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PROFESSOR: Over the past
several lectures, we've
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00:00:57,270 --> 00:01:01,880
developed a representation for
linear time-invariant systems.
10
00:01:01,880 --> 00:01:05,500
A particularly important set of
systems, which are linear
11
00:01:05,500 --> 00:01:08,700
and time-invariant, are those
that are represented by linear
12
00:01:08,700 --> 00:01:11,240
constant-coefficient
differential equations in
13
00:01:11,240 --> 00:01:14,950
continuous time or linear
constant-coefficient
14
00:01:14,950 --> 00:01:17,630
difference equations
in discrete time.
15
00:01:17,630 --> 00:01:20,780
For example, electrical circuits
that are built, let's
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00:01:20,780 --> 00:01:23,300
say, out of resistors,
inductors, and capacitors,
17
00:01:23,300 --> 00:01:26,780
perhaps with op-amps, correspond
to systems
18
00:01:26,780 --> 00:01:29,150
described by differential
equations.
19
00:01:29,150 --> 00:01:31,690
Mechanical systems with
springs and dashpots,
20
00:01:31,690 --> 00:01:35,010
likewise, are described by
differential equations.
21
00:01:35,010 --> 00:01:39,300
And in the discrete-time case,
things such as moving average
22
00:01:39,300 --> 00:01:44,470
filters, digital filters, and
most simple kinds of data
23
00:01:44,470 --> 00:01:47,840
smoothing are all linear
constant-coefficient
24
00:01:47,840 --> 00:01:50,110
difference equations.
25
00:01:50,110 --> 00:01:55,180
Now, presumably in a previous
course, you've had some
26
00:01:55,180 --> 00:02:00,260
exposure to differential
equations for continuous time,
27
00:02:00,260 --> 00:02:04,000
and their solution using
notions like particular
28
00:02:04,000 --> 00:02:06,220
solution, and homogeneous
solution, initial
29
00:02:06,220 --> 00:02:09,380
conditions, et cetera.
30
00:02:09,380 --> 00:02:13,760
Later on in the course, when
we've developed the concept of
31
00:02:13,760 --> 00:02:17,450
the Fourier transform after
that, the Laplace transform,
32
00:02:17,450 --> 00:02:21,500
we'll see some very efficient
and useful ways of generating
33
00:02:21,500 --> 00:02:23,160
solutions, both for
differential
34
00:02:23,160 --> 00:02:25,010
and difference equations.
35
00:02:25,010 --> 00:02:29,620
At this point, however, I'd like
to just introduce linear
36
00:02:29,620 --> 00:02:33,020
constant-coefficient
differential equations and
37
00:02:33,020 --> 00:02:35,130
their discrete-time
counterpart.
38
00:02:35,130 --> 00:02:38,890
And address, among other things,
the issue of when they
39
00:02:38,890 --> 00:02:43,650
do and don't correspond to
linear time-invariant systems.
40
00:02:43,650 --> 00:02:48,420
Well, let's first consider
what I refer to as an
41
00:02:48,420 --> 00:02:51,010
nth-order linear
constant-coefficient
42
00:02:51,010 --> 00:02:54,270
differential equation, as
I've indicated here.
43
00:02:54,270 --> 00:02:57,350
And what it consists of is
a linear combination of
44
00:02:57,350 --> 00:03:01,560
derivatives of the system
output, y(t), equal to a
45
00:03:01,560 --> 00:03:04,660
linear combination of
derivatives of the system
46
00:03:04,660 --> 00:03:07,300
input x(t).
47
00:03:07,300 --> 00:03:12,470
And it's referred to as a
constant-coefficient equation,
48
00:03:12,470 --> 00:03:15,550
of course, because the
coefficients are constant.
49
00:03:15,550 --> 00:03:18,670
In other words, not assumed
to be time-varying.
50
00:03:18,670 --> 00:03:25,200
And it's referred to as linear
because it corresponds to a
51
00:03:25,200 --> 00:03:30,820
linear combination of these
derivatives, not because it
52
00:03:30,820 --> 00:03:32,620
corresponds to a
linear system.
53
00:03:32,620 --> 00:03:36,910
And, in fact, as we'll see,
or as I'll indicate, this
54
00:03:36,910 --> 00:03:39,340
equation may or may
not, in fact,
55
00:03:39,340 --> 00:03:42,410
correspond to a linear system.
56
00:03:42,410 --> 00:03:47,500
In the discrete-time case, the
corresponding equation is a
57
00:03:47,500 --> 00:03:50,710
linear constant-coefficient
difference equation.
58
00:03:50,710 --> 00:03:55,650
And that corresponds to, again,
a linear combination of
59
00:03:55,650 --> 00:04:00,370
delayed versions of the output
equal to a linear combination
60
00:04:00,370 --> 00:04:02,630
of delayed versions
of the input.
61
00:04:02,630 --> 00:04:04,590
This equation is referred to an
62
00:04:04,590 --> 00:04:07,290
nth-order difference equation.
63
00:04:07,290 --> 00:04:10,750
The n referring to the number
of delays of the output
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00:04:10,750 --> 00:04:15,690
involved, just as an Nth-order
differential equation, the n
65
00:04:15,690 --> 00:04:18,230
or the order of the equation
refers to the number of
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00:04:18,230 --> 00:04:21,050
derivatives of the output.
67
00:04:21,050 --> 00:04:25,150
Now, let's first begin with
linear constant-coefficient
68
00:04:25,150 --> 00:04:27,170
differential equations.
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00:04:27,170 --> 00:04:34,290
And the basic point of the
solution for the differential
70
00:04:34,290 --> 00:04:40,300
equations is the fact that if
we've generated some solution,
71
00:04:40,300 --> 00:04:44,600
which I refer to here as y_p(t),
some solution to the
72
00:04:44,600 --> 00:04:49,890
equation for a given input,
then, in fact, we can add to
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00:04:49,890 --> 00:04:55,750
that solution any other solution
which satisfies
74
00:04:55,750 --> 00:04:59,910
what's referred to as the
homogeneous equation.
75
00:04:59,910 --> 00:05:06,620
So in fact, this differential
equation by itself is not a
76
00:05:06,620 --> 00:05:10,300
unique specification
of the system.
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00:05:10,300 --> 00:05:15,360
If I have any solution, then I
can add to that solution any
78
00:05:15,360 --> 00:05:18,950
other solution which satisfies
the homogeneous equation, and
79
00:05:18,950 --> 00:05:22,140
the sum of those two will
likewise be a solution.
80
00:05:22,140 --> 00:05:24,630
And that's very straightforward
to verify.
81
00:05:24,630 --> 00:05:28,770
By simply substituting into the
differential equation, the
82
00:05:28,770 --> 00:05:33,910
sum of a particular and the
homogeneous solution, and what
83
00:05:33,910 --> 00:05:38,180
you'll see is that the
homogeneous contribution, in
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00:05:38,180 --> 00:05:41,950
fact, goes to zero by definition
of what we mean by
85
00:05:41,950 --> 00:05:44,670
the homogeneous equation.
86
00:05:44,670 --> 00:05:48,430
Now, the homogeneous solution
for a linear
87
00:05:48,430 --> 00:05:53,290
constant-coefficient
differential equation is of
88
00:05:53,290 --> 00:05:58,030
the form that I indicate
at the bottom.
89
00:05:58,030 --> 00:06:07,230
And it typically consists of a
sum of N complex exponentials.
90
00:06:07,230 --> 00:06:11,650
And the constants are
undetermined by
91
00:06:11,650 --> 00:06:14,000
the equation itself.
92
00:06:14,000 --> 00:06:17,240
And this form for the
homogeneous solution, in
93
00:06:17,240 --> 00:06:23,860
essence, drops out of examining
the homogeneous
94
00:06:23,860 --> 00:06:28,890
equation, where if we assume
that the form of the
95
00:06:28,890 --> 00:06:34,210
homogeneous solution is a
complex exponential with some
96
00:06:34,210 --> 00:06:38,570
unspecified amplitude and
unspecified exponent.
97
00:06:38,570 --> 00:06:43,580
If we substitute this into our
homogeneous equation, we end
98
00:06:43,580 --> 00:06:46,510
up with the equation that
I've indicated here.
99
00:06:46,510 --> 00:06:50,780
The factor A and the e^(st) can
in fact be canceled out,
100
00:06:50,780 --> 00:06:56,110
and we find that that equation
is satisfied
101
00:06:56,110 --> 00:07:02,160
for N values of s.
102
00:07:02,160 --> 00:07:06,540
And that's true no matter what
choice is made for these
103
00:07:06,540 --> 00:07:07,890
coefficients.
104
00:07:07,890 --> 00:07:12,510
And the essential consequence
of all of that is that the
105
00:07:12,510 --> 00:07:16,660
homogeneous solution is of
the form that I indicated
106
00:07:16,660 --> 00:07:18,160
previously.
107
00:07:18,160 --> 00:07:26,010
Namely it consists of a sum of
N complex exponentials, where
108
00:07:26,010 --> 00:07:29,720
the coefficients, the N
coefficients, attach to each
109
00:07:29,720 --> 00:07:33,320
of those complex exponential
is undetermined or
110
00:07:33,320 --> 00:07:34,980
unspecified.
111
00:07:34,980 --> 00:07:39,710
So what this says is that in
order to obtain the solution
112
00:07:39,710 --> 00:07:41,360
for a linear
constant-coefficient
113
00:07:41,360 --> 00:07:44,750
differential equation, we need
some kind of auxiliary
114
00:07:44,750 --> 00:07:49,760
information that tells us
is how to obtain these N
115
00:07:49,760 --> 00:07:51,690
undetermined constants.
116
00:07:51,690 --> 00:07:55,550
And there are variety of ways
of specifying this auxiliary
117
00:07:55,550 --> 00:07:57,930
information or auxiliary
conditions.
118
00:07:57,930 --> 00:08:02,250
For example, in addition to
the differential equation,
119
00:08:02,250 --> 00:08:10,000
what I can tell you is the value
of the output and N - 1
120
00:08:10,000 --> 00:08:15,300
of its derivatives, at some
specified time, t_0.
121
00:08:15,300 --> 00:08:20,590
And so the differential equation
together with the
122
00:08:20,590 --> 00:08:25,020
auxiliary information, the
initial conditions, then lets
123
00:08:25,020 --> 00:08:28,760
you determine the total
solution, which namely lets
124
00:08:28,760 --> 00:08:31,750
you determine these
previously-unspecified
125
00:08:31,750 --> 00:08:35,130
coefficients in the homogeneous
solution.
126
00:08:35,130 --> 00:08:39,120
Now, depending on how the
auxiliary information is
127
00:08:39,120 --> 00:08:42,350
stated or what auxiliary
information is available, the
128
00:08:42,350 --> 00:08:47,080
system may or may not correspond
to a linear system,
129
00:08:47,080 --> 00:08:49,360
and may or may not correspond
to a linear
130
00:08:49,360 --> 00:08:51,880
time-invariant system.
131
00:08:51,880 --> 00:08:55,180
One essential condition for it
to correspond to a linear
132
00:08:55,180 --> 00:09:01,110
system is that the initial
conditions must be 0.
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00:09:01,110 --> 00:09:06,550
And one can see the reason for
that, if we refer back to the
134
00:09:06,550 --> 00:09:10,090
previous lecture in which we saw
that for a linear system,
135
00:09:10,090 --> 00:09:13,090
if we put 0 in, we get 0 out.
136
00:09:13,090 --> 00:09:17,560
So if x(t), the input, is
0, the output must be 0.
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00:09:17,560 --> 00:09:20,710
And so that, in essence, tells
us that, at least for the
138
00:09:20,710 --> 00:09:26,220
system to be linear, these
initial conditions must be 0.
139
00:09:26,220 --> 00:09:33,380
Now, beyond that, if we want
the system to be causal and
140
00:09:33,380 --> 00:09:38,890
linear and time-invariant, then
what's required on the
141
00:09:38,890 --> 00:09:42,840
initial conditions is that they
be consistent with what's
142
00:09:42,840 --> 00:09:46,520
referred to as initial rest.
143
00:09:46,520 --> 00:09:53,770
Initial rest says that the
output must be 0 up until the
144
00:09:53,770 --> 00:09:56,910
time that the input
becomes non-zero.
145
00:09:56,910 --> 00:09:59,590
And we can see, of course, that
that's consistent with
146
00:09:59,590 --> 00:10:03,150
the notion of causality,
as we talked about in
147
00:10:03,150 --> 00:10:04,730
the previous lecture.
148
00:10:04,730 --> 00:10:10,400
And it's relatively
straightforward to see that if
149
00:10:10,400 --> 00:10:14,920
the system is causal and linear
and time-invariant,
150
00:10:14,920 --> 00:10:18,810
that will require
initial rest.
151
00:10:18,810 --> 00:10:23,600
It's somewhat more difficult
to see that if we specify
152
00:10:23,600 --> 00:10:27,410
initial rest, then that, in
fact, is sufficient to
153
00:10:27,410 --> 00:10:31,040
determine that the system is
both causal and linear and
154
00:10:31,040 --> 00:10:32,100
time invariant.
155
00:10:32,100 --> 00:10:35,730
But the essential point then
is that it requires initial
156
00:10:35,730 --> 00:10:40,500
rest for both linearity
and causality.
157
00:10:40,500 --> 00:10:46,360
OK, well, let's look at an
example, and let's take the
158
00:10:46,360 --> 00:10:52,580
example of a first-order
differential equation, as I've
159
00:10:52,580 --> 00:10:54,470
indicated here.
160
00:10:54,470 --> 00:10:58,623
So we have a first-order
differential equation, dy(t) /
161
00:10:58,623 --> 00:11:03,350
dt + ay(t) is the input x(t).
162
00:11:03,350 --> 00:11:07,360
And let's first look at what
the homogeneous solution of
163
00:11:07,360 --> 00:11:08,500
this equation is.
164
00:11:08,500 --> 00:11:13,370
And so, we consider the
homogeneous equation, namely
165
00:11:13,370 --> 00:11:17,600
the equation that specifies
solutions, which would
166
00:11:17,600 --> 00:11:20,650
correspond to 0 input.
167
00:11:20,650 --> 00:11:27,180
We, in essence, guess, or impose
a solution of the form.
168
00:11:27,180 --> 00:11:31,590
The homogeneous solution is an
amplitude factor times a
169
00:11:31,590 --> 00:11:33,450
complex exponential.
170
00:11:33,450 --> 00:11:38,390
Substituting this into the
homogeneous equation, we then
171
00:11:38,390 --> 00:11:40,860
get the equation that
I've indicated here.
172
00:11:40,860 --> 00:11:45,780
What you can see is that in this
equation, I can cancel
173
00:11:45,780 --> 00:11:50,530
out the amplitude factor and
this complex exponential.
174
00:11:50,530 --> 00:11:56,590
So let's just cancel those out
on both sides of the equation.
175
00:11:56,590 --> 00:12:02,340
And what we're left with is an
equation that specifies what
176
00:12:02,340 --> 00:12:04,760
the complex exponent must be.
177
00:12:04,760 --> 00:12:08,110
In particular, for the
homogeneous solution, s must
178
00:12:08,110 --> 00:12:12,060
be equal to -a, and so finally,
our homogeneous
179
00:12:12,060 --> 00:12:16,380
solution is as I've
indicated here.
180
00:12:19,050 --> 00:12:23,150
Now, let's look at the solution
for a specific input.
181
00:12:23,150 --> 00:12:25,990
Let's consider, for example,
an input which is
182
00:12:25,990 --> 00:12:28,000
a scaled unit step.
183
00:12:28,000 --> 00:12:31,530
And although I won't work out
the solution in detail and
184
00:12:31,530 --> 00:12:36,510
perhaps using what you've worked
on previously, you know
185
00:12:36,510 --> 00:12:39,805
how to carry out the
solution for that.
186
00:12:39,805 --> 00:12:44,950
A solution with a step input is
what I've indicated here: a
187
00:12:44,950 --> 00:12:47,060
scalar, 1 minus an exponential,
188
00:12:47,060 --> 00:12:48,540
times a unit step.
189
00:12:48,540 --> 00:12:51,430
And you can verify that simply
by substituting into the
190
00:12:51,430 --> 00:12:53,170
differential equation.
191
00:12:53,170 --> 00:12:56,930
Now, we know that there's
a family of solutions.
192
00:12:56,930 --> 00:13:02,620
In other words, any solution
with a homogeneous solution
193
00:13:02,620 --> 00:13:04,990
added to it, is, again,
a solution.
194
00:13:04,990 --> 00:13:08,670
And so if we consider the
solution that I just
195
00:13:08,670 --> 00:13:13,160
indicated, we generate the
entire family of solutions by
196
00:13:13,160 --> 00:13:17,440
adding a homogeneous solution
to it, and so this then
197
00:13:17,440 --> 00:13:21,400
corresponds to the entire family
of solutions, where the
198
00:13:21,400 --> 00:13:27,640
constant A is unspecified so
far, and needs to be specified
199
00:13:27,640 --> 00:13:31,470
through some type of auxiliary
conditions.
200
00:13:31,470 --> 00:13:36,640
Now, a class of auxiliary
conditions is the condition of
201
00:13:36,640 --> 00:13:41,310
initial rest, which as I
indicated before, is
202
00:13:41,310 --> 00:13:44,830
equivalent to the statement that
the system is causal and
203
00:13:44,830 --> 00:13:46,900
linear and time-invariant.
204
00:13:46,900 --> 00:13:52,080
And in that case, for the
initial rest condition, we
205
00:13:52,080 --> 00:13:56,660
would then require in this
equation above that this
206
00:13:56,660 --> 00:13:59,710
constant be equal to 0.
207
00:13:59,710 --> 00:14:04,860
And so finally, the response
to a scaled step--
208
00:14:04,860 --> 00:14:07,730
if the system is to correspond
to a causal linear
209
00:14:07,730 --> 00:14:12,760
time-invariant system, is then
just this term, namely, a
210
00:14:12,760 --> 00:14:18,870
constant times 1 minus an
exponential, times the step.
211
00:14:18,870 --> 00:14:23,060
Now, if the system is a linear
time-invariant system, it can,
212
00:14:23,060 --> 00:14:26,260
as we know, be described through
its impulse response.
213
00:14:26,260 --> 00:14:30,420
And as you've worked out
previously in the video course
214
00:14:30,420 --> 00:14:34,730
manual, for a linear
time-invariant system, the
215
00:14:34,730 --> 00:14:36,480
impulse response is the
216
00:14:36,480 --> 00:14:38,750
derivative of the step response.
217
00:14:38,750 --> 00:14:41,730
And just to quickly remind
you of where that
218
00:14:41,730 --> 00:14:43,960
result comes from.
219
00:14:43,960 --> 00:14:48,020
In essence, we can consider
two linear time-invariant
220
00:14:48,020 --> 00:14:51,770
systems in cascade, one a
differentiator, the other the
221
00:14:51,770 --> 00:14:55,060
system that we're talking
about described by the
222
00:14:55,060 --> 00:14:56,850
differential equation.
223
00:14:56,850 --> 00:15:01,700
And a step in here then
generates an impulse into our
224
00:15:01,700 --> 00:15:04,600
system, and out comes the
impulse response.
225
00:15:04,600 --> 00:15:07,100
Well, just using the fact that
these are both linear
226
00:15:07,100 --> 00:15:10,840
time-invariant systems, and they
can be cascaded in either
227
00:15:10,840 --> 00:15:15,970
order, then means that if we
have the step response to our
228
00:15:15,970 --> 00:15:20,380
system, and that goes through
the differentiator, what must
229
00:15:20,380 --> 00:15:24,570
come out, again, is the
impulse response.
230
00:15:24,570 --> 00:15:27,380
So differentiating the
step response, we
231
00:15:27,380 --> 00:15:29,310
get the impulse response.
232
00:15:29,310 --> 00:15:33,240
Here again, is the step response
as we just worked it
233
00:15:33,240 --> 00:15:35,990
out, this time for
a unit step.
234
00:15:35,990 --> 00:15:40,760
If we differentiate, we have
then, since the step response
235
00:15:40,760 --> 00:15:44,120
is the product of two terms, the
derivative of a product is
236
00:15:44,120 --> 00:15:45,940
the sum of the derivatives.
237
00:15:45,940 --> 00:15:49,640
And carrying that algebra
through, and using the fact
238
00:15:49,640 --> 00:15:54,350
that the derivative of the step
is an impulse, finally we
239
00:15:54,350 --> 00:15:59,180
come down to this statement
for the impulse response.
240
00:15:59,180 --> 00:16:01,820
And then recognizing
that this is a time
241
00:16:01,820 --> 00:16:03,970
function times an impulse.
242
00:16:03,970 --> 00:16:07,800
And we know that if a time
function times an impulse
243
00:16:07,800 --> 00:16:14,160
takes on the value at the time
that the impulse occurs, then
244
00:16:14,160 --> 00:16:18,390
this term is simply 0.
245
00:16:18,390 --> 00:16:22,140
And the impulse response
then finally is an
246
00:16:22,140 --> 00:16:24,420
exponential of this form.
247
00:16:24,420 --> 00:16:28,390
And this is the decaying
exponential for a-positive,
248
00:16:28,390 --> 00:16:31,640
it's a growing exponential
for a-negative.
249
00:16:31,640 --> 00:16:38,480
And recall that, as we talked
about previously, a linear
250
00:16:38,480 --> 00:16:42,400
time-invariant system is stable
if its impulse response
251
00:16:42,400 --> 00:16:44,610
is absolutely integrable.
252
00:16:44,610 --> 00:16:49,760
For this particular case, this
impulse response is absolutely
253
00:16:49,760 --> 00:16:56,250
integrable provided that the
exponential factor a is
254
00:16:56,250 --> 00:16:59,290
greater than 0.
255
00:16:59,290 --> 00:17:04,700
Okay, so what we've seen then is
the impulse response for a
256
00:17:04,700 --> 00:17:07,230
system described by a linear
constant-coefficient
257
00:17:07,230 --> 00:17:10,829
differential equation, where in
addition, we would impose
258
00:17:10,829 --> 00:17:15,069
causality, linearity, and
time-invariance, essentially,
259
00:17:15,069 --> 00:17:19,619
through the initial conditions
of initial rest.
260
00:17:19,619 --> 00:17:23,609
Now, pretty much the same kinds
of things happen with
261
00:17:23,609 --> 00:17:27,079
difference equations as we've
gone through with
262
00:17:27,079 --> 00:17:29,740
differential equations.
263
00:17:29,740 --> 00:17:33,820
In particular, again, let me
remind you of the form of an
264
00:17:33,820 --> 00:17:35,640
Nth order linear constant
265
00:17:35,640 --> 00:17:37,760
coefficient difference equation.
266
00:17:37,760 --> 00:17:40,820
It's as I indicate here.
267
00:17:40,820 --> 00:17:44,390
And, again, a linear
combination, this time of
268
00:17:44,390 --> 00:17:48,220
delayed versions of the output
equal to a linear combination
269
00:17:48,220 --> 00:17:51,520
of delayed versions
of the input.
270
00:17:51,520 --> 00:17:57,210
Once again, difference equation
is not a complete
271
00:17:57,210 --> 00:18:00,310
specification of the system
because we can add to the
272
00:18:00,310 --> 00:18:03,820
response any homogeneous
solution.
273
00:18:03,820 --> 00:18:07,500
In other words, any solution
that satisfies the homogeneous
274
00:18:07,500 --> 00:18:12,230
equation and the sum of those
will also satisfy the original
275
00:18:12,230 --> 00:18:13,910
difference equation.
276
00:18:13,910 --> 00:18:21,500
So if we have a particular
response that satisfies the
277
00:18:21,500 --> 00:18:27,070
difference equation, then adding
to that any response
278
00:18:27,070 --> 00:18:32,680
that is a solution to the
homogeneous equation will also
279
00:18:32,680 --> 00:18:36,640
be a solution to the
total equation.
280
00:18:36,640 --> 00:18:41,470
The homogeneous solution, again,
is of the form of a
281
00:18:41,470 --> 00:18:44,650
linear combination
of exponentials.
282
00:18:44,650 --> 00:18:48,300
Here, we have the homogeneous
equation.
283
00:18:48,300 --> 00:18:53,390
As with differential equations,
we can guess or
284
00:18:53,390 --> 00:18:58,010
impose solutions of the form
A times an exponential.
285
00:18:58,010 --> 00:19:01,780
When we substitute this into the
homogeneous equation, we
286
00:19:01,780 --> 00:19:05,830
then end up with the equation
that I've indicated here.
287
00:19:05,830 --> 00:19:15,580
We recognize again that A, the
amplitude, and z^n, this
288
00:19:15,580 --> 00:19:19,310
exponential factor,
cancel out.
289
00:19:19,310 --> 00:19:27,000
And so this equation is
satisfied for any values of z
290
00:19:27,000 --> 00:19:30,310
that satisfy this equation.
291
00:19:30,310 --> 00:19:34,310
And there are N roots,
z_1 through z_N.
292
00:19:34,310 --> 00:19:38,830
And so finally, the form for the
homogeneous solution is a
293
00:19:38,830 --> 00:19:44,240
linear combination of capital N
exponentials, where capital
294
00:19:44,240 --> 00:19:46,930
N is the order of
the equation.
295
00:19:46,930 --> 00:19:49,400
With each of those exponentials,
the amplitude
296
00:19:49,400 --> 00:19:53,220
factor is undetermined and needs
to be determined in some
297
00:19:53,220 --> 00:19:58,630
way through the imposition of
appropriate initial conditions
298
00:19:58,630 --> 00:19:59,880
or boundary conditions.
299
00:20:01,990 --> 00:20:07,370
So the general form then for the
solution to the difference
300
00:20:07,370 --> 00:20:13,590
equation is a sum of
exponentials plus any
301
00:20:13,590 --> 00:20:15,830
particular solution.
302
00:20:15,830 --> 00:20:20,730
It's through auxiliary
conditions that we determine
303
00:20:20,730 --> 00:20:23,150
these coefficients.
304
00:20:23,150 --> 00:20:28,420
We have N undetermined
coefficients and, so we
305
00:20:28,420 --> 00:20:31,850
require N auxiliary
conditions.
306
00:20:31,850 --> 00:20:37,520
For example, some set of values
of the output at N
307
00:20:37,520 --> 00:20:41,210
distinct instance of time.
308
00:20:41,210 --> 00:20:44,200
Now this was the same as with
differential equations.
309
00:20:44,200 --> 00:20:46,640
In the case of differential
equations, we talked about
310
00:20:46,640 --> 00:20:50,510
specifying the value of the
output and its derivatives.
311
00:20:50,510 --> 00:20:53,270
And there, we indicated
that for
312
00:20:53,270 --> 00:20:56,890
linearity, what we required--
313
00:20:56,890 --> 00:21:00,740
for linearity, what we required
is that the auxiliary
314
00:21:00,740 --> 00:21:02,580
conditions be 0.
315
00:21:02,580 --> 00:21:05,830
And the same thing applies
here for the same reason.
316
00:21:05,830 --> 00:21:09,350
Namely, if the system is to be
linear, then the response, if
317
00:21:09,350 --> 00:21:13,440
there's no input, must
be equal to 0.
318
00:21:13,440 --> 00:21:19,200
In addition, what we may want
to impose on the system is
319
00:21:19,200 --> 00:21:22,290
that it be causal, and
in addition to linear
320
00:21:22,290 --> 00:21:28,320
time-invariant, and what that
requires, again, is that the
321
00:21:28,320 --> 00:21:32,220
auxiliary conditions be
consistent with initial rest,
322
00:21:32,220 --> 00:21:37,690
namely, that if the input is 0
prior to some time, then the
323
00:21:37,690 --> 00:21:42,680
output is 0 prior to
the same time.
324
00:21:42,680 --> 00:21:47,240
So we've seen a very direct
parallel so far between
325
00:21:47,240 --> 00:21:52,210
differential equations and
difference equations.
326
00:21:52,210 --> 00:21:55,930
In fact, one difference between
them that, in some
327
00:21:55,930 --> 00:22:00,140
sense, makes difference
equations easier to deal with
328
00:22:00,140 --> 00:22:04,000
in some situations, is that in
contrast to a differential
329
00:22:04,000 --> 00:22:07,980
equation, a difference equation,
if we assume
330
00:22:07,980 --> 00:22:13,580
causality, in fact is an
explicit input-output
331
00:22:13,580 --> 00:22:15,080
relationship for the system.
332
00:22:15,080 --> 00:22:18,670
Now, let me show you
what I mean.
333
00:22:18,670 --> 00:22:22,130
Let's consider the nth-order
difference equation, as I've
334
00:22:22,130 --> 00:22:23,490
indicated here.
335
00:22:23,490 --> 00:22:28,590
And let's assume that we're
imposing causality so that the
336
00:22:28,590 --> 00:22:33,000
output can only depend on prior
values of the input, and
337
00:22:33,000 --> 00:22:35,730
therefore, aren't prior
values of the output.
338
00:22:35,730 --> 00:22:41,450
Well, we can simply rearrange
this equation solving for y[n]
339
00:22:41,450 --> 00:22:44,680
the leading term, with k = 0.
340
00:22:44,680 --> 00:22:47,660
Taking all of the other terms
over to the right side of the
341
00:22:47,660 --> 00:22:54,740
equation, and we then have a
recursive equation, namely, an
342
00:22:54,740 --> 00:22:59,070
equation that expresses the
output in terms of prior
343
00:22:59,070 --> 00:23:03,850
values of the input, which is
this term, and prior values of
344
00:23:03,850 --> 00:23:05,640
the output.
345
00:23:05,640 --> 00:23:10,050
And so if, in fact, we have this
equation running, then
346
00:23:10,050 --> 00:23:13,800
once it started, we know how to
compute the output for the
347
00:23:13,800 --> 00:23:16,740
next time instant.
348
00:23:16,740 --> 00:23:19,170
Well, how do we get
it started?
349
00:23:19,170 --> 00:23:22,240
The way we get it started,
of course, is through the
350
00:23:22,240 --> 00:23:26,240
appropriate set of initial
conditions or boundary
351
00:23:26,240 --> 00:23:27,300
conditions.
352
00:23:27,300 --> 00:23:30,760
And, if for example, we assume
initial rest corresponding to
353
00:23:30,760 --> 00:23:35,360
a causal linear time-invariant
system, then if the input is 0
354
00:23:35,360 --> 00:23:39,560
up until some time, the output
must be 0 up until that time.
355
00:23:39,560 --> 00:23:43,390
And that, in essence, helps us
get the equation started.
356
00:23:43,390 --> 00:23:48,150
Well, let's look at this
specifically in the context of
357
00:23:48,150 --> 00:23:50,160
a first-order difference
equation.
358
00:23:50,160 --> 00:23:55,510
So let's take a first-order
difference equation, as I've
359
00:23:55,510 --> 00:23:57,560
indicated here.
360
00:23:57,560 --> 00:24:00,650
And so, we have an equation
that tells us that y[n]
361
00:24:00,650 --> 00:24:03,920
- ay[n-1]
362
00:24:03,920 --> 00:24:06,750
= x[n].
363
00:24:06,750 --> 00:24:10,970
Now, if we want this to
correspond to a causal linear
364
00:24:10,970 --> 00:24:16,780
time-invariant system, we impose
initial rest on it.
365
00:24:16,780 --> 00:24:21,550
We can rewrite the first-order
difference equation by taking
366
00:24:21,550 --> 00:24:24,280
the term involving y[n-1]
367
00:24:24,280 --> 00:24:25,945
over two the right hand
side of the equation.
368
00:24:28,450 --> 00:24:33,560
Now, this gives us a recursive
equation that expresses the
369
00:24:33,560 --> 00:24:38,500
output in terms of the input and
past values of the output.
370
00:24:38,500 --> 00:24:43,610
And since we've imposed
causality, and if we're
371
00:24:43,610 --> 00:24:46,820
talking about a linear
time-invariant system, we can
372
00:24:46,820 --> 00:24:50,350
now inquire as to what the
impulse response is.
373
00:24:50,350 --> 00:24:53,120
And we know, of course, the
impulse response tells us
374
00:24:53,120 --> 00:24:56,910
everything that we need to
know about the system.
375
00:24:56,910 --> 00:25:02,990
So let's choose an input,
which is an impulse.
376
00:25:02,990 --> 00:25:07,070
So the impulse response
is delta[n]
377
00:25:07,070 --> 00:25:09,040
corresponding to the x[n]
378
00:25:09,040 --> 00:25:13,490
up here, plus a delta of--
379
00:25:13,490 --> 00:25:15,880
this should be h[n-1].
380
00:25:15,880 --> 00:25:17,970
And let me just correct that.
381
00:25:17,970 --> 00:25:20,640
This is h[n-1].
382
00:25:20,640 --> 00:25:24,270
So we have the impulse
response as delta[n]
383
00:25:24,270 --> 00:25:25,520
plus a times h[n-1].
384
00:25:28,470 --> 00:25:33,235
Now, from initial rest, we know
that since the input,
385
00:25:33,235 --> 00:25:37,620
namely an impulse, is 0 for
n less than 0, the impulse
386
00:25:37,620 --> 00:25:42,410
response is likewise 0
for n less than 0.
387
00:25:42,410 --> 00:25:45,620
And now, let's work
out what h[0]
388
00:25:45,620 --> 00:25:46,870
is.
389
00:25:46,870 --> 00:25:52,900
Well, h of 0, with n
= 0, is delta[0]
390
00:25:52,900 --> 00:25:55,940
plus a times h[n-1].
391
00:25:55,940 --> 00:25:56,950
h[n-1]
392
00:25:56,950 --> 00:25:58,010
is 0.
393
00:25:58,010 --> 00:25:59,600
And so, h[0]
394
00:25:59,600 --> 00:26:01,440
is equal to 1.
395
00:26:05,490 --> 00:26:09,030
Now that we have h[0], we
can figure out h[1]
396
00:26:09,030 --> 00:26:12,340
by running this recursive
equation.
397
00:26:12,340 --> 00:26:14,640
So h[1]
398
00:26:14,640 --> 00:26:21,360
is delta[1], which is 0, plus
a times h[0], which we just
399
00:26:21,360 --> 00:26:22,820
figured out is 1.
400
00:26:22,820 --> 00:26:24,410
So, h[1]
401
00:26:24,410 --> 00:26:26,120
is equal to a.
402
00:26:26,120 --> 00:26:29,380
And if we carry this through,
we'll have h[2]
403
00:26:29,380 --> 00:26:34,180
equal to a^2, and this
will continue on.
404
00:26:34,180 --> 00:26:37,660
And in fact, what we can
recognize by looking at this
405
00:26:37,660 --> 00:26:41,860
and how we would expect these
terms to build, we would see
406
00:26:41,860 --> 00:26:46,120
that the impulse response, h[n],
in fact, is of the form
407
00:26:46,120 --> 00:26:50,630
a^n times u[n].
408
00:26:50,630 --> 00:26:56,080
And we also can recognize then
that this corresponds to a
409
00:26:56,080 --> 00:27:02,890
stable system, if and only if
the impulse response, which is
410
00:27:02,890 --> 00:27:05,640
what we just figured out, if
and only if the impulse
411
00:27:05,640 --> 00:27:08,150
response is absolutely
summable.
412
00:27:08,150 --> 00:27:12,820
And what that will require is
that the magnitude of a be
413
00:27:12,820 --> 00:27:14,070
less than 1.
414
00:27:17,400 --> 00:27:21,110
Now, we imposed causality,
linearity, and
415
00:27:21,110 --> 00:27:25,750
time-invariance, and generated
a solution recursively.
416
00:27:25,750 --> 00:27:29,590
And now, of course, if we want
to generate the more general
417
00:27:29,590 --> 00:27:33,880
set of solutions to this
difference equation, we can do
418
00:27:33,880 --> 00:27:38,650
that by adding all of the
homogeneous solutions, namely,
419
00:27:38,650 --> 00:27:42,860
all the solutions that satisfy
the homogeneous equation.
420
00:27:42,860 --> 00:27:46,230
So here, we have the
causal linear
421
00:27:46,230 --> 00:27:49,330
time-invariant impulse response.
422
00:27:49,330 --> 00:27:55,220
In fact, with an impulse input,
all of the possible
423
00:27:55,220 --> 00:28:00,380
solutions are that impulse
response plus
424
00:28:00,380 --> 00:28:02,500
the homogeneous solutions.
425
00:28:02,500 --> 00:28:06,620
The homogeneous solution is the
solution that satisfies
426
00:28:06,620 --> 00:28:09,660
the homogeneous equation.
427
00:28:09,660 --> 00:28:14,960
That will, in general, be of the
form an amplitude factor
428
00:28:14,960 --> 00:28:19,950
times an exponential factor, and
if we substitute this into
429
00:28:19,950 --> 00:28:24,610
this equation, then we see that
the homogeneous equation
430
00:28:24,610 --> 00:28:30,480
is satisfied for any values of
a and any values of z that
431
00:28:30,480 --> 00:28:33,780
satisfy this equation.
432
00:28:33,780 --> 00:28:36,180
Again, as we did with
differential equations, the
433
00:28:36,180 --> 00:28:39,960
factor A cancels out.
434
00:28:39,960 --> 00:28:43,870
And also, in fact, I can
cancel out a z^n
435
00:28:43,870 --> 00:28:48,100
there and a z^n here.
436
00:28:48,100 --> 00:28:52,420
And so what we're left with is a
statement that tells us then
437
00:28:52,420 --> 00:28:58,890
that the homogeneous solution is
of the form a(z^n), for any
438
00:28:58,890 --> 00:29:05,180
value of A and any value of z
that satisfies this equation.
439
00:29:05,180 --> 00:29:10,200
And that value of z, in
particular, is z equal to a.
440
00:29:10,200 --> 00:29:14,620
So the homogeneous solution
then is any exponential of
441
00:29:14,620 --> 00:29:17,080
this form with any
amplitude factor.
442
00:29:17,080 --> 00:29:20,670
And so, the family of solutions,
with an impulse
443
00:29:20,670 --> 00:29:25,800
input is the solution
corresponding to the system
444
00:29:25,800 --> 00:29:29,810
being causal, linear, and
time-invariant, plus the
445
00:29:29,810 --> 00:29:32,010
homogeneous term.
446
00:29:32,010 --> 00:29:36,950
If we impose causality,
linearity, and time-invariance
447
00:29:36,950 --> 00:29:41,430
on the system, then of course,
that additional exponential
448
00:29:41,430 --> 00:29:43,170
factor will be 0.
449
00:29:43,170 --> 00:29:45,400
In other words, A
is equal to 0.
450
00:29:48,970 --> 00:29:54,450
Now, we've seen the differential
equations and
451
00:29:54,450 --> 00:29:59,880
difference equations in terms
of the fact that there are
452
00:29:59,880 --> 00:30:04,040
families of solutions, and in
order to get causality,
453
00:30:04,040 --> 00:30:07,350
linearity, and time-invariance
requires imposing a particular
454
00:30:07,350 --> 00:30:10,930
set of initial conditions,
namely, imposing initial rest
455
00:30:10,930 --> 00:30:12,480
on the system.
456
00:30:12,480 --> 00:30:17,310
Let's now look at the difference
equation and then
457
00:30:17,310 --> 00:30:20,540
later, the differential
equation, interpreted in block
458
00:30:20,540 --> 00:30:23,190
diagram terms.
459
00:30:23,190 --> 00:30:27,790
Now, the difference equation,
as I just simply repeated
460
00:30:27,790 --> 00:30:30,780
here, is y[n]
461
00:30:30,780 --> 00:30:31,690
= x[n]
462
00:30:31,690 --> 00:30:37,360
+ ay[n-1], where I've taken the
delayed term over to the
463
00:30:37,360 --> 00:30:39,440
right hand side of
the equation.
464
00:30:39,440 --> 00:30:42,200
So, in effect, what
I'm imposing on
465
00:30:42,200 --> 00:30:44,460
this system is causality.
466
00:30:44,460 --> 00:30:47,990
I'm assuming that if I know the
past history of the input
467
00:30:47,990 --> 00:30:52,190
and the output, I can determine
the next value of
468
00:30:52,190 --> 00:30:53,860
the output.
469
00:30:53,860 --> 00:30:58,600
Well, we can, in fact, draw a
block diagram that represents
470
00:30:58,600 --> 00:31:00,130
that equation.
471
00:31:00,130 --> 00:31:03,140
The equations says,
we take x[n]
472
00:31:03,140 --> 00:31:07,890
for any given value of n, say
n0, whatever value we're
473
00:31:07,890 --> 00:31:09,980
computing the output for.
474
00:31:09,980 --> 00:31:15,850
We take the input at that time,
and add to it the factor
475
00:31:15,850 --> 00:31:21,390
a times the output value that
we calculated last time.
476
00:31:21,390 --> 00:31:27,880
So if we have x[n], which is
our input, and if we have
477
00:31:27,880 --> 00:31:34,160
y[n], which is our output, we,
in fact, can get y of n by
478
00:31:34,160 --> 00:31:41,290
taking the last value of y[n],
indicated here by putting y[n]
479
00:31:41,290 --> 00:31:52,070
through a delay, multiplying
that by the factor a, and then
480
00:31:52,070 --> 00:31:59,930
adding that result
to the input.
481
00:31:59,930 --> 00:32:05,430
And the result of doing
that is y[n].
482
00:32:05,430 --> 00:32:09,750
So the way this block diagram
might be interpreted, for
483
00:32:09,750 --> 00:32:16,720
example, as an algorithm is to
say that we take x[n], add to
484
00:32:16,720 --> 00:32:20,680
it a times the previous
value the output.
485
00:32:20,680 --> 00:32:24,160
That sum gives us the current
value of the output, which we
486
00:32:24,160 --> 00:32:29,440
then put out of the system,
and also put into a delay
487
00:32:29,440 --> 00:32:33,630
element, or basically, into a
storage register, to use on
488
00:32:33,630 --> 00:32:35,980
the next iteration
or recursion.
489
00:32:35,980 --> 00:32:37,580
Now, how do we get
this started?
490
00:32:37,580 --> 00:32:40,680
Well, we know that the
difference equation requires
491
00:32:40,680 --> 00:32:42,500
initial conditions.
492
00:32:42,500 --> 00:32:47,560
And, in fact, the initial
conditions correspond to what
493
00:32:47,560 --> 00:32:52,210
we store in the delay register
when this block diagram or
494
00:32:52,210 --> 00:32:55,300
equation initially starts up.
495
00:32:55,300 --> 00:33:01,380
OK Now, let's look at this for
the case of difference
496
00:33:01,380 --> 00:33:03,275
equations more generally.
497
00:33:06,650 --> 00:33:13,360
So, what we've said is that we
can calculate the output by
498
00:33:13,360 --> 00:33:18,460
having previous values of the
input, previous values of the
499
00:33:18,460 --> 00:33:22,270
output, and forming the
appropriate linear
500
00:33:22,270 --> 00:33:23,850
combination.
501
00:33:23,850 --> 00:33:27,550
So let's just build up the more
general block diagram
502
00:33:27,550 --> 00:33:28,960
that would correspond to this.
503
00:33:31,870 --> 00:33:35,900
And what it says is that we want
to have a mechanism for
504
00:33:35,900 --> 00:33:40,130
storing past values of the
input, and a mechanism for
505
00:33:40,130 --> 00:33:41,930
storing past values
of the output.
506
00:33:41,930 --> 00:33:47,400
And I've indicated that on
this figure, so far, by a
507
00:33:47,400 --> 00:33:55,610
chain of delay elements,
indicating that what the
508
00:33:55,610 --> 00:34:02,090
output of each delay is, is the
input delayed by one time
509
00:34:02,090 --> 00:34:05,050
instant or interval.
510
00:34:05,050 --> 00:34:09,560
And so what we see down this
chain of delays are delayed
511
00:34:09,560 --> 00:34:13,010
replications of the input.
512
00:34:13,010 --> 00:34:19,159
And what we see on the other
chain is delayed replications
513
00:34:19,159 --> 00:34:20,409
of the output.
514
00:34:22,960 --> 00:34:26,139
Now, the difference equation
says that we want to take
515
00:34:26,139 --> 00:34:31,150
these, and multiply them by the
appropriate coefficients,
516
00:34:31,150 --> 00:34:33,909
the coefficients in the
difference equation, and so,
517
00:34:33,909 --> 00:34:37,690
we can do that as I've
indicated here.
518
00:34:37,690 --> 00:34:42,880
So now, we have these delay
elements, each multiplied by
519
00:34:42,880 --> 00:34:46,090
the appropriate coefficients
on the input, and by
520
00:34:46,090 --> 00:34:50,139
appropriate coefficients
on the output.
521
00:34:50,139 --> 00:34:56,300
Those are then summed together,
and so we now will
522
00:34:56,300 --> 00:35:00,330
sum these and will sum these.
523
00:35:00,330 --> 00:35:03,990
After we've summed these, we
want to add those together.
524
00:35:03,990 --> 00:35:07,200
And there's a factor of
1 / a_0 that comes in.
525
00:35:07,200 --> 00:35:11,300
And so that then generates
our output.
526
00:35:11,300 --> 00:35:16,690
And so this, in fact, then
represents a block diagram,
527
00:35:16,690 --> 00:35:20,370
which is a general block diagram
for implementing or
528
00:35:20,370 --> 00:35:22,670
representing a linear
constant-coefficient
529
00:35:22,670 --> 00:35:25,060
difference equation.
530
00:35:25,060 --> 00:35:28,190
Now, if you think about what
it means in terms of, let's
531
00:35:28,190 --> 00:35:31,530
say, a computer algorithm or a
piece of hardware, in fact,
532
00:35:31,530 --> 00:35:34,670
this block diagram is a recipe
or algorithm for doing the
533
00:35:34,670 --> 00:35:36,410
implementation.
534
00:35:36,410 --> 00:35:41,680
But it's important to recognize,
even at this point,
535
00:35:41,680 --> 00:35:45,980
that it's only one of many
possible algorithms or
536
00:35:45,980 --> 00:35:48,990
implementations for this
difference equation.
537
00:35:48,990 --> 00:35:57,280
Just for example, I can consider
that equation for
538
00:35:57,280 --> 00:35:59,320
that block diagram.
539
00:35:59,320 --> 00:36:03,290
And here, I've re-drawn it.
540
00:36:03,290 --> 00:36:05,630
So here, are once again.
541
00:36:05,630 --> 00:36:09,140
I have the same block diagram
that we just saw.
542
00:36:09,140 --> 00:36:13,810
And I can recognize, for
example, that this, in
543
00:36:13,810 --> 00:36:19,620
essence, corresponds to two
linear time-invariant systems
544
00:36:19,620 --> 00:36:22,180
in cascade.
545
00:36:22,180 --> 00:36:26,540
Now, that assumes, of course,
that my initial conditions are
546
00:36:26,540 --> 00:36:30,020
such that the system is,
in fact, linear.
547
00:36:30,020 --> 00:36:33,480
And that, in turn, requires that
we're assuming initial
548
00:36:33,480 --> 00:36:35,300
rests, namely, before
the input does
549
00:36:35,300 --> 00:36:37,880
anything other than 0.
550
00:36:37,880 --> 00:36:41,030
There are just 0 values stored
in the registers.
551
00:36:41,030 --> 00:36:43,710
But assuming that it corresponds
to a linear
552
00:36:43,710 --> 00:36:48,640
time-invariant system, this
is a cascade of two linear
553
00:36:48,640 --> 00:36:50,540
time-invariant invriant
systems.
554
00:36:50,540 --> 00:36:52,990
We know that two linear
time-invariant systems can be
555
00:36:52,990 --> 00:36:55,160
cascaded in either order.
556
00:36:55,160 --> 00:36:59,090
So, in particular, I can
consider breaking this cascade
557
00:36:59,090 --> 00:37:03,820
here, and moving this block
over to the other side.
558
00:37:03,820 --> 00:37:05,520
And so let's just do that.
559
00:37:08,610 --> 00:37:16,160
And when I do, I then have this
combination of systems,
560
00:37:16,160 --> 00:37:19,370
and, of course, you can ask
what advantage there is to
561
00:37:19,370 --> 00:37:24,570
doing that, and the advantage
arises because of the fact
562
00:37:24,570 --> 00:37:31,990
that in this form, exactly what
is stored in these delays
563
00:37:31,990 --> 00:37:35,120
is also stored in these
delay registers.
564
00:37:35,120 --> 00:37:38,270
In other words, it's this
intermediate variable--
565
00:37:38,270 --> 00:37:39,500
whatever it is--
566
00:37:39,500 --> 00:37:42,850
down this chain of the delays
and down this chain of delays,
567
00:37:42,850 --> 00:37:47,150
and so, in fact, I can collapse
those delays into a
568
00:37:47,150 --> 00:37:49,000
single chain of delays.
569
00:37:49,000 --> 00:37:53,790
And the network that I'm left
with is the network that I
570
00:37:53,790 --> 00:37:57,930
indicate on this view graph,
where what I've done is to
571
00:37:57,930 --> 00:38:01,980
simply collapse that double
chain of delays into a single
572
00:38:01,980 --> 00:38:04,580
change of delays.
573
00:38:04,580 --> 00:38:10,180
Now, one can ask, well, what's
the advantage to doing that?
574
00:38:10,180 --> 00:38:13,020
And one advantage, simply
stated, is that when you think
575
00:38:13,020 --> 00:38:16,160
in terms of an implementation
of a difference equation, a
576
00:38:16,160 --> 00:38:21,340
delay corresponds to a storage
register, a memory location,
577
00:38:21,340 --> 00:38:24,980
and by simply using the fact
that we can interchange the
578
00:38:24,980 --> 00:38:27,200
order in which linear
time-invariant systems are
579
00:38:27,200 --> 00:38:30,030
cascaded, we can reduce
the amount of memory
580
00:38:30,030 --> 00:38:31,280
by a factor of 2.
581
00:38:35,280 --> 00:38:40,570
Now, an essentially similar
procedure can also be used for
582
00:38:40,570 --> 00:38:44,910
differential equations, in terms
of implementation using
583
00:38:44,910 --> 00:38:47,590
block diagrams or the
interpretation of
584
00:38:47,590 --> 00:38:50,350
implementations using
block diagrams.
585
00:38:50,350 --> 00:38:52,380
And let me first do that--
586
00:38:52,380 --> 00:38:56,840
rather than in general-- let me
first do it in the context
587
00:38:56,840 --> 00:39:00,870
of a specific example.
588
00:39:00,870 --> 00:39:06,830
So let's consider a linear
constant-coefficient
589
00:39:06,830 --> 00:39:10,790
differential equation, as I've
indicated here, and I have
590
00:39:10,790 --> 00:39:13,930
terms on the left side and
terms on the right side.
591
00:39:13,930 --> 00:39:19,870
And with the differential
equation, let's consider
592
00:39:19,870 --> 00:39:22,800
taking all the terms over
to the right side of the
593
00:39:22,800 --> 00:39:26,030
equation, except
for the highest
594
00:39:26,030 --> 00:39:28,950
derivative in the output.
595
00:39:28,950 --> 00:39:33,250
Next, we integrate both sides
of the equation so that when
596
00:39:33,250 --> 00:39:35,540
we're done, we end up with
on the left side of the
597
00:39:35,540 --> 00:39:37,480
equation with y(t).
598
00:39:37,480 --> 00:39:40,480
On the right side of the
equation with the appropriate
599
00:39:40,480 --> 00:39:42,700
number of integrations.
600
00:39:42,700 --> 00:39:46,170
And so the integral equation
that we'll get for this
601
00:39:46,170 --> 00:39:52,410
example y(t), the output, is
x(t) plus b, the scale factor
602
00:39:52,410 --> 00:39:56,370
times the integral of the input,
and minus a, that scale
603
00:39:56,370 --> 00:39:59,480
factor, times the integral
of the output.
604
00:39:59,480 --> 00:40:05,600
So to form the output in the
block diagram terms, we form a
605
00:40:05,600 --> 00:40:09,140
linear combination of the input,
a scaled integral of
606
00:40:09,140 --> 00:40:12,970
the input, and a scaled integral
of the output, all of
607
00:40:12,970 --> 00:40:14,820
that added together.
608
00:40:14,820 --> 00:40:20,540
So we need, in addition to
the input, we need the
609
00:40:20,540 --> 00:40:22,070
integral of the input.
610
00:40:22,070 --> 00:40:25,250
And so this box indicates
an integrator.
611
00:40:25,250 --> 00:40:28,260
In addition to the output,
we need the
612
00:40:28,260 --> 00:40:30,670
integral of the output.
613
00:40:30,670 --> 00:40:38,490
And now, to form y(t), we
multiply the integrated input
614
00:40:38,490 --> 00:40:41,540
by the scale factor, b.
615
00:40:41,540 --> 00:40:49,860
And add that to x(t), and we
take the integrated output,
616
00:40:49,860 --> 00:40:57,330
multiply it by -a, and add
to that the result of the
617
00:40:57,330 --> 00:41:00,530
previous addition, and according
to the integral
618
00:41:00,530 --> 00:41:05,180
equation, then that
forms the output.
619
00:41:05,180 --> 00:41:09,580
So just as we did with the
difference equation, we've
620
00:41:09,580 --> 00:41:12,330
converted the differential
equation to an integral
621
00:41:12,330 --> 00:41:17,200
equation, and we have a block
diagram form very similar to
622
00:41:17,200 --> 00:41:20,100
what we had in the case of
the difference equation.
623
00:41:20,100 --> 00:41:24,160
Now, the initial conditions,
of course, are tied up in,
624
00:41:24,160 --> 00:41:27,430
again, how these integrators
are initialized.
625
00:41:27,430 --> 00:41:30,980
Assuming that we impose initial
rest on the system, we
626
00:41:30,980 --> 00:41:33,710
can think of the overall
system as a linear
627
00:41:33,710 --> 00:41:37,740
time-invariant system, and it's
a cascade of one linear
628
00:41:37,740 --> 00:41:40,470
time-invariant system
with a second.
629
00:41:40,470 --> 00:41:47,190
So we can, in fact, break
this, and consider
630
00:41:47,190 --> 00:41:49,620
interchanging the order
in which these
631
00:41:49,620 --> 00:41:51,520
two systems are cascaded.
632
00:41:51,520 --> 00:41:54,580
And so I've indicated
that down below.
633
00:41:54,580 --> 00:41:58,840
Here, I've simply taken
the top block diagram,
634
00:41:58,840 --> 00:42:01,920
interchanged the order
in which the
635
00:42:01,920 --> 00:42:05,590
two systems are cascaded.
636
00:42:05,590 --> 00:42:09,490
And here, again, we can ask what
the advantages to this,
637
00:42:09,490 --> 00:42:12,090
as opposed to the
previous one.
638
00:42:12,090 --> 00:42:14,830
And what you can see, just as
we saw with the difference
639
00:42:14,830 --> 00:42:18,790
equation, is that now,
the integrators--
640
00:42:18,790 --> 00:42:19,780
both integrators--
641
00:42:19,780 --> 00:42:21,730
are integrating the
same thing.
642
00:42:21,730 --> 00:42:27,170
In particular, the input to this
integrator and the input
643
00:42:27,170 --> 00:42:30,210
to this integrator
are identical.
644
00:42:30,210 --> 00:42:34,660
So in fact, rather than using
this one, we can simply tap
645
00:42:34,660 --> 00:42:36,970
off from here.
646
00:42:36,970 --> 00:42:40,660
We can, in fact, remove this
integrator, break this
647
00:42:40,660 --> 00:42:45,610
connection, and tap
in at this point.
648
00:42:45,610 --> 00:42:49,910
And so what we've done then, by
interchanging the order in
649
00:42:49,910 --> 00:42:53,420
which the systems are cascaded,
is reduced the
650
00:42:53,420 --> 00:42:56,730
implementation to the
implementation with a single
651
00:42:56,730 --> 00:42:58,060
integrator.
652
00:42:58,060 --> 00:43:01,830
Very much similar to what we
talked about in the case of
653
00:43:01,830 --> 00:43:03,630
the difference equation.
654
00:43:03,630 --> 00:43:07,600
Now, let's just, again, with the
integral equation or the
655
00:43:07,600 --> 00:43:13,270
differential equation, look at
this somewhat more generally.
656
00:43:13,270 --> 00:43:17,360
Again, if we take the
differential equation, the
657
00:43:17,360 --> 00:43:20,480
general differential equation,
integrate it a sufficient
658
00:43:20,480 --> 00:43:23,740
number of times to convert it
to an integral equation.
659
00:43:23,740 --> 00:43:28,320
We would then have this
cascade of systems.
660
00:43:28,320 --> 00:43:33,000
And again, if we assume initial
rest, so that these
661
00:43:33,000 --> 00:43:36,520
are both linear time-invariant
systems, we can interchange
662
00:43:36,520 --> 00:43:38,610
the order in which
they're cascaded.
663
00:43:38,610 --> 00:43:45,990
Namely, take the second system,
and move it to precede
664
00:43:45,990 --> 00:43:48,220
the first system.
665
00:43:48,220 --> 00:43:51,890
And then what we recognize is
that the input to this chain
666
00:43:51,890 --> 00:43:54,530
of integrators and this
chain of integrators
667
00:43:54,530 --> 00:43:56,220
is exactly the same.
668
00:43:56,220 --> 00:44:00,320
And so, in fact we can collapse
these together using
669
00:44:00,320 --> 00:44:02,310
only one chain of integrators.
670
00:44:02,310 --> 00:44:07,890
And the system that we're left
with then is a system that
671
00:44:07,890 --> 00:44:09,730
looks as I've indicated here.
672
00:44:09,730 --> 00:44:14,890
So we have now just a single
chains of integrators instead
673
00:44:14,890 --> 00:44:18,330
of the two sets of
integrators.
674
00:44:18,330 --> 00:44:21,670
So we've seen that the situation
is very similar here
675
00:44:21,670 --> 00:44:24,510
as it was in the case of the
difference equation.
676
00:44:24,510 --> 00:44:26,900
Again, why do we want
to cut the number of
677
00:44:26,900 --> 00:44:28,310
integrators in half?
678
00:44:28,310 --> 00:44:32,140
Well, one reason is because
integrators, in effect,
679
00:44:32,140 --> 00:44:33,710
represent hardware.
680
00:44:33,710 --> 00:44:38,155
And if we have half as many
integrators, then we're using
681
00:44:38,155 --> 00:44:39,410
half as much hardware.
682
00:44:41,940 --> 00:44:44,440
Well, let me just conclude by
683
00:44:44,440 --> 00:44:47,550
summarizing a number of points.
684
00:44:47,550 --> 00:44:50,340
I indicated at the beginning
that linear
685
00:44:50,340 --> 00:44:52,690
constant-coefficient
differential equations and
686
00:44:52,690 --> 00:44:57,060
difference equations will play
an important role as linear
687
00:44:57,060 --> 00:45:01,490
time-invariant systems
throughout this course and
688
00:45:01,490 --> 00:45:04,160
throughout this set
of lectures.
689
00:45:04,160 --> 00:45:09,210
I also stressed the fact that
differential or difference
690
00:45:09,210 --> 00:45:13,990
equations, by themselves, are
not a complete specification
691
00:45:13,990 --> 00:45:18,190
of the system because of the
fact that we can add to any
692
00:45:18,190 --> 00:45:21,970
solution a homogeneous
solution.
693
00:45:21,970 --> 00:45:26,080
How do we specify the
appropriate initial conditions
694
00:45:26,080 --> 00:45:28,270
to ensure--
695
00:45:28,270 --> 00:45:30,260
how do we specify the
appropriate initial conditions
696
00:45:30,260 --> 00:45:34,250
to ensure that the system is
linear and time-invariant?
697
00:45:34,250 --> 00:45:39,110
Well, the auxiliary information,
namely, the
698
00:45:39,110 --> 00:45:45,100
initial conditions associated
with the system being causal,
699
00:45:45,100 --> 00:45:47,790
linear, and time-invariant
are the
700
00:45:47,790 --> 00:45:50,630
conditions of initial rest.
701
00:45:50,630 --> 00:45:54,130
And, in fact, for most of the
course, what we'll be
702
00:45:54,130 --> 00:45:57,200
interested in are systems that
are in fact, causal, linear,
703
00:45:57,200 --> 00:45:58,420
and time-invariant.
704
00:45:58,420 --> 00:46:01,150
And so we will, in fact, be
assuming initial rest
705
00:46:01,150 --> 00:46:03,640
conditions.
706
00:46:03,640 --> 00:46:09,670
Now, as I also indicated,
there are a variety of
707
00:46:09,670 --> 00:46:12,580
efficient procedures for solving
differential and
708
00:46:12,580 --> 00:46:15,560
difference equations that we
haven't yet addressed.
709
00:46:15,560 --> 00:46:18,990
And beginning with the next
set of lectures, we'll be
710
00:46:18,990 --> 00:46:22,510
talking about the Fourier
Transform and much later in
711
00:46:22,510 --> 00:46:26,470
the course, what's referred to
as the Laplace Transform for
712
00:46:26,470 --> 00:46:30,070
continuous time and the
Z-transform for discrete time.
713
00:46:30,070 --> 00:46:34,030
And what we'll see is that with
the Fourier Transform and
714
00:46:34,030 --> 00:46:37,300
later with the Laplace and
Z-transform, we'll have a
715
00:46:37,300 --> 00:46:42,480
number of efficient and very
useful ways of generating the
716
00:46:42,480 --> 00:46:48,000
solution for differential and
difference equations under the
717
00:46:48,000 --> 00:46:50,705
assumption that the system
is causal, linear, and
718
00:46:50,705 --> 00:46:52,510
time-invariant.
719
00:46:52,510 --> 00:46:56,020
Also, we'll see in addition
to the block diagram
720
00:46:56,020 --> 00:46:58,510
implementations of these systems
that we've talked
721
00:46:58,510 --> 00:47:04,170
about so far, we'll see a
number of other useful
722
00:47:04,170 --> 00:47:07,880
implementations that exploit
a variety of properties
723
00:47:07,880 --> 00:47:10,870
associated with Fourier and
Laplace Transforms.
724
00:47:10,870 --> 00:47:12,120
Thank you.