1
00:00:45,079 --> 00:00:52,079
Before we start, before we start I just like
to point out a small slip, in the last class
2
00:00:53,579 --> 00:00:58,870
there should have been the derivative that
you take for multiple poles; this should have
3
00:00:58,870 --> 00:01:05,870
been multiplied by m, please make the corrections,
z minus p K to the power m. In the example,
4
00:01:06,399 --> 00:01:13,399
you have used this, it was a small slip. We
are discussing in the last class; the inverse
5
00:01:23,560 --> 00:01:30,560
DTFT from the given frequency domain representation,
as given by this expression.
6
00:01:36,490 --> 00:01:43,140
So, x n will be this discrete line. Now, how
do you know, that this is going to give you
7
00:01:43,140 --> 00:01:50,140
the same sequence, you get my point? Given
a sequence x n, if I take its Fourier transform
8
00:01:53,780 --> 00:02:00,780
by this integral, do you get back the same
sequence? So, let us see let us verify this
9
00:02:02,580 --> 00:02:09,580
relation 1 by 2 p i minus p i to plus p I;
this is nothing but sigma x k e to the power
10
00:02:15,830 --> 00:02:22,830
minus j omega k into e to the power j omega
n, d omega, okay.
11
00:02:27,580 --> 00:02:34,580
You may write this as, sigma x k integral
minus p i to plus p i, 1 by 2 p i e to the
12
00:02:46,549 --> 00:02:53,549
power j omega n minus k d omega, okay. So,
that gives me if we integrate this, we will
13
00:03:03,340 --> 00:03:10,340
get after substitution of these values sin
n minus k p i by n minus k into p i. And this
14
00:03:24,090 --> 00:03:31,090
is; this particular function will be equal
to 1 when n is equal to k and equal to 0 otherwise.
15
00:03:39,760 --> 00:03:46,760
So, this gives me sigma sin n. So, basically
that gives you the series as x n, basically;
16
00:03:52,959 --> 00:03:59,959
this is a delta function n is equal to k and
n not equal to k, this function is basically
17
00:04:01,330 --> 00:04:08,330
delta n minus k. So, x k multiplied by delta
n minus k is x n itself, okay. Now, we take
18
00:04:16,150 --> 00:04:22,550
some difference equations and see what will
be the solution in the normal time domain.
19
00:04:22,550 --> 00:04:28,160
How to compute by the normal classical method?
20
00:04:28,160 --> 00:04:35,160
You have say a 0 x n plus y n is equal to
a 0 x n plus a 1 x n minus 1 plus and so on,
21
00:04:49,759 --> 00:04:56,759
A K x n minus k, all right. Now, the output
of the see for a sequence like this; suppose
22
00:05:11,580 --> 00:05:18,580
you have a term like a minus 1 x n plus 1
and so on, suppose the x n sequence is considered
23
00:05:19,129 --> 00:05:26,129
in the left hand side of the time domain,
can you have an output? Can you have an output
24
00:05:28,039 --> 00:05:35,039
which will be a function of the previous inputs,
as well as inputs which are yet to come?
25
00:05:36,089 --> 00:05:43,089
It is not possible, inputs which have not
yet come how can the system response, identify
26
00:05:43,289 --> 00:05:50,289
that, all right? So, that will be a non-causal
system, okay. So, causal system means, this
27
00:05:56,900 --> 00:06:03,900
kind of a term will not be present; for y
n I cannot have on the right hand side, the
28
00:06:04,719 --> 00:06:11,719
dependence of y n on x n plus 1 n plus 2,
okay.
29
00:06:13,389 --> 00:06:19,909
Sometimes we try find out, some kind of a
relationship between the present value and
30
00:06:19,909 --> 00:06:26,650
the future values; that is possible only when
you are having offline delta that is, the
31
00:06:26,650 --> 00:06:31,289
history is already available, so what is going
to come say after one hour that is already
32
00:06:31,289 --> 00:06:33,969
known to you, that record is there.
33
00:06:33,969 --> 00:06:39,029
So, can you correlate with the present value,
if it is possible? There are now causal systems
34
00:06:39,029 --> 00:06:46,029
also; you take an image all right, you focus
a light here, now it is not time domain function,
35
00:06:48,999 --> 00:06:55,059
it is phase domain function, means phase domain
this is plus all right. So, a light here will
36
00:06:55,059 --> 00:07:01,159
be influencing the illumination here, here,
here, it will also be influencing the illumination
37
00:07:01,159 --> 00:07:03,949
on this side, all right.
38
00:07:03,949 --> 00:07:10,949
So, if I consider in-space these as positive
side then this as negative side, then in that
39
00:07:11,389 --> 00:07:18,389
case at any instant of time; we are not considering
time as such, the illumination of a particular
40
00:07:20,849 --> 00:07:27,849
of a particular point will be depending also
on its previous values, I mean the location
41
00:07:29,439 --> 00:07:32,089
can be on both sides, that what I mean.
42
00:07:32,089 --> 00:07:39,089
So, there we can have functions with z to
the power plus 1, if you consider just one
43
00:07:39,129 --> 00:07:46,129
dimensional; say one line of a particular
object, of an image, then if this is z to
44
00:07:47,339 --> 00:07:51,839
the power minus 1, the normal backward shift
operator then this will be forward shift operator
45
00:07:51,839 --> 00:07:58,839
z to the power of plus 1, so both are possible.
So, this is that way something like a non-causal
46
00:08:00,499 --> 00:08:06,429
system, otherwise all natural processes are
causal systems.
47
00:08:06,429 --> 00:08:13,429
Now, suppose we have a difference equation;
k y on this side which is the summation of
48
00:08:19,240 --> 00:08:26,240
weighted values like this, equal to b k x
n minus k. k varying from say 1 to M, k varying
49
00:08:34,000 --> 00:08:41,000
from say, 0 to M and here it is 0 to N, okay.
You can write y solution y n, as b k x n minus
50
00:08:54,010 --> 00:09:01,010
k summation minus k varying from 0 to M minus
a k y n minus k, k varying from 1 to N, okay.
51
00:09:15,830 --> 00:09:22,830
So, when you have a general expression like
this; that is y n depends on the previous
52
00:09:30,380 --> 00:09:36,830
values of y, y n minus 1 n minus 2 because
k starts from 1 n minus 1 n minus 2 and so
53
00:09:36,830 --> 00:09:43,830
on and some values of the inputs then that
is known as auto-regressive process, okay.
54
00:09:53,580 --> 00:10:00,580
Auto-regressive and moving average, moving
average means; it is a weighted mean of the
55
00:10:05,560 --> 00:10:10,550
inputs average of the input, average in some
sense.
56
00:10:10,550 --> 00:10:16,910
You give more weightage to the present value
of the input, a little less weightage to the
57
00:10:16,910 --> 00:10:21,340
previous input and so on, but the previous
inputs also have an influence in the present
58
00:10:21,340 --> 00:10:27,340
on the present output. So, this is a moving
average part and it is regressing on itself,
59
00:10:27,340 --> 00:10:30,150
its previous values, so it is autoregressive.
60
00:10:30,150 --> 00:10:37,150
So, it is auto-regressive and moving average
more commonly known as, ARMA process, all
61
00:10:43,470 --> 00:10:50,470
right. ARMA, auto-regressive part what is
the order? N, ARMA N, moving average, what
62
00:10:53,740 --> 00:11:00,740
is the order? M. So, we write ARMA N M, it
is an ARMA, M N, N M process or somebody may
63
00:11:04,590 --> 00:11:10,630
choose P Q, ARMA, P Q process.
64
00:11:10,630 --> 00:11:17,630
When you have y n, as just one term of x n
and other terms of y; we call it simple auto
65
00:11:35,050 --> 00:11:42,050
regressive process, there no moving average,
just one input is that okay. When you have
66
00:11:50,160 --> 00:11:57,160
y n dependent on only inputs, it is called
moving average process, all right; moving
67
00:12:11,850 --> 00:12:18,850
average of order M, this is auto regressive
of A R process of order N and this also called
68
00:12:26,560 --> 00:12:33,560
FIR okay, Finite impulse response.
69
00:12:34,260 --> 00:12:41,260
Suppose we have, suppose we have
a system like this; will come to this very
soon, a system like this, what will be say
70
00:12:50,420 --> 00:12:57,420
in the Z domain, what will be the transfer
function? Y z by X z, it will be summation
71
00:13:06,390 --> 00:13:13,390
b k z to the power minus k, k varying from
0 to M, okay. So, this is H z, is it not?
72
00:13:20,050 --> 00:13:27,050
So, what will be h K impulse response? It
will be b 0, b 1 up to b K, b M okay.
73
00:13:32,390 --> 00:13:39,390
So, if I excite the system if excite the system,
with an impulse then this system; if I give
74
00:13:42,530 --> 00:13:48,460
an impulse as input then it will give you
output up to b M and after that it will be
75
00:13:48,460 --> 00:13:55,460
0. So, it is an infinite length, so that is
that is why it is called finite impulse response
76
00:13:56,360 --> 00:14:02,120
system, FIR system. It is same as, in mathematics
we called it M A, moving average module, it
77
00:14:02,120 --> 00:14:03,040
one and the same thing.
78
00:14:03,040 --> 00:14:10,040
If on the other hand, if you have a system
like this that is Y z by X z comes out as
79
00:14:22,430 --> 00:14:29,430
1 by some A z, that is basically 1 by 1 plus
a 1 z to the power minus 1 plus a 2 z to the
80
00:14:40,230 --> 00:14:46,390
power minus 2 and so on, the polynomial a
N z to the power minus N, okay.
81
00:14:46,390 --> 00:14:53,390
If you get something of this type then it
is called an IIR function, infinite impulse
82
00:15:01,730 --> 00:15:08,730
response. Let us see, what ii means infinite
impulse response. Let us take just one term,
83
00:15:14,930 --> 00:15:21,930
suppose H z equal to 1 by 1 minus 0.8 z inverse,
it can be plus or minus, it really does not
84
00:15:29,930 --> 00:15:36,930
matter. Then Y z is equal to X z by 1 minus
0.8 z inverse okay or multiplied by this Y
85
00:15:48,730 --> 00:15:55,730
K minus 0.8 Y k minus 1 is equal to x k, if
I multiply by this and then take the inverse,
86
00:16:04,190 --> 00:16:11,190
this will be the difference equation.
So, Y k is a solution, I can write 0.8 Y k
87
00:16:13,970 --> 00:16:20,970
minus 1 plus X k, this is a difference equation.
Now, I want to see its impulse response; that
88
00:16:22,860 --> 00:16:28,839
is X k is delta k, what will be the response,
okay.
89
00:16:28,839 --> 00:16:35,839
So, X k is delta k, that is x 0 is 1 then
x 1, x 2 etcetera are all 0’s, is it not?
90
00:16:46,550 --> 00:16:51,180
You are giving just an impulse at k equal
to 0 and after that it is all zero, so what
91
00:16:51,180 --> 00:16:58,180
will be y 0? If the system is initially unexcited
then there is, if it is a causal system; before
92
00:17:00,120 --> 00:17:07,120
the input is given there was no output, so
y at minus 1 is 0, okay. So, y 0 will be 0.8
93
00:17:12,500 --> 00:17:19,500
into y at minus 1 which is 0 plus x at 0,
which means 0 plus x at 0 is 1. So, y 0 will
94
00:17:22,860 --> 00:17:23,159
be 1.
95
00:17:23,159 --> 00:17:30,159
What will be y 1? 0.8, it will be now y 0
which is 1 plus x at 1 is 0, so that will
96
00:17:40,950 --> 00:17:47,950
be 0.8. What will be y at 2? It will be 0.8
into again 0.8 plus 0. So, 0.64, so it will
97
00:17:53,039 --> 00:18:00,039
continue to give you output which will be
gradually diminishing in magnitude but it
98
00:18:03,679 --> 00:18:10,679
continues theoretically, it continues upto
infinite. It does not stop at a definite point.
99
00:18:10,830 --> 00:18:17,759
You can approximate, you can approximate truncate
it somewhere, but otherwise it is an infinite
100
00:18:17,759 --> 00:18:24,759
sequence; so that is why it is called IIR,
this kind of response is infinite impulse
101
00:18:25,769 --> 00:18:32,769
response. Now, what will be the response of
any system, in any difference equation that
102
00:18:42,690 --> 00:18:49,690
we have written of this kind? A difference
equation of this type, causal system; what
103
00:18:49,759 --> 00:18:56,759
will be the response, due to any general input?
So, the input whatever we give will decide
104
00:18:58,960 --> 00:19:05,590
the output but apart from that like you have
done in mathematics for a differential equation,
105
00:19:05,590 --> 00:19:12,590
there are two parts two components or solution,
a particular integral and a general, okay.
106
00:19:14,879 --> 00:19:21,879
So, y n will have y particular plus y complimentary,
okay. So, y complimentary when there is no
107
00:19:41,059 --> 00:19:48,059
input, when x n is 0 there is no input; there
is no input you calculate a K y n minus k
108
00:20:03,159 --> 00:20:10,159
to the right hand, it is all 0. So, this solution
is the complimentary solution and why particular
109
00:20:19,440 --> 00:20:25,999
there is particular solution will be due to
x n itself.
110
00:20:25,999 --> 00:20:32,999
So we again start with an assume solution
here. So, let assume a solution y n is equal
111
00:20:41,570 --> 00:20:48,570
to lambda to the power n, okay. So, substitute
it here. So, a K lambda n minus k summation,
112
00:20:58,379 --> 00:21:05,379
all right which will be a 0 plus a 1 lambda
plus a 2 lambda squared and so on, all right.
113
00:21:17,940 --> 00:21:24,940
And that is equal to 0, you get the solution
for lambda, is this all right, or other way?
114
00:21:25,330 --> 00:21:32,330
It is lambda to the power n, so okay a 0 lambda
to the power n, so it will be a 0 lambda to
115
00:21:38,929 --> 00:21:45,929
the power N; it should be capital N because
it is going upto a certain, see this n is
116
00:21:48,720 --> 00:21:55,399
a variable n and this N is a finite length,
since we are considering a finite length of
117
00:21:55,399 --> 00:22:02,399
this sequence, 1 to capital N, so I should
have taken N
minus k, okay.
118
00:22:11,570 --> 00:22:18,570
So, it will be lambda N minus 1 and so on
up to a N, that is equal to 0; so you can
119
00:22:25,230 --> 00:22:32,230
calculate the roots of lambda, there are N
roots all right. So, once you know the roots,
120
00:22:33,519 --> 00:22:36,730
linear combination of the roots will be the
solution.
121
00:22:36,730 --> 00:22:43,730
So, C 1 lambda 1 plus C 2 lambda 2 plus C
3 lambda 3 and so on; C N lambda N will be
122
00:22:50,220 --> 00:22:57,220
solution Y complimentary, lambda 1 to the
power n lambda 2 to the power n lambda 3 to
123
00:23:07,499 --> 00:23:14,499
the power n, because our assumed solution
was lambda to the power n as y n, okay. Y
124
00:23:20,710 --> 00:23:27,710
particular integral, particular solution
will be depending on the type of input, if
you give a sinusoidal input correspondingly
125
00:23:33,470 --> 00:23:39,850
output will also be sinusoidal, if you give
a constant input like an impulse, like a step
126
00:23:39,850 --> 00:23:43,840
then output will also be a constant, okay.
127
00:23:43,840 --> 00:23:50,499
So, you once again you assume a solution corresponding
to the type of input that you give; if it
128
00:23:50,499 --> 00:23:57,499
is not a regular input then of course it becomes
a little difficult. So, Y will be depending
129
00:24:03,610 --> 00:24:10,610
on, say we give an input of step x n is a
some k times u n then correspondingly; we
130
00:24:21,600 --> 00:24:28,600
can assume a solution alpha because it is
a constant, so output corresponding to this
131
00:24:37,869 --> 00:24:40,659
part will be a constant.
132
00:24:40,659 --> 00:24:47,659
So from the equation, that you have already
got say; a 0 we wrote a k, all right, a k
133
00:24:50,059 --> 00:24:57,059
a k plus a sorry a k y n plus a k minus 1
y n minus 1 and so on. This will be all alpha
134
00:25:17,580 --> 00:25:24,580
plus alpha plus alpha, so on because they
are all constant, n terms should be equal
135
00:25:27,470 --> 00:25:34,470
to the given value k into u n k, okay. So,
N times alpha is k, so alpha you have got
136
00:25:39,609 --> 00:25:40,369
k by N.
137
00:25:40,369 --> 00:25:47,369
The input has been given as an constant magnitude,
step function, all right; so all the time
138
00:25:51,179 --> 00:25:58,179
it will be equal to k. So in the difference
equation, this is given as x n which is k
139
00:26:00,899 --> 00:26:07,899
times u n, so you get with of course with
sorry sorry a k minus 1 etcetera, I am sorry.
140
00:26:13,600 --> 00:26:20,600
So, it will be summation a k into alpha is
equal to k, so alpha is summation a k, all
141
00:26:24,669 --> 00:26:31,669
right. Let us workout a small example, it
will be clear. Once you have got this, try
142
00:26:33,450 --> 00:26:38,820
to use it with the initial conditions, try
to match with the initial conditions, you
143
00:26:38,820 --> 00:26:43,450
can calculate all the constants.
144
00:26:43,450 --> 00:26:50,450
We have a system, y n plus 5 into y n minus
1 plus 4 into y n minus 2 is equal to x n,
145
00:27:03,379 --> 00:27:10,379
all right. Y at minus 1 is equal to 2; y at
minus 2 is equal to 1, determine the response
146
00:27:27,190 --> 00:27:34,190
corresponding to x n is equal to 4 times u
n, okay. So, let us first consider when x
147
00:27:41,659 --> 00:27:48,659
n is 0 that is a complimentary solution. Y
complimentary will be, we assume lambda to
148
00:27:59,299 --> 00:28:06,299
the power n, so the solution for that will
be lambda square plus 5 lambda plus 4 lambda
149
00:28:06,570 --> 00:28:13,570
square, sorry lambda square plus 5 lambda
plus 4, okay equal to 0.
150
00:28:15,759 --> 00:28:22,759
This is also known as characteristic equation;
lambda is equal to minus 4 or minus 1, okay.
151
00:28:25,529 --> 00:28:32,529
So, the solution
will be some C 1 times minus 4 to the power
n plus C 2 times minus 1 to the power n into
152
00:28:47,679 --> 00:28:54,309
of course u n, okay.
153
00:28:54,309 --> 00:29:01,309
Next, we take y particular solution as some
constant alpha because you are exciting it
154
00:29:12,149 --> 00:29:19,149
by, a step function then alpha plus 5 into
alpha plus 4 into alpha will be equal to 4,
155
00:29:27,220 --> 00:29:34,220
is it not? So, alpha plus 5 into alpha plus
4 into alpha is equal to 4, alpha is equal
156
00:29:38,639 --> 00:29:45,639
to 4 by 10, all right. Now, y n is equal to
C 1 into minus 4 to the power n plus C 2 minus
157
00:29:58,359 --> 00:30:05,359
1 to the power n plus 0.4, because it is a
complimentary solution plus particular solution,
158
00:30:09,259 --> 00:30:11,830
all right.
159
00:30:11,830 --> 00:30:18,269
Substitute the initial values, initial condition
are initial conditions are y at minus 1, what
160
00:30:18,269 --> 00:30:25,269
be the value at minus 1? C 1 minus 4 to the
power minus 1, N is equal to minus 1 plus
161
00:30:29,389 --> 00:30:36,389
C 2 minus 1 to the power minus 1 plus 0.4
and that is given as 2; this is minus of 4,
162
00:30:41,190 --> 00:30:48,190
so minus C 1 by 4 minus C 2 is equal to 2
minus 0.4, that is 1.6, okay.
163
00:30:54,179 --> 00:31:01,179
If you simplify C 1 plus 4, C 2 is equal to
minus 6.4, okay this is one equation. We substitute
164
00:31:09,470 --> 00:31:12,379
in the second equation, second condition.
165
00:31:12,379 --> 00:31:19,379
So, y at minus 2 is similarly minus 4 to the
power minus 2; so it will be 1 by 16 plus
166
00:31:26,519 --> 00:31:33,519
1 by 16 minus 1 to the power minus 2, is again
plus 1. So, C 1 by 16 I am just writing the
167
00:31:37,909 --> 00:31:44,909
final values 0.4 and that is equal to 1 or
from this C 1 plus 16 C 2 is equal to 9.6,
168
00:31:50,849 --> 00:31:57,849
0.4 if I bring it to this side, it becomes
0.6 0.6 into 16.9.6, this a second equation.
169
00:31:58,599 --> 00:32:05,599
First equation was C 1 plus 4 C 2 equal to
minus 6.4 okay; this is was the first equation.
170
00:32:07,509 --> 00:32:12,619
So, if you subtract one from the other, C
1 will get cancelled. So, 12 C 2 will give
171
00:32:12,619 --> 00:32:19,619
me 9.6 plus 6.4. So, C 2 becomes 4 by 3, checks
whether, is it all right, okay. And then this
172
00:32:26,080 --> 00:32:33,080
is approximately 11.733, okay with a negative
sign, C 1 is negative.
173
00:32:35,580 --> 00:32:42,580
So the solution will be y n equal minus 11.733
minus 4 to the power n plus 1.33 minus 1 to
174
00:32:54,609 --> 00:33:01,609
the power n plus 0.4, this is a general solution
for n greater than equal to 0, is it all right.
175
00:33:04,849 --> 00:33:11,849
Whether same thing you can solve by Z transform
method also, that we have already seen.
176
00:33:16,479 --> 00:33:23,479
Sometimes when you study signals, sometimes
you want to compare one signal with other;
177
00:33:24,009 --> 00:33:29,720
especially in communication. Some signals
are coming, you want to check whether it is
178
00:33:29,720 --> 00:33:36,720
matching with some other signal or not. So,
you have got the sample values, you try correlate
179
00:33:38,869 --> 00:33:45,869
them, so it define a correlation function
between two signals. If it is between two
180
00:33:46,970 --> 00:33:53,349
different signals say, x n and Y n we call
it cross correlation. If it is a signal with
181
00:33:53,349 --> 00:33:56,340
it itself then it is auto-correlation
182
00:33:56,340 --> 00:34:03,340
So, cross correlation we define r x y of lag
l as, sigma x n y n minus l; l equal to 0
183
00:34:24,250 --> 00:34:31,250
plus minus 1 plus minus 2 and so on, I can
have lag positive or negative.
184
00:34:38,800 --> 00:34:45,800
r y x l will be sigma Y n x n minus l, okay.
This n is over this range minus infinite plus
185
00:35:04,910 --> 00:35:11,910
infinite. So, this will be if you change the
variables; you will get r x y minus l, this
186
00:35:28,750 --> 00:35:35,750
is something very similar to which term, very
similar to convolution, is it not? You try
187
00:35:39,150 --> 00:35:44,330
to find out the certain difference between
convolution of a sequence up to sequences
188
00:35:44,330 --> 00:35:51,150
and cross correlation, there is a difference,
all right.
189
00:35:51,150 --> 00:35:58,150
So, here it is n, n minus l the other one
is l minus n. Now, auto-correlation
of a sequence x, it r x x l is equal to x
190
00:36:18,030 --> 00:36:25,030
n into x n minus l, okay. r x x 0, will be
X square n x n into x n, it is also called
191
00:36:39,730 --> 00:36:46,730
the energy of the signal. Now this is an even
function, r x x l is an even function, can
192
00:37:02,900 --> 00:37:09,900
I tell me why, this is say x n this is also
x n all right; now if I give it a shift by
193
00:37:14,460 --> 00:37:20,790
l all right or if I give shift to this one,
it is one at the same thing.
194
00:37:20,790 --> 00:37:27,790
Means, if I shifted forward or backward, it
is symmetrical, it is a same sequence, all
195
00:37:28,990 --> 00:37:35,990
right. So, it is an even function, is it not?
For example, if you have hundred points; x
196
00:37:36,630 --> 00:37:43,400
zero, x one, x two, x three, x hundred then
if I take a lag of 1, so it will be x 0 into
197
00:37:43,400 --> 00:37:50,110
x one, x one into x two, x two into x three
and so on, x ninety-nine to x hundred.
198
00:37:50,110 --> 00:37:57,080
If I take a lag on the other side; it will
be x one into x zero, x two into x one, it
199
00:37:57,080 --> 00:38:04,080
is just x one and x two; I mean the the two
sequences are reversed all right, but they
200
00:38:04,400 --> 00:38:11,400
give you the same product, so it is an even
function…..Now, let us before we go to any
201
00:38:38,330 --> 00:38:45,330
other discussions; let us define a positive
definite matrix, definite matrix or positive
202
00:39:00,050 --> 00:39:02,300
definite function.
203
00:39:02,300 --> 00:39:09,300
What is a positive definite function? Okay,
let us take a vector alpha 1, alpha 2, alpha
204
00:39:11,040 --> 00:39:18,040
3 and a vector vector beta as beta 1, beta
2 beta 3, okay. So, what is alpha transposed
205
00:39:25,110 --> 00:39:32,110
beta; it is an inner product which will be
alpha 1, alpha 2, alpha 3 into beta 1, beta
206
00:39:38,070 --> 00:39:45,070
2, beta 3, is alpha 1 beta 1 plus alpha 2
beta 2 plus alpha 3 beta 3, it is same as
207
00:39:48,710 --> 00:39:51,450
beta transposed alpha, is it not?
208
00:39:51,450 --> 00:39:55,040
If I would have transposed these and then
multiplied by this, I would have got the same
209
00:39:55,040 --> 00:40:02,040
product, is that so. Now, let us take y as
alpha transposed x okay. A vector x is converted
210
00:40:19,280 --> 00:40:26,280
to a is transformed to a new vector y, so
y 1, y 2, y 3 is a vector which is alpha 1
211
00:40:33,510 --> 00:40:40,510
x 1 plus alpha 2 x 2 plus alpha 3 x 3, okay
where alpha is alpha 1, alpha 2, sorry sorry
212
00:40:50,950 --> 00:40:57,950
sorry I am sorry, this not the inner product.
213
00:40:59,970 --> 00:41:06,970
y is equal to, okay y is equal to A x, all
right, any matrix A. So, x 1, x 2, x 3 multiplied
214
00:41:17,890 --> 00:41:24,890
by A will give me vector Y, all right. The
elements of x may be positive or negative,
215
00:41:30,360 --> 00:41:37,360
I don not know all right. What is alpha transposed
alpha? Any vector pre multiplied by its transpose
216
00:41:42,440 --> 00:41:47,430
will give the will give me alpha 1 square
plus alpha 2 square plus alpha 3 squared which
217
00:41:47,430 --> 00:41:51,120
is always positive, okay.
218
00:41:51,120 --> 00:41:58,120
So, what is y transposed Y? It will be A X
transpose A X, all right which is X transpose
219
00:42:10,180 --> 00:42:17,180
A X all right. This A transpose A is this
entire thing is always positive, because Y
220
00:42:24,890 --> 00:42:31,210
transpose y is always positive. I have shown
for any vector alpha, alpha transpose alpha
221
00:42:31,210 --> 00:42:37,720
is always positive; so any vector Y will be
always giving you, Y transpose y as a positive
222
00:42:37,720 --> 00:42:44,720
quantity, okay. So, this is a positive quantity
irrespective of the vector x.
223
00:42:45,200 --> 00:42:52,200
So when you get such a function of X, that
is a vector x 1, x 2, x 3 etcetera multiplied
224
00:42:54,510 --> 00:43:01,510
by a matrix P which is transpose A is a square
matrix okay; post multiplied by the vector
225
00:43:05,250 --> 00:43:12,250
again, x 1, x 2, x 3 are the elements. If
it is always positive then this is called
226
00:43:13,290 --> 00:43:20,290
a positive definite matrix, okay and this
function is a positive definite function of
227
00:43:26,910 --> 00:43:33,780
X, whatever be the value of x the elements
may be positive or negative; it is always
228
00:43:33,780 --> 00:43:33,860
positive, okay.
229
00:43:33,860 --> 00:43:40,860
It is something like an energy function, say
if move my hand in this direction there is
230
00:43:41,160 --> 00:43:47,590
lots of frictional energy all right, if I
move it backward then also it will be positive,
231
00:43:47,590 --> 00:43:54,040
heat will be dissipated whether I move it
this way or that way, all right. So, it is
232
00:43:54,040 --> 00:44:01,040
something like energy which is always positive
so positive definite matrix and positive definite
233
00:44:01,410 --> 00:44:03,350
functions is defined.
234
00:44:03,350 --> 00:44:09,220
Positive definite function is when it is always
positive, if it is sometimes positive, sometimes
235
00:44:09,220 --> 00:44:14,720
zero, it is called positive semi-definite.
Similarly, if some function is always negative
236
00:44:14,720 --> 00:44:20,560
just put a negative sign on this, this becomes
a negative definite or negative semi-definite.
237
00:44:20,560 --> 00:44:27,560
If
238
00:44:30,710 --> 00:44:37,710
you have a finite energy sequence, X n and
a combination of a x n plus another sequence
239
00:45:00,890 --> 00:45:07,890
with
a with a lag, what will be the energy content
of this signal?
240
00:45:16,150 --> 00:45:23,150
That means, you have a linear combination
of two signals, all right you are measuring
241
00:45:23,890 --> 00:45:30,890
the energy content of that signal; if x n
is a finite energy, y n is a finite energy
242
00:45:32,930 --> 00:45:39,720
then it will also be a finite energy sequence,
all right. And let us see, what will be the
243
00:45:39,720 --> 00:45:46,720
energy of this sequence. Energy of this sequence
is summation of this, which will be a squared
244
00:45:55,740 --> 00:46:02,740
x n squared plus twice a sigma x n y n minus
l plus sigma y n minus l square, equal to
245
00:46:23,220 --> 00:46:30,220
a squared this is nothing but r x x 0 plus
2 a, this is nothing but r x y l and this
246
00:46:39,180 --> 00:46:46,180
is r y y 0, okay because there all varying
over minus infinituve to plus infinitive.
247
00:46:50,540 --> 00:46:57,540
We are assuming the sequence to be, extended
in both the directions all right. And that
248
00:46:58,720 --> 00:47:05,720
must be always greater than equal to 0, because
this is finite energy, this is finite energy.
249
00:47:06,880 --> 00:47:13,880
So, we can write this as a 1 r x x 0, r x
y l r x y l, r y y 0, this matrix into a 1
250
00:47:40,620 --> 00:47:47,620
that is greater than equal to 0; so this is
a positive semi-definite form matrix, positive
251
00:47:49,400 --> 00:47:52,680
semi-definite matrix.
252
00:47:52,680 --> 00:47:59,680
If
it is positive semi-definite, basically for
a two by two matrix it is very simple, the
253
00:48:05,330 --> 00:48:11,000
determinant must be always greater than equal
to 0, determinant must be always greater than
254
00:48:11,000 --> 00:48:12,650
equal to 0.
255
00:48:12,650 --> 00:48:19,650
That means r x x 0 into r y y 0 minus r x
y l squared, should be greater than equal
256
00:48:35,680 --> 00:48:42,680
to 0 or r x y should be less than equal to
r x x 0 r y y 0 under root; that is under
257
00:48:52,630 --> 00:48:59,630
root of E x energy of signal x, energy of
signal y, okay. So by the same logic when
258
00:49:08,210 --> 00:49:15,210
we are considering the same signal x, therefore
r x x at any lag should be less than r x x,
259
00:49:20,330 --> 00:49:27,330
this is l; into r x x under root will be r
x x 0 or that is equal to E x.
260
00:49:39,520 --> 00:49:46,520
So, if you take a signal; if you take its
autocorrelation with a lag say, 1 or 2 or
261
00:49:48,200 --> 00:49:54,150
3 that will be always less than or equal to
at the most; the value of the autocorrelation
262
00:49:54,150 --> 00:50:00,520
when you are taking at n equal to 0, when
you are multiplying a quantity with itself
263
00:50:00,520 --> 00:50:05,140
and then take the product sum, that is giving
the maximum value.
264
00:50:05,140 --> 00:50:12,140
So, we will stop discussion on this, at this
point. Next time, we would like to take up
265
00:50:14,110 --> 00:50:21,110
discrete Fourier transform, in most of the
books they do not distinguish between distinct
266
00:50:22,370 --> 00:50:28,950
between discrete Fourier transform, discrete
Fourier series and discrete time Fourier transform.
267
00:50:28,950 --> 00:50:35,950
So, we would like to bring out the certain
difference between these terms first and then
268
00:50:36,090 --> 00:50:40,520
we will go over to the competition of discrete
Fourier transform.
269
00:50:40,520 --> 00:50:47,280
Discrete Fourier transform is the most widely
used transform in the signal processing and
270
00:50:47,280 --> 00:50:54,260
its numerical; the algorithmic version is
fast Fourier transform which is a basically
271
00:50:54,260 --> 00:51:01,260
competition of DFT, very fast algorithm for competition purpose. So, we will take up DFT
272
00:51:02,580 --> 00:51:05,070
and FFT in the next class. Thank you very much.