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[MUSIC PLAYING]

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ALAN OPPENHEIM: Last
time, we introduced

10
00:00:50,150 --> 00:00:52,430
the topic of digital
filter design,

11
00:00:52,430 --> 00:00:55,790
and we discussed two
design techniques which

12
00:00:55,790 --> 00:00:59,900
permitted us to obtain digital
filter designs from analog

13
00:00:59,900 --> 00:01:02,360
or continuous time designs.

14
00:01:02,360 --> 00:01:06,770
In particular, the first
technique that we discussed

15
00:01:06,770 --> 00:01:09,860
was basically
mapping differentials

16
00:01:09,860 --> 00:01:13,375
in a linear constant coefficient
differential equation

17
00:01:13,375 --> 00:01:17,000
to differences in a linear
constant coefficient difference

18
00:01:17,000 --> 00:01:18,620
equation.

19
00:01:18,620 --> 00:01:21,320
We found that that
corresponded basically

20
00:01:21,320 --> 00:01:25,070
to a mapping from the
s plane to the z plane.

21
00:01:25,070 --> 00:01:30,110
But we saw also that it
was not a good technique

22
00:01:30,110 --> 00:01:32,190
for two basic reasons.

23
00:01:32,190 --> 00:01:36,740
One is that it didn't correspond
to mapping the frequency

24
00:01:36,740 --> 00:01:41,960
characteristic from the j
omega axis to the unit circle.

25
00:01:41,960 --> 00:01:45,230
And second of all, and
perhaps even more serious,

26
00:01:45,230 --> 00:01:49,370
it didn't result in a mapping
of stable analog filters

27
00:01:49,370 --> 00:01:52,700
to stable digital filters.

28
00:01:52,700 --> 00:01:55,040
The second technique
that we discussed

29
00:01:55,040 --> 00:01:57,830
was the technique of
impulse and invariance,

30
00:01:57,830 --> 00:02:01,160
where basically, the
discrete time filter

31
00:02:01,160 --> 00:02:05,810
is designed by
sampling periodically

32
00:02:05,810 --> 00:02:09,650
the impulse response of
the continuous time filter.

33
00:02:09,650 --> 00:02:13,280
And we saw that what
this corresponded to

34
00:02:13,280 --> 00:02:16,430
was basically a
mapping of the system

35
00:02:16,430 --> 00:02:20,690
function for the continuous
time filter to a system

36
00:02:20,690 --> 00:02:26,330
function for the digital filter,
which preserved the residues

37
00:02:26,330 --> 00:02:32,060
and mapped the poles of the
continuous time filter at poles

38
00:02:32,060 --> 00:02:35,810
at s sub k to poles of
the digital filter at e

39
00:02:35,810 --> 00:02:38,570
to the s sub k capital T.

40
00:02:38,570 --> 00:02:43,040
Well, let me remind you
again of the advantages

41
00:02:43,040 --> 00:02:45,440
and disadvantages
of this technique

42
00:02:45,440 --> 00:02:47,450
by illustrating it
with a simple example.

43
00:02:50,420 --> 00:02:55,690
Let's take the case of a simple,
second order analog filter,

44
00:02:55,690 --> 00:02:58,720
let's say with one
0 and two poles,

45
00:02:58,720 --> 00:03:01,060
so that the analog
system function

46
00:03:01,060 --> 00:03:05,820
is given by S plus a divided
by S plus a squared plus

47
00:03:05,820 --> 00:03:07,900
b squared.

48
00:03:07,900 --> 00:03:10,660
For the impulse invariant
design procedure,

49
00:03:10,660 --> 00:03:15,550
we want to expand this into
a partial fraction expansion.

50
00:03:15,550 --> 00:03:20,350
And in that form, this system
function becomes a 1/2--

51
00:03:20,350 --> 00:03:22,090
1/2 as the residu3--

52
00:03:22,090 --> 00:03:25,120
divided by S plus a plus jb.

53
00:03:25,120 --> 00:03:31,600
The pole is then at minus a
minus jb, plus a residue of 1/2

54
00:03:31,600 --> 00:03:35,050
over S plus a minus
jb, in other words,

55
00:03:35,050 --> 00:03:39,530
a pole at minus a plus jb.

56
00:03:39,530 --> 00:03:40,030
All right.

57
00:03:40,030 --> 00:03:44,860
Well, mapping this analog system
function to the digital system

58
00:03:44,860 --> 00:03:49,060
function, we
maintain the residue

59
00:03:49,060 --> 00:03:53,350
and map the pole
at minus a minus jb

60
00:03:53,350 --> 00:03:56,890
to a pole in the z plane
at e to the minus a capital

61
00:03:56,890 --> 00:04:00,760
T e to the minus
jb capital T. So

62
00:04:00,760 --> 00:04:04,240
the resulting digital
transfer function

63
00:04:04,240 --> 00:04:07,390
has these same
residues, and it has

64
00:04:07,390 --> 00:04:13,330
poles which are at e to the
minus aT e to the minus jgT,

65
00:04:13,330 --> 00:04:18,399
and e to the minus
aT e to the plus jbT.

66
00:04:18,399 --> 00:04:23,320
And this system function,
we can now recombine.

67
00:04:23,320 --> 00:04:25,510
We can recombine the two terms.

68
00:04:25,510 --> 00:04:28,870
And the resulting
digital transfer function

69
00:04:28,870 --> 00:04:31,490
is as I've indicated here.

70
00:04:31,490 --> 00:04:35,380
Now, this illustrates one of
the important points, which

71
00:04:35,380 --> 00:04:40,150
is the fact that the
poles got mapped according

72
00:04:40,150 --> 00:04:43,270
to a mapping which
is of the form z

73
00:04:43,270 --> 00:04:47,290
equals e to the s
capital T. But notice

74
00:04:47,290 --> 00:04:50,770
that, in fact, the 0 didn't
get mapped in the same way.

75
00:04:50,770 --> 00:04:55,450
The 0 at minus a got mapped,
for this particular example,

76
00:04:55,450 --> 00:05:00,070
to a 0 at e to the
minus aT t cosine bT.

77
00:05:00,070 --> 00:05:03,820
So in fact, it's not
a mapping of S to z,

78
00:05:03,820 --> 00:05:07,420
although the poles of the
analog system function

79
00:05:07,420 --> 00:05:10,180
get mapped to poles of the
digital system function

80
00:05:10,180 --> 00:05:13,790
according to the mapping
z equals e to the sT.

81
00:05:13,790 --> 00:05:21,640
Well, what this example looks
like in the frequency domain,

82
00:05:21,640 --> 00:05:23,500
I've indicated here.

83
00:05:23,500 --> 00:05:26,920
Here is the s plane
plot of the two poles

84
00:05:26,920 --> 00:05:30,250
and the 0 from the
analog system function.

85
00:05:30,250 --> 00:05:35,320
And the result of the
mapping or the result

86
00:05:35,320 --> 00:05:40,180
of applying the impulse
invariant procedure

87
00:05:40,180 --> 00:05:46,830
are the poles and 0s, which are
indicated here in the z plane.

88
00:05:46,830 --> 00:05:50,400
Well, to see what happens
to the frequency response,

89
00:05:50,400 --> 00:05:53,340
for this particular
analog example,

90
00:05:53,340 --> 00:05:57,900
we have a resonance at a
frequency corresponding

91
00:05:57,900 --> 00:06:01,020
to the vertical
position of these poles.

92
00:06:01,020 --> 00:06:03,630
And that's the resonance
that we have here

93
00:06:03,630 --> 00:06:07,640
in the magnitude of the
analog system function.

94
00:06:07,640 --> 00:06:12,000
And one of the important
aspects of the impulse invariant

95
00:06:12,000 --> 00:06:16,860
procedure, which I pointed
out last time, is that,

96
00:06:16,860 --> 00:06:19,980
except for aliasing-- and
that's a big except for--

97
00:06:19,980 --> 00:06:24,750
except for aliasing,
the frequency response

98
00:06:24,750 --> 00:06:28,470
has the same general shape
in the discrete domain

99
00:06:28,470 --> 00:06:31,020
as it does in the
continuous domain.

100
00:06:31,020 --> 00:06:35,400
So the corresponding discrete
time frequency response

101
00:06:35,400 --> 00:06:38,070
is as I've indicated here.

102
00:06:38,070 --> 00:06:40,380
We have a resonance
corresponding

103
00:06:40,380 --> 00:06:43,470
to these pole locations.

104
00:06:43,470 --> 00:06:46,380
But whereas this
frequency response

105
00:06:46,380 --> 00:06:52,320
drops off asymptotically to 0,
this frequency response drops

106
00:06:52,320 --> 00:06:55,380
off to some value,
non-zero value,

107
00:06:55,380 --> 00:06:57,270
and then continues
back up again,

108
00:06:57,270 --> 00:07:02,100
basically because this frequency
response has to be periodic.

109
00:07:02,100 --> 00:07:06,870
But in particular, the value
which this frequency response

110
00:07:06,870 --> 00:07:15,140
decays to is dictated in part
by the effect of aliasing.

111
00:07:15,140 --> 00:07:18,830
And you can see here that
in comparing this value

112
00:07:18,830 --> 00:07:21,260
to this one, this is higher.

113
00:07:21,260 --> 00:07:23,000
And the reason it's
higher is because

114
00:07:23,000 --> 00:07:25,040
of the effect of aliasing.

115
00:07:25,040 --> 00:07:28,610
So there is the advantage
to impulse invariance

116
00:07:28,610 --> 00:07:33,260
that the frequency axis
is a scaled replica

117
00:07:33,260 --> 00:07:35,630
of the analog frequency axis.

118
00:07:35,630 --> 00:07:38,750
But then there's a
distortion that's introduced,

119
00:07:38,750 --> 00:07:42,460
which is the effect of aliasing.

120
00:07:42,460 --> 00:07:45,280
There's another
technique which is

121
00:07:45,280 --> 00:07:47,770
a very important
and commonly used

122
00:07:47,770 --> 00:07:52,420
technique for designing
digital filters, which

123
00:07:52,420 --> 00:07:55,210
avoids the problem the
problem of aliasing,

124
00:07:55,210 --> 00:07:58,480
but it avoids that
problem at a price.

125
00:07:58,480 --> 00:08:05,350
That technique is referred to
as the bilinear transformation

126
00:08:05,350 --> 00:08:10,960
and basically involves mapping
from the s plane to the z plane

127
00:08:10,960 --> 00:08:14,770
by applying-- by
relating s and z

128
00:08:14,770 --> 00:08:17,830
according to the bilinear
transformation, a bilinear

129
00:08:17,830 --> 00:08:20,110
transformation.

130
00:08:20,110 --> 00:08:22,930
There, in fact, is
an interesting way

131
00:08:22,930 --> 00:08:25,780
of deriving this
technique, which

132
00:08:25,780 --> 00:08:28,750
is somewhat similar
to the technique

133
00:08:28,750 --> 00:08:31,660
involving mapping
differentials to differences.

134
00:08:31,660 --> 00:08:34,090
But of course, the
bilinear transformation

135
00:08:34,090 --> 00:08:37,940
will turn out to be a much
more useful technique.

136
00:08:37,940 --> 00:08:39,340
This technique can be derived--

137
00:08:39,340 --> 00:08:41,200
and we won't be going
through this in detail

138
00:08:41,200 --> 00:08:45,040
here, although we
do in the text--

139
00:08:45,040 --> 00:08:49,390
this can be derived
essentially by beginning

140
00:08:49,390 --> 00:08:53,830
with a continuous time
differential equation,

141
00:08:53,830 --> 00:08:58,160
integrating the equation to
obtain an integral equation,

142
00:08:58,160 --> 00:09:01,570
and then approximating
the integrals,

143
00:09:01,570 --> 00:09:03,250
using the trapezoidal rule.

144
00:09:03,250 --> 00:09:06,250
In other words, replacing
the integrals by sums,

145
00:09:06,250 --> 00:09:09,010
using the trapezoidal
rule for integration,

146
00:09:09,010 --> 00:09:13,300
the resulting equation is
then a difference equation.

147
00:09:13,300 --> 00:09:16,090
And that difference
equation will then

148
00:09:16,090 --> 00:09:20,230
represent the digital filter,
and that digital filter,

149
00:09:20,230 --> 00:09:23,320
as it turns out, can be related
back to the analog filter.

150
00:09:23,320 --> 00:09:25,570
The system functions
can be related

151
00:09:25,570 --> 00:09:28,970
through a transformation
from s to z,

152
00:09:28,970 --> 00:09:32,910
which is, in fact, the
bilinear transformation.

153
00:09:32,910 --> 00:09:37,130
So the technique
then is basically

154
00:09:37,130 --> 00:09:40,520
to begin with an
analog system function,

155
00:09:40,520 --> 00:09:44,900
convert that to a
digital system function,

156
00:09:44,900 --> 00:09:51,230
and do that by replacing s
by a bilinear transformation,

157
00:09:51,230 --> 00:09:53,060
where the bilinear
transformation

158
00:09:53,060 --> 00:09:55,700
is as I've indicated here.

159
00:09:55,700 --> 00:09:59,480
There's a parameter, capital
T, that I've introduced,

160
00:09:59,480 --> 00:10:02,720
and it's a parameter
that comes in, basically,

161
00:10:02,720 --> 00:10:06,950
because of what's involved
in applying the integration

162
00:10:06,950 --> 00:10:10,370
or applying the trapezoidal
rule to implementing

163
00:10:10,370 --> 00:10:12,380
the integration.

164
00:10:12,380 --> 00:10:17,300
But this then is the general
form of the transformation.

165
00:10:17,300 --> 00:10:21,380
Or expressing z instead
as a function of s,

166
00:10:21,380 --> 00:10:25,340
we obtain z equals
1 plus sT over 2

167
00:10:25,340 --> 00:10:28,980
divided by 1 minus sT over 2.

168
00:10:28,980 --> 00:10:31,950
So one important
aspect of this then,

169
00:10:31,950 --> 00:10:35,130
which, in fact, is different
than the impulse invariant

170
00:10:35,130 --> 00:10:39,720
procedure, is that it,
in fact, does correspond

171
00:10:39,720 --> 00:10:42,240
to a mapping from s to z.

172
00:10:42,240 --> 00:10:45,120
It's a mapping of the s
plane to the z plane, which

173
00:10:45,120 --> 00:10:48,760
is not quite what the
impulse invariant method is.

174
00:10:48,760 --> 00:10:53,970
Well, recalling that there are
two guidelines that we set down

175
00:10:53,970 --> 00:10:58,930
last time, we'd like to
verify, first of all,

176
00:10:58,930 --> 00:11:02,430
whether the bilinear
transformation in fact

177
00:11:02,430 --> 00:11:06,210
preserves the characteristics,
the frequency characteristics--

178
00:11:06,210 --> 00:11:09,090
that is, whether it maps
from the j omega axis

179
00:11:09,090 --> 00:11:11,250
to the unit circle, first.

180
00:11:11,250 --> 00:11:15,000
And second of all, does it
map stable analog filters

181
00:11:15,000 --> 00:11:17,730
to stable digital filters?

182
00:11:17,730 --> 00:11:21,150
Well, let's look at the
first question first.

183
00:11:21,150 --> 00:11:24,510
Let's look at the unit
circle and inquire as to what

184
00:11:24,510 --> 00:11:29,580
part of the s plane the
unit circle came from.

185
00:11:29,580 --> 00:11:35,380
So we have, then, z equals e to
the j omega on the unit circle.

186
00:11:35,380 --> 00:11:42,245
And so we want to inquire
as to what values of s

187
00:11:42,245 --> 00:11:46,610
land on the unit circle through
the bilinear transformation.

188
00:11:46,610 --> 00:11:49,210
And we do that simply
by substituting

189
00:11:49,210 --> 00:11:54,110
z equals e to the
j omega, so that s

190
00:11:54,110 --> 00:11:57,060
is equal to 2 over
capital T times 1 minus e

191
00:11:57,060 --> 00:12:02,030
to the minus j omega divided by
1 plus e to the minus j omega.

192
00:12:02,030 --> 00:12:09,270
And now we can change
this term and likewise

193
00:12:09,270 --> 00:12:13,065
the denominator term
by factoring out e

194
00:12:13,065 --> 00:12:18,630
to the minus j omega over 2.

195
00:12:18,630 --> 00:12:21,980
And then this
numerator term becomes

196
00:12:21,980 --> 00:12:27,260
e to the plus j
omega over 2 minus

197
00:12:27,260 --> 00:12:32,190
e to the minus j omega over 2.

198
00:12:32,190 --> 00:12:34,920
And we recognize
this numerator then

199
00:12:34,920 --> 00:12:38,490
as 2j times sine omega over 2.

200
00:12:38,490 --> 00:12:42,360
Well, likewise, we can carry
out a similar manipulation

201
00:12:42,360 --> 00:12:44,580
with the denominator term.

202
00:12:44,580 --> 00:12:48,240
The result then is that
this can be rewritten

203
00:12:48,240 --> 00:12:54,090
as 2 over capital T times j
times the tangent of omega

204
00:12:54,090 --> 00:12:55,980
over 2.

205
00:12:55,980 --> 00:13:00,840
Well, since omega
ranges from 0 to 2 pi,

206
00:13:00,840 --> 00:13:05,460
this is always a real
number, tangent omega over 2.

207
00:13:05,460 --> 00:13:10,290
And consequently, we can rewrite
this as j times a real number

208
00:13:10,290 --> 00:13:12,310
capital Omega.

209
00:13:12,310 --> 00:13:12,810
All right.

210
00:13:12,810 --> 00:13:14,080
Well, what does that say?

211
00:13:14,080 --> 00:13:18,000
It says that if z is
on the unit circle,

212
00:13:18,000 --> 00:13:23,040
then s is going to be of the
form j times a real number.

213
00:13:23,040 --> 00:13:26,490
And that has to have been,
then, the j omega axis.

214
00:13:26,490 --> 00:13:33,090
So s equal to j omega means z
equals e to the j omega, where

215
00:13:33,090 --> 00:13:37,650
capital Omega and little
omega are related by capital

216
00:13:37,650 --> 00:13:41,310
Omega equal to 2
over capital T times

217
00:13:41,310 --> 00:13:44,820
the tangent of
little omega over 2.

218
00:13:44,820 --> 00:13:48,990
Well, that says that,
indeed, the j omega axis

219
00:13:48,990 --> 00:13:52,260
gets mapped to the unit
circle, and vice versa.

220
00:13:52,260 --> 00:13:57,090
And in fact, the
entire j omega axis

221
00:13:57,090 --> 00:14:01,300
gets mapped to exactly once
around the unit circle.

222
00:14:01,300 --> 00:14:06,850
Well, let's look at the mapping
in a little more detail.

223
00:14:06,850 --> 00:14:12,970
Here we have the s plane,
and here we have the z plane.

224
00:14:12,970 --> 00:14:17,710
We have the relationship between
the j omega axis and the unit

225
00:14:17,710 --> 00:14:24,010
circle, that the j omega axis
gets mapped to the unit circle.

226
00:14:24,010 --> 00:14:29,680
And in fact, as capital Omega
runs from minus infinity

227
00:14:29,680 --> 00:14:39,090
to plus infinity, we map exactly
once around the unit circle.

228
00:14:39,090 --> 00:14:42,800
Now, that's an important
point for two reasons.

229
00:14:42,800 --> 00:14:46,760
One is that, obviously, there's
no aliasing involved anymore,

230
00:14:46,760 --> 00:14:51,740
because the entire j
omega axis ended up once

231
00:14:51,740 --> 00:14:52,760
around the unit circle.

232
00:14:52,760 --> 00:14:54,218
That is, there was
nothing that got

233
00:14:54,218 --> 00:14:57,230
added on to something else that
got added on to something else,

234
00:14:57,230 --> 00:15:01,500
as it did with the
impulse invariant method.

235
00:15:01,500 --> 00:15:07,200
The second point is that
that axis is infinitely long.

236
00:15:07,200 --> 00:15:11,020
The unit circle only has
a finite circumference.

237
00:15:11,020 --> 00:15:13,650
So clearly, something
has to happen.

238
00:15:13,650 --> 00:15:17,520
Something has to be distorted
in order to fit all of that

239
00:15:17,520 --> 00:15:19,590
onto just this little bit.

240
00:15:19,590 --> 00:15:23,430
In particular,
the mapping, if we

241
00:15:23,430 --> 00:15:27,060
draw it, between little
omega, angular frequency

242
00:15:27,060 --> 00:15:30,420
around the unit circle, and
distance along the j omega

243
00:15:30,420 --> 00:15:34,680
axis, looks as I've indicated
here, which, of course, is

244
00:15:34,680 --> 00:15:37,250
a nonlinear curve.

245
00:15:37,250 --> 00:15:40,810
Well, let me return
to that in a second.

246
00:15:40,810 --> 00:15:43,850
An additional point about
the bilinear transformation,

247
00:15:43,850 --> 00:15:49,000
which I'll state simply and
leave it to you to verify,

248
00:15:49,000 --> 00:15:51,460
is that the bilinear
transformation has

249
00:15:51,460 --> 00:15:55,930
the property that the
left 1/2 of the s plane

250
00:15:55,930 --> 00:15:59,500
gets mapped to the interior
of the unit circle,

251
00:15:59,500 --> 00:16:01,630
and the right half
of the s plane

252
00:16:01,630 --> 00:16:05,450
gets mapped to the exterior
of the unit circle.

253
00:16:05,450 --> 00:16:08,900
Well, that's good,
because that says

254
00:16:08,900 --> 00:16:11,840
that, if we have poles or 0s--

255
00:16:11,840 --> 00:16:14,630
but poles is what we really
care about right now--

256
00:16:14,630 --> 00:16:17,570
we have poles in the
left 1/2 of the s plane,

257
00:16:17,570 --> 00:16:20,190
they'll end up inside
the unit circle.

258
00:16:20,190 --> 00:16:23,030
So if we have a
stable analog filter,

259
00:16:23,030 --> 00:16:26,120
we'll end up with a
stable digital filter.

260
00:16:26,120 --> 00:16:28,550
If we have an unstable
analog filter,

261
00:16:28,550 --> 00:16:31,640
corresponding to poles on
the right 1/2 of the s plane,

262
00:16:31,640 --> 00:16:35,090
then we'll end up with an
unstable digital filter,

263
00:16:35,090 --> 00:16:39,110
corresponding to poles
outside the unit circle.

264
00:16:39,110 --> 00:16:39,610
All right.

265
00:16:39,610 --> 00:16:45,790
Well, now, returning to the
mapping of the j omega axis

266
00:16:45,790 --> 00:16:50,110
to digital frequency, the
mapping from analog frequency

267
00:16:50,110 --> 00:16:53,770
to digital frequency, one
of the important aspects

268
00:16:53,770 --> 00:16:58,600
of the bilinear transformation
is that that mapping ends up

269
00:16:58,600 --> 00:17:02,340
as a nonlinear mapping.

270
00:17:02,340 --> 00:17:04,800
Well, what are the
consequences of that?

271
00:17:04,800 --> 00:17:09,540
Let's take a look at it
in a little more detail.

272
00:17:09,540 --> 00:17:12,899
And in addition to pointing
out some of the problems

273
00:17:12,899 --> 00:17:17,550
that that raises, this
offers us an opportunity

274
00:17:17,550 --> 00:17:21,990
to indicate how, in
fact, this distortion can

275
00:17:21,990 --> 00:17:25,680
be taken into account in
implementing a digital filter

276
00:17:25,680 --> 00:17:27,490
design.

277
00:17:27,490 --> 00:17:34,060
So here is the frequency mapping
of the bilinear transformation.

278
00:17:34,060 --> 00:17:35,940
This is digital frequency.

279
00:17:35,940 --> 00:17:38,280
Here is analog frequency.

280
00:17:38,280 --> 00:17:42,300
And this curve, of course,
goes out to infinity in capital

281
00:17:42,300 --> 00:17:46,530
Omega and asymptotically
approaches pi, as it does that,

282
00:17:46,530 --> 00:17:47,820
in little omega.

283
00:17:47,820 --> 00:17:51,000
That says that the top
1/2 of the j omega axis

284
00:17:51,000 --> 00:17:55,260
gets mapped to the upper
1/2 of the unit circle.

285
00:17:55,260 --> 00:18:00,780
Well, suppose that we had,
say, an analog filter frequency

286
00:18:00,780 --> 00:18:06,720
response that was, let's
say, linear in frequency.

287
00:18:06,720 --> 00:18:10,830
Would the corresponding
digital frequency response

288
00:18:10,830 --> 00:18:13,150
also be linear in frequency?

289
00:18:13,150 --> 00:18:16,650
Well, the answer to that,
unfortunately, is no.

290
00:18:16,650 --> 00:18:22,800
In other words, a frequency
characteristic along this axis

291
00:18:22,800 --> 00:18:25,750
will get distorted
as it gets reflected

292
00:18:25,750 --> 00:18:28,500
through this nonlinear curve.

293
00:18:28,500 --> 00:18:32,670
For example, let's
look specifically

294
00:18:32,670 --> 00:18:39,120
at a linear analog
frequency characteristic,

295
00:18:39,120 --> 00:18:42,450
as I've indicated here
which might, for example, be

296
00:18:42,450 --> 00:18:44,820
the frequency response
of a differentiator.

297
00:18:44,820 --> 00:18:47,610
That, in fact, is the
kind of frequency response

298
00:18:47,610 --> 00:18:50,700
that a differentiator
might have.

299
00:18:50,700 --> 00:18:55,140
As we reflect that through
this nonlinear curve,

300
00:18:55,140 --> 00:18:57,750
because of the fact
that that's nonlinear,

301
00:18:57,750 --> 00:19:02,640
we don't maintain the linearity
of the analog frequency

302
00:19:02,640 --> 00:19:04,050
response.

303
00:19:04,050 --> 00:19:06,990
Now, if it happened that this
was the kind of frequency

304
00:19:06,990 --> 00:19:10,080
response that we wanted,
we could get that

305
00:19:10,080 --> 00:19:12,880
from an analog differentiator.

306
00:19:12,880 --> 00:19:17,160
But more typically, what
we might be interested in

307
00:19:17,160 --> 00:19:20,310
is a digital
differentiator which

308
00:19:20,310 --> 00:19:23,740
would have, say, a linear
frequency characteristic.

309
00:19:23,740 --> 00:19:28,200
And one of the important aspects
of the bilinear transformation

310
00:19:28,200 --> 00:19:33,540
is that we couldn't design that
by designing-- by mapping over

311
00:19:33,540 --> 00:19:38,610
an analog differentiator to
a digital differentiator,

312
00:19:38,610 --> 00:19:40,770
using the bilinear
transformation.

313
00:19:40,770 --> 00:19:43,800
Now, that's in contrast to
the impulse invariant design

314
00:19:43,800 --> 00:19:49,050
procedure, where, in fact,
for an impulse invariant

315
00:19:49,050 --> 00:19:53,220
design, if we have a
band-limited filter that

316
00:19:53,220 --> 00:19:56,700
is linear over some part
of the frequency band,

317
00:19:56,700 --> 00:19:58,710
impulse invariance
would maintain

318
00:19:58,710 --> 00:20:01,690
the linearity of the
frequency characteristics.

319
00:20:01,690 --> 00:20:05,070
So that's an important
distinction between these two

320
00:20:05,070 --> 00:20:09,890
methods and, in fact, a drawback
to the bilinear transformation.

321
00:20:09,890 --> 00:20:12,560
Well, where might we
be able to tolerate

322
00:20:12,560 --> 00:20:14,700
this kind of a
nonlinear distortion

323
00:20:14,700 --> 00:20:16,760
in the frequency axis?

324
00:20:16,760 --> 00:20:23,830
One place we could is when
the desired digital filter

325
00:20:23,830 --> 00:20:27,490
is piecewise constant,
in which case

326
00:20:27,490 --> 00:20:31,390
that will get reflected into
a piecewise constant analog

327
00:20:31,390 --> 00:20:32,350
filter.

328
00:20:32,350 --> 00:20:34,630
And let me indicate
that more explicitly.

329
00:20:42,130 --> 00:20:47,970
Let's suppose that we were
interested in designing

330
00:20:47,970 --> 00:20:52,150
a digital filter, which
was a low pass filter.

331
00:20:52,150 --> 00:20:56,260
Here's the magnitude axis,
and here's the frequency axis.

332
00:20:56,260 --> 00:21:00,310
And the digital low
pass filter was unity,

333
00:21:00,310 --> 00:21:06,080
say, in the passband, a rapid
cutoff at the passband edge,

334
00:21:06,080 --> 00:21:07,810
and then 0.

335
00:21:07,810 --> 00:21:12,750
Well, clearly, this
part being constant

336
00:21:12,750 --> 00:21:16,530
gets reflected through this
nonlinear frequency curve

337
00:21:16,530 --> 00:21:22,590
into a constant analog
frequency characteristic.

338
00:21:22,590 --> 00:21:25,750
Similarly, the
stopband gets reflected

339
00:21:25,750 --> 00:21:30,930
into a constant amplitude,
in this case, 0,

340
00:21:30,930 --> 00:21:33,090
in the analog
frequency response.

341
00:21:33,090 --> 00:21:37,680
And what's the effect of
this nonlinear distortion

342
00:21:37,680 --> 00:21:39,540
in the frequency axis?

343
00:21:39,540 --> 00:21:44,280
The only effect that it has
is that the cutoff frequency

344
00:21:44,280 --> 00:21:49,140
of the filter or, in the more
general case, the frequencies

345
00:21:49,140 --> 00:21:54,810
at which the piecewise constant
characteristic changes from one

346
00:21:54,810 --> 00:22:00,870
piece to another gets changed
according to this frequency

347
00:22:00,870 --> 00:22:02,560
warping characteristic.

348
00:22:02,560 --> 00:22:08,070
So this cutoff frequency gets
converted into a new cutoff

349
00:22:08,070 --> 00:22:11,670
frequency, where the
analog cutoff frequency is

350
00:22:11,670 --> 00:22:14,370
related to the digital
cutoff frequency

351
00:22:14,370 --> 00:22:17,620
according to that curve.

352
00:22:17,620 --> 00:22:23,540
Now-- and so one of the
things that that says is that

353
00:22:23,540 --> 00:22:31,040
for the filter design, if
we wanted to design an ideal

354
00:22:31,040 --> 00:22:35,330
digital filter, an ideal
digital low pass filter,

355
00:22:35,330 --> 00:22:39,140
then to decide on the
specifications for the ideal

356
00:22:39,140 --> 00:22:42,590
analog filter, we
would simply pre-warp--

357
00:22:42,590 --> 00:22:47,210
what's called pre-warping--
pre-warp the cutoff frequency

358
00:22:47,210 --> 00:22:51,080
to an analog cutoff frequency,
which would get unraveled back

359
00:22:51,080 --> 00:22:55,310
to the right cutoff frequency
when we go through the bilinear

360
00:22:55,310 --> 00:22:57,410
transformation.

361
00:22:57,410 --> 00:23:00,700
Now, we know that there
aren't ideal filters.

362
00:23:00,700 --> 00:23:03,230
And so, in fact,
we're, in general,

363
00:23:03,230 --> 00:23:07,310
faced with the problem of
approximating a characteristic

364
00:23:07,310 --> 00:23:08,540
like that.

365
00:23:08,540 --> 00:23:11,750
But if the specifications
for the filter

366
00:23:11,750 --> 00:23:16,410
are still piecewise
constant, then, in fact,

367
00:23:16,410 --> 00:23:21,690
we can still concentrate
just simply on a pre-warping

368
00:23:21,690 --> 00:23:24,060
of the critical frequencies.

369
00:23:24,060 --> 00:23:28,560
And the piecewise
constant portions

370
00:23:28,560 --> 00:23:30,480
of the frequency
response will then

371
00:23:30,480 --> 00:23:32,490
fall in the correct places.

372
00:23:32,490 --> 00:23:38,400
And let me illustrate that by
showing a more realistic kind

373
00:23:38,400 --> 00:23:40,120
of filter frequency response.

374
00:23:53,250 --> 00:23:55,740
Well, that's pretty close.

375
00:23:55,740 --> 00:23:56,370
All right.

376
00:23:56,370 --> 00:24:02,580
Here we have what would be more
typical of the digital filter

377
00:24:02,580 --> 00:24:05,940
that we would design
and implement--

378
00:24:05,940 --> 00:24:09,070
some allowed tolerance
in the passband,

379
00:24:09,070 --> 00:24:13,680
some allowed tolerance in the
stopband, a passband cutoff

380
00:24:13,680 --> 00:24:17,680
frequency, and a stopband
cutoff frequency.

381
00:24:17,680 --> 00:24:21,900
Well, if that was our
desired digital filter,

382
00:24:21,900 --> 00:24:23,700
then how would
those specifications

383
00:24:23,700 --> 00:24:26,500
translate to an analog filter?

384
00:24:26,500 --> 00:24:30,780
Well, if the deviation
in the digital filter

385
00:24:30,780 --> 00:24:34,860
was between 1 and 1
minus delta sub p,

386
00:24:34,860 --> 00:24:38,880
then that would get
converted into specifications

387
00:24:38,880 --> 00:24:41,970
on the analog filter,
likewise between 1 and 1

388
00:24:41,970 --> 00:24:44,880
minus delta sub p.

389
00:24:44,880 --> 00:24:47,820
The passband cutoff
edge would get

390
00:24:47,820 --> 00:24:50,840
reflected through
this nonlinear curve

391
00:24:50,840 --> 00:24:54,390
to a new analog
cutoff frequency.

392
00:24:54,390 --> 00:24:58,350
Likewise, the transition
region would get reflected

393
00:24:58,350 --> 00:25:00,780
into a new transition region.

394
00:25:00,780 --> 00:25:03,540
If this frequency
characteristic was, let's say,

395
00:25:03,540 --> 00:25:06,210
linear in the transition
region, of course,

396
00:25:06,210 --> 00:25:09,120
it wouldn't end up linear
in this transition region.

397
00:25:09,120 --> 00:25:11,430
But who cares?

398
00:25:11,430 --> 00:25:14,160
The stopband edge
would get reflected

399
00:25:14,160 --> 00:25:16,170
into a new stopband edge.

400
00:25:16,170 --> 00:25:19,950
And finally, the
stopband specifications,

401
00:25:19,950 --> 00:25:22,200
let's say, delta
sub s, would get

402
00:25:22,200 --> 00:25:26,710
mapped to the same stopband
specifications, delta sub s.

403
00:25:26,710 --> 00:25:30,090
Furthermore, notice
that if, let's say,

404
00:25:30,090 --> 00:25:32,580
we had an optimum
filter which turns out

405
00:25:32,580 --> 00:25:36,450
to be equal ripple for
the digital case, then

406
00:25:36,450 --> 00:25:40,590
that would correspond to
an equal ripple filter

407
00:25:40,590 --> 00:25:42,000
in the analog case.

408
00:25:42,000 --> 00:25:45,420
Or said another way, if
we were able to design

409
00:25:45,420 --> 00:25:47,850
an equal ripple
analog filter, then

410
00:25:47,850 --> 00:25:50,220
map through the bilinear
transformation, that

411
00:25:50,220 --> 00:25:55,480
would likewise result in an
equal ripple digital filter.

412
00:25:55,480 --> 00:26:01,050
So the important point
here then is that we can,

413
00:26:01,050 --> 00:26:05,280
for piecewise constant
frequency characteristics

414
00:26:05,280 --> 00:26:09,000
and specifications, it really
is only the critical frequencies

415
00:26:09,000 --> 00:26:11,280
that we care about, the edge
of the passband, the edge

416
00:26:11,280 --> 00:26:13,960
of the stopband, and in the
multiple band case, of course,

417
00:26:13,960 --> 00:26:15,990
there will be a number of those.

418
00:26:15,990 --> 00:26:20,670
By pre-warping those through
this arctangent curve

419
00:26:20,670 --> 00:26:25,770
into the corresponding
analog critical frequencies,

420
00:26:25,770 --> 00:26:30,600
if this filter is then designed
and mapped to a digital filter,

421
00:26:30,600 --> 00:26:33,060
using the bilinear
transformation,

422
00:26:33,060 --> 00:26:35,910
then the passband
and stopband edges

423
00:26:35,910 --> 00:26:39,240
will fall at the
frequencies that we want.

424
00:26:39,240 --> 00:26:43,680
And this point again
will be emphasized

425
00:26:43,680 --> 00:26:51,750
in the next lecture, when I
show a design example comparing

426
00:26:51,750 --> 00:26:55,060
impulse invariance in the
bilinear transformation.

427
00:26:55,060 --> 00:27:01,110
So the key point then about
the bilinear transformation

428
00:27:01,110 --> 00:27:05,460
is that it avoids the problem
of aliasing at a price.

429
00:27:05,460 --> 00:27:08,130
The price is that there's a
distortion in the frequency

430
00:27:08,130 --> 00:27:09,090
axis.

431
00:27:09,090 --> 00:27:13,650
And so consequently, we can only
use the bilinear transformation

432
00:27:13,650 --> 00:27:17,460
to design filters that can
tolerate that distortion.

433
00:27:17,460 --> 00:27:21,450
Typically, that means designing
piecewise constant digital

434
00:27:21,450 --> 00:27:21,950
filters.

435
00:27:24,410 --> 00:27:24,910
All right.

436
00:27:24,910 --> 00:27:32,260
Well, that basically concludes
the discussion of the design

437
00:27:32,260 --> 00:27:38,080
techniques that correspond to
mapping analog filter designs

438
00:27:38,080 --> 00:27:41,110
to digital filter designs.

439
00:27:41,110 --> 00:27:44,590
There is one final
class of techniques

440
00:27:44,590 --> 00:27:49,450
that I would like to just
comment on somewhat briefly.

441
00:27:49,450 --> 00:27:53,380
And that is the class of
techniques that correspond

442
00:27:53,380 --> 00:27:55,780
to algorithmic design.

443
00:27:55,780 --> 00:28:00,730
Now, there are algorithmic
design procedures

444
00:28:00,730 --> 00:28:02,690
for analog filters.

445
00:28:02,690 --> 00:28:06,970
And the same
algorithmic procedures

446
00:28:06,970 --> 00:28:09,760
can likewise be used
for digital filters.

447
00:28:09,760 --> 00:28:12,770
Since we have to go through
an algorithmic procedure--

448
00:28:12,770 --> 00:28:16,830
in fact, there is no
advantage, in that case,

449
00:28:16,830 --> 00:28:19,330
to first designing
the analog filter

450
00:28:19,330 --> 00:28:22,150
and then mapping that over
to the digital filter.

451
00:28:22,150 --> 00:28:24,650
Rather, for these
algorithmic techniques,

452
00:28:24,650 --> 00:28:27,490
it makes more sense to
just simply apply them

453
00:28:27,490 --> 00:28:31,460
directly to the design
of a digital filter.

454
00:28:31,460 --> 00:28:34,030
Well, there are a number
of techniques, actually

455
00:28:34,030 --> 00:28:39,940
a fairly long list, many of them
simple modifications of others.

456
00:28:39,940 --> 00:28:44,650
There are only two that I want
to point to in this lecture.

457
00:28:44,650 --> 00:28:47,590
When we talk about the design
of finite impulse response

458
00:28:47,590 --> 00:28:51,190
filters, there are some
additional algorithmic design

459
00:28:51,190 --> 00:28:54,400
procedures that we'll point to.

460
00:28:54,400 --> 00:28:57,910
Well, the first algorithmic
design procedure,

461
00:28:57,910 --> 00:29:01,930
which I am restricting,
in my discussion here,

462
00:29:01,930 --> 00:29:06,260
to the design of infinite
impulse response filters,

463
00:29:06,260 --> 00:29:10,630
the first is simply
designing the filter

464
00:29:10,630 --> 00:29:14,800
to minimize the mean square
error between the desired

465
00:29:14,800 --> 00:29:17,410
frequency characteristic,
the desired frequency

466
00:29:17,410 --> 00:29:22,340
response, and the actual
frequency response.

467
00:29:22,340 --> 00:29:25,690
So let's assume that we
have some desired frequency

468
00:29:25,690 --> 00:29:29,080
response, H sub d
of e to the j omega.

469
00:29:29,080 --> 00:29:33,070
We would like to match
that with the frequency

470
00:29:33,070 --> 00:29:39,100
response of the rational
transfer function.

471
00:29:39,100 --> 00:29:43,360
It, of course, isn't
possible to attempt

472
00:29:43,360 --> 00:29:47,180
to do that at an arbitrary
number of frequencies.

473
00:29:47,180 --> 00:29:52,450
So we can pick out some set
of frequencies, omega sub

474
00:29:52,450 --> 00:29:56,710
i, at which we would
like to minimize

475
00:29:56,710 --> 00:29:59,530
the error between
the desired frequency

476
00:29:59,530 --> 00:30:04,100
response and the actual
frequency response.

477
00:30:04,100 --> 00:30:07,310
The error, then,
which we'll be using,

478
00:30:07,310 --> 00:30:10,240
or one error which
is used, is the error

479
00:30:10,240 --> 00:30:13,310
that corresponds to minimization
of the mean square error

480
00:30:13,310 --> 00:30:18,320
is the mean square difference
between the magnitude

481
00:30:18,320 --> 00:30:23,150
of the actual transfer
function and the magnitude

482
00:30:23,150 --> 00:30:26,670
of the desired
transfer function.

483
00:30:26,670 --> 00:30:31,880
So this error, then, summed over
the frequency points at which

484
00:30:31,880 --> 00:30:36,740
we specify the desired
frequency response, this error,

485
00:30:36,740 --> 00:30:39,320
m is to be minimized.

486
00:30:39,320 --> 00:30:42,260
Well, how do we pick H?

487
00:30:42,260 --> 00:30:47,190
Well, the general form for H
is as a rational function--

488
00:30:47,190 --> 00:30:49,370
that's what we've been
restricting ourselves

489
00:30:49,370 --> 00:30:51,560
to throughout the discussion--

490
00:30:51,560 --> 00:30:56,660
the rational function of z,
and expressed, for example,

491
00:30:56,660 --> 00:31:00,830
in cascade form, involving
second order sections,

492
00:31:00,830 --> 00:31:05,060
that is, two coefficients
for the 0s and two

493
00:31:05,060 --> 00:31:09,290
coefficients for the poles,
with capital K sections

494
00:31:09,290 --> 00:31:11,240
all together.

495
00:31:11,240 --> 00:31:13,970
So basically what the
procedure then consists

496
00:31:13,970 --> 00:31:16,370
of-- it's somewhat brute force--

497
00:31:16,370 --> 00:31:20,720
is to specify the
order of the filter.

498
00:31:20,720 --> 00:31:23,240
We then have a set
of coefficients

499
00:31:23,240 --> 00:31:26,030
which we need to solve for.

500
00:31:26,030 --> 00:31:34,130
We have a set of equations
which result from minimizing

501
00:31:34,130 --> 00:31:36,960
this error function.

502
00:31:36,960 --> 00:31:40,050
And by solving that
set of equations,

503
00:31:40,050 --> 00:31:46,130
by minimizing this
error, what results then

504
00:31:46,130 --> 00:31:48,350
are values for the
parameters, which

505
00:31:48,350 --> 00:31:50,750
provide the parameters
for the filter

506
00:31:50,750 --> 00:31:53,520
that we want to implement.

507
00:31:53,520 --> 00:31:57,060
Well, one of the issues
with this method,

508
00:31:57,060 --> 00:32:00,600
one of the difficulties with
it, of course, is that--

509
00:32:00,600 --> 00:32:02,340
and you can kind of
imagine that this

510
00:32:02,340 --> 00:32:04,020
is what's going to happen--

511
00:32:04,020 --> 00:32:09,570
is that, if you take
this rational function

512
00:32:09,570 --> 00:32:14,430
and put it into this equation
and minimize this error,

513
00:32:14,430 --> 00:32:16,020
you end up with a
set of equations

514
00:32:16,020 --> 00:32:19,500
for the a sub k's, b sub
k, c sub k, and d sub k,

515
00:32:19,500 --> 00:32:22,870
but it's a set of
nonlinear equations.

516
00:32:22,870 --> 00:32:25,740
And basically what the
algorithmic part of this design

517
00:32:25,740 --> 00:32:29,640
procedure amounts to
is algorithmically

518
00:32:29,640 --> 00:32:33,910
solving this nonlinear
set of equations.

519
00:32:33,910 --> 00:32:38,160
Well, let me just illustrate,
with one simple example,

520
00:32:38,160 --> 00:32:43,710
how this procedure-- what
type of filter this procedure

521
00:32:43,710 --> 00:32:44,580
might lead to.

522
00:32:44,580 --> 00:32:47,730
I don't want to, obviously,
go through all the details

523
00:32:47,730 --> 00:32:49,780
of solving the equations.

524
00:32:49,780 --> 00:32:53,650
But let's just quickly
look at an example.

525
00:32:53,650 --> 00:32:59,730
This is an example incidentally
that is taken from a paper

526
00:32:59,730 --> 00:33:03,870
by Professor Kenneth Steiglitz,
at Princeton University.

527
00:33:03,870 --> 00:33:06,600
And the objective
here was to design

528
00:33:06,600 --> 00:33:09,000
what's called a
format filter, which

529
00:33:09,000 --> 00:33:11,610
is used for speech synthesis.

530
00:33:11,610 --> 00:33:15,990
And the frequency
characteristic--

531
00:33:15,990 --> 00:33:18,120
the aspects of the
frequency characteristic

532
00:33:18,120 --> 00:33:23,640
that were important were to
have peaks at this frequency

533
00:33:23,640 --> 00:33:26,455
and this frequency
and this frequency--

534
00:33:26,455 --> 00:33:28,080
those are the peaks
that, incidentally,

535
00:33:28,080 --> 00:33:31,140
correspond to the resonances
of the vocal tract--

536
00:33:31,140 --> 00:33:35,580
and to have valleys in
between those peaks.

537
00:33:35,580 --> 00:33:39,360
Well, one of the things that you
can see from the kind of filter

538
00:33:39,360 --> 00:33:42,390
that we're talking
about designing here

539
00:33:42,390 --> 00:33:46,750
is that this isn't
piecewise constant.

540
00:33:46,750 --> 00:33:49,410
And in fact, it's
not band-limited.

541
00:33:49,410 --> 00:33:52,770
And it tends to have-- it's
not arbitrary, of course,

542
00:33:52,770 --> 00:33:56,700
but it's a much more general
frequency characteristic

543
00:33:56,700 --> 00:33:59,790
shape than we had
been talking about

544
00:33:59,790 --> 00:34:03,270
in discussing the previous
methods, in particular,

545
00:34:03,270 --> 00:34:05,960
than we had been talking about
in discussing the bilinear

546
00:34:05,960 --> 00:34:08,310
transformation.

547
00:34:08,310 --> 00:34:10,920
Well, the procedure
that was used, then,

548
00:34:10,920 --> 00:34:17,760
was to specify the values of
the desired frequency response

549
00:34:17,760 --> 00:34:20,100
and to try to specify
them in such a way

550
00:34:20,100 --> 00:34:23,100
that the frequency
characteristic was more

551
00:34:23,100 --> 00:34:27,929
or less forced to go
through the proper values

552
00:34:27,929 --> 00:34:29,980
at the important frequencies.

553
00:34:29,980 --> 00:34:32,940
So what was specified,
then, was a point here.

554
00:34:32,940 --> 00:34:34,620
The frequency
response was specified

555
00:34:34,620 --> 00:34:38,670
to have that value
there, one here to get

556
00:34:38,670 --> 00:34:40,889
us heading toward
the resonance, three

557
00:34:40,889 --> 00:34:46,250
points at the peak of the
resonance there, three points

558
00:34:46,250 --> 00:34:49,770
to force the frequency
response back down again,

559
00:34:49,770 --> 00:34:51,270
another three to
bring it back up

560
00:34:51,270 --> 00:34:55,530
for the second resonance, a few
points distributed along here

561
00:34:55,530 --> 00:35:00,420
to provide a general
characteristic to tend

562
00:35:00,420 --> 00:35:04,080
toward the next resonance,
three points at this resonance,

563
00:35:04,080 --> 00:35:07,110
and then a few points to have
the frequency response taper

564
00:35:07,110 --> 00:35:09,350
off.

565
00:35:09,350 --> 00:35:14,990
Well, what was used was a total
of three second order sections.

566
00:35:14,990 --> 00:35:18,860
And the resulting frequency
response that was obtained

567
00:35:18,860 --> 00:35:23,910
is what is shown
here in solid lines.

568
00:35:23,910 --> 00:35:25,940
And you can see
that, in fact, that's

569
00:35:25,940 --> 00:35:30,530
a reasonably good fit to the
points that were specified.

570
00:35:30,530 --> 00:35:35,570
This particular example I
chose to illustrate primarily,

571
00:35:35,570 --> 00:35:39,890
because it is one of the
harder examples involving

572
00:35:39,890 --> 00:35:42,500
the minimization of
mean square error.

573
00:35:42,500 --> 00:35:48,830
And in fact, in terms of
computer time on an IBM 36065,

574
00:35:48,830 --> 00:35:52,280
this particular example
required approximately 2 and 1/2

575
00:35:52,280 --> 00:35:53,280
minutes.

576
00:35:53,280 --> 00:35:57,030
So that's a reasonable
amount of computation,

577
00:35:57,030 --> 00:36:00,200
but one of the
important aspects to it

578
00:36:00,200 --> 00:36:04,400
is that it allows the design,
and it allowed the design,

579
00:36:04,400 --> 00:36:08,240
of a somewhat general
frequency shape,

580
00:36:08,240 --> 00:36:10,970
frequency characteristic.

581
00:36:10,970 --> 00:36:11,470
All right.

582
00:36:11,470 --> 00:36:15,670
So that is one
algorithmic method.

583
00:36:15,670 --> 00:36:18,580
And it, in fact,
is a good method.

584
00:36:18,580 --> 00:36:21,520
It's one of the methods
that is commonly used,

585
00:36:21,520 --> 00:36:23,050
particularly when
the frequency--

586
00:36:23,050 --> 00:36:24,670
when some of the other methods--

587
00:36:24,670 --> 00:36:27,550
the bilinear transformation
or impulse invariance

588
00:36:27,550 --> 00:36:32,530
can't be used, because the
filter isn't of a simple form.

589
00:36:32,530 --> 00:36:36,700
But one of the major
disadvantages to it

590
00:36:36,700 --> 00:36:40,840
is the fact that the
equations that result

591
00:36:40,840 --> 00:36:42,790
are not linear equations.

592
00:36:42,790 --> 00:36:45,610
And in general, solving
nonlinear equations

593
00:36:45,610 --> 00:36:49,970
is a fairly difficult
algorithmic job.

594
00:36:49,970 --> 00:36:53,300
Well, there's an
alternative to that,

595
00:36:53,300 --> 00:36:58,250
which is a method which is
commonly referred to as least

596
00:36:58,250 --> 00:37:01,790
squares inverse design.

597
00:37:01,790 --> 00:37:06,900
And in that design
procedure, we can

598
00:37:06,900 --> 00:37:12,540
imagine specifying a
desired impulse response.

599
00:37:12,540 --> 00:37:19,490
And in this case, we restrict
the filter transfer function

600
00:37:19,490 --> 00:37:24,430
to have no 0s and only poles.

601
00:37:24,430 --> 00:37:27,510
Well, it turns out-- and
I don't want to elaborate

602
00:37:27,510 --> 00:37:28,890
on the details--

603
00:37:28,890 --> 00:37:32,490
but basically, the way
this procedure is set up

604
00:37:32,490 --> 00:37:39,120
is to consider a system
whose system function is

605
00:37:39,120 --> 00:37:45,320
the inverse of H of z, so that
that system is in all 0 system,

606
00:37:45,320 --> 00:37:48,500
and then choose as an
input to that system

607
00:37:48,500 --> 00:37:51,290
the desired unit
sample response.

608
00:37:51,290 --> 00:37:56,660
Now, we have the desired unit
sample response is the input.

609
00:37:56,660 --> 00:38:00,140
We've got the
reciprocal of the filter

610
00:38:00,140 --> 00:38:06,950
that we're going to implement
in this as the system function.

611
00:38:06,950 --> 00:38:11,900
And if, in fact, the
realized unit sample response

612
00:38:11,900 --> 00:38:15,920
was exactly equal to the desired
unit sample response, what

613
00:38:15,920 --> 00:38:18,360
would the output be?

614
00:38:18,360 --> 00:38:21,180
Well, it would be a unit
sample, because this

615
00:38:21,180 --> 00:38:25,590
would be exactly the inverse
of this unit sample response.

616
00:38:25,590 --> 00:38:29,670
Well, let's think of
the output as g of n.

617
00:38:29,670 --> 00:38:36,390
And ideally, g of n should
be an impulse, a unit sample.

618
00:38:36,390 --> 00:38:40,560
Well, in general, of course,
it won't be, since H sub d of n

619
00:38:40,560 --> 00:38:44,530
we won't be able to fit exactly
by a model of that form.

620
00:38:44,530 --> 00:38:47,610
So let's consider
the error, which

621
00:38:47,610 --> 00:38:50,165
is the mean square
difference between g

622
00:38:50,165 --> 00:38:53,280
of n and a unit sample.

623
00:38:53,280 --> 00:38:57,150
And that then is the
error that we minimize.

624
00:38:57,150 --> 00:39:00,930
So the idea behind the
least squares inverse design

625
00:39:00,930 --> 00:39:07,410
is basically to design the
inverse to the desired unit

626
00:39:07,410 --> 00:39:10,590
sample response, do
that by minimizing

627
00:39:10,590 --> 00:39:13,710
the mean square error
between the output

628
00:39:13,710 --> 00:39:16,590
of this inverse system
and the desired output.

629
00:39:16,590 --> 00:39:21,030
And the desired output, of
course, is a unit sample.

630
00:39:21,030 --> 00:39:24,300
Well, there's the
legitimate question

631
00:39:24,300 --> 00:39:28,680
as to whether that is a
reasonable error criterion

632
00:39:28,680 --> 00:39:29,730
to use.

633
00:39:29,730 --> 00:39:33,360
It's, of course, a possible
error criterion to use.

634
00:39:33,360 --> 00:39:35,310
But one could ask,
why in the world

635
00:39:35,310 --> 00:39:38,160
would you want to set
up the problem that way?

636
00:39:38,160 --> 00:39:40,200
Well, there's a
very good reason.

637
00:39:40,200 --> 00:39:43,250
The reason is that,
if you do that,

638
00:39:43,250 --> 00:39:48,820
minimize this error in
terms of these coefficients,

639
00:39:48,820 --> 00:39:52,360
the equations that result
are linear equations.

640
00:39:52,360 --> 00:39:55,990
And linear equations are
simple equations to solve,

641
00:39:55,990 --> 00:39:58,340
as opposed to
nonlinear equations.

642
00:39:58,340 --> 00:40:01,180
So this is a technique
that, in fact, turns out

643
00:40:01,180 --> 00:40:06,190
to be a good technique for
a wide variety of filters.

644
00:40:06,190 --> 00:40:09,100
It has the important
advantage that it

645
00:40:09,100 --> 00:40:12,340
leads to a set of
linear equations,

646
00:40:12,340 --> 00:40:16,330
as opposed to the minimization
of mean square error,

647
00:40:16,330 --> 00:40:19,430
which leads to a set
of nonlinear equations.

648
00:40:19,430 --> 00:40:21,730
Well, this is just
a very quick glimpse

649
00:40:21,730 --> 00:40:25,210
at two algorithmic
design procedures.

650
00:40:25,210 --> 00:40:28,000
As I said previously,
there are some others,

651
00:40:28,000 --> 00:40:30,070
which we won't be going
into, some of which

652
00:40:30,070 --> 00:40:31,840
are described in the text.

653
00:40:31,840 --> 00:40:36,610
And there are a few related to
finite impulse response filters

654
00:40:36,610 --> 00:40:39,820
that we'll be discussing
in a number of lectures.

655
00:40:39,820 --> 00:40:42,940
In the next lecture,
what I would like to do

656
00:40:42,940 --> 00:40:48,790
is review the impulse invariant
and bilinear design procedures,

657
00:40:48,790 --> 00:40:51,670
since those are the
two major design

658
00:40:51,670 --> 00:40:55,240
procedures for infinite
impulse response filters,

659
00:40:55,240 --> 00:40:58,800
review those two procedures
through a design example.

660
00:40:58,800 --> 00:40:59,300
Thank you.

661
00:41:04,100 --> 00:41:07,150
[MUSIC PLAYING]