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[MUSIC PLAYING]

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ALAN OPPENHEIM: In the
last several lectures,

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00:00:57,270 --> 00:01:00,970
we've considered a
number of procedures

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00:01:00,970 --> 00:01:06,610
for designing digital
filters for the class

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00:01:06,610 --> 00:01:09,730
of infinite impulse
response filters.

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00:01:09,730 --> 00:01:12,880
And the two primary
design techniques

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that we considered, in
fact, the two useful design

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00:01:16,120 --> 00:01:19,240
techniques that we developed,
were the techniques

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00:01:19,240 --> 00:01:24,190
which we refer to as
impulse invariance, and also

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00:01:24,190 --> 00:01:27,400
the bilinear transformation.

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00:01:27,400 --> 00:01:31,180
You recall, hopefully, that
both of these techniques

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00:01:31,180 --> 00:01:37,560
are techniques that correspond
to mapping a continuous time

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00:01:37,560 --> 00:01:42,120
or analog filter to
a digital filter.

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00:01:42,120 --> 00:01:43,660
What I'd like to
do in this lecture

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00:01:43,660 --> 00:01:47,860
is consider an
example of the design

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00:01:47,860 --> 00:01:52,180
of a digital low-pass filter
using each of these techniques,

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00:01:52,180 --> 00:01:55,900
using impulse invariance and
the bilinear transformation.

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00:01:55,900 --> 00:01:59,740
And hopefully this will give
you a better feeling for how

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00:01:59,740 --> 00:02:03,080
both of these techniques work.

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00:02:03,080 --> 00:02:09,050
The class of analog filters
that I would like to use

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00:02:09,050 --> 00:02:13,200
are the class of so-called
analog Butterworth filters,

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00:02:13,200 --> 00:02:16,970
which are a useful
class of lowpass analog

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00:02:16,970 --> 00:02:20,525
and digital filters, and also
a relatively simple class.

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00:02:23,120 --> 00:02:28,360
To first of all, define
the class of analog filters

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00:02:28,360 --> 00:02:31,270
that we're considering,
they are the class

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00:02:31,270 --> 00:02:33,550
of analog Butterworth filters.

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00:02:33,550 --> 00:02:39,970
And this class of filters is
defined by the analog system

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00:02:39,970 --> 00:02:45,820
function H sub a of j omega
magnitude squared of the form

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00:02:45,820 --> 00:02:51,910
1 over 1 plus j omega divided
by j omega sub c to the 2N.

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00:02:51,910 --> 00:02:55,750
Now, there are two parameters
in this expression.

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00:02:55,750 --> 00:03:00,370
One is what's referred to often
as the cutoff frequency, omega

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00:03:00,370 --> 00:03:01,720
sub c.

40
00:03:01,720 --> 00:03:04,600
And the second is the
order of the filter,

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00:03:04,600 --> 00:03:07,720
namely N. Let me
remind you that we're

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00:03:07,720 --> 00:03:10,660
looking at here an expression
for the squared magnitude

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00:03:10,660 --> 00:03:11,970
function.

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00:03:11,970 --> 00:03:15,910
And so we have a 2N here.

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00:03:15,910 --> 00:03:18,610
But in fact, N is
what corresponds

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00:03:18,610 --> 00:03:21,490
to the order of the filter.

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00:03:21,490 --> 00:03:24,000
Well, if we look at
this frequency response

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00:03:24,000 --> 00:03:30,390
characteristic, we observe
that at omega equal to 0,

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00:03:30,390 --> 00:03:35,180
the gain or the magnitude
squared is equal to unity.

50
00:03:35,180 --> 00:03:41,810
At omega equal to omega sub c,
the j's of course cancel out.

51
00:03:41,810 --> 00:03:46,610
At omega equal to omega sub c,
this term then is equal to 1,

52
00:03:46,610 --> 00:03:50,660
and we have for the magnitude
squared function, a half.

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00:03:50,660 --> 00:03:53,210
Or for the magnitude
function, which

54
00:03:53,210 --> 00:03:56,810
is what we're looking
at here, 1 divided

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00:03:56,810 --> 00:03:58,910
by the square root of 2.

56
00:03:58,910 --> 00:04:01,820
So at omega equal
to omega sub c,

57
00:04:01,820 --> 00:04:06,410
an analog Butterworth filter has
a magnitude which is down by 1

58
00:04:06,410 --> 00:04:08,570
over the square root of 2.

59
00:04:08,570 --> 00:04:11,510
And the other important
characteristic

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00:04:11,510 --> 00:04:17,000
is that the frequency response
is a monotonic function

61
00:04:17,000 --> 00:04:18,589
of capital Omega.

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00:04:18,589 --> 00:04:23,720
So that as capital Omega
increases from 0 to infinity,

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00:04:23,720 --> 00:04:26,180
the magnitude or the
frequency response

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00:04:26,180 --> 00:04:28,990
is a monotonic function.

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00:04:28,990 --> 00:04:32,250
Well, we observed that the
cutoff frequency, the parameter

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00:04:32,250 --> 00:04:39,500
omega sub c, adjusts in a
sense where we might divide

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00:04:39,500 --> 00:04:43,480
the frequency characteristic
into passband, transition band,

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00:04:43,480 --> 00:04:45,530
and stopband.

69
00:04:45,530 --> 00:04:48,090
We have another parameter,
which is the parameter

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00:04:48,090 --> 00:04:50,540
N, the order of the filter.

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00:04:50,540 --> 00:04:55,220
And basically the effect
that the parameter N has

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00:04:55,220 --> 00:05:00,500
is to affect the shape of this
frequency response in the sense

73
00:05:00,500 --> 00:05:06,590
that if capital N is larger,
the frequency response tends

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00:05:06,590 --> 00:05:12,650
to be flatter longer, so that
for a larger N than what I've

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00:05:12,650 --> 00:05:15,590
drawn with the black curve,
the frequency response might

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00:05:15,590 --> 00:05:21,150
look more like this and
then drop off sharper.

77
00:05:21,150 --> 00:05:25,500
Whereas if capital N
is smaller, then it

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00:05:25,500 --> 00:05:28,350
will be less flat
in the passband

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00:05:28,350 --> 00:05:30,430
and drop off more slowly.

80
00:05:30,430 --> 00:05:34,760
So in fact, the way
that I've drawn it here,

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00:05:34,760 --> 00:05:38,480
this is the behavior of the
frequency response curve

82
00:05:38,480 --> 00:05:44,780
as the parameter
capital N increases.

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00:05:44,780 --> 00:05:47,630
So, the higher the
order of the filter,

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00:05:47,630 --> 00:05:52,040
the sharper the
drop from what we

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00:05:52,040 --> 00:05:56,270
could consider as a passband
region into a stopband region.

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00:05:56,270 --> 00:05:59,870
And obviously then, the
sharper the filter we want,

87
00:05:59,870 --> 00:06:04,430
the higher order
filter we would design.

88
00:06:04,430 --> 00:06:07,880
Now, to look at
the pole 0 pattern

89
00:06:07,880 --> 00:06:13,470
or the 0 pattern really
of the Butterworth filter,

90
00:06:13,470 --> 00:06:19,070
we can convert this expression
into an expression relating

91
00:06:19,070 --> 00:06:22,160
the magnitude squared
function expressed

92
00:06:22,160 --> 00:06:26,240
in terms of the Laplace
transform variable s.

93
00:06:26,240 --> 00:06:31,250
And so this expression
corresponds to the product

94
00:06:31,250 --> 00:06:36,500
H sub a of s times H sub a
of minus s, evaluated at s

95
00:06:36,500 --> 00:06:41,840
equals j omega, in
which case the j omega

96
00:06:41,840 --> 00:06:46,755
is what corresponds to s, and
so H sub a of s times H sub

97
00:06:46,755 --> 00:06:51,590
a of minus s is of
the form 1 over 1 plus

98
00:06:51,590 --> 00:06:58,650
s divided by j omega sub
c raised to the 2N power.

99
00:06:58,650 --> 00:07:05,040
Well, what we recognize is
that this expression has poles,

100
00:07:05,040 --> 00:07:07,200
if we solve this equation.

101
00:07:07,200 --> 00:07:10,470
This expression has poles
at of course the roots

102
00:07:10,470 --> 00:07:15,690
of the denominator polynomial
or at s sub p equal to minus 1

103
00:07:15,690 --> 00:07:19,830
to the over 2N
times j omega sub c.

104
00:07:19,830 --> 00:07:23,040
That's simply solving the
equation with this denominator

105
00:07:23,040 --> 00:07:26,830
polynomial equal to 0.

106
00:07:26,830 --> 00:07:30,460
Well, we can see
more specifically

107
00:07:30,460 --> 00:07:34,990
where these poles
lie in the s plane

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00:07:34,990 --> 00:07:40,870
by recognizing that these
are the 2N roots of minus 1.

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00:07:40,870 --> 00:07:44,020
And then we have a
j, which corresponds

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00:07:44,020 --> 00:07:48,700
to e to the j pi over 2.

111
00:07:48,700 --> 00:07:52,810
So we can rewrite this
expression for the poles

112
00:07:52,810 --> 00:07:59,450
as s sub p equal to e
to the j pi over 2N.

113
00:08:03,336 --> 00:08:09,450
That's one of the roots of
minus 1, plus k times 2 pi

114
00:08:09,450 --> 00:08:11,380
where k is an integer.

115
00:08:11,380 --> 00:08:14,790
This generates the
roots of minus 1.

116
00:08:14,790 --> 00:08:17,310
We need to multiply by
j, and we can do that

117
00:08:17,310 --> 00:08:20,220
by multiplying by e to
the j pi over 2, which

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00:08:20,220 --> 00:08:25,680
is equal to j times omega sub
c, which is the cutoff frequency

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00:08:25,680 --> 00:08:28,680
parameter for the
Butterworth filter.

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00:08:28,680 --> 00:08:32,370
So this then specifies
where the pole locations

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00:08:32,370 --> 00:08:34,470
are for the Butterworth filter.

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00:08:34,470 --> 00:08:38,940
For example, if we
consider the case,

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00:08:38,940 --> 00:08:45,570
let's say with n equal to
3, k let's say equal to 0.

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00:08:45,570 --> 00:08:49,105
That says then that
we have a pole at e

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00:08:49,105 --> 00:08:56,300
to the j pi divided by 6,
e to the j pi over 6, times

126
00:08:56,300 --> 00:08:58,320
e to the j pi over 2.

127
00:08:58,320 --> 00:09:01,670
In other words, we have
a pole in the s plane

128
00:09:01,670 --> 00:09:05,570
at an angle of pi
over 6 plus pi over 2,

129
00:09:05,570 --> 00:09:08,870
which corresponds
to this location.

130
00:09:08,870 --> 00:09:14,750
Then we'll have poles
distributed in angle

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00:09:14,750 --> 00:09:18,410
around a circle in the
s plane, distributed

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00:09:18,410 --> 00:09:22,400
in an angle where the
angular spacing is pi over 3.

133
00:09:22,400 --> 00:09:28,010
The pi over 3 arising from
a consideration of 2 pi k

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00:09:28,010 --> 00:09:32,930
divided by 2N, where in
this case, N is equal to 6.

135
00:09:32,930 --> 00:09:39,100
So these poles are distributed
equally spaced in angle

136
00:09:39,100 --> 00:09:42,070
on a circle in the s plane.

137
00:09:42,070 --> 00:09:45,810
For N equal to 3, the
angular spacing is pi over 3.

138
00:09:45,810 --> 00:09:50,500
For k equal to 0 that first
pole corresponds to this one,

139
00:09:50,500 --> 00:09:53,040
and then we continue around.

140
00:09:53,040 --> 00:09:56,640
Now these are poles
distributed on a circle, which

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00:09:56,640 --> 00:09:59,910
tends to look like the z plane.

142
00:09:59,910 --> 00:10:03,360
But I stress that what
we're talking about here

143
00:10:03,360 --> 00:10:04,410
is the s plane.

144
00:10:04,410 --> 00:10:07,830
In particular, we're talking
about the analog frequency

145
00:10:07,830 --> 00:10:12,730
response design as the class
of analog Butterworth filters,

146
00:10:12,730 --> 00:10:15,630
which we're then going to
map to digital filters.

147
00:10:15,630 --> 00:10:18,840
So this is a circle
in the s plane.

148
00:10:18,840 --> 00:10:21,870
It's often referred to as
the Butterworth circle.

149
00:10:21,870 --> 00:10:28,320
And it has a radius which
is equal to omega sub c.

150
00:10:28,320 --> 00:10:33,240
In designing an analog
Butterworth filter then,

151
00:10:33,240 --> 00:10:38,720
the procedure is to
appropriately choose

152
00:10:38,720 --> 00:10:41,490
the parameters omega
sub c and capital

153
00:10:41,490 --> 00:10:45,840
N. Those are the parameters of
the analog Butterworth filter.

154
00:10:45,840 --> 00:10:52,530
And then from the poles
distributed on the Butterworth

155
00:10:52,530 --> 00:10:58,240
circle, this gives us the pole
locations for the product,

156
00:10:58,240 --> 00:11:04,970
H sub a of s times
H sub a of minus s.

157
00:11:04,970 --> 00:11:09,500
We can then simply associate
with the transfer function

158
00:11:09,500 --> 00:11:11,180
H sub a of s.

159
00:11:11,180 --> 00:11:13,490
The poles in the
left half plane,

160
00:11:13,490 --> 00:11:18,200
to generate a stable filter,
and then the ones which

161
00:11:18,200 --> 00:11:21,740
are contributed by the
factor H sub a of minus s

162
00:11:21,740 --> 00:11:26,200
are the corresponding image
poles in the right half plane.

163
00:11:26,200 --> 00:11:31,800
The procedure then is to design
a squared magnitude function.

164
00:11:31,800 --> 00:11:33,810
Talking about a
Butterworth filter,

165
00:11:33,810 --> 00:11:36,375
we choose these
parameters omega sub c

166
00:11:36,375 --> 00:11:39,300
and N. That tells us where
the poles are located

167
00:11:39,300 --> 00:11:42,150
on the Butterworth circle.

168
00:11:42,150 --> 00:11:46,320
From the pole locations for
the squared magnitude function,

169
00:11:46,320 --> 00:11:50,700
then we can factor that into
H sub a of s times H sub

170
00:11:50,700 --> 00:11:53,760
a of minus s, by
choosing the poles that

171
00:11:53,760 --> 00:11:56,820
fall in the left half
plane, to give us

172
00:11:56,820 --> 00:12:01,631
a causal, stable analog filter.

173
00:12:01,631 --> 00:12:02,130
All right.

174
00:12:02,130 --> 00:12:05,970
Well, that is an analog filter.

175
00:12:05,970 --> 00:12:07,830
What we would like
to talk about is

176
00:12:07,830 --> 00:12:10,500
the design of a digital filter.

177
00:12:10,500 --> 00:12:13,140
And the procedure that
we're going to go through

178
00:12:13,140 --> 00:12:18,600
is to choose a set of specs
for the digital filter,

179
00:12:18,600 --> 00:12:22,230
and then decide on what the
corresponding specifications

180
00:12:22,230 --> 00:12:25,170
should be for the analog filter,
depending on whether we're

181
00:12:25,170 --> 00:12:29,130
using impulse invariance or
the bilinear transformation,

182
00:12:29,130 --> 00:12:32,640
design the analog
filter, and then map

183
00:12:32,640 --> 00:12:36,300
the result to a
digital filter using

184
00:12:36,300 --> 00:12:39,300
one of those two techniques.

185
00:12:39,300 --> 00:12:44,070
So let's choose a set of
digital filter specifications

186
00:12:44,070 --> 00:12:50,490
for a digital lowpass
filter with a passband which

187
00:12:50,490 --> 00:12:55,710
occurs for frequencies
now in the digital domain,

188
00:12:55,710 --> 00:13:03,150
for frequencies below 0.2 pi,
and a stopband region that

189
00:13:03,150 --> 00:13:07,050
occurs for frequencies
greater than 0.3 pi,

190
00:13:07,050 --> 00:13:10,110
and the transition band then,
of course, is the region

191
00:13:10,110 --> 00:13:13,370
in between those.

192
00:13:13,370 --> 00:13:20,060
So, we would like to design
a digital filter, which

193
00:13:20,060 --> 00:13:24,020
stays within some limits
in this passband region,

194
00:13:24,020 --> 00:13:28,670
drops below some limit, delta
sub s in the stopband region.

195
00:13:28,670 --> 00:13:32,810
And we'll choose as the
limits in the passband,

196
00:13:32,810 --> 00:13:37,100
unity for the upper limit and 1
minus delta sub p for the lower

197
00:13:37,100 --> 00:13:38,390
limit.

198
00:13:38,390 --> 00:13:42,860
Where delta 1 minus
delta sub p might not

199
00:13:42,860 --> 00:13:47,100
and in fact it won't turn out to
be 1 over the square root of 2,

200
00:13:47,100 --> 00:13:50,390
in other words, this
cutoff frequency

201
00:13:50,390 --> 00:13:54,725
will not necessarily correspond
to the parameter omega sub

202
00:13:54,725 --> 00:13:57,850
c for the Butterworth filter.

203
00:13:57,850 --> 00:13:58,350
All right.

204
00:13:58,350 --> 00:14:01,950
Well, so let's choose a
set of specifications.

205
00:14:01,950 --> 00:14:05,250
The ones that I've chosen
are the requirement

206
00:14:05,250 --> 00:14:11,970
that the frequency response
characteristic be within 1db

207
00:14:11,970 --> 00:14:14,340
of unity in the passband.

208
00:14:14,340 --> 00:14:17,160
In other words, the requirement
that 1 minus deltas sub

209
00:14:17,160 --> 00:14:20,970
p be greater than or
equal to minus 1db.

210
00:14:20,970 --> 00:14:26,080
So this deviation
must be less than 1db.

211
00:14:26,080 --> 00:14:28,800
And then the requirement
that in the stopband

212
00:14:28,800 --> 00:14:31,260
the frequency response
characteristic

213
00:14:31,260 --> 00:14:35,550
is below minus 15db.

214
00:14:35,550 --> 00:14:38,580
In other words, the
stopband attenuation

215
00:14:38,580 --> 00:14:45,090
for frequencies greater than 0.3
pi is greater than minus 15db.

216
00:14:45,090 --> 00:14:48,270
It's further down
than minus 15db.

217
00:14:48,270 --> 00:14:52,500
So delta sub s is less than
or equal to minus 15db.

218
00:14:52,500 --> 00:14:55,980
The one other specification
that we'll impose,

219
00:14:55,980 --> 00:14:59,220
or in fact, it's really
a consequence that

220
00:14:59,220 --> 00:15:02,070
will result from using
a Butterworth filter,

221
00:15:02,070 --> 00:15:07,800
is the fact that the frequency
characteristic is monotonic.

222
00:15:07,800 --> 00:15:12,030
So we have the passband
edge, the stopband edge,

223
00:15:12,030 --> 00:15:15,330
the passband ripple
or deviation,

224
00:15:15,330 --> 00:15:18,700
and the stopband
ripple or deviation.

225
00:15:18,700 --> 00:15:24,240
And we can, of course,
rewrite these specifications

226
00:15:24,240 --> 00:15:30,240
in terms of requirements on the
transfer function or frequency

227
00:15:30,240 --> 00:15:36,420
response, the magnitude
of the frequency response.

228
00:15:36,420 --> 00:15:41,760
The requirement that the
deviation be less than 1db

229
00:15:41,760 --> 00:15:45,510
in the passband is
the requirement then

230
00:15:45,510 --> 00:15:49,680
that 20 log to the base 10 of
the magnitude of the frequency

231
00:15:49,680 --> 00:15:51,030
response--

232
00:15:51,030 --> 00:15:53,270
that's the deviation db--

233
00:15:53,270 --> 00:15:58,425
be greater than or equal to
minus 1, or that the magnitude

234
00:15:58,425 --> 00:16:03,650
be greater than or equal
to 10 to the minus 0.05.

235
00:16:03,650 --> 00:16:06,960
And of course, we wanted
also to be less than unity,

236
00:16:06,960 --> 00:16:09,110
and that's an implied condition.

237
00:16:09,110 --> 00:16:11,210
So it must be less
than unity, and it

238
00:16:11,210 --> 00:16:14,740
must be greater than or
equal to 10 to the minus 0.05

239
00:16:14,740 --> 00:16:19,100
for frequencies
from 0 to 0.2 pi.

240
00:16:19,100 --> 00:16:21,560
So in particular,
since it's monotonic,

241
00:16:21,560 --> 00:16:28,280
we'll impose that condition
at the frequency point 0.2 pi,

242
00:16:28,280 --> 00:16:30,520
corresponding to the
end of the stopband

243
00:16:30,520 --> 00:16:33,590
and the beginning of
the transition band.

244
00:16:33,590 --> 00:16:36,770
The second condition
that we impose

245
00:16:36,770 --> 00:16:40,980
is the condition
that by the time

246
00:16:40,980 --> 00:16:45,340
we get to omega
equal to 0.3 pi, we

247
00:16:45,340 --> 00:16:49,780
would like the frequency
response to be down at least 15

248
00:16:49,780 --> 00:16:52,810
db, and more if possible.

249
00:16:52,810 --> 00:16:56,300
But the condition
is that the 20 log

250
00:16:56,300 --> 00:16:59,980
to the base 10 of the magnitude
at omega equal to 0.3 pi

251
00:16:59,980 --> 00:17:04,310
must be less than or
equal to minus 15.

252
00:17:04,310 --> 00:17:07,720
So this says we must be
further down than 15 db,

253
00:17:07,720 --> 00:17:11,433
or we can translate that into
a condition on the magnitude,

254
00:17:11,433 --> 00:17:13,599
so that the magnitude of
that frequency is less than

255
00:17:13,599 --> 00:17:18,790
or equal to 10 to
the minus 0.75.

256
00:17:18,790 --> 00:17:22,319
Well, let's consider two--

257
00:17:22,319 --> 00:17:25,589
we want to consider
two design procedures.

258
00:17:25,589 --> 00:17:30,360
Let's first consider the impulse
invariant design procedure,

259
00:17:30,360 --> 00:17:34,590
in which case we would like to
convert these specifications

260
00:17:34,590 --> 00:17:40,800
on the digital filter to
corresponding specifications

261
00:17:40,800 --> 00:17:44,050
on the analog filter.

262
00:17:44,050 --> 00:17:47,500
Now, we know that
from discussions

263
00:17:47,500 --> 00:17:52,120
that we've had previously that
if we have an analog frequency

264
00:17:52,120 --> 00:17:57,130
response and we apply impulse
invariance to the impulse

265
00:17:57,130 --> 00:18:01,630
response, the digital
frequency response consists

266
00:18:01,630 --> 00:18:05,920
of the analog frequency
response scaled in frequency so

267
00:18:05,920 --> 00:18:09,940
that capital Omega is equal
to little omega divided

268
00:18:09,940 --> 00:18:16,775
by capital T. And then
that scaled replica of H

269
00:18:16,775 --> 00:18:23,950
sub a added to itself delayed
by little omega equal to 2 pi, 4

270
00:18:23,950 --> 00:18:24,790
pi, et cetera.

271
00:18:24,790 --> 00:18:28,210
That is multiples of 2 pi.

272
00:18:28,210 --> 00:18:36,220
So, that means then that
the digital filter is scaled

273
00:18:36,220 --> 00:18:39,310
repeated replicas of
the analog filter,

274
00:18:39,310 --> 00:18:44,380
and so we would choose
the analog filter

275
00:18:44,380 --> 00:18:48,760
to have similar specifications
to the digital filter,

276
00:18:48,760 --> 00:18:53,830
but with the corresponding
frequency parameters dictated

277
00:18:53,830 --> 00:18:57,610
by this transformation, capital
Omega equal to little omega,

278
00:18:57,610 --> 00:19:05,260
divided by capital T. Well, that
of course neglects aliasing,

279
00:19:05,260 --> 00:19:08,980
and in fact as we go through
the impulse invariant design,

280
00:19:08,980 --> 00:19:11,890
we'll initially
neglect aliasing.

281
00:19:11,890 --> 00:19:17,800
And a rationale for doing this
or a reason why, in fact, it's

282
00:19:17,800 --> 00:19:20,770
not unreasonable to
do that, will emerge

283
00:19:20,770 --> 00:19:22,439
as we go through the example.

284
00:19:22,439 --> 00:19:23,980
So for the time
being, in fact, we're

285
00:19:23,980 --> 00:19:27,310
going to neglect
aliasing, so that we'll

286
00:19:27,310 --> 00:19:31,030
think of the analog and
digital frequency responses

287
00:19:31,030 --> 00:19:34,450
as simply related by this
substitution of frequency

288
00:19:34,450 --> 00:19:37,210
variable, capital Omega
equal to little omega

289
00:19:37,210 --> 00:19:40,480
divided by capital T.

290
00:19:40,480 --> 00:19:49,230
Well, that leads us then to
the analog frequency response

291
00:19:49,230 --> 00:19:52,720
which I've depicted here.

292
00:19:52,720 --> 00:19:57,220
We had a factor of 1 over
capital T in the amplitude

293
00:19:57,220 --> 00:20:00,230
of this expression.

294
00:20:00,230 --> 00:20:03,340
So for the digital
filter to result

295
00:20:03,340 --> 00:20:06,970
with an amplitude
of unity at dc,

296
00:20:06,970 --> 00:20:11,425
we would scale the analog
frequency response by capital

297
00:20:11,425 --> 00:20:18,620
T. So the important values
then in capital Omega

298
00:20:18,620 --> 00:20:23,870
are the ones corresponding to
little omega equal to 0.2 pi,

299
00:20:23,870 --> 00:20:27,290
or since capital Omega is
equal to little omega divided

300
00:20:27,290 --> 00:20:33,170
by capital T. We have a passband
edge occurring at capital

301
00:20:33,170 --> 00:20:36,310
Omega equal to 0.2
pi over capital T,

302
00:20:36,310 --> 00:20:42,250
and a stopband edge equal
to 0.3 pi over capital T.

303
00:20:42,250 --> 00:20:46,660
At this passband edge,
according to the specifications

304
00:20:46,660 --> 00:20:49,600
on the digital
filter that we want,

305
00:20:49,600 --> 00:20:53,530
we would like the frequency
response for frequencies

306
00:20:53,530 --> 00:20:54,500
below this--

307
00:20:54,500 --> 00:20:56,560
In other words,
in the passband--

308
00:20:56,560 --> 00:21:00,130
to be between capital T
and capital T times 10

309
00:21:00,130 --> 00:21:02,080
to the minus 0.05.

310
00:21:02,080 --> 00:21:04,630
That's the guarantee that
we have less than 1 db

311
00:21:04,630 --> 00:21:06,610
ripple in the passband.

312
00:21:06,610 --> 00:21:10,000
And for frequencies
greater than 0.3 pi,

313
00:21:10,000 --> 00:21:12,940
we would like to be
below capital T times 10

314
00:21:12,940 --> 00:21:14,950
to the minus 0.75.

315
00:21:14,950 --> 00:21:19,330
That's the guarantee that
we're down 15 db or more

316
00:21:19,330 --> 00:21:21,880
in the stopband.

317
00:21:21,880 --> 00:21:26,620
The way I've drawn it here,
I've implied that I will exactly

318
00:21:26,620 --> 00:21:29,980
meet the specifications
at the passband edge,

319
00:21:29,980 --> 00:21:31,690
and exactly meet
the specifications

320
00:21:31,690 --> 00:21:33,550
at the stopband edge.

321
00:21:33,550 --> 00:21:39,550
You can imagine, of course,
that if I did that and then

322
00:21:39,550 --> 00:21:44,350
converted to a digital filter
using impulse invariance

323
00:21:44,350 --> 00:21:48,340
that there would be an
effect due to aliasing, which

324
00:21:48,340 --> 00:21:51,610
in fact would result
in my not meeting

325
00:21:51,610 --> 00:21:55,450
exactly the specifications
at the stopband edge.

326
00:21:55,450 --> 00:21:59,870
And that's a point that we'll
deal with in a few minutes.

327
00:21:59,870 --> 00:22:02,740
All right, well now we
have our specifications.

328
00:22:02,740 --> 00:22:06,220
Let me incidentally
point out that there's

329
00:22:06,220 --> 00:22:10,990
this parameter capital T
that we're carrying around.

330
00:22:10,990 --> 00:22:15,730
And as I'll justify partly
on the basis of this example,

331
00:22:15,730 --> 00:22:20,510
the parameter capital T is
an irrelevant parameter.

332
00:22:20,510 --> 00:22:22,810
It's a parameter
that will disappear.

333
00:22:22,810 --> 00:22:26,800
We'll carry it through with
us for this one example.

334
00:22:26,800 --> 00:22:29,800
But for other
examples and problems

335
00:22:29,800 --> 00:22:33,340
that you work through
the study guide,

336
00:22:33,340 --> 00:22:35,320
you should have
absorbed the fact

337
00:22:35,320 --> 00:22:38,080
that capital T is irrelevant,
and in fact capital

338
00:22:38,080 --> 00:22:40,720
T can be chosen equal to unity.

339
00:22:40,720 --> 00:22:43,120
But let's carry it
through in this example,

340
00:22:43,120 --> 00:22:47,300
and just see how it disappears.

341
00:22:47,300 --> 00:22:47,800
All right.

342
00:22:47,800 --> 00:22:53,710
Well, so we want the
analog Butterworth filter

343
00:22:53,710 --> 00:22:56,360
meaning these specifications.

344
00:22:56,360 --> 00:22:59,830
And we want it at
omega equal to 0

345
00:22:59,830 --> 00:23:03,460
to have a gain magnitude squared
function to be T squared,

346
00:23:03,460 --> 00:23:06,100
so that the magnitude
is T. So here we

347
00:23:06,100 --> 00:23:11,590
have the form then of the
analog Butterworth filter.

348
00:23:11,590 --> 00:23:15,310
And in order to meet
these specifications

349
00:23:15,310 --> 00:23:20,930
at this frequency and this one,
we then have two equations.

350
00:23:20,930 --> 00:23:24,650
One equation says
that at capital Omega

351
00:23:24,650 --> 00:23:28,680
equal to 0.2 pi
over capital T, we

352
00:23:28,680 --> 00:23:35,450
would like the value of
the magnitude function

353
00:23:35,450 --> 00:23:38,480
to be T times 10
to the minus 0.05,

354
00:23:38,480 --> 00:23:41,450
which implies that
this denominator should

355
00:23:41,450 --> 00:23:46,438
be equal to 10 to the 0.1.

356
00:23:46,438 --> 00:23:50,160
Second of all, at the
stopband frequency,

357
00:23:50,160 --> 00:23:54,510
0.3 pi divided by
capital T, the fact

358
00:23:54,510 --> 00:24:00,480
that we want the magnitude to
be T times 10 to the minus 0.75

359
00:24:00,480 --> 00:24:05,410
implies the magnitude squared
is of course the square of that.

360
00:24:05,410 --> 00:24:12,300
And we want then the denominator
to be equal to 10 to the 1.5.

361
00:24:12,300 --> 00:24:15,500
So we have two points that
we're trying to nail down.

362
00:24:15,500 --> 00:24:20,020
We have two parameters,
omega sub c and N.

363
00:24:20,020 --> 00:24:22,330
From these two points that
we're trying to nail down,

364
00:24:22,330 --> 00:24:24,520
we get two equations.

365
00:24:24,520 --> 00:24:28,510
And we can now solve these
equations for our parameters,

366
00:24:28,510 --> 00:24:31,690
omega sub c and capital N.

367
00:24:31,690 --> 00:24:35,200
Well, if we do that, and it's
a little tedious to do it,

368
00:24:35,200 --> 00:24:37,150
let me point out
incidentally since we're

369
00:24:37,150 --> 00:24:39,940
trying to keep track
of this capital T,

370
00:24:39,940 --> 00:24:44,080
that capital T drops
down into the numerator

371
00:24:44,080 --> 00:24:49,300
and so, in fact, we can solve
for capital N and omega sub

372
00:24:49,300 --> 00:24:51,910
c times capital
T. In other words,

373
00:24:51,910 --> 00:24:55,570
we can just move this
down here algebraically.

374
00:24:55,570 --> 00:24:58,990
And so we have then instead
of the parameter omega sub c,

375
00:24:58,990 --> 00:25:03,730
we have the parameter omega
sub c times capital T.

376
00:25:03,730 --> 00:25:07,090
If you grind through
these two equations,

377
00:25:07,090 --> 00:25:12,640
what results is a value for N
equal to 5.8858 and probably

378
00:25:12,640 --> 00:25:14,170
some other digits.

379
00:25:14,170 --> 00:25:20,130
And omega sub c capital
T equal to 0.70474.

380
00:25:20,130 --> 00:25:24,220
And we could imagine
that at this point

381
00:25:24,220 --> 00:25:31,130
we're done, except
that the parameter N

382
00:25:31,130 --> 00:25:35,410
corresponding to the order
of the filter, of course,

383
00:25:35,410 --> 00:25:38,000
has to be an integer.

384
00:25:38,000 --> 00:25:43,000
So, what that means is that
we can't use the value of N

385
00:25:43,000 --> 00:25:44,920
as I've indicated here.

386
00:25:44,920 --> 00:25:47,790
We need an integer value.

387
00:25:47,790 --> 00:25:51,500
Also if we were to round
this down or truncate

388
00:25:51,500 --> 00:25:55,860
it down to a choice
of N equal to 5,

389
00:25:55,860 --> 00:25:58,410
then since the
order is lower you

390
00:25:58,410 --> 00:26:00,480
can imagine that the
filter characteristics

391
00:26:00,480 --> 00:26:02,200
won't be as good.

392
00:26:02,200 --> 00:26:06,540
In other words, we're not going
to meet these specifications.

393
00:26:06,540 --> 00:26:12,600
So the other choice then is to
use instead of this value of N,

394
00:26:12,600 --> 00:26:15,540
use a value of N equal to 6.

395
00:26:15,540 --> 00:26:21,190
In other words, round up
to the next integer value.

396
00:26:21,190 --> 00:26:26,550
And if we do that then
these two equations

397
00:26:26,550 --> 00:26:30,010
will no longer be satisfied.

398
00:26:30,010 --> 00:26:32,590
We're now using a
higher order filter

399
00:26:32,590 --> 00:26:35,550
than these equations
would dictate.

400
00:26:35,550 --> 00:26:37,380
Of course, we're using
an integer order,

401
00:26:37,380 --> 00:26:39,120
because that's what we need.

402
00:26:39,120 --> 00:26:44,710
We're using a
sixth order filter.

403
00:26:44,710 --> 00:26:51,040
And what that means is that
if we choose N equal to 6,

404
00:26:51,040 --> 00:26:56,130
we can satisfy this equation,
or we can satisfy this equation,

405
00:26:56,130 --> 00:26:57,850
but we can't satisfy
both of them.

406
00:27:00,480 --> 00:27:03,570
Well, which one
should we try to meet

407
00:27:03,570 --> 00:27:07,440
and which one should we exceed?

408
00:27:07,440 --> 00:27:09,840
Well, we know that
with impulse invariance

409
00:27:09,840 --> 00:27:14,460
there is the issue of aliasing.

410
00:27:14,460 --> 00:27:17,850
And so in fact, in
impulse invariant design

411
00:27:17,850 --> 00:27:22,530
it will often happen that
although we design exactly

412
00:27:22,530 --> 00:27:25,370
to meet the stopband
specifications,

413
00:27:25,370 --> 00:27:27,030
the stopband
specifications won't

414
00:27:27,030 --> 00:27:29,400
be met because of aliasing.

415
00:27:29,400 --> 00:27:33,500
Consequently, given the
flexibility that we have,

416
00:27:33,500 --> 00:27:36,870
due to the fact that we can
actually implement a higher

417
00:27:36,870 --> 00:27:38,610
order filter than
these equations

418
00:27:38,610 --> 00:27:42,420
dictate, to compensate
for aliasing

419
00:27:42,420 --> 00:27:47,580
we can exceed the
stopband specifications

420
00:27:47,580 --> 00:27:51,840
and meet exactly the
passband specifications.

421
00:27:51,840 --> 00:27:56,940
So if we substitute
for N, N equals 6

422
00:27:56,940 --> 00:28:03,280
into this equation
that then requires

423
00:28:03,280 --> 00:28:07,180
that omega sub c T be
such that we exactly

424
00:28:07,180 --> 00:28:11,950
meet the passband
specifications, in which case

425
00:28:11,950 --> 00:28:13,600
this equation won't be met.

426
00:28:13,600 --> 00:28:15,970
In fact, what will
happen at the stopband

427
00:28:15,970 --> 00:28:19,240
edge is that the
stopband specifications

428
00:28:19,240 --> 00:28:21,490
will be exceeded.

429
00:28:21,490 --> 00:28:27,580
If we do this, then we
have for the equation

430
00:28:27,580 --> 00:28:31,210
that we solve for
omega sub c T from,

431
00:28:31,210 --> 00:28:37,150
the equation which specified the
passband edge, which is where

432
00:28:37,150 --> 00:28:39,310
we're going to meet the specs.

433
00:28:39,310 --> 00:28:45,370
And what results then is for
omega sub c T, this value,

434
00:28:45,370 --> 00:28:48,490
rather than the previous
value that had been obtained

435
00:28:48,490 --> 00:28:51,670
from the two equations.

436
00:28:51,670 --> 00:28:55,490
Well, from these
parameters then,

437
00:28:55,490 --> 00:29:00,060
we have that omega sub c is
0.7032 divided by capital T.

438
00:29:00,060 --> 00:29:05,610
We can substitute these
parameters into the form

439
00:29:05,610 --> 00:29:09,285
for the squared magnitude
function, H sub a of s times H

440
00:29:09,285 --> 00:29:12,660
sub a of minus s,
and actually it's

441
00:29:12,660 --> 00:29:16,620
this raised to the 2N,
which is the 2 times

442
00:29:16,620 --> 00:29:19,530
sixth power, which
is the 12th power.

443
00:29:19,530 --> 00:29:24,420
So this then is the form for
the analog squared magnitude

444
00:29:24,420 --> 00:29:27,100
function.

445
00:29:27,100 --> 00:29:30,700
Now notice that the
parameter capital T still

446
00:29:30,700 --> 00:29:32,050
hasn't disappeared.

447
00:29:32,050 --> 00:29:32,770
We still have it.

448
00:29:32,770 --> 00:29:35,020
We have it there,
and we have it here.

449
00:29:35,020 --> 00:29:39,270
But notice that when we
substitute in for omega sub c,

450
00:29:39,270 --> 00:29:41,590
this number divided
by capital T if you

451
00:29:41,590 --> 00:29:45,340
track through that
algebra, capital T

452
00:29:45,340 --> 00:29:50,200
reappears in this expression
as a multiplier on s.

453
00:29:50,200 --> 00:29:53,770
That is we had 0.7032
divided by capital T.

454
00:29:53,770 --> 00:29:56,620
That gets reflected back
up into this numerator,

455
00:29:56,620 --> 00:29:59,650
so that we have s
times capital T. Well,

456
00:29:59,650 --> 00:30:02,800
let's see what that means.

457
00:30:02,800 --> 00:30:06,500
Suppose that capital
T were equal to 1.

458
00:30:06,500 --> 00:30:12,060
And we got an analog impulse
response, H sub a of T,

459
00:30:12,060 --> 00:30:14,790
or the system
function H sub a of s.

460
00:30:17,470 --> 00:30:21,960
If instead we
reinserted capital T

461
00:30:21,960 --> 00:30:24,880
and considered H
sub a of s times

462
00:30:24,880 --> 00:30:31,610
capital T multiplied
by capital T

463
00:30:31,610 --> 00:30:34,760
that impulse response would
be related to this one

464
00:30:34,760 --> 00:30:37,046
by simply a time scaling.

465
00:30:37,046 --> 00:30:38,420
In other words,
this one would be

466
00:30:38,420 --> 00:30:44,660
replaced by H sub a of t
divided by capital T. Well,

467
00:30:44,660 --> 00:30:49,280
let's look at this and ask
what happens when we sample it

468
00:30:49,280 --> 00:30:52,910
at samples spaced by capital T.

469
00:30:52,910 --> 00:30:55,850
So, we look at this
impulse response

470
00:30:55,850 --> 00:30:59,230
with T replaced
by n times capital

471
00:30:59,230 --> 00:31:02,932
T, as we've specified here.

472
00:31:02,932 --> 00:31:05,015
And we observe that no
matter what we pick capital

473
00:31:05,015 --> 00:31:08,750
T to be that
parameter disappears.

474
00:31:08,750 --> 00:31:11,990
Because little t
equal to n times

475
00:31:11,990 --> 00:31:15,380
capital T gives
us samples H sub a

476
00:31:15,380 --> 00:31:19,730
of n, which are exactly the
same samples that we would

477
00:31:19,730 --> 00:31:23,600
get no matter how we chose
the parameter capital T,

478
00:31:23,600 --> 00:31:26,240
or equivalently whether or
not we chose the parameter

479
00:31:26,240 --> 00:31:30,020
capital T equal to unity.

480
00:31:30,020 --> 00:31:32,570
I think that it requires
a little reflection

481
00:31:32,570 --> 00:31:35,240
to absorb this point.

482
00:31:35,240 --> 00:31:39,980
It's basically tied to the fact
that the frequency response

483
00:31:39,980 --> 00:31:42,440
is specified for
the digital filter,

484
00:31:42,440 --> 00:31:44,420
not for the analog filter.

485
00:31:44,420 --> 00:31:46,940
And specifying the
frequency response

486
00:31:46,940 --> 00:31:49,160
for the digital
filter, in effect,

487
00:31:49,160 --> 00:31:55,340
specifies capital Omega
times capital T. Well,

488
00:31:55,340 --> 00:31:59,480
I leave it to you to
think about this a little.

489
00:31:59,480 --> 00:32:03,880
But the important point is
that, in fact this parameter

490
00:32:03,880 --> 00:32:07,940
T, if we specify
the filter in terms

491
00:32:07,940 --> 00:32:11,270
of digital specifications,
this parameter capital T

492
00:32:11,270 --> 00:32:14,760
is really irrelevant.

493
00:32:14,760 --> 00:32:15,260
All right.

494
00:32:15,260 --> 00:32:20,190
Well, so here is
our analog design.

495
00:32:20,190 --> 00:32:26,190
And we can obtain H
sub a of s from this,

496
00:32:26,190 --> 00:32:28,320
according to the
method that we talked

497
00:32:28,320 --> 00:32:30,720
about previously taking
the left half plane

498
00:32:30,720 --> 00:32:33,120
poles of this expression.

499
00:32:33,120 --> 00:32:35,730
Converting those
to a digital filter

500
00:32:35,730 --> 00:32:39,300
by using the transformation
that we discussed

501
00:32:39,300 --> 00:32:42,030
in the previous two
lectures, expanding,

502
00:32:42,030 --> 00:32:45,300
for example, in terms
of residues and poles,

503
00:32:45,300 --> 00:32:49,380
and then applying the impulse
invariant transformation.

504
00:32:49,380 --> 00:32:59,160
And if we do that, what results
is a digital lowpass filter,

505
00:32:59,160 --> 00:33:02,280
which was the lowpass filter
that we set out to design.

506
00:33:02,280 --> 00:33:06,330
And let me just show how
those specifications,

507
00:33:06,330 --> 00:33:09,510
or how that design comes out.

508
00:33:09,510 --> 00:33:14,040
So this, for the example that
we've been talking about,

509
00:33:14,040 --> 00:33:20,220
is a graph of the frequency
response characteristics.

510
00:33:20,220 --> 00:33:24,570
Here is the magnitude as a
function of digital frequency

511
00:33:24,570 --> 00:33:25,920
omega.

512
00:33:25,920 --> 00:33:29,130
And here is the same
frequency response

513
00:33:29,130 --> 00:33:32,180
plotted in terms of db.

514
00:33:32,180 --> 00:33:37,340
And we see that we had
in the specifications

515
00:33:37,340 --> 00:33:45,400
a passband edge at 0.2 pi,
a stopband edge at 0.3 pi.

516
00:33:45,400 --> 00:33:51,190
At 0.2 pi we asked that
the gain be down by 1db.

517
00:33:51,190 --> 00:33:55,840
And of course, it's hard to tell
exactly that we're down 1db,

518
00:33:55,840 --> 00:33:57,640
but in fact we are.

519
00:33:57,640 --> 00:34:01,930
At omega equal to 0.3
pi, our specification

520
00:34:01,930 --> 00:34:07,900
was that we wanted
to be down by 15 db.

521
00:34:07,900 --> 00:34:10,810
In fact, we overcompensated,
because of the fact

522
00:34:10,810 --> 00:34:14,650
that we had to round the order
of the filter up to an integer

523
00:34:14,650 --> 00:34:15,860
value.

524
00:34:15,860 --> 00:34:20,360
And you can see, in fact, that
if you look at this carefully,

525
00:34:20,360 --> 00:34:25,929
we exceed the specifications
of 0.3 pi by a little bit.

526
00:34:25,929 --> 00:34:30,460
In other words, we're down
by a little more than 15 db.

527
00:34:30,460 --> 00:34:32,620
In fact, for this
particular filter

528
00:34:32,620 --> 00:34:35,980
you can imagine that the
effective aliasing is, in fact,

529
00:34:35,980 --> 00:34:37,239
rather minimal.

530
00:34:37,239 --> 00:34:42,340
Because this frequency response,
by the time we get out to pi

531
00:34:42,340 --> 00:34:43,870
is essentially 0.

532
00:34:43,870 --> 00:34:48,014
Here we're down to minus 75 db.

533
00:34:48,014 --> 00:34:49,389
And that's, of
course, the amount

534
00:34:49,389 --> 00:34:52,040
that would get aliased back in.

535
00:34:52,040 --> 00:34:55,960
So this is the resulting filter
frequency characteristics

536
00:34:55,960 --> 00:34:59,560
for the impulse
invariant design where

537
00:34:59,560 --> 00:35:06,840
we exceed slightly the
stopband edge specifications.

538
00:35:06,840 --> 00:35:10,050
Now, the next thing
that I'd like to go onto

539
00:35:10,050 --> 00:35:14,430
is the corresponding
design, the design

540
00:35:14,430 --> 00:35:17,100
or the corresponding
Butterworth filter,

541
00:35:17,100 --> 00:35:19,770
now using the bilinear
transformation

542
00:35:19,770 --> 00:35:23,280
rather than using
impulse invariance.

543
00:35:23,280 --> 00:35:29,160
I remind you that the basic
idea with the bilinear

544
00:35:29,160 --> 00:35:34,770
transformation, as we've
discussed in the last lectures,

545
00:35:34,770 --> 00:35:44,790
is that we basically choose
the analog specifications

546
00:35:44,790 --> 00:35:48,330
by pre-warping the
critical frequencies

547
00:35:48,330 --> 00:35:53,730
for the digital filter through
the nonlinear distortion

548
00:35:53,730 --> 00:35:57,970
curve corresponding to the
bilinear transformation.

549
00:35:57,970 --> 00:36:01,839
So, we would start here
for the digital filter.

550
00:36:01,839 --> 00:36:04,380
Of course, here we're talking
about a monotonic filter rather

551
00:36:04,380 --> 00:36:05,940
than one that ripples.

552
00:36:05,940 --> 00:36:09,420
But we have a passband
edge at 0.2 pi,

553
00:36:09,420 --> 00:36:12,120
a stopband edge at 0.3 pi.

554
00:36:12,120 --> 00:36:14,850
We reflect those
critical frequencies

555
00:36:14,850 --> 00:36:18,090
through the arctangent
curve to obtain

556
00:36:18,090 --> 00:36:21,090
the corresponding
critical frequencies

557
00:36:21,090 --> 00:36:25,320
for the analog
Butterworth filter.

558
00:36:25,320 --> 00:36:27,990
When the analog Butterworth
filter is designed,

559
00:36:27,990 --> 00:36:31,770
we then obtain the
digital Butterworth filter

560
00:36:31,770 --> 00:36:35,700
by transforming this filter
back through the bilinear

561
00:36:35,700 --> 00:36:38,390
transformation.

562
00:36:38,390 --> 00:36:47,070
So, let's look at how
this design works then.

563
00:36:47,070 --> 00:36:52,590
We have, as we've just
reflected on that curve,

564
00:36:52,590 --> 00:36:58,680
we begin with the digital
frequency specifications,

565
00:36:58,680 --> 00:37:02,310
reflect those through the
bilinear warping curve

566
00:37:02,310 --> 00:37:07,590
to the analog specifications,
and consequently

567
00:37:07,590 --> 00:37:09,840
the critical frequencies--

568
00:37:09,840 --> 00:37:13,650
and let me emphasize
first of all that

569
00:37:13,650 --> 00:37:16,500
just as with impulse
invariance, you

570
00:37:16,500 --> 00:37:18,600
can show that the
parameter capital

571
00:37:18,600 --> 00:37:21,090
T is an irrelevant parameter.

572
00:37:21,090 --> 00:37:24,690
We could, in fact, carry
it all the way through.

573
00:37:24,690 --> 00:37:28,680
And what would happen is that
we would find that no matter

574
00:37:28,680 --> 00:37:33,600
what we chose capital T to
be, the digital filter design

575
00:37:33,600 --> 00:37:35,920
that resulted would
be exactly the same.

576
00:37:35,920 --> 00:37:39,330
So in fact, for this example
I'll work through it simply

577
00:37:39,330 --> 00:37:41,900
with capital T equal to 1.

578
00:37:41,900 --> 00:37:42,400
All right.

579
00:37:42,400 --> 00:37:47,070
With capital T equal
to 1, we reflect

580
00:37:47,070 --> 00:37:51,390
these critical frequencies
into this passband frequency

581
00:37:51,390 --> 00:37:54,060
edge and this stopband edge.

582
00:37:54,060 --> 00:37:57,900
And consequently,
the analog filter--

583
00:37:57,900 --> 00:38:01,290
this is the analog filter--

584
00:38:01,290 --> 00:38:05,640
the analog filter that
we design is one that

585
00:38:05,640 --> 00:38:08,340
meets these specifications.

586
00:38:08,340 --> 00:38:14,100
And consequently we have,
again, the two equations

587
00:38:14,100 --> 00:38:19,840
that say that at
the passband edge,

588
00:38:19,840 --> 00:38:25,730
we want the magnitude to
be within 1db of unity.

589
00:38:25,730 --> 00:38:28,930
And at the stopband edge,
we want the magnitude

590
00:38:28,930 --> 00:38:34,940
to be less than, down
by more than 15 db.

591
00:38:34,940 --> 00:38:36,830
And the two
equations that result

592
00:38:36,830 --> 00:38:40,580
are these, exactly the same as
we had in the impulse invariant

593
00:38:40,580 --> 00:38:45,150
case, except that the
frequencies are different.

594
00:38:45,150 --> 00:38:49,820
So now substituting in for the
form of the analog Butterworth

595
00:38:49,820 --> 00:38:54,710
filter into these expressions,
and setting this up

596
00:38:54,710 --> 00:38:59,210
for equality in
these expressions,

597
00:38:59,210 --> 00:39:03,650
what results then are
the two equations which

598
00:39:03,650 --> 00:39:08,390
are identical to
the two equations

599
00:39:08,390 --> 00:39:11,030
that we had for the
impulse invariant design,

600
00:39:11,030 --> 00:39:17,930
except that the frequencies at
which the passband and stopband

601
00:39:17,930 --> 00:39:23,620
edges of the analog Butterworth
filter fall are different.

602
00:39:23,620 --> 00:39:27,050
They're different because of the
warping of the bilinear curve,

603
00:39:27,050 --> 00:39:29,670
or the pre-warping
that we have to apply.

604
00:39:29,670 --> 00:39:32,660
So, these are the two equations.

605
00:39:32,660 --> 00:39:40,700
And if we solve these two
equations for capital N,

606
00:39:40,700 --> 00:39:45,840
the resulting value for capital
N is 5.3 and some digits.

607
00:39:45,840 --> 00:39:48,350
Notice incidentally
that it's less

608
00:39:48,350 --> 00:39:51,020
than it was for the
impulse invariant design.

609
00:39:51,020 --> 00:39:53,270
And in general that's
a characteristic

610
00:39:53,270 --> 00:39:55,310
of the bilinear transformation.

611
00:39:55,310 --> 00:40:00,452
But so it's less, but
it is not an integer.

612
00:40:00,452 --> 00:40:01,910
So the first thing
we can ask again

613
00:40:01,910 --> 00:40:04,940
is, could we just simply
truncate this down or round it

614
00:40:04,940 --> 00:40:07,850
down to the nearest integer
value, in other words,

615
00:40:07,850 --> 00:40:09,380
N equals 5.

616
00:40:09,380 --> 00:40:11,750
But of course if
we did that, then

617
00:40:11,750 --> 00:40:14,390
we wouldn't be able
to pick omega sub c,

618
00:40:14,390 --> 00:40:17,510
so that these equations
were satisfied or so

619
00:40:17,510 --> 00:40:20,970
that the specifications
were met or exceeded.

620
00:40:20,970 --> 00:40:26,510
So in fact, we need to choose,
we have to choose N equal to 6.

621
00:40:26,510 --> 00:40:30,830
We have to choose the next
highest integer value.

622
00:40:30,830 --> 00:40:34,550
Choosing N equal to 6,
we then have the choice

623
00:40:34,550 --> 00:40:38,990
of exactly satisfying
this equation,

624
00:40:38,990 --> 00:40:42,440
or exactly satisfying
this equation, or in fact

625
00:40:42,440 --> 00:40:44,150
not satisfying
either one of them,

626
00:40:44,150 --> 00:40:47,780
but playing back and
forth between the two.

627
00:40:47,780 --> 00:40:51,050
Well let's do it in
a particular way.

628
00:40:51,050 --> 00:40:55,250
We can, for example,
exactly meet

629
00:40:55,250 --> 00:40:57,830
the stopband specifications.

630
00:40:57,830 --> 00:41:01,110
In other words, we can use
N equals 6 in this equation,

631
00:41:01,110 --> 00:41:04,400
and solve for omega
sub c, in which case

632
00:41:04,400 --> 00:41:08,540
the passband specifications
will be exceeded.

633
00:41:08,540 --> 00:41:10,770
Recall that in the
impulse invariant design

634
00:41:10,770 --> 00:41:12,630
we did it the other way around.

635
00:41:12,630 --> 00:41:16,580
We met exactly the passband
specifications, in which case

636
00:41:16,580 --> 00:41:19,250
the stopband specs
were exceeded.

637
00:41:19,250 --> 00:41:21,380
But there we had a good
reason for doing that.

638
00:41:21,380 --> 00:41:24,920
The reason that we did that
is because we knew in advance

639
00:41:24,920 --> 00:41:29,270
that in any case the passband
specs wouldn't be met exactly,

640
00:41:29,270 --> 00:41:31,340
because of the
effect of aliasing.

641
00:41:31,340 --> 00:41:33,110
For the bilinear
transformation, we

642
00:41:33,110 --> 00:41:35,180
don't have a problem
with aliasing.

643
00:41:35,180 --> 00:41:40,130
So, in fact, we're free to play
with this flexibility in a more

644
00:41:40,130 --> 00:41:41,881
general way.

645
00:41:41,881 --> 00:41:42,380
All right.

646
00:41:42,380 --> 00:41:46,750
Well, in any case then if
we meet the stopband specs,

647
00:41:46,750 --> 00:41:49,320
substituting N equals
6 in this equation,

648
00:41:49,320 --> 00:41:53,030
and solving for omega
sub c; we end up

649
00:41:53,030 --> 00:41:58,910
with a value for omega
sub c, which is 0.76622.

650
00:41:58,910 --> 00:42:03,590
And so this then gives us
the parameters omega sub c

651
00:42:03,590 --> 00:42:09,530
and N for the analog
Butterworth filter.

652
00:42:09,530 --> 00:42:12,770
We can then use
those specifications,

653
00:42:12,770 --> 00:42:15,790
again to generate
the pole pattern

654
00:42:15,790 --> 00:42:20,060
for the squared magnitude
function H sub a

655
00:42:20,060 --> 00:42:23,480
of s times H sub a of minus s.

656
00:42:26,470 --> 00:42:30,820
Factor this by choosing the left
half plane poles in s plane,

657
00:42:30,820 --> 00:42:34,810
and then mapping this function
to a digital filter using

658
00:42:34,810 --> 00:42:36,760
the bilinear transformation.

659
00:42:36,760 --> 00:42:42,340
And let's now look at how this
filter comes out as compared

660
00:42:42,340 --> 00:42:49,060
with the digital Butterworth
filter that we obtained

661
00:42:49,060 --> 00:42:52,090
by using impulse invariance.

662
00:42:52,090 --> 00:42:59,710
Well again, I have a display of
the magnitude of the frequency

663
00:42:59,710 --> 00:43:03,105
response and also
the gain in db.

664
00:43:05,770 --> 00:43:08,020
From what you may
recall from the curve

665
00:43:08,020 --> 00:43:12,010
that I showed previously,
this looks roughly the same.

666
00:43:12,010 --> 00:43:16,900
It's a monotonic
filter and drops off.

667
00:43:16,900 --> 00:43:19,780
It's down by some
amount at 0.2 pi,

668
00:43:19,780 --> 00:43:24,820
and the stopband is at 0.3 pi.

669
00:43:24,820 --> 00:43:27,250
In particular, according
to the specifications,

670
00:43:27,250 --> 00:43:32,860
we wanted this to be down by
no more than 1db at 0.2 pi.

671
00:43:32,860 --> 00:43:38,710
Well, it's hard again to say
that this is 1db or less.

672
00:43:38,710 --> 00:43:41,560
But if we look at
0.3 pi, we can see

673
00:43:41,560 --> 00:43:48,160
that this is in fact
exactly down by 15 db.

674
00:43:48,160 --> 00:43:51,700
And that's the way
that we designed it.

675
00:43:51,700 --> 00:43:55,750
We designed it to exactly
meet the stopband cutoff

676
00:43:55,750 --> 00:43:59,280
and exceed the passband cutoff.

677
00:43:59,280 --> 00:44:03,960
And then the frequency response
continues on from there.

678
00:44:03,960 --> 00:44:06,570
It's interesting, in fact,
to compare the design

679
00:44:06,570 --> 00:44:11,130
that we got from the bilinear
transformation with the design

680
00:44:11,130 --> 00:44:14,860
that we got previously
with impulse invariance.

681
00:44:14,860 --> 00:44:20,350
And let me do that by
superimposing the two of them.

682
00:44:20,350 --> 00:44:22,500
So that the one that
I'm putting on now

683
00:44:22,500 --> 00:44:26,400
is the impulse invariant
design, and let

684
00:44:26,400 --> 00:44:30,372
me just lay them on
top of each other.

685
00:44:34,300 --> 00:44:37,006
OK.

686
00:44:37,006 --> 00:44:40,030
You can see incidentally
that looking at the frequency

687
00:44:40,030 --> 00:44:43,540
response on a magnitude
plot there, in fact,

688
00:44:43,540 --> 00:44:44,830
are some differences.

689
00:44:44,830 --> 00:44:47,170
This line has gotten
a little thicker.

690
00:44:47,170 --> 00:44:50,140
That is, the two don't
exactly superimpose.

691
00:44:50,140 --> 00:44:53,350
But you can't really
see the differences.

692
00:44:53,350 --> 00:44:58,060
Whereas you can when we look
at this on a logarithmic plot.

693
00:44:58,060 --> 00:45:01,810
In particular, this is
the frequency response

694
00:45:01,810 --> 00:45:05,720
that corresponded to the
bilinear transformation.

695
00:45:05,720 --> 00:45:07,960
This is the frequency
response that corresponded

696
00:45:07,960 --> 00:45:10,270
to impulse invariance.

697
00:45:10,270 --> 00:45:14,050
And this is, in general,
a characteristic

698
00:45:14,050 --> 00:45:17,610
of the use of the
bilinear transformation.

699
00:45:17,610 --> 00:45:21,280
It's a result essentially
of the frequency distortion,

700
00:45:21,280 --> 00:45:25,360
because we've
taken the frequency

701
00:45:25,360 --> 00:45:29,590
axis and the continuous
time frequency axis.

702
00:45:29,590 --> 00:45:32,470
As you run along it
at a constant rate,

703
00:45:32,470 --> 00:45:39,400
it gets compressed more and
more along the digital frequency

704
00:45:39,400 --> 00:45:40,120
axis.

705
00:45:40,120 --> 00:45:41,140
That wasn't well said.

706
00:45:41,140 --> 00:45:44,920
But what I really mean to
say is as we walk along here

707
00:45:44,920 --> 00:45:48,850
at a constant rate, the farther
we walk, the faster we're

708
00:45:48,850 --> 00:45:51,200
running along the j omega axis.

709
00:45:51,200 --> 00:45:57,970
And that's why, in fact, the use
of the bilinear transformation

710
00:45:57,970 --> 00:46:03,580
on a lowpass filter will
give better attenuation

711
00:46:03,580 --> 00:46:07,030
at the higher frequencies
than impulse invariance will.

712
00:46:07,030 --> 00:46:09,220
And I don't know
if you can really

713
00:46:09,220 --> 00:46:18,400
see the fact that the
passband specs were exceeded

714
00:46:18,400 --> 00:46:20,990
for the bilinear design.

715
00:46:20,990 --> 00:46:24,250
In fact, when you look at these
view graphs in the study guide,

716
00:46:24,250 --> 00:46:28,420
I think you'll see that
for impulse invariance

717
00:46:28,420 --> 00:46:32,440
the curve is slightly lower
than it is for the bilinear

718
00:46:32,440 --> 00:46:34,810
transformation

719
00:46:34,810 --> 00:46:36,400
OK well, look.

720
00:46:36,400 --> 00:46:43,750
This is a simple example that
hopefully at least illustrates

721
00:46:43,750 --> 00:46:46,960
for you the basic
procedure involved

722
00:46:46,960 --> 00:46:51,460
in carrying through a design
of an infinite impulse response

723
00:46:51,460 --> 00:46:55,600
filter using impulse
invariance and the bilinear

724
00:46:55,600 --> 00:46:57,610
transformation.

725
00:46:57,610 --> 00:47:01,840
There are other examples that
are considered in the text.

726
00:47:01,840 --> 00:47:08,900
There is a discussion of
digital Chebyshev filters,

727
00:47:08,900 --> 00:47:11,660
in other words digital
designs beginning

728
00:47:11,660 --> 00:47:13,850
from an analog Chebyshev filter.

729
00:47:13,850 --> 00:47:15,350
And you'll also
have an opportunity

730
00:47:15,350 --> 00:47:20,235
to carry through some
designs in the study guide.

731
00:47:20,235 --> 00:47:24,040
A point that I would
like to mention briefly

732
00:47:24,040 --> 00:47:26,540
is that most of the
discussion, in fact

733
00:47:26,540 --> 00:47:29,260
all of the discussion
that we've gone through,

734
00:47:29,260 --> 00:47:33,410
has focused entirely
on lowpass filters.

735
00:47:33,410 --> 00:47:36,310
In fact, the
discussion generalizes

736
00:47:36,310 --> 00:47:41,030
in a fairly straightforward
way to bandpass, bandstop,

737
00:47:41,030 --> 00:47:43,000
highpass, et cetera filters.

738
00:47:43,000 --> 00:47:47,530
In particular, as is discussed
in some detail in the text,

739
00:47:47,530 --> 00:47:51,400
there are a class of
frequency transformations

740
00:47:51,400 --> 00:47:54,430
that allow lowpass
designs to get

741
00:47:54,430 --> 00:48:00,010
converted to bandstop, bandpass,
highpass, et cetera designs.

742
00:48:00,010 --> 00:48:05,980
So this concludes our
discussion of design procedures

743
00:48:05,980 --> 00:48:10,380
for recursive or
infinite impulse response

744
00:48:10,380 --> 00:48:11,860
digital filters.

745
00:48:11,860 --> 00:48:17,470
In the next lecture I'll discuss
the design of finite impulse

746
00:48:17,470 --> 00:48:20,830
response filters,
and in particular

747
00:48:20,830 --> 00:48:25,330
introduce a number of
designs, one of which

748
00:48:25,330 --> 00:48:28,020
corresponds to a class
of algorithmic designs.

749
00:48:28,020 --> 00:48:30,500
Thank you.

750
00:48:30,500 --> 00:48:33,250
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