 --> 00:35:30
PROFESSOR STRANG: Right.
639
00:35:30 --> 00:35:32
It's just convention.
640
00:35:32 --> 00:35:34
The difference was that
that was the Fourier
641
00:35:34 --> 00:35:36
integral problem.
642
00:35:36 --> 00:35:39
Where we put, you might have
noticed that the 2pi goes on
643
00:35:39 --> 00:35:46
the other - and this was the
Fourier series problem.
644
00:35:46 --> 00:35:47
That was its only difference.
645
00:35:47 --> 00:35:50
So it's just a convention.
646
00:35:50 --> 00:35:54
And maybe other people would
choose a different convention.
647
00:35:54 --> 00:36:01
So that is a slight wiggle in
the presentation that the 2pi
648
00:36:01 --> 00:36:08
in the Fourier integral section
got moved to the transform
649
00:36:08 --> 00:36:09
in the other direction.
650
00:36:09 --> 00:36:12
Yeah, good.
651
00:36:12 --> 00:36:13
Yes, thanks.
652
00:36:13 --> 00:36:16
AUDIENCE: [INAUDIBLE]
653
00:36:16 --> 00:36:19
PROFESSOR STRANG: Oh, OK.
654
00:36:19 --> 00:36:24
I'm really just going by this
rule that if there is a jump
655
00:36:24 --> 00:36:29
in the function, and nothing
worse, then the decay rate is
656
00:36:29 --> 00:36:35
1/k, and if there's a jump
in the derivative, as
657
00:36:35 --> 00:36:37
there would be here.
658
00:36:37 --> 00:36:40
So that has a jump in slope,
then the decay rate is one
659
00:36:40 --> 00:36:42
over k squared and so on.
660
00:36:42 --> 00:36:44
But those are the main cases.
661
00:36:44 --> 00:36:50
Yeah, so I'm not using
anything deep there.
662
00:36:50 --> 00:36:54
And by the way, I realize here
that one MATLAB problem that I
663
00:36:54 --> 00:36:59
didn't assign this semester but
it's quite fun, is to actually
664
00:36:59 --> 00:37:03
see the Gibbs phenomenon.
665
00:37:03 --> 00:37:09
We know the terms in the
Fourier series, with a jump.
666
00:37:09 --> 00:37:16
And if you computed - make it
the periodic case, if you've
667
00:37:16 --> 00:37:20
computed the terms in the
Fourier series they would stay
668
00:37:20 --> 00:37:27
near here and then I'd see
the famous Gibbs stuff here.
669
00:37:27 --> 00:37:32
So that - you know, it's quite
pleasant to see the printout
670
00:37:32 --> 00:37:37
of the Gibbs phenomenon.
671
00:37:37 --> 00:37:42
Not getting any shorter, just
moving closer to the jump as
672
00:37:42 --> 00:37:44
you increase the
number of terms.
673
00:37:44 --> 00:37:48
Yeah, it's quite interesting.
674
00:37:48 --> 00:37:50
But couldn't do everything.
675
00:37:50 --> 00:37:51
Yeah, ready for more.
676
00:37:51 --> 00:37:54
Any questions?
677
00:37:54 --> 00:37:55
Well, these are good.
678
00:37:55 --> 00:37:57
Yes, thanks.
679
00:37:57 --> 00:37:58
AUDIENCE: [INAUDIBLE]
680
00:37:58 --> 00:37:59
PROFESSOR STRANG: With
cyclic convolutions.
681
00:37:59 --> 00:38:13
OK, what could I do with
cyclic convolutions?
682
00:38:13 --> 00:38:17
Let's see.
683
00:38:17 --> 00:38:20
What should I do with
cyclic convolutions?
684
00:38:20 --> 00:38:30
I better make a little space.
685
00:38:30 --> 00:38:31
Let me make a little
space and think.
686
00:38:31 --> 00:38:43
Anybody got a suggested problem
for a cyclic convolution?
687
00:38:43 --> 00:38:50
I mean, one type of problem
would certainly be if I gave
688
00:38:50 --> 00:39:02
you a cyclic convolution and
asked for - I mean, the direct
689
00:39:02 --> 00:39:06
way would be to say OK, what's
the cyclic convolution of
690
00:39:06 --> 00:39:14
(1, 4, 2), cyclically with
(2, 1, 3) or something.
691
00:39:14 --> 00:39:17
But you could - that's
just a calculation, so
692
00:39:17 --> 00:39:19
you would get that.
693
00:39:19 --> 00:39:22
I won't discuss that further.
694
00:39:22 --> 00:39:28
Now suppose I ask it with
the unknown, the thing
695
00:39:28 --> 00:39:30
that I don't know here?
696
00:39:30 --> 00:39:31
So let me do that.
697
00:39:31 --> 00:39:40
Shall I say three,
yeah. (3, 1, 1).
698
00:39:40 --> 00:39:49
Cyclically convolved with some
unknown (x_0, x_1, x_2) equals
699
00:39:49 --> 00:39:51
some right-hand side, what
should we take for the
700
00:39:51 --> 00:39:57
right-hand side? (1,
1, 1), for example?
701
00:39:57 --> 00:40:05
OK, I'll erase the top one.
702
00:40:05 --> 00:40:14
OK, how would you go with that?
703
00:40:14 --> 00:40:17
So that's three by
three, three unknowns.
704
00:40:17 --> 00:40:20
It's linear, so somehow I
could write that as three
705
00:40:20 --> 00:40:23
equations in three unknowns.
706
00:40:23 --> 00:40:28
The question is how do I get
some insight into this.
707
00:40:28 --> 00:40:31
And when I'm seeing a
convolution, what's my
708
00:40:31 --> 00:40:34
immediate thought?
709
00:40:34 --> 00:40:36
Transform.
710
00:40:36 --> 00:40:39
That's what this
month is all about.
711
00:40:39 --> 00:40:42
November.
712
00:40:42 --> 00:40:46
So I'll take the
DFT of everything.
713
00:40:46 --> 00:40:51
So what's the discrete Fourier
transform of (1, 1, 1)?
714
00:40:51 --> 00:40:54
Up there is going to
be the transform.
715
00:40:54 --> 00:40:59
So let me call it x_0
hat, x_1 hat, x_2 hat.
716
00:40:59 --> 00:41:01
Just to emphasize.
717
00:41:01 --> 00:41:07
So n is three here,
in this problem.
718
00:41:07 --> 00:41:12
Or maybe c_0, c_1, c_2
you might prefer.
719
00:41:12 --> 00:41:13
Whatever.
720
00:41:13 --> 00:41:17
And now this is going to be
a multiplication, so it's
721
00:41:17 --> 00:41:20
in MATLAB notation will
be a dot star, right?
722
00:41:20 --> 00:41:25
And so now I have to - these
are the known ones, so what's
723
00:41:25 --> 00:41:28
the discrete Fourier transform
of (1, 1, 1), the cyclic
724
00:41:28 --> 00:41:31
convolution of that?
725
00:41:31 --> 00:41:33
It's (1, 0, 0) is it?
726
00:41:33 --> 00:41:35
Or it's (3, 0, 0)?
727
00:41:35 --> 00:41:43
Is it three - I guess when I
asked that question, you're
728
00:41:43 --> 00:41:44
open to two answers.
729
00:41:44 --> 00:41:51
You could either be going from
x space to frequency space
730
00:41:51 --> 00:41:52
or frequency to the other.
731
00:41:52 --> 00:41:57
Because I haven't told
you what space we're in.
732
00:41:57 --> 00:41:59
OK, let's go with that one.
733
00:41:59 --> 00:42:07
How did you get to there?
734
00:42:07 --> 00:42:10
You multiplied by the three
by three Fourier matrix,
735
00:42:10 --> 00:42:12
is that what you did?
736
00:42:12 --> 00:42:15
Aright, so now can
you do that here?
737
00:42:15 --> 00:42:18
So separately I have to do a
little multiplication of the
738
00:42:18 --> 00:42:20
three by three Fourier matrix.
739
00:42:20 --> 00:42:22
One, one - sorry, let
me get it right.
740
00:42:22 --> 00:42:27
So I'm going to transform this.
741
00:42:27 --> 00:42:28
Make a little space here.
742
00:42:28 --> 00:42:29
Transform that.
743
00:42:29 --> 00:42:37
So it's 1, 1, 1, 1, 1 and this
is w, w squared, w squared and
744
00:42:37 --> 00:42:40
w to the fourth, which
is the same as w.
745
00:42:40 --> 00:42:44
OK, and now I want to
multiply that by (3, 1, 1).
746
00:42:44 --> 00:42:47
And get the answer there.
747
00:42:47 --> 00:42:50
OK, so what do I have?
748
00:42:50 --> 00:42:52
Five.
749
00:42:52 --> 00:42:53
And what is that other one?
750
00:42:53 --> 00:42:54
Oh, boy.
751
00:42:54 --> 00:42:59
What is three plus
w plus w squared?
752
00:42:59 --> 00:43:02
Anybody know that one?
753
00:43:02 --> 00:43:07
Well, I do know what one
plus w plus w squared is.
754
00:43:07 --> 00:43:09
It is zero.
755
00:43:09 --> 00:43:12
Why is one plus w
plus w squared zero?
756
00:43:12 --> 00:43:15
So I claim that if I have
three plus, I think the
757
00:43:15 --> 00:43:17
answer there is two.
758
00:43:17 --> 00:43:20
And here I think I've also got
three, w, w squared, I think if
759
00:43:20 --> 00:43:24
it was a one the answer
would be a zero.
760
00:43:24 --> 00:43:26
But it's a three, so I
still have two there.
761
00:43:26 --> 00:43:29
I think that's probably right.
762
00:43:29 --> 00:43:32
Can I just remind you why?
763
00:43:32 --> 00:43:38
There's one, there's w,
and there's w squared
764
00:43:38 --> 00:43:40
and they add to zero.
765
00:43:40 --> 00:43:42
Yeah, yeah.
766
00:43:42 --> 00:43:47
For many reasons.
767
00:43:47 --> 00:43:51
That's a very, very,
handy property.
768
00:43:51 --> 00:43:57
That the sum of the nth roots
of one add to zero, OK,
769
00:43:57 --> 00:43:58
so now where I am I?
770
00:43:58 --> 00:44:04
So I got (5, 2, 2).
771
00:44:04 --> 00:44:05
So I've transformed.
772
00:44:05 --> 00:44:09
Now what do I do next?
773
00:44:09 --> 00:44:11
I divide.
774
00:44:11 --> 00:44:18
So this is telling me that this
is three separate equations.
775
00:44:18 --> 00:44:23
There's three, two,
x_1 hat is zero.
776
00:44:23 --> 00:44:24
Right?
777
00:44:24 --> 00:44:26
No, x_2 hat.
778
00:44:26 --> 00:44:28
And 2x_2 hat.
779
00:44:28 --> 00:44:31
Sorry, one was right.
780
00:44:31 --> 00:44:35
And 2x_2 hat is zero.
781
00:44:35 --> 00:44:40
I'm just doing the
multiplication as
782
00:44:40 --> 00:44:41
I'm supposed to.
783
00:44:41 --> 00:44:45
In the other space it's
at each, for each
784
00:44:45 --> 00:44:46
component separately.
785
00:44:46 --> 00:44:52
So now I know x_0 hat is 3/5,
and the others are zero.
786
00:44:52 --> 00:44:55
So now I know what
these numbers are.
787
00:44:55 --> 00:45:01
Oh, yeah. (3/5, 0, 0).
788
00:45:01 --> 00:45:08
And now what?
789
00:45:08 --> 00:45:11
I mean, that statement's
clearly true, right?
790
00:45:11 --> 00:45:14
Point star means five times
this gives me three.
791
00:45:14 --> 00:45:17
Two times zero gives me zero,
two times zero gives zero.
792
00:45:17 --> 00:45:21
So I'm golden, but
I'm not finished.
793
00:45:21 --> 00:45:26
So that's not x,
that's its transform.
794
00:45:26 --> 00:45:29
So what do I have to do?
795
00:45:29 --> 00:45:30
Go back.
796
00:45:30 --> 00:45:36
OK, so now can you
tell me what this is?
797
00:45:36 --> 00:45:41
Up to a factor of 1/n, or
n, which we may have to
798
00:45:41 --> 00:45:45
figure out separately.
799
00:45:45 --> 00:45:50
So apart from this possibly
appearing, and probably
800
00:45:50 --> 00:45:56
appearing, give me an idea of
what's the inverse transform of
801
00:45:56 --> 00:46:01
that? (3/5, 3/5, 3/5), good.
802
00:46:01 --> 00:46:06
3/5, all three times.
803
00:46:06 --> 00:46:16
OK, and now I can check
to see if it's right.
804
00:46:16 --> 00:46:22
And again, I have not attempted
to get the 1/n's right, I've
805
00:46:22 --> 00:46:23
just waited to the end.
806
00:46:23 --> 00:46:26
So now I'll just do this
convolution and see if I
807
00:46:26 --> 00:46:28
really do get (1, 1, 1).
808
00:46:28 --> 00:46:30
So can we do that convolution?
809
00:46:30 --> 00:46:36
What's the typical,
say, this guy.
810
00:46:36 --> 00:46:40
Where does that come from
in the convolution?
811
00:46:40 --> 00:46:45
It comes from this, times what?
812
00:46:45 --> 00:46:46
This.
813
00:46:46 --> 00:46:48
That's the constant, right?
814
00:46:48 --> 00:46:49
And that's the constant.
815
00:46:49 --> 00:46:51
Contributes to the constant.
816
00:46:51 --> 00:46:55
And what other products
contribute to the constant?
817
00:46:55 --> 00:47:00
So we're really seeing -
thank you for this question.
818
00:47:00 --> 00:47:13
So this one multiplies
which one?
819
00:47:13 --> 00:47:16
The answer is, this one
multiplies this guy.
820
00:47:16 --> 00:47:18
You may say don't worry
me about it because
821
00:47:18 --> 00:47:19
they're the same.
822
00:47:19 --> 00:47:21
But it's nice to know.
823
00:47:21 --> 00:47:23
So this one
multiplies this one.
824
00:47:23 --> 00:47:25
And this guy
multiplies this guy.
825
00:47:25 --> 00:47:27
And we'll remember
in a minute why.
826
00:47:27 --> 00:47:30
And then you add them up,
that's what convolution is.
827
00:47:30 --> 00:47:38
So I'm getting 9/5+3/5+3/5
is 15/5, so that's three.
828
00:47:38 --> 00:47:43
So what's your conclusion here?
829
00:47:43 --> 00:47:45
I should divide
by three, right?
830
00:47:45 --> 00:47:46
So I should divide by three.
831
00:47:46 --> 00:47:48
So that's easy to do.
832
00:47:48 --> 00:47:49
Those will all be one.
833
00:47:49 --> 00:47:52
Yeah, OK.
834
00:47:52 --> 00:47:54
So that's the right
answer, right?
835
00:47:54 --> 00:47:59
And then I check
that it's right.
836
00:47:59 --> 00:48:04
So let me just close by
remembering why was it -
837
00:48:04 --> 00:48:08
so I'm doing up just a
forward convolution here.
838
00:48:08 --> 00:48:11
Solving the equation
was a deconvolution,
839
00:48:11 --> 00:48:12
figuring out that x.
840
00:48:12 --> 00:48:16
But now that I've got it and
I'm just checking, that's just
841
00:48:16 --> 00:48:19
do it forward and see if
you got the right answer.
842
00:48:19 --> 00:48:24
And so we're just remembering
what it means to
843
00:48:24 --> 00:48:30
do a convolution.
844
00:48:30 --> 00:48:34
So you remember the point
about convolutions.
845
00:48:34 --> 00:48:41
That if I'm convolving (a, b,
c) cyclically with (d, e,
846
00:48:41 --> 00:48:45
f), that's a constant term.
847
00:48:45 --> 00:48:48
Just tell me again, now, what's
the constant terms in this?
848
00:48:48 --> 00:48:55
It's ad plus b times what?
849
00:48:55 --> 00:49:01
This is the first term
in the convolution.
850
00:49:01 --> 00:49:06
You remember that, the day
we brought convolutions
851
00:49:06 --> 00:49:07
into the course?
852
00:49:07 --> 00:49:13
What does b multiply? f, why f.
853
00:49:13 --> 00:49:21
Because b is the coefficient of
w, if I'm taking polynomials,
854
00:49:21 --> 00:49:25
this is a plus bw plus cw
squared, and this corresponds
855
00:49:25 --> 00:49:29
to d plus ew plus fw squared.
856
00:49:29 --> 00:49:32
And I'm looking
for the constant.
857
00:49:32 --> 00:49:37
So I get an a times ad gives
me a constant. bw times
858
00:49:37 --> 00:49:43
what? times f w squared
gives me the w cubed.
859
00:49:43 --> 00:49:47
Which is the one so that bf
here and then there's a ce.
860
00:49:48 --> 00:49:55
Because c w squared multiplying
ew gives me the w cubed,
861
00:49:55 --> 00:49:57
the one which folds in.
862
00:49:57 --> 00:50:01
Yeah, OK, so anyway.
863
00:50:01 --> 00:50:04
That's a quick reminder
of cyclic convolutions.
864
00:50:04 --> 00:50:07
That was a good
question to ask, yeah.
865
00:50:07 --> 00:50:12
You know, I fully realize this
especially this third topic,
866
00:50:12 --> 00:50:14
lots of things like this.
867
00:50:14 --> 00:50:17
Straightforward if you
have lots of time, but
868
00:50:17 --> 00:50:20
we had to keep moving.
869
00:50:20 --> 00:50:26
So remembering what cyclic
convolution was is a
870
00:50:26 --> 00:50:29
good chance to do it.