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[MUSIC PLAYING]

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PROFESSOR: In the
last lecture, we

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introduced an alternative
set of algorithms

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for computation of the
fast Fourier transform

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00:00:57,690 --> 00:01:00,930
or rather, computation of the
discrete Fourier transform.

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And we refer to these as
the decimation in frequency

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form of the algorithm.

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00:01:06,990 --> 00:01:10,900
As you recall, we
developed the decimation

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in frequency form
of the algorithm

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by essentially organizing
the computation of the even

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numbered DFT points and the odd
numbered DFT points separately

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00:01:24,390 --> 00:01:27,820
and then proceeded
in a similar manner.

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00:01:27,820 --> 00:01:31,290
This corresponded to breaking
the computation into two

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00:01:31,290 --> 00:01:35,400
and over two point DFT's
proceeded in a similar manner

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00:01:35,400 --> 00:01:38,910
to decompose the computation
of these separately

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00:01:38,910 --> 00:01:41,937
into their even numbered points
and odd numbered points, et

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00:01:41,937 --> 00:01:42,660
cetera.

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00:01:42,660 --> 00:01:48,690
And the flow graph that resulted
from this procedure was--

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00:01:48,690 --> 00:01:51,460
as I've shown here--

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00:01:51,460 --> 00:01:55,540
which is one form,
then, of the decimation

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00:01:55,540 --> 00:02:01,890
in frequency form of the fast
Fourier transform algorithm.

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00:02:01,890 --> 00:02:08,699
It begins with the data sorted
in normal order and proceeds

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00:02:08,699 --> 00:02:12,090
to the data sorted in
big reversed order.

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00:02:12,090 --> 00:02:14,940
And as I stressed
last time, it also

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corresponds to an in-place
computation because of the way

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00:02:19,560 --> 00:02:22,410
in which the butterflies
are laid out.

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00:02:22,410 --> 00:02:26,190
In particular, the output
nodes for each butterfly

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00:02:26,190 --> 00:02:30,300
are horizontally adjacent
to the input nodes.

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00:02:30,300 --> 00:02:35,160
Now this is similar in
a number of respects

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00:02:35,160 --> 00:02:41,070
to one of the decimation in
time forms of the FFT algorithm.

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00:02:41,070 --> 00:02:44,440
Although, as I also
indicated last time,

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00:02:44,440 --> 00:02:47,280
the basic structure of
the butterfly computation

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00:02:47,280 --> 00:02:48,570
is somewhat different.

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00:02:48,570 --> 00:02:53,510
In particular, for
this set of algorithms,

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00:02:53,510 --> 00:02:56,520
the multiplication
by powers of W

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00:02:56,520 --> 00:02:59,850
is implemented at the
output of the butterfly,

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00:02:59,850 --> 00:03:02,970
as opposed to the
decimation in time forms,

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00:03:02,970 --> 00:03:05,430
where the multiplication
by powers of W

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00:03:05,430 --> 00:03:09,300
are implemented at the
input to the butterfly.

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00:03:09,300 --> 00:03:12,900
However, there is a
very close relationship

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00:03:12,900 --> 00:03:17,850
between this flow graph
and the original flow graph

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00:03:17,850 --> 00:03:23,160
that we implemented for the
decimation in time algorithm.

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00:03:23,160 --> 00:03:25,620
In particular, the
two flow graphs

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00:03:25,620 --> 00:03:28,630
are transposes of each other.

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00:03:28,630 --> 00:03:32,280
Now I remind you that in
one of the early lectures,

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00:03:32,280 --> 00:03:36,360
when we introduced the notation
of flow graphs and some

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00:03:36,360 --> 00:03:39,090
of the properties
of flow graphs,

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00:03:39,090 --> 00:03:42,780
one of the properties of flow
graphs that I mentioned--

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00:03:42,780 --> 00:03:44,400
although we didn't derive it--

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00:03:44,400 --> 00:03:48,120
was the fact that
transposition of a flow graph

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00:03:48,120 --> 00:03:51,670
doesn't change the input
output characteristics.

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00:03:51,670 --> 00:03:56,550
So then in fact, if we took
this flow graph, changed

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00:03:56,550 --> 00:03:59,550
the direction of
all of the arrows,

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00:03:59,550 --> 00:04:02,640
treated this set of
nodes as the input nodes

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00:04:02,640 --> 00:04:05,820
and this set of nodes
as the output nodes,

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00:04:05,820 --> 00:04:09,660
then since the flow graph,
as it's indicated here,

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computes the discrete
Fourier transform,

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00:04:12,510 --> 00:04:15,120
the transposed flow
graph would also

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00:04:15,120 --> 00:04:17,709
compute the discrete
Fourier transform.

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00:04:17,709 --> 00:04:21,269
And in fact, the transposition
of this flow graph

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00:04:21,269 --> 00:04:25,820
is identical to the
first flow graph--

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00:04:25,820 --> 00:04:28,260
FFT flow graph--
that we introduced,

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00:04:28,260 --> 00:04:32,070
which was the initial
decimation in time form.

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00:04:32,070 --> 00:04:37,410
And that I refer
you to again, here.

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00:04:37,410 --> 00:04:41,670
So this is the
decimation in time form

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00:04:41,670 --> 00:04:44,490
of the algorithm with
the input bit reversed

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and the output in normal order.

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00:04:47,220 --> 00:04:55,710
This is the transposition
of this flow graph, which

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is the decimation in frequency
form of the algorithm

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00:04:59,340 --> 00:05:02,880
with the input in normal
order and the output in bit

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00:05:02,880 --> 00:05:04,830
reversed order.

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00:05:04,830 --> 00:05:08,430
The fact that these are
related through transposition

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00:05:08,430 --> 00:05:12,420
of the flow graph isn't a
particularly important point

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from a practical point of view.

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But it is a useful
piece of insight

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into the relationship between
these various algorithms.

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00:05:22,321 --> 00:05:25,860
All right, now just as we
did with the decimation

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00:05:25,860 --> 00:05:29,580
in time forms of
the algorithm, we

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00:05:29,580 --> 00:05:34,590
can rearrange the decimation in
frequency form of the algorithm

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to either arrange the output to
be in normal order, as opposed

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00:05:41,790 --> 00:05:44,040
to having the input
in normal order,

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00:05:44,040 --> 00:05:47,010
or rearrange it so that both
the input and output are

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00:05:47,010 --> 00:05:50,910
in normal order or
rearrange it as we will,

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00:05:50,910 --> 00:05:54,090
so that we can utilize
sequential memory rather

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00:05:54,090 --> 00:05:56,880
than random access memory.

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00:05:56,880 --> 00:06:00,210
To rearrange the flow
graph so that the output is

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00:06:00,210 --> 00:06:03,330
in normal order, we
would, again, simply

95
00:06:03,330 --> 00:06:07,920
take horizontal lines
and re-sort them

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00:06:07,920 --> 00:06:13,680
so that the output nodes are
sorted in a normal order.

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00:06:13,680 --> 00:06:18,150
And the flow graph
that results is

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00:06:18,150 --> 00:06:24,120
a decimation in frequency
form of the algorithm,

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00:06:24,120 --> 00:06:30,030
for which the outputs
are in normal order.

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00:06:30,030 --> 00:06:35,520
But the inputs are in
bit reversed order.

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00:06:35,520 --> 00:06:38,880
Again, as this flow
graph has been sorted,

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00:06:38,880 --> 00:06:42,180
it is an in-place computation.

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00:06:42,180 --> 00:06:45,450
In-place, again, because
the butterflies are laid out

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00:06:45,450 --> 00:06:50,280
in such a way that they involve
horizontally adjacent nodes.

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00:06:50,280 --> 00:06:52,950
So this is an
in-place computation

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00:06:52,950 --> 00:06:55,860
with the input in
bit reversed order

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00:06:55,860 --> 00:06:59,730
and the output in normal order.

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00:06:59,730 --> 00:07:10,170
Again this is a transposition
of one of the flow graphs

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00:07:10,170 --> 00:07:12,480
that we discussed in
the previous lecture.

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00:07:12,480 --> 00:07:16,860
In particular, it is a
transposition of the flow graph

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00:07:16,860 --> 00:07:20,880
for which, if you imagine this
flow graph being transposed

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00:07:20,880 --> 00:07:23,430
so that the direction of
the arrows are reversed

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00:07:23,430 --> 00:07:26,400
and this is the input,
then it corresponds

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00:07:26,400 --> 00:07:29,880
to the decimation in time form
of the algorithm for which

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00:07:29,880 --> 00:07:34,410
the input is in normal order
and the output comes out

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00:07:34,410 --> 00:07:36,280
in bit reversed order.

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00:07:36,280 --> 00:07:42,720
And that is the flow graph
that I have indicated here.

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00:07:42,720 --> 00:07:47,550
So this is a decimation in time
form of the algorithm, which

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00:07:47,550 --> 00:07:52,710
is the transposition of
the decimation in frequency

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00:07:52,710 --> 00:07:58,230
form of the algorithm
that we just looked at.

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00:07:58,230 --> 00:08:05,320
All right, that corresponds
then to two possible decimation

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00:08:05,320 --> 00:08:07,560
in frequency algorithms.

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00:08:07,560 --> 00:08:10,060
One in which the input
is in normal order,

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00:08:10,060 --> 00:08:11,850
the output is bit reversed.

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00:08:11,850 --> 00:08:14,970
This one in which the input
is bit reversed and the output

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00:08:14,970 --> 00:08:17,190
is in normal order.

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00:08:17,190 --> 00:08:21,000
We can also rearrange
this so that the input is

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00:08:21,000 --> 00:08:24,300
in normal order and the
output is in normal order.

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00:08:24,300 --> 00:08:28,110
And the flow graph that
results when we implement that

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00:08:28,110 --> 00:08:35,500
is similar to the corresponding
decimation in time flow graph.

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00:08:35,500 --> 00:08:40,530
In fact, it is a transposition
of that flow graph.

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00:08:40,530 --> 00:08:45,300
And just as we found in the
decimation in time algorithm,

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00:08:45,300 --> 00:08:49,440
for this case, although the
input is sorted in normal order

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00:08:49,440 --> 00:08:52,390
and the output comes
out in normal order,

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00:08:52,390 --> 00:08:56,190
the indexing is
generally complicated

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00:08:56,190 --> 00:08:58,920
as we proceed from
stage to stage.

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00:08:58,920 --> 00:09:01,980
And furthermore, it
does not correspond

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00:09:01,980 --> 00:09:05,820
to an in-place computation,
because of the fact

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00:09:05,820 --> 00:09:09,990
that the outputs
of the butterflies

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00:09:09,990 --> 00:09:14,040
are no longer horizontally
adjacent to the inputs

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00:09:14,040 --> 00:09:15,810
to the butterfly.

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00:09:15,810 --> 00:09:19,050
So generally, this flow
graph and its decimation

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00:09:19,050 --> 00:09:26,500
in time counterpart are not
practically very important.

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00:09:26,500 --> 00:09:30,180
A final rearrangement
of the flow graph

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00:09:30,180 --> 00:09:36,390
is the rearrangement
which permits

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00:09:36,390 --> 00:09:41,490
the use of sequential memory
rather than random access

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00:09:41,490 --> 00:09:42,570
memory.

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00:09:42,570 --> 00:09:46,350
And this, then,
is the flow graph

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00:09:46,350 --> 00:09:49,020
that is the counterpart
to the flow graph

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00:09:49,020 --> 00:09:51,390
that we discussed last
time, which I referred to

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00:09:51,390 --> 00:09:56,460
as the Singleton algorithm,
or I attributed to Singleton.

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00:09:56,460 --> 00:10:02,210
This is a variation of
that type of algorithm,

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00:10:02,210 --> 00:10:07,350
in which now, again, we
recognize that the indexing is

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00:10:07,350 --> 00:10:12,120
identical from stage to stage.

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00:10:12,120 --> 00:10:14,370
The input is in normal order.

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00:10:14,370 --> 00:10:17,950
And the output comes out
in bit reversed order.

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00:10:17,950 --> 00:10:21,270
And let me just
compare this for you

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00:10:21,270 --> 00:10:23,700
with the corresponding
flow graph

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00:10:23,700 --> 00:10:29,370
that we had last time,
which I've indicated here.

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00:10:29,370 --> 00:10:32,040
That has the input
in bit reversed

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00:10:32,040 --> 00:10:34,200
order, the output
in normal order.

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00:10:34,200 --> 00:10:37,440
In fact, the transposition
of this flow graph

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00:10:37,440 --> 00:10:40,410
is the decimation
in frequency form

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00:10:40,410 --> 00:10:44,400
of the algorithm, for which
the input is in normal order,

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00:10:44,400 --> 00:10:47,790
and the output is in
bit reversed order.

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00:10:47,790 --> 00:10:52,020
All right, so the indexing is
identical from stage to stage.

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00:10:52,020 --> 00:10:55,200
And just as we
had previously, we

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00:10:55,200 --> 00:10:59,010
can implement this
algorithm by utilizing

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00:10:59,010 --> 00:11:02,610
sequential memory as opposed
to random access memory.

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00:11:02,610 --> 00:11:06,570
Although the organization of
the memory for the decimation

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00:11:06,570 --> 00:11:10,020
in frequency form of
the Singleton algorithm

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00:11:10,020 --> 00:11:13,230
is somewhat different
than the organization

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00:11:13,230 --> 00:11:16,800
of the memory for the decimation
in time form of the Singleton

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00:11:16,800 --> 00:11:18,420
algorithm.

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00:11:18,420 --> 00:11:22,170
In particular, last time when
we discussed the decimation

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00:11:22,170 --> 00:11:28,680
in time algorithm, we had four
memories, as we will this time.

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00:11:28,680 --> 00:11:30,700
The first half of
the data in memory A,

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00:11:30,700 --> 00:11:32,740
the second half of
the memory in memory

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00:11:32,740 --> 00:11:37,170
B. We went, first of all,
through all of memory A,

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00:11:37,170 --> 00:11:41,640
then through all of memory B,
storing results alternatively

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00:11:41,640 --> 00:11:44,430
in memories C and D.

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00:11:44,430 --> 00:11:48,180
In this case, the strategy
in utilizing the memory

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00:11:48,180 --> 00:11:50,440
is somewhat different.

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00:11:50,440 --> 00:11:57,910
In this case, we store the first
half of the data points, again,

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00:11:57,910 --> 00:12:00,790
in memory A, the
second half of the data

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00:12:00,790 --> 00:12:04,810
points, again, in memory
B. But notice that now

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00:12:04,810 --> 00:12:11,470
in the computation utilizing
the computation for this point,

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00:12:11,470 --> 00:12:16,930
for example, we use the first
output point from memory A

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00:12:16,930 --> 00:12:20,140
and the first output
point from memory b.

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00:12:20,140 --> 00:12:26,230
We store the result in the
first register in memory C

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00:12:26,230 --> 00:12:30,100
and in the second
register in memory C.

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00:12:30,100 --> 00:12:32,710
Then, to compute
the next two points,

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00:12:32,710 --> 00:12:36,386
we take the next
point from memory A

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00:12:36,386 --> 00:12:42,550
and the next point
from memory B,

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00:12:42,550 --> 00:12:45,340
do the appropriate addition
and subtraction, multiplication

196
00:12:45,340 --> 00:12:52,360
by powers of W, and store
those in the next two registers

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00:12:52,360 --> 00:12:55,420
in memory C. For this
particular example,

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00:12:55,420 --> 00:12:58,300
namely n equals eight,
that fills up memory C.

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00:12:58,300 --> 00:13:02,500
And we proceed, likewise,
accessing the input data,

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00:13:02,500 --> 00:13:05,650
alternating between
the two input memories

201
00:13:05,650 --> 00:13:10,570
and storing data first in all
of one of the output memories

202
00:13:10,570 --> 00:13:14,230
and then in all of the second
of the output memories.

203
00:13:14,230 --> 00:13:18,670
And then we proceed in that
fashion from stage to stage.

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00:13:18,670 --> 00:13:21,040
But again, one of
the important aspects

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00:13:21,040 --> 00:13:23,770
is that it utilizes
sequential memory,

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00:13:23,770 --> 00:13:28,180
for example, disk or drum or
tape or shift register memory.

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00:13:32,340 --> 00:13:42,090
OK, so this then is
essentially a picture

208
00:13:42,090 --> 00:13:45,510
of a variety of the
algorithms, which

209
00:13:45,510 --> 00:13:49,620
can be used for the computation
of the discrete Fourier

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00:13:49,620 --> 00:13:53,430
transform, some of
which were derived

211
00:13:53,430 --> 00:13:56,250
on the basis of a
decimation in time argument.

212
00:13:56,250 --> 00:14:00,720
Some were derived on the basis
of a decimation in frequency

213
00:14:00,720 --> 00:14:01,740
argument.

214
00:14:01,740 --> 00:14:05,310
But we saw, in fact, that the
decimation in frequency forms

215
00:14:05,310 --> 00:14:08,580
of the algorithm are,
in terms of flow graph

216
00:14:08,580 --> 00:14:11,610
notation or
interpretation, transposes

217
00:14:11,610 --> 00:14:15,750
of the decimation in time form.

218
00:14:15,750 --> 00:14:20,340
What I would like
to discuss now are

219
00:14:20,340 --> 00:14:23,820
a few of the
computational issues that

220
00:14:23,820 --> 00:14:28,980
are involved in the implementing
the fast Fourier transform

221
00:14:28,980 --> 00:14:35,850
algorithm and return, also, to
a brief discussion of at least

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00:14:35,850 --> 00:14:39,120
one situation in which
there is a preference

223
00:14:39,120 --> 00:14:45,660
for using decimation in time or
using decimation in frequency.

224
00:14:45,660 --> 00:14:47,910
After we've discussed
some of the computational

225
00:14:47,910 --> 00:14:53,520
considerations, I will then
just very briefly discuss

226
00:14:53,520 --> 00:14:58,310
the generalization of the fast
Fourier transform algorithm

227
00:14:58,310 --> 00:15:01,350
to situations in which
the number of data points

228
00:15:01,350 --> 00:15:04,530
is not a power of
two, but is a more

229
00:15:04,530 --> 00:15:07,260
arbitrary composite number.

230
00:15:07,260 --> 00:15:11,220
Well first of all,
then, let me introduce

231
00:15:11,220 --> 00:15:14,280
a few of the
computational issues

232
00:15:14,280 --> 00:15:18,280
that I would like to
draw your attention to.

233
00:15:18,280 --> 00:15:24,450
The first is-- and it's a point
that I raised in the first

234
00:15:24,450 --> 00:15:27,630
lecture, in which I introduced
the fast Fourier transform

235
00:15:27,630 --> 00:15:28,920
algorithm--

236
00:15:28,920 --> 00:15:32,730
that whereas our discussion
has been directed entirely

237
00:15:32,730 --> 00:15:36,930
toward the computation
of the DFT,

238
00:15:36,930 --> 00:15:40,470
it applies also, in a
very straightforward way,

239
00:15:40,470 --> 00:15:45,120
to a computation
of the inverse DFT.

240
00:15:45,120 --> 00:15:49,110
In particular, we know that
the inverse discrete Fourier

241
00:15:49,110 --> 00:15:54,750
transform relationship is given
by 1 over N times the sum of x

242
00:15:54,750 --> 00:15:58,200
of k W sub-N to the minus NK.

243
00:15:58,200 --> 00:16:02,130
In contrast to the discrete
Fourier transform--

244
00:16:02,130 --> 00:16:04,240
that is the forward transform--

245
00:16:04,240 --> 00:16:07,200
which does not have
the factor of 1 over N

246
00:16:07,200 --> 00:16:14,080
and involves multiplication by
W sub-N to the plus NK, rather

247
00:16:14,080 --> 00:16:18,630
than W sub-N to the minus NK.

248
00:16:18,630 --> 00:16:23,280
To use the algorithms
that we have just

249
00:16:23,280 --> 00:16:27,090
been discussing
for the computation

250
00:16:27,090 --> 00:16:30,690
of the inverse discrete
Fourier transform,

251
00:16:30,690 --> 00:16:34,150
the modification is
relatively straightforward.

252
00:16:34,150 --> 00:16:36,180
First of all, it
involves a factor of 1

253
00:16:36,180 --> 00:16:40,420
over N, which we can
accommodate in a number of ways.

254
00:16:40,420 --> 00:16:43,920
One is, of course, by
shifting the output

255
00:16:43,920 --> 00:16:50,190
or by applying scaling at each
stage of the FFT, for example.

256
00:16:50,190 --> 00:16:54,930
So this factor of 1 over
N is easily accommodated.

257
00:16:54,930 --> 00:17:01,140
And the other difference
is the inclusion

258
00:17:01,140 --> 00:17:05,550
of a minus sign in the
exponent in powers of W

259
00:17:05,550 --> 00:17:09,060
rather than a plus sign
for the forward transform.

260
00:17:09,060 --> 00:17:14,099
That means, in essence,
that these coefficients

261
00:17:14,099 --> 00:17:18,450
are the complex conjugate
of these coefficients.

262
00:17:18,450 --> 00:17:27,720
Because W sub-capital N is e
of the J two pi over capital N.

263
00:17:27,720 --> 00:17:32,340
This minus sign represents
a conjugation of W.

264
00:17:32,340 --> 00:17:36,870
And so one procedure that we
can follow for implementing

265
00:17:36,870 --> 00:17:41,100
the inverse discrete Fourier
transform using all of the flow

266
00:17:41,100 --> 00:17:46,650
graphs that we've talked
about is simply conjugate

267
00:17:46,650 --> 00:17:51,960
the powers of W that are
involved in the computation.

268
00:17:51,960 --> 00:17:55,860
An alternative procedure,
which is basically equivalent,

269
00:17:55,860 --> 00:18:02,070
is to use the flow graphs
as they stand, in which case

270
00:18:02,070 --> 00:18:06,120
the output data,
which is obtained,

271
00:18:06,120 --> 00:18:13,110
is the desired result with
little n replaced by minus n.

272
00:18:13,110 --> 00:18:17,910
So we can either
compute the inverse DFT

273
00:18:17,910 --> 00:18:22,650
from an algorithm, which is
aimed at the forward DFT,

274
00:18:22,650 --> 00:18:26,130
by conjugating the
coefficients or equivalently

275
00:18:26,130 --> 00:18:33,360
by essentially flipping the
output, modulo capital N. So

276
00:18:33,360 --> 00:18:37,140
all of the algorithms that
we have talked about, then,

277
00:18:37,140 --> 00:18:39,390
relate in a very
straightforward manner

278
00:18:39,390 --> 00:18:42,360
to the computation of the
inverse discrete Fourier

279
00:18:42,360 --> 00:18:44,940
transform.

280
00:18:44,940 --> 00:18:47,040
The second issue
is that as we've

281
00:18:47,040 --> 00:18:50,640
seen in essentially
all of the algorithms

282
00:18:50,640 --> 00:18:56,500
of practical interest, they
involve either bit reversal

283
00:18:56,500 --> 00:19:01,300
at the input to the flow graph
or a bit reversal at the output

284
00:19:01,300 --> 00:19:03,496
of the flow graph.

285
00:19:03,496 --> 00:19:09,490
Bit reversal, again, is a
relatively straightforward

286
00:19:09,490 --> 00:19:12,120
thing to implement.

287
00:19:12,120 --> 00:19:17,880
First of all, it is important to
recognize that bit reversal is

288
00:19:17,880 --> 00:19:22,590
an in-place computation, if we
think of it as a computation.

289
00:19:22,590 --> 00:19:29,760
In particular, suppose that we
had the seven data indices zero

290
00:19:29,760 --> 00:19:31,690
through seven.

291
00:19:31,690 --> 00:19:36,990
And we want to rearrange
this data so that the data is

292
00:19:36,990 --> 00:19:40,200
arranged in bit reversed order.

293
00:19:40,200 --> 00:19:45,990
Well that means that, of course,
zero bit reversed is zero,

294
00:19:45,990 --> 00:19:48,300
one bit reversed is four.

295
00:19:48,300 --> 00:19:52,210
But of course, four
bit reversed is one.

296
00:19:52,210 --> 00:19:57,480
So in fact, we would
want to store data

297
00:19:57,480 --> 00:20:04,350
with whose index is one
in storage location four.

298
00:20:04,350 --> 00:20:07,920
But we will also want to
store data whose data index

299
00:20:07,920 --> 00:20:11,700
is four in storage location one.

300
00:20:11,700 --> 00:20:15,240
So in fact, implementing
the bit reversal

301
00:20:15,240 --> 00:20:21,210
can be accomplished in place
by essentially swapping

302
00:20:21,210 --> 00:20:23,260
these two pieces of data.

303
00:20:23,260 --> 00:20:26,440
Similarly of course,
two bit reverses two.

304
00:20:26,440 --> 00:20:28,080
And so that wouldn't move.

305
00:20:28,080 --> 00:20:30,310
Three bit reversed is six.

306
00:20:30,310 --> 00:20:32,760
But six bit reversed is three.

307
00:20:32,760 --> 00:20:36,660
And so again, we can
carry out that interchange

308
00:20:36,660 --> 00:20:38,700
as an in-place computation.

309
00:20:38,700 --> 00:20:43,140
So bit reversal, then, can
be implemented in place.

310
00:20:43,140 --> 00:20:47,760
And consequently, as
we restore the data

311
00:20:47,760 --> 00:20:50,670
in a bit reversed
order, we don't

312
00:20:50,670 --> 00:20:55,050
require double the storage,
since we can carry that out

313
00:20:55,050 --> 00:20:58,080
as an in-place operation.

314
00:20:58,080 --> 00:21:02,100
Second of all, to implement
bit reversal, of course,

315
00:21:02,100 --> 00:21:06,240
to implement bit reversal
in hardware, to obtain a bit

316
00:21:06,240 --> 00:21:08,910
reversed index and
hardware is very simple.

317
00:21:08,910 --> 00:21:13,560
We just simply rearrange
the order of the wires.

318
00:21:13,560 --> 00:21:20,620
To implement a bit reversed
index register or an index

319
00:21:20,620 --> 00:21:25,130
algorithmically is a little
more difficult. But in fact--

320
00:21:25,130 --> 00:21:28,570
and this is discussed in some
more detail in the text--

321
00:21:28,570 --> 00:21:31,450
one of the most
straightforward procedures,

322
00:21:31,450 --> 00:21:36,790
normally, is to
implement a bit reversed

323
00:21:36,790 --> 00:21:42,910
counter so that as we
proceed along with an index

324
00:21:42,910 --> 00:21:47,170
register accessing through
an index in normal order,

325
00:21:47,170 --> 00:21:51,730
we can also run a counter that
runs in bit reversed order

326
00:21:51,730 --> 00:21:56,110
and use, essentially, those two
as index registers to tell us

327
00:21:56,110 --> 00:21:57,910
how to swap the data.

328
00:21:57,910 --> 00:22:00,760
So a bit reversal, in
fact, algorithmically

329
00:22:00,760 --> 00:22:03,160
is relatively straightforward.

330
00:22:03,160 --> 00:22:07,600
It's an interesting point that
it is a somewhat inefficient

331
00:22:07,600 --> 00:22:14,860
procedure, as you'll see if you
try to program bit reversal.

332
00:22:14,860 --> 00:22:17,020
But algorithmically
and conceptually,

333
00:22:17,020 --> 00:22:21,920
it's a fairly straightforward
procedure to implement.

334
00:22:21,920 --> 00:22:24,750
A third computational
issue, which

335
00:22:24,750 --> 00:22:26,970
I would like to draw
your attention to

336
00:22:26,970 --> 00:22:34,830
is the question of obtaining the
coefficients to use in the FFT

337
00:22:34,830 --> 00:22:36,360
computation.

338
00:22:36,360 --> 00:22:39,360
And there are basically
two procedures

339
00:22:39,360 --> 00:22:42,600
that are commonly used.

340
00:22:42,600 --> 00:22:48,270
One, of course, is to store
the coefficients in a table

341
00:22:48,270 --> 00:22:53,220
and then simply access
them as they're needed.

342
00:22:53,220 --> 00:22:56,490
A second, which saves
storage but requires

343
00:22:56,490 --> 00:23:01,590
some computational time, is
to generate the coefficients

344
00:23:01,590 --> 00:23:03,090
recursively.

345
00:23:03,090 --> 00:23:04,950
That is essentially
using an oscillator--

346
00:23:04,950 --> 00:23:06,780
programming an oscillator--

347
00:23:06,780 --> 00:23:10,020
and generating the
coefficients as they're needed.

348
00:23:10,020 --> 00:23:12,810
Along those lines, it's
interesting to observe

349
00:23:12,810 --> 00:23:16,600
that in both the decimation
in time and decimation

350
00:23:16,600 --> 00:23:19,110
in frequency forms
of the algorithm,

351
00:23:19,110 --> 00:23:22,560
as you proceed from
stage to stage,

352
00:23:22,560 --> 00:23:28,170
the powers of W that are
involved in the computation

353
00:23:28,170 --> 00:23:34,380
are powers of a basic
power of W that doubles--

354
00:23:34,380 --> 00:23:35,970
or at least that changes--

355
00:23:35,970 --> 00:23:39,300
in each stage as you go
through the computation.

356
00:23:39,300 --> 00:23:43,380
And this has
implications as you think

357
00:23:43,380 --> 00:23:48,360
of either storing a table of
coefficients and accessing them

358
00:23:48,360 --> 00:23:53,850
or as you think of generating
the coefficients recursively.

359
00:23:53,850 --> 00:23:59,010
The fact that the powers of W
are related from stage to stage

360
00:23:59,010 --> 00:24:04,140
suggests some fairly efficient
procedures for doing this.

361
00:24:04,140 --> 00:24:10,850
One additional
consideration, though,

362
00:24:10,850 --> 00:24:18,950
in obtaining the
coefficients is that

363
00:24:18,950 --> 00:24:21,900
in some of the forms
of the FFT algorithm,

364
00:24:21,900 --> 00:24:29,040
the coefficients are naturally
accessed in a normal order,

365
00:24:29,040 --> 00:24:33,030
whereas in some other forms,
they are naturally accessed

366
00:24:33,030 --> 00:24:35,320
in a bit reversed order.

367
00:24:35,320 --> 00:24:41,250
For example, if we think of
the decimation in frequency

368
00:24:41,250 --> 00:24:45,750
form of the algorithm,
here is the decimation

369
00:24:45,750 --> 00:24:47,910
in frequency form
of the algorithm

370
00:24:47,910 --> 00:24:50,250
with the input in normal
order and the output in bit

371
00:24:50,250 --> 00:24:52,480
reversed order.

372
00:24:52,480 --> 00:24:59,430
Notice that the powers of W
that we have here occur in what

373
00:24:59,430 --> 00:25:01,440
looks like normal order.

374
00:25:01,440 --> 00:25:03,660
At least these powers
are in normal order.

375
00:25:07,830 --> 00:25:13,440
And this is normal input order,
bit reversed output order.

376
00:25:13,440 --> 00:25:17,690
Whereas if we took,
let's say, the decimation

377
00:25:17,690 --> 00:25:20,910
in time form of the algorithm--

378
00:25:20,910 --> 00:25:27,240
where we have normally ordered
input and bit reversed output--

379
00:25:27,240 --> 00:25:32,460
in that case, if you look
at these powers of W,

380
00:25:32,460 --> 00:25:33,960
they're not in normal order.

381
00:25:33,960 --> 00:25:37,860
In fact, what they're in
is bit reversed order.

382
00:25:37,860 --> 00:25:41,490
So in fact, in addition
to the consideration

383
00:25:41,490 --> 00:25:44,850
as to whether the
input is bit reversed

384
00:25:44,850 --> 00:25:49,020
or the output is bit reversed,
in some forms of the algorithm,

385
00:25:49,020 --> 00:25:54,300
the coefficients would tend
to be stored in normal order,

386
00:25:54,300 --> 00:25:57,090
whereas in some other
forms of the algorithm,

387
00:25:57,090 --> 00:25:58,770
the coefficients
would tend to be

388
00:25:58,770 --> 00:26:01,620
stored in a bit reversed order.

389
00:26:01,620 --> 00:26:04,860
Or if we think about
generating the coefficients,

390
00:26:04,860 --> 00:26:11,100
clearly it is simpler to think
of generating coefficients

391
00:26:11,100 --> 00:26:13,200
in normal order
than it is to think

392
00:26:13,200 --> 00:26:17,740
of them as being generated
in bit reversed order.

393
00:26:17,740 --> 00:26:23,920
Now with regard to the question
of the input and output

394
00:26:23,920 --> 00:26:29,590
being bit reversed, one
important area in which

395
00:26:29,590 --> 00:26:31,960
the computation of the DFT--

396
00:26:31,960 --> 00:26:34,300
in other words, the
FFT algorithms--

397
00:26:34,300 --> 00:26:40,540
play a role is in implementing
convolution or correlation.

398
00:26:40,540 --> 00:26:44,150
In that case, we
compute a transform,

399
00:26:44,150 --> 00:26:47,660
multiply by something-- in
the case of a convolution,

400
00:26:47,660 --> 00:26:54,080
we multiply by the transform
of the impulse response--

401
00:26:54,080 --> 00:26:56,680
and then implement
an inverse transform.

402
00:26:56,680 --> 00:27:00,010
Well the fact that there
are two transforms involved

403
00:27:00,010 --> 00:27:05,110
suggests the possibility that
we can organize the computation

404
00:27:05,110 --> 00:27:08,980
in such a way that we
completely avoid bit reversal.

405
00:27:08,980 --> 00:27:12,370
For example, we can
choose an algorithm

406
00:27:12,370 --> 00:27:15,730
for the direct
transform, which utilizes

407
00:27:15,730 --> 00:27:20,020
the input in normal order and
generates the transform in bit

408
00:27:20,020 --> 00:27:21,730
reversed order.

409
00:27:21,730 --> 00:27:24,070
We then simply
have the transform

410
00:27:24,070 --> 00:27:27,880
of the impulse response stored
in bit reversed order, carry

411
00:27:27,880 --> 00:27:31,210
out the multiplication,
and then choose

412
00:27:31,210 --> 00:27:35,410
a form of the algorithm for
the inverse transform, which

413
00:27:35,410 --> 00:27:39,190
has bit reversed
input and results

414
00:27:39,190 --> 00:27:42,200
in a normally ordered output.

415
00:27:42,200 --> 00:27:45,550
So in that case, what we would
have is the forward transform,

416
00:27:45,550 --> 00:27:48,080
normally ordered data
being transformed

417
00:27:48,080 --> 00:27:49,990
to bit reversed data.

418
00:27:49,990 --> 00:27:51,730
And then as the
inverse transform,

419
00:27:51,730 --> 00:27:55,280
bit reversed input and
normally ordered output.

420
00:27:55,280 --> 00:27:59,510
Well even given that, there are
a number of options available.

421
00:27:59,510 --> 00:28:03,190
For example, we
have an algorithm--

422
00:28:03,190 --> 00:28:06,590
the decimation in
time algorithm--

423
00:28:06,590 --> 00:28:09,445
which is normally ordered input.

424
00:28:12,070 --> 00:28:18,550
I'm sorry, normally ordered
input and bit reversed

425
00:28:18,550 --> 00:28:22,330
output, which we could use
as our forward transform.

426
00:28:22,330 --> 00:28:26,230
Although as we just
illustrated, it

427
00:28:26,230 --> 00:28:29,660
involves bit reversed
coefficients.

428
00:28:29,660 --> 00:28:32,680
Well we could think of storing
the coefficients in bit

429
00:28:32,680 --> 00:28:37,210
reversed order and then using
the companion decimation

430
00:28:37,210 --> 00:28:45,010
in time form of the algorithm,
which has bit reversed input

431
00:28:45,010 --> 00:28:51,790
and normally ordered output, to
achieve the inverse transform.

432
00:28:51,790 --> 00:28:54,250
However, in this
case the coefficients

433
00:28:54,250 --> 00:28:56,480
are in normal order.

434
00:28:56,480 --> 00:28:59,920
So if we match up the
algorithms that way,

435
00:28:59,920 --> 00:29:01,690
then we're faced
with the problem

436
00:29:01,690 --> 00:29:04,750
that in the direct transform,
the coefficients would

437
00:29:04,750 --> 00:29:07,870
be stored in bit
reversed order, whereas

438
00:29:07,870 --> 00:29:09,820
in the inverse
transform they would

439
00:29:09,820 --> 00:29:13,030
be stored in normal order.

440
00:29:13,030 --> 00:29:15,960
Well you can ask whether
for the decimation

441
00:29:15,960 --> 00:29:22,360
of frequency form of the
algorithm, we can avoid that.

442
00:29:22,360 --> 00:29:24,910
But in fact, the same
problem arises there.

443
00:29:24,910 --> 00:29:27,010
If we have the
decimation in time

444
00:29:27,010 --> 00:29:30,270
in frequency form of the
algorithm with normal input,

445
00:29:30,270 --> 00:29:34,390
bit reversed output,
then the coefficients

446
00:29:34,390 --> 00:29:37,360
are stored in normal order.

447
00:29:37,360 --> 00:29:44,170
But for the inverse decimation
in frequency transform with bit

448
00:29:44,170 --> 00:29:47,440
reversed input and
normally ordered output,

449
00:29:47,440 --> 00:29:52,060
the coefficients would be
stored in bit reversed order.

450
00:29:52,060 --> 00:29:55,550
The solution is to
match up a decimation

451
00:29:55,550 --> 00:29:59,800
in time form of the algorithm
with a decimation in frequency

452
00:29:59,800 --> 00:30:05,620
form of the algorithm
so that, for example,

453
00:30:05,620 --> 00:30:11,200
we can choose as the
forward transform

454
00:30:11,200 --> 00:30:18,220
the decimation in frequency form
of the algorithm with the input

455
00:30:18,220 --> 00:30:21,880
in normal order, the output
in bit reversed order,

456
00:30:21,880 --> 00:30:23,980
and the coefficients
normally ordered,

457
00:30:23,980 --> 00:30:27,340
which is generally
more convenient.

458
00:30:27,340 --> 00:30:32,650
And then choose for the inverse
transform not the decimation

459
00:30:32,650 --> 00:30:37,690
in frequency algorithm, but the
decimation in time algorithm,

460
00:30:37,690 --> 00:30:41,980
for which the input
is now bit reversed,

461
00:30:41,980 --> 00:30:47,050
the output is in normal
order, and the coefficients

462
00:30:47,050 --> 00:30:49,930
are also in normal order.

463
00:30:49,930 --> 00:30:55,930
So in fact, this suggests
that if we are implementing

464
00:30:55,930 --> 00:31:00,070
a forward transform and
an inverse transform,

465
00:31:00,070 --> 00:31:04,300
generally there are
advantages to matching up

466
00:31:04,300 --> 00:31:08,080
the decimation in time form
and the decimation in frequency

467
00:31:08,080 --> 00:31:12,190
form so that we have similarly
accessed coefficients

468
00:31:12,190 --> 00:31:14,200
in both cases.

469
00:31:14,200 --> 00:31:18,130
Well there are, of
course, a large variety

470
00:31:18,130 --> 00:31:20,350
of other computational
issues to be

471
00:31:20,350 --> 00:31:24,190
considered in implementing
the fast Fourier transform

472
00:31:24,190 --> 00:31:25,390
algorithms.

473
00:31:25,390 --> 00:31:28,240
Some of these are
discussed in the text.

474
00:31:28,240 --> 00:31:32,290
Many of them you will
discover for yourselves

475
00:31:32,290 --> 00:31:35,450
as you try to program
the algorithm.

476
00:31:35,450 --> 00:31:39,400
But hopefully, this
discussion provides

477
00:31:39,400 --> 00:31:44,320
at least some indication of
what some of the strategies are

478
00:31:44,320 --> 00:31:47,560
and some of the issues
are that are involved

479
00:31:47,560 --> 00:31:51,370
in computation of
the FFT of the forms

480
00:31:51,370 --> 00:31:53,790
that we've been talking about.

481
00:31:53,790 --> 00:31:57,830
Now all of this
discussion has been

482
00:31:57,830 --> 00:32:02,490
related to the computation
of the discrete Fourier

483
00:32:02,490 --> 00:32:06,770
transform when the number of
data points is a power of two.

484
00:32:06,770 --> 00:32:10,640
And these algorithms are
referred to as the radix two

485
00:32:10,640 --> 00:32:13,850
forms of the FFT algorithm.

486
00:32:13,850 --> 00:32:17,150
As I indicated in the first
lecture, in which we discussed

487
00:32:17,150 --> 00:32:21,860
the FFT, the FFT, in
fact, is more general

488
00:32:21,860 --> 00:32:25,940
in that it generally
applies when

489
00:32:25,940 --> 00:32:29,360
N is a highly composite number.

490
00:32:29,360 --> 00:32:32,780
We chose it to be
composed of powers of two.

491
00:32:32,780 --> 00:32:36,500
But in fact, there are
a variety of other forms

492
00:32:36,500 --> 00:32:40,350
of the FFT algorithm
for different radices.

493
00:32:40,350 --> 00:32:43,730
And in some cases, there
are some advantages

494
00:32:43,730 --> 00:32:47,690
to be found in using not
a power of two algorithm,

495
00:32:47,690 --> 00:32:51,290
but a different algorithm.

496
00:32:51,290 --> 00:32:53,390
On the other hand,
in many situations,

497
00:32:53,390 --> 00:32:56,810
the disadvantages of that
outweigh the advantages.

498
00:32:56,810 --> 00:33:00,380
However, what I would like to
do is conclude the discussion

499
00:33:00,380 --> 00:33:03,370
of the fast Fourier
transform algorithm

500
00:33:03,370 --> 00:33:12,320
by just very briefly outlining
what the structure of the FFT

501
00:33:12,320 --> 00:33:14,900
algorithm is more generally.

502
00:33:14,900 --> 00:33:19,940
And again, in the text
there are a number

503
00:33:19,940 --> 00:33:24,920
of examples of FFT
algorithms for radices

504
00:33:24,920 --> 00:33:28,350
other than a power of two and
a more complete discussion

505
00:33:28,350 --> 00:33:32,400
than I feel that is appropriate
to go through right now.

506
00:33:32,400 --> 00:33:37,400
However, let me just outline
what some of the issues

507
00:33:37,400 --> 00:33:44,930
are in discussing a more
general radix FFT algorithm.

508
00:33:44,930 --> 00:33:48,620
Generally the computation
of the discrete Fourier

509
00:33:48,620 --> 00:33:51,830
transform using this
class of algorithms

510
00:33:51,830 --> 00:33:55,430
is directed toward
capitalizing on the fact

511
00:33:55,430 --> 00:33:59,120
that the size transform
to be implemented

512
00:33:59,120 --> 00:34:02,840
is a product of numbers.

513
00:34:02,840 --> 00:34:07,880
And it turns out that
the more numbers--

514
00:34:07,880 --> 00:34:10,670
the more terms in
the decomposition--

515
00:34:10,670 --> 00:34:13,580
the greater the efficiency
that can be obtained

516
00:34:13,580 --> 00:34:16,600
in implementing the transform.

517
00:34:16,600 --> 00:34:20,420
In that case, generally one
would think of these numbers

518
00:34:20,420 --> 00:34:23,179
as primes, since
that is the biggest

519
00:34:23,179 --> 00:34:26,030
decomposition of any number
that we can carry out.

520
00:34:26,030 --> 00:34:30,469
But in fact, for the derivation,
it is not a requirement

521
00:34:30,469 --> 00:34:32,960
that the p's be primes.

522
00:34:32,960 --> 00:34:36,170
So let's think of N, then,
as decomposed as a product

523
00:34:36,170 --> 00:34:41,900
of p1 times p2, et
cetera, through p sub-Nu,

524
00:34:41,900 --> 00:34:47,630
which we can alternatively
write as p1 times q1,

525
00:34:47,630 --> 00:34:52,670
where q1 is then represented
by the product of the remaining

526
00:34:52,670 --> 00:34:54,889
terms.

527
00:34:54,889 --> 00:35:01,220
And we can proceed,
essentially, along a root

528
00:35:01,220 --> 00:35:06,830
similar to the decimation in
time algorithms or along a root

529
00:35:06,830 --> 00:35:09,890
similar to the decimation
in frequency algorithms.

530
00:35:09,890 --> 00:35:13,190
Let me just indicate
the strategy paralleling

531
00:35:13,190 --> 00:35:16,640
the decimation in time
form of the algorithm,

532
00:35:16,640 --> 00:35:20,490
as we discussed that
for N, a power of two.

533
00:35:20,490 --> 00:35:25,320
In that case, we decomposed
p1 is equal to two,

534
00:35:25,320 --> 00:35:28,140
and q1 is equal to N over two.

535
00:35:28,140 --> 00:35:31,560
And we decompose the
original sequence

536
00:35:31,560 --> 00:35:34,620
into two sub-sequences,
one consisting

537
00:35:34,620 --> 00:35:37,050
of the even numbered
points, the other consisting

538
00:35:37,050 --> 00:35:39,180
of the odd numbered points.

539
00:35:39,180 --> 00:35:43,170
More generally, we would
decompose the sequence

540
00:35:43,170 --> 00:35:51,820
into a set of sub-sequences
consisting of every p1th point.

541
00:35:51,820 --> 00:35:53,870
So how many sequences are there?

542
00:35:53,870 --> 00:35:56,950
Well, there are p1 sequences.

543
00:35:56,950 --> 00:36:00,370
And the length of
each sequence is q1.

544
00:36:00,370 --> 00:36:05,110
For example, if p1 was equal
to two, and q1 was N over two,

545
00:36:05,110 --> 00:36:07,284
this would be every other point.

546
00:36:07,284 --> 00:36:08,950
That's choosing the
even numbered points

547
00:36:08,950 --> 00:36:10,750
and the odd numbered points.

548
00:36:10,750 --> 00:36:13,720
There would be
two sub-sequences,

549
00:36:13,720 --> 00:36:17,820
each sub-sequence of
length N over two.

550
00:36:17,820 --> 00:36:22,920
More generally, then, if
we had, say, N equal to 12,

551
00:36:22,920 --> 00:36:32,220
and p1 equal to three, we would
generate three sub-sequences,

552
00:36:32,220 --> 00:36:38,970
each consisting of four points
chosen by selecting every p1th,

553
00:36:38,970 --> 00:36:40,590
or every third point.

554
00:36:40,590 --> 00:36:44,700
So one sub sequence, which
I've denoted here by A,

555
00:36:44,700 --> 00:36:48,520
would be this point and this
one and this one and that one.

556
00:36:48,520 --> 00:36:52,080
The second sub-sequence
would be the sub-sequence B,

557
00:36:52,080 --> 00:36:56,320
which is comprised of this
point, this, this, and that.

558
00:36:56,320 --> 00:36:59,640
And the third sub-sequence
would be sub-sequence

559
00:36:59,640 --> 00:37:02,220
C, which is this point,
this point, that point,

560
00:37:02,220 --> 00:37:04,830
and that point.

561
00:37:04,830 --> 00:37:08,190
All right, so we decompose
the original sequence, then,

562
00:37:08,190 --> 00:37:13,020
into p1 sequences,
each of length q1.

563
00:37:13,020 --> 00:37:17,190
Given that, we can
organize the sum involved

564
00:37:17,190 --> 00:37:21,240
in the discrete
Fourier transform

565
00:37:21,240 --> 00:37:26,040
by decomposing the sum
into separate sums,

566
00:37:26,040 --> 00:37:32,600
involving each one of
these p1 sub-sequences.

567
00:37:32,600 --> 00:37:38,360
In particular, if we think
of the argument p1 times

568
00:37:38,360 --> 00:37:46,235
r, as r ranges from
zero to q1 minus one,

569
00:37:46,235 --> 00:37:50,930
we're selecting, with this
argument, every p1th point,

570
00:37:50,930 --> 00:37:54,470
starting with the
zero-eth point.

571
00:37:54,470 --> 00:37:57,890
If we think of the
argument p1 r plus one,

572
00:37:57,890 --> 00:38:00,740
as r runs from zero
to q1 minus one,

573
00:38:00,740 --> 00:38:04,640
we're then collecting
together the terms,

574
00:38:04,640 --> 00:38:08,680
which are every p1th point,
starting with the first point.

575
00:38:08,680 --> 00:38:12,740
Et cetera, we can
proceed to decompose this

576
00:38:12,740 --> 00:38:20,030
into a set of p1 sub-sums,
as I've indicated here.

577
00:38:20,030 --> 00:38:22,460
Or when we combine
these together,

578
00:38:22,460 --> 00:38:27,740
then, we can express
x of k, the DFT,

579
00:38:27,740 --> 00:38:36,000
as a double sum, where we
have the terms involving

580
00:38:36,000 --> 00:38:41,130
the separate sub sequences
and then the multiplication

581
00:38:41,130 --> 00:38:46,260
by a power of W to
combine the computation

582
00:38:46,260 --> 00:38:50,430
on these sub-sequences
together to obtain the DFT.

583
00:38:50,430 --> 00:38:53,940
These terms, of course,
are obtained by the fact

584
00:38:53,940 --> 00:38:57,730
that this power of
W is p1 r plus one.

585
00:38:57,730 --> 00:39:01,200
In the next sum, it will be p1
r plus two, p1 r plus three,

586
00:39:01,200 --> 00:39:02,220
et cetera.

587
00:39:02,220 --> 00:39:07,050
We decompose that into a
product of two powers of W.

588
00:39:07,050 --> 00:39:10,620
And then one of those terms--

589
00:39:10,620 --> 00:39:12,390
since it doesn't
depend on r-- can

590
00:39:12,390 --> 00:39:17,051
be removed from the sum on r
and shows up in this second sum.

591
00:39:17,051 --> 00:39:18,630
Well I would suggest--

592
00:39:18,630 --> 00:39:20,640
that's a somewhat
rapid treatment--

593
00:39:20,640 --> 00:39:22,740
I would suggest
just simply working

594
00:39:22,740 --> 00:39:24,010
through that for an example.

595
00:39:24,010 --> 00:39:28,290
And I think it will be clear how
this original sum is decomposed

596
00:39:28,290 --> 00:39:31,180
into this one.

597
00:39:31,180 --> 00:39:35,140
All right, well now, just as
we did in both the decimation

598
00:39:35,140 --> 00:39:37,480
in time and decimation
in frequency forms

599
00:39:37,480 --> 00:39:41,710
of the algorithm
for powers of two,

600
00:39:41,710 --> 00:39:47,320
we can recognize
this power of W sub-N

601
00:39:47,320 --> 00:39:53,830
as a different power of a W
with a different subscript.

602
00:39:53,830 --> 00:40:00,760
In particular, W sub-N to
the p1 rk, as I have here,

603
00:40:00,760 --> 00:40:06,730
is equal to e to the j two pi
over capital-N times p1 rk.

604
00:40:06,730 --> 00:40:12,310
But N is equal to p1 times q1.

605
00:40:12,310 --> 00:40:15,670
That's the way we
originally decomposed it.

606
00:40:15,670 --> 00:40:19,210
And the p1's cancel out.

607
00:40:19,210 --> 00:40:22,690
And we're left with e to
the j two pi over q1 times

608
00:40:22,690 --> 00:40:29,680
rk or equivalently,
W sub-q1 to the rk.

609
00:40:29,680 --> 00:40:36,490
So substituting this
for W sub-N to the p1 r

610
00:40:36,490 --> 00:40:42,490
k in the previous expression,
what results is then x of k

611
00:40:42,490 --> 00:40:46,960
expressed as a sum from l
equals zero, p1 minus one,

612
00:40:46,960 --> 00:40:50,650
of these powers of
W sub-N. But then

613
00:40:50,650 --> 00:40:56,620
the important thing is that
the inner sum involves q1 point

614
00:40:56,620 --> 00:40:59,240
sequences.

615
00:40:59,240 --> 00:41:02,710
This factor is W
sub-q1 to the rk.

616
00:41:02,710 --> 00:41:09,730
And in fact, this entire
summation is a q1 point DFT.

617
00:41:09,730 --> 00:41:14,790
So pursuing this, then,
what we basically decomposed

618
00:41:14,790 --> 00:41:24,100
the computation into is the
computation of p1 q1 point

619
00:41:24,100 --> 00:41:26,890
DFT's.

620
00:41:26,890 --> 00:41:30,970
We obtained the q1 point
DFT's and then combined them,

621
00:41:30,970 --> 00:41:34,900
according to the
expression as we have here.

622
00:41:34,900 --> 00:41:38,380
This would then lead to the
generalization of the butterfly

623
00:41:38,380 --> 00:41:43,780
computation as we had it for
the radix two algorithms.

624
00:41:43,780 --> 00:41:47,290
And if you count up the
number of multiplies and adds,

625
00:41:47,290 --> 00:41:51,130
again, you'll see that there is
some computational efficiency

626
00:41:51,130 --> 00:41:53,380
that results from doing this.

627
00:41:53,380 --> 00:41:57,580
And then, just as we did with
the radix two algorithms,

628
00:41:57,580 --> 00:42:02,590
we can continue to
decompose these transforms

629
00:42:02,590 --> 00:42:09,310
so that we have q1 is given
by the product p2 times p3

630
00:42:09,310 --> 00:42:11,560
through p sub-Nu.

631
00:42:11,560 --> 00:42:16,990
This product we denote as q2.

632
00:42:16,990 --> 00:42:22,030
And then we can proceed to
decompose these transforms--

633
00:42:22,030 --> 00:42:25,270
these q1 point transforms--

634
00:42:25,270 --> 00:42:30,610
into p2 q2 point transforms.

635
00:42:30,610 --> 00:42:34,570
Then we can decompose the q2
point transforms, et cetera.

636
00:42:34,570 --> 00:42:37,330
If you do this, then
what you'll find--

637
00:42:37,330 --> 00:42:40,030
and the counting is done
a little more carefully

638
00:42:40,030 --> 00:42:42,160
in the text--

639
00:42:42,160 --> 00:42:46,360
what you'll find is that the
number of multiplies and adds

640
00:42:46,360 --> 00:42:50,440
involved in computing the
discrete Fourier transform

641
00:42:50,440 --> 00:42:59,080
this way is proportional to
N times a sum of the factors

642
00:42:59,080 --> 00:43:03,660
involved in the
decomposition of N.

643
00:43:03,660 --> 00:43:07,630
I've indicated an additional
factor here of minus Nu.

644
00:43:07,630 --> 00:43:11,620
Depending on how you choose
to count multiplies and adds,

645
00:43:11,620 --> 00:43:15,130
this factor of minus Nu
either shows up or not.

646
00:43:15,130 --> 00:43:20,660
In fact, to make this
expression consistent with what

647
00:43:20,660 --> 00:43:25,340
we obtained for the
power of two algorithms,

648
00:43:25,340 --> 00:43:27,470
the minus Nu should be in here.

649
00:43:27,470 --> 00:43:30,680
And it's simply a
question of whether you

650
00:43:30,680 --> 00:43:34,730
count or don't count some
multiplications by unity.

651
00:43:34,730 --> 00:43:39,110
Well OK, that's a very quick
treatment-- discussion--

652
00:43:39,110 --> 00:43:43,520
of the generalization of
the radix two algorithms

653
00:43:43,520 --> 00:43:45,800
to more arbitrary radix.

654
00:43:45,800 --> 00:43:49,490
And there, as I indicated,
are some examples

655
00:43:49,490 --> 00:43:52,730
of this, which are
given in the text.

656
00:43:52,730 --> 00:43:55,580
Although, in fact,
what you'll find

657
00:43:55,580 --> 00:43:58,850
is that in most
practical contexts,

658
00:43:58,850 --> 00:44:05,480
it is a radix two algorithm that
is the most efficient to use.

659
00:44:05,480 --> 00:44:09,620
And I indicate,
incidentally, that if you

660
00:44:09,620 --> 00:44:13,370
are faced with the problem
of transforming data, which

661
00:44:13,370 --> 00:44:16,610
is of a length which is not
a power of two, it of course

662
00:44:16,610 --> 00:44:20,180
can always be made to be
of length power of two

663
00:44:20,180 --> 00:44:24,470
by simply augmenting
the sequence with zeros.

664
00:44:24,470 --> 00:44:29,480
Well this concludes, then,
the discussion of the FFT

665
00:44:29,480 --> 00:44:35,120
algorithms, the computation of
the discrete Fourier transform.

666
00:44:35,120 --> 00:44:40,280
And there are, I hope,
a number of issues

667
00:44:40,280 --> 00:44:43,550
which you will
have an opportunity

668
00:44:43,550 --> 00:44:46,700
to dwell on some as you
read through the text

669
00:44:46,700 --> 00:44:50,330
and as your attention will be
drawn to in the study guide.

670
00:44:50,330 --> 00:44:50,870
Thank you.

671
00:44:54,770 --> 00:44:56,920
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