1
00:00:44,969 --> 00:00:56,899
Discussing about inverse inverse Z transform, sorry. The last method that is the residue
2
00:00:56,899 --> 00:01:11,399
method of competition, we will discuss today.
x n the sequence is obtained by evaluating,
3
00:01:11,399 --> 00:01:18,399
this integral z to the power n minus 1, X
z, d z. The contour C is taken to include
4
00:01:28,189 --> 00:01:35,189
all the poles, to include all the poles of
X z; you choose any contour which can include
5
00:01:44,849 --> 00:01:46,610
all the poles.
6
00:01:46,610 --> 00:01:53,610
So, x n can be written as by Cautious theorem;
sum of the residues of z to the power n minus
7
00:02:05,009 --> 00:02:12,009
1 into X z at all the poles, okay. And for
multiple poles at a particular location, for
8
00:02:25,409 --> 00:02:32,409
multiple poles the residue is calculated,
residue F z at p K where F z is this product,
9
00:02:43,580 --> 00:02:49,549
residue of say F z equal to z to the power
n minus 1 into X z.
10
00:02:49,549 --> 00:02:56,549
So, the residue of F z will be 1 by m minus
1 factorial, d m minus 1 by d z m minus 1,
11
00:03:03,230 --> 00:03:10,230
z minus p K into F z, evaluated at z is equal
to p K, okay. Let us see this with an example,
12
00:03:20,060 --> 00:03:25,959
how this can be applied.
13
00:03:25,959 --> 00:03:32,959
Let us take X z, the same example; z inverse
by 1 minus 0.25, z inverse minus 0.375 z to
14
00:03:41,550 --> 00:03:48,550
the power minus 2 which we took last time.
So, this will be z by I can write; 1 minus
15
00:03:49,830 --> 00:03:56,830
okay, z minus 0.75 these factors we got last
time, z plus 0.5, okay. So, F z will be z
16
00:04:08,300 --> 00:04:14,920
to the power n minus 1 into z which is z to
the power n minus 1 into z; that will give
17
00:04:14,920 --> 00:04:21,920
me z to the power n by z minus 0.75 into z
plus 0.5, okay.
18
00:04:21,930 --> 00:04:28,930
So, you can
evaluate x n, by that method of residue once
you have got F z there are two poles, one
19
00:04:37,169 --> 00:04:44,169
at minus 0.5 the other one is at plus 0.5.
So, evaluate it, it comes as x n. So, okay
20
00:04:46,169 --> 00:04:53,169
let me write the residues first; x n will
be residue of the first pole that is z to
21
00:05:02,630 --> 00:05:09,630
the power n, if I multiplied by the first
pole, so z plus not five I get in the denominator
22
00:05:11,979 --> 00:05:18,979
this is to be evaluated at z is equal to 0.75
plus residue the second one is z to the power
23
00:05:23,880 --> 00:05:30,880
n by z minus 0.75, evaluated at z is equal
to 0.5, is it all right, minus 0.5, thank
24
00:05:35,530 --> 00:05:38,000
you.
25
00:05:38,000 --> 00:05:45,000
So that gives me, if I put z is equal to 0.75,
so it will be 0.75 to the power n, okay; 0.75
26
00:05:54,680 --> 00:06:01,680
will make it 1 by 0.5 plus 0.75, 1.25 and
minus 0.5 and the denominator will be minus
27
00:06:05,940 --> 00:06:12,940
1.25 in the second case, and minus 0.5 to
the power n, so whole thing multiplied by
28
00:06:15,479 --> 00:06:22,479
u n will be x n, it is so simple. This is
what; we got by the other method, 1 by 1.25
29
00:06:24,900 --> 00:06:31,900
was 0.8 so I can take that as common.
30
00:06:32,919 --> 00:06:39,919
So, that gives you 0.8 into 0.75 to the power
n minus 0.5 u n. Let us, take another example
31
00:06:53,460 --> 00:07:00,460
with multiple roots. X z equal to z square
by z minus 0.5 into z minus 1 whole squared,
32
00:07:12,050 --> 00:07:19,050
okay. So, you have to differentiate and then
find out the residue. So, residue for the
33
00:07:20,280 --> 00:07:27,280
first part, before that let me compute F z;
which will be z to the power n minus 1 into
34
00:07:30,960 --> 00:07:37,960
X z which will be z to the power n plus 1,
z square into z n minus 1, all right.
35
00:07:44,389 --> 00:07:51,389
So, x n will be residue at the first pole,
it will be z to the power n plus 1 by z minus
36
00:07:59,620 --> 00:08:06,620
1 whole square, 0.5 evaluated at this pole,
is the first one, okay. I will call this as
37
00:08:14,880 --> 00:08:21,880
say x 1, this will be 0.5 to the power n plus
1 by this is 0.5 minus 1, so minus 0.5 squared;
38
00:08:28,520 --> 00:08:35,520
so 0.52, so that gives me okay 0.5 to the
power n minus 1, I may write as 2 into 0.5
39
00:08:44,839 --> 00:08:50,990
to the power n if you permit me, I can write
like this. Similarly, the second one I am
40
00:08:50,990 --> 00:08:57,990
writing separately x 1, and x 2 and add them
together, is it all right.
41
00:09:00,069 --> 00:09:07,069
x 2 n for the second one, I will take residue
F z at this pole; which will be if you differentiate
42
00:09:23,730 --> 00:09:30,730
here, this one will be z to the power n plus
1 by z minus 0.5 d d z of, this one is it
43
00:09:44,730 --> 00:09:51,730
not? You have to multiply by z to the power,
z minus 1 whole squared and then calculate
44
00:09:56,079 --> 00:10:03,079
the function, okay. So that gives me, z to
the power; okay n plus 1 into z to the power
45
00:10:08,579 --> 00:10:15,579
n into z minus 0.5 minus z to the power n
plus 1 into z minus 0.5; the derivative is
46
00:10:22,360 --> 00:10:29,360
1, divided by z minus 0.5 squared and this
is to be evaluated at z is equal to 1.
47
00:10:33,089 --> 00:10:40,089
So that gives me, if you put 1 n plus 1 into
0.5, okay n plus 1 into 0.5 minus 1 divided
48
00:10:46,410 --> 00:10:53,410
by 0.5 squared. So, that gives me 2 into n
minus 1, okay. Therefore, x n is 2 into n
49
00:11:12,670 --> 00:11:19,670
minus 1 plus 0.5 squared into u n. It was
true; sorry to the power n not square, sorry
50
00:11:41,489 --> 00:11:48,489
is that all right. This is to the power n,
x 1. Now, there are certain important properties
51
00:11:53,049 --> 00:12:00,049
of Z transform, let me just list out some
of them, will take up one or two for proof.
52
00:12:00,929 --> 00:12:07,929
One is very helpful linearity. a into x 1
n plus b into x 2 n gives me, a into corresponding
53
00:12:23,009 --> 00:12:30,009
Z transforms X 1 z and X 2 z multiplied by
a and b, okay. This we also observed for frequency
54
00:12:36,869 --> 00:12:43,869
domain representation, of frequency transform
a into X 1 e to the power j omega plus b X
55
00:12:46,829 --> 00:12:53,829
2 e to the power j omega, okay where we discussed
about D T F T, discrete time Fourier transform,
56
00:12:55,949 --> 00:13:02,949
we will get back to that very soon.
Then delay or shift, this also you observed
57
00:13:11,399 --> 00:13:18,399
earlier; if x n is having a transform X z
then x n minus say capital N, will be z to
58
00:13:29,619 --> 00:13:36,619
the power minus N X z, okay. Convolution,
y n if it is convolved in the discrete time
59
00:13:51,559 --> 00:13:58,559
domain, x K and h K; if they convolved then
in the frequency domain in the Z domain, they
60
00:14:04,709 --> 00:14:11,709
will be in a product form. These proofs are
very similar to Laplace transforms, where
61
00:14:15,730 --> 00:14:22,730
you have seen convolution, a delay, linearity,
is a very simple relation.
62
00:14:29,629 --> 00:14:36,629
Differentiation, x n is X z, x n is paired
with X z then n times x n is minus z, d x
63
00:14:50,939 --> 00:14:57,939
z, by d z. Let us take up this one for example;
x n by definition is a sequence x 0, x 1 and
64
00:15:05,910 --> 00:15:12,910
so on, okay so X z is x 0 plus x 1 z inverse
and so on, all right. So, what is d x z by
65
00:15:23,019 --> 00:15:30,019
d z will be x 0 derivative is 0 then x 1 and
z 1 if I take the, derivative it will be minus
66
00:15:37,009 --> 00:15:44,009
z to the power minus 2 plus minus z to the
power minus 3 into 2 into x 2 and so on.
67
00:15:49,239 --> 00:15:56,239
If, I take z to the power minus 1 common it
will be minus z to the power minus 1, minus
68
00:15:58,879 --> 00:16:05,879
z common, okay. Sorry, z to minus 1 common,
sorry z to the power minus 1 common will be
69
00:16:07,839 --> 00:16:14,839
x 1 z to the power minus 1 plus 2 into x 2
z to the power minus 2, 3 into x 3 z to the
70
00:16:20,879 --> 00:16:27,879
power minus 3 and so on; which is nothing
but z inverse, this is a Z transform of what,
71
00:16:34,529 --> 00:16:41,529
n x n is it not. As, if 0 into x 0 plus 1
into x 1, 2 into x 2, 3 into x 3 these are
72
00:16:43,489 --> 00:16:48,970
appearing, the 0 term is missing because it
is multiplied with 0, okay.
73
00:16:48,970 --> 00:16:55,970
So, it is Z transform of n times x n series.
So, if I transfer this to this side, we get
74
00:16:56,779 --> 00:17:03,779
the result minus z into d x z by d z is equal
to z transform of n x n, okay.
75
00:17:12,169 --> 00:17:19,169
Then modulation, that is y n is x n into h
n, all right. Then Y z will be 1 by 2 p i
76
00:17:39,450 --> 00:17:46,450
j, G v, H z by v v to the power minus 1 d
v. The ROC, region of convergence includes
77
00:18:03,350 --> 00:18:10,350
R g at R h, okay. We will be taking up, some
simple application of Z transform in a real
78
00:18:23,809 --> 00:18:30,809
systems, okay. Then we will go back to discrete
time domain and frequency domain presentation
79
00:18:30,809 --> 00:18:37,509
once again and try to find out, the common
link between Z transform and frequency domain
80
00:18:37,509 --> 00:18:43,320
transform, okay.
81
00:18:43,320 --> 00:18:50,320
Say, we are given x n, x n as 0.5, 0.5 to
the power n, u n. h n, 0.2 to the power n
82
00:19:18,889 --> 00:19:25,889
minus 1 u, n minus 1. What will be y n which
is convolution of x n and h n? Sometimes we
83
00:19:42,370 --> 00:19:49,370
use a circle around the star for denoting
convolution, because star is quite often used
84
00:19:49,610 --> 00:19:56,610
for complex sequences also complex conjugates,
so to avoid confusions, okay we use this.
85
00:19:57,669 --> 00:20:04,669
So, let us use that relationship for convolution.
Y z will be therefore; X z into H z and what
86
00:20:09,809 --> 00:20:16,809
is X z for this one? 1 by 1 minus 0.5 z inverse,
is it not? For 0.5 to the power n and H z,
87
00:20:23,769 --> 00:20:30,409
this is delayed by one step, so it will be
z to the power minus 1 and 0.2 to the power
88
00:20:30,409 --> 00:20:37,409
n; Z transform is 0.2 z inverse, is that all
right? So, you know the product make partial
89
00:20:43,750 --> 00:20:45,750
fractions, okay.
90
00:20:45,750 --> 00:20:52,750
Let me use this paper okay. So, Y z is
z to the power minus 1 by 0.3, 1 by 1 minus
0.5 z inverse minus 1 by 1 minus 0.2 z inverse,
91
00:21:18,879 --> 00:21:25,879
seen such cases it is very simple. If you
have the product like, 1 by 1 minus a z inverse
92
00:21:28,779 --> 00:21:34,110
into 1 by 1 minus b z inverse then you straight
away take the partial fraction as 1 minus
93
00:21:34,110 --> 00:21:40,179
a z inverse minus 1 by 1 minus b z inverse,
okay. Once you will get cancelled, so what
94
00:21:40,179 --> 00:21:45,909
you will be left with; a minus b in the numerator,
so divide by a minus b. That is what I have
95
00:21:45,909 --> 00:21:49,730
done; the difference between 0.5 and 0.2 is
0.3, so divided by 0.3.
96
00:21:49,730 --> 00:21:56,730
And there was z inverse in the numerator,
so I have taken it out. So that is approximately
97
00:21:57,830 --> 00:22:04,830
3.33 z inverse into this. So, what will be
Y K? It will be 3.33 or you can write 10 by
98
00:22:10,200 --> 00:22:17,200
3 either way, solution for this to be delayed
by one, inverse of this to be delayed by one
99
00:22:19,000 --> 00:22:26,000
step. So this is 0.5 to the power n minus
1 u, n minus 1 minus 0.2 to the power n minus
100
00:22:32,370 --> 00:22:39,370
1 u, n minus 1, okay. So, this is a solution.
Yes, please?
101
00:22:45,419 --> 00:22:52,419
z inverse will come as n minus 1, partial
fractions; 1 will get cancelled, there is
102
00:22:58,639 --> 00:23:04,019
a z inverse already coming in so this will
not be there, so thank you very much. So,
103
00:23:04,019 --> 00:23:11,019
I need not have z inverse, this itself will
give me the z inverse, thank you. So, it will
104
00:23:14,929 --> 00:23:21,929
be just n no then 0.2 u n, yes it will be
sorry it will be this; z inverse will automatically
105
00:23:35,409 --> 00:23:42,409
come from the subtraction.
Now, let us take an interesting application,
106
00:23:48,779 --> 00:23:55,779
radar tracking system. In a radar tracking
system in a radar tracking system, what do
107
00:24:03,779 --> 00:24:10,039
you do, you have a source of signals, you
are sending pulses all right. Suppose, you
108
00:24:10,039 --> 00:24:15,490
are tracking a moving object all right, so
the pulses will be going at the velocity of
109
00:24:15,490 --> 00:24:22,490
light of course and then that will reflected
back, you are sensing the reflected signal,
110
00:24:22,950 --> 00:24:29,610
you are finding out the time delay that divided
by 2 will be giving you the distance of the
111
00:24:29,610 --> 00:24:34,059
object, all right. Now you keep on measuring
it, all right.
112
00:24:34,059 --> 00:24:39,399
So, if you are sending signals at regular
intervals and receiving signals at regular
113
00:24:39,399 --> 00:24:46,399
intervals then you try to compute, you try
to estimate its position and also its velocity.
114
00:24:48,230 --> 00:24:54,499
If you know its position at a certain instant
of time and its velocity; you can predict
115
00:24:54,499 --> 00:24:59,620
after say one second, what will be its possible
location. You know its direction velocity
116
00:24:59,620 --> 00:25:04,970
and position, so you know its direction; so
along that direction, you know after one second
117
00:25:04,970 --> 00:25:06,460
it will be in this position.
118
00:25:06,460 --> 00:25:13,460
So you fire the next shot, all right you fire
the shots to the object, to the enemy object,
119
00:25:16,139 --> 00:25:23,139
you know the velocity of the missile and then
you can destroy that, okay that is the idea.
120
00:25:23,779 --> 00:25:30,779
So, let us see very simple mathematical model,
for a radar technique system.
121
00:25:30,960 --> 00:25:37,960
So, I will show you a very simple, schematic,
diagram. So, this is an enemy plane and this
122
00:25:49,850 --> 00:25:56,850
is radar. So, you are sending a signal, this
is object, c meters per second; it is a velocity
123
00:25:58,940 --> 00:26:05,940
of light, the signal is sent and you are sending
the pulses like this.
124
00:26:10,019 --> 00:26:17,019
So, this is T at regular intervals, all right.
The received signal ideally that any instant,
125
00:26:34,929 --> 00:26:41,929
this is delta t, this is the received signal,
of course this delta t will keep on changing
126
00:26:48,429 --> 00:26:54,129
because it is changing its position. So, the
received signal, the time occupy a time taken
127
00:26:54,129 --> 00:26:59,620
by the pulse to return back will not be the
same throughout, that is changing so this
128
00:26:59,620 --> 00:27:03,350
gap is changing, okay.
129
00:27:03,350 --> 00:27:10,259
But the actual signal is contaminated. The
received signal, actual received signal will
130
00:27:10,259 --> 00:27:17,259
be somewhat like this. So, there is threshold
that we take and this is our actual time we
131
00:27:23,809 --> 00:27:30,809
are accounting; okay, it will be contaminated
by noise and there is a filtering effect of
132
00:27:33,399 --> 00:27:40,399
the space. So, this is the actual received
signal and this is the ideal received signal,
133
00:27:47,169 --> 00:27:53,490
this is a sending pulses.
134
00:27:53,490 --> 00:28:00,490
Let us take X K, as a measurement
of the range, measurement of the range from
the K th pulse okay, so what will be X K?
135
00:28:37,210 --> 00:28:44,210
It will be delta t k by 2; delta t k is a
time for the K th pulse to return, all right.
136
00:28:46,749 --> 00:28:53,749
So, delta t k by 2 into the velocity of light,
so many meters, okay. Y K is an estimate of
137
00:29:07,509 --> 00:29:14,509
the range after processing K th pulse, okay
that means; when I ask you to make a measurement,
138
00:29:35,419 --> 00:29:39,009
measurement does not give you any measurement
for that matter, measurement does not give
139
00:29:39,009 --> 00:29:39,960
the exact value.
140
00:29:39,960 --> 00:29:45,960
For example, in your high school physics experiment
or even here; you take a measurement ten times
141
00:29:45,960 --> 00:29:51,149
and then take the flat average sometime all
right. In schools, your you learn to take
142
00:29:51,149 --> 00:29:56,360
only the arithmetic average, later on you
must have observed, you take the most probable
143
00:29:56,360 --> 00:30:03,360
value that means the value which is occurring
more often is taken as the estimate, estimate.
144
00:30:03,909 --> 00:30:08,600
You can always make an estimate, now there
will be always some measurement error all
145
00:30:08,600 --> 00:30:11,730
right, personal error and so on, so instrumental
error.
146
00:30:11,730 --> 00:30:18,730
So, depending on the complexity of the experiment,
you make an estimate of any item any variable
147
00:30:20,080 --> 00:30:24,379
from the set of measurement, okay.
148
00:30:24,379 --> 00:30:31,379
Then y v K is an estimate of the objects velocity,
and y p K is a prediction of the objects range.
149
00:31:01,799 --> 00:31:07,240
When you whenever you are making a prediction,
prediction is based on the previous measurements
150
00:31:07,240 --> 00:31:14,240
all right; so after processing K minus 1 th
pulse, so on the basis of K minus 1 readings
151
00:31:25,049 --> 00:31:30,220
you make, a forecast, you make a prediction
what will be its position at the K the instant.
152
00:31:30,220 --> 00:31:31,970
And you have made an actual measurement.
153
00:31:31,970 --> 00:31:38,460
So, that gives you an error, prediction error.
So, you take that error as a feedback and
154
00:31:38,460 --> 00:31:45,460
let us see; how these estimates are modified
with the help of that feedback. So, y p K
155
00:31:49,610 --> 00:31:56,610
the predicted value, how much is a predicted
value? Say, up to this instant of time I have
156
00:32:00,559 --> 00:32:07,559
got, y K minus 1; y K minus 1 as the estimate,
what will be my prediction in the next instant?
157
00:32:10,460 --> 00:32:16,440
If I know its velocity, if I have an estimate
of its velocity then velocity multiplied by
158
00:32:16,440 --> 00:32:23,440
that time gap capital T that we are considering,
so multiplied by that time gap will be the
159
00:32:23,519 --> 00:32:30,519
additional increase in the range, so T into
Y v K minus 1.
160
00:32:32,610 --> 00:32:39,409
Suppose, this is K minus 1th instant, so at
K minus 1th instant; if I have estimated the
161
00:32:39,409 --> 00:32:46,119
value of the velocity of the object, if I
multiplied by the time interval, I will get
162
00:32:46,119 --> 00:32:53,119
the increase in the value delta y, okay. So,
that multiplied by the present estimate, that
163
00:32:54,389 --> 00:33:01,389
will be giving me the prediction. Similarly,
y K whatever the estimate at the next instant;
164
00:33:05,769 --> 00:33:12,769
that is after possessing K number of measurements,
what will be its value? y K, how will you
165
00:33:12,980 --> 00:33:18,009
estimate?
166
00:33:18,009 --> 00:33:25,009
What is y K? It is the predicted value, whatever
was predicted to be the value at this moment
167
00:33:25,499 --> 00:33:32,499
from the previous instant, y p K minus 1 plus
we take a correction factor into the error,
168
00:33:34,669 --> 00:33:41,669
all right. And what is the error measurement,
and using sorry third bracket; so let me be
169
00:33:46,210 --> 00:33:53,210
consistent, minus, is that all right? This
was the prediction error.
170
00:33:59,950 --> 00:34:06,950
So, prediction error multiplied by a constant,
normally less than 1 plus the prediction met
171
00:34:10,070 --> 00:34:16,740
in the previous instant, it should be; prediction
is made at this instant, remind you prediction
172
00:34:16,740 --> 00:34:23,740
is made at these instant from the previous
data. This is prediction error, okay and y
173
00:34:31,520 --> 00:34:38,520
v K, what will be the estimate of the velocity?
It will be whatever velocity was estimated
174
00:34:43,280 --> 00:34:50,280
in the previous instant plus the prediction
error from the prediction error, I will estimate
175
00:34:50,570 --> 00:34:54,790
the prediction, prediction error in velocity.
176
00:34:54,790 --> 00:35:01,790
So, beta times 1 by T times the same prediction
error, is that all right. Whatever is the
177
00:35:11,549 --> 00:35:18,369
error in the position that divided by that
time T, gives me the error in the estimate
178
00:35:18,369 --> 00:35:25,299
of the velocity. So, that multiplied by a
factor beta, if I add with the previous estimate
179
00:35:25,299 --> 00:35:32,299
of the velocity; that will give me the present
estimate of the velocity, oaky. This is known
180
00:35:33,369 --> 00:35:40,369
as, an alpha beta tracker. Now, let us take
Z transform of these.
181
00:35:42,700 --> 00:35:49,700
So, from here if you permit me, these three
equations I will write, for say the for example;
182
00:36:01,079 --> 00:36:05,400
the second equation what will be the Z transform
of this?
183
00:36:05,400 --> 00:36:12,400
Y z is equal to Y p z, Y p z plus alpha times
x K minus 1. So, x z into z inverse
minus y p z, is that all right? Similarly,
184
00:36:37,210 --> 00:36:44,210
first equation here will give me y p z, okay
let me arrange these it will be Y z minus
185
00:36:47,880 --> 00:36:54,880
Y p z into 1 minus alpha, okay, is it all
right, Y p z this one, okay. On this side
186
00:37:21,599 --> 00:37:28,599
alpha z inverse X z, is it so, is this all
right?
187
00:37:32,040 --> 00:37:39,040
Next one, y p K is equal to y K minus 1 plus
T y v K minus 1. So, if I take the Z transform,
188
00:37:44,160 --> 00:37:51,160
Y p z equal to z inverse y z plus T z inverse
y v z or in other words z inverse into y z
189
00:38:02,230 --> 00:38:09,230
minus z inverse into y z minus y p z plus
T z inverse y v z equal to 0. And the third,
190
00:38:24,250 --> 00:38:31,250
third equation this one gives me, y v z okay,
equal z inverse y v z plus beta by T, z inverse
191
00:38:44,319 --> 00:38:51,319
X z minus y p z or in other words; I can write
it 0 into y z, there is no y z term here,
192
00:39:02,339 --> 00:39:09,339
okay. Plus beta by T y p z plus 1 minus z
inverse into y v z is equal beta by T z inverse
193
00:39:22,040 --> 00:39:26,960
X z, this is equation three.
194
00:39:26,960 --> 00:39:33,549
I have tried to arrange the three equations
in such a way; that you can write a very simple
195
00:39:33,549 --> 00:39:39,930
form, a matrix equation.
196
00:39:39,930 --> 00:39:46,930
I would request you to fill up the blanks
Y z, Y v z, Y p z, and Y v z will give you
197
00:40:02,559 --> 00:40:09,559
something into X z, these elements you fill
up from these equations. So, it is very simple,
198
00:40:13,680 --> 00:40:20,680
matrix equation. You can find out, Y z by
X z, you can find out Y v z by X z, you can
199
00:40:29,130 --> 00:40:36,130
find out Y p z by X z; try at home, what will
be the transfer functions?
200
00:40:38,440 --> 00:40:45,440
That means, the input signals the input measurements
from the measurements not from the signals,
201
00:40:47,540 --> 00:40:54,540
X z is a measurement so with the measurement,
how are you making estimates of the velocities,
202
00:40:57,369 --> 00:41:04,369
position and the prediction, okay. Now, if
you have a general equation of this type;
203
00:41:13,400 --> 00:41:20,400
say, a 0, y K plus a 1 y K minus 1 plus a
2 y K minus 2 and so on, up to a N y K minus
204
00:41:36,119 --> 00:41:38,799
N.
205
00:41:38,799 --> 00:41:45,799
On this side, you have b 0, X K plus b 1 x
K minus 1 and so on, up to b M x K minus M.
206
00:42:00,480 --> 00:42:07,480
This is obtained is a generalized form of
a difference equation, obtained from differential
207
00:42:08,589 --> 00:42:15,230
equation. Once earlier we have discussed,
how from a differential equation you can approximate
208
00:42:15,230 --> 00:42:22,039
and you can make certain approximations about
the derivatives and you can get a difference
209
00:42:22,039 --> 00:42:22,859
equation.
210
00:42:22,859 --> 00:42:29,859
Now, difference equations will not be identical,
for the same differential equation; if you
211
00:42:30,390 --> 00:42:36,780
change the step length, if you because the
factor T squared T, T Q, would depending on
212
00:42:36,780 --> 00:42:41,930
the order of the derivatives. So, it all depends
on the step length that is a sampling time
213
00:42:41,930 --> 00:42:46,440
that you choose, okay. If you change the sampling
time, the difference equation coefficients
214
00:42:46,440 --> 00:42:51,020
will also be different, okay. This is something
very important.
215
00:42:51,020 --> 00:42:58,020
Now, such difference equations can be converted
into, in the cab be converted into Z domain.
216
00:42:58,559 --> 00:43:05,559
So, you get a transformed equation. So, you
get a 0, Y z if you assume zero initial conditions
217
00:43:08,690 --> 00:43:15,690
to start with; that will be z inverse for
Y z, a N Z to the power minus N Y z, it will
218
00:43:22,250 --> 00:43:29,250
be b 0, X z plus b 1 z inverse X z and so
on, up to b M z to the power minus M X z.
219
00:43:38,569 --> 00:43:45,569
I can take Y z common, similarly on this side;
I can take X z common and Y by X will be b
220
00:43:49,440 --> 00:43:56,440
0 plus b 1 z inverse b M z to the power minus
M divided by a 0 plus a 1 z inverse and so
221
00:44:03,319 --> 00:44:10,319
on, a N z to the power minus N. We call this
as transfer function in the discrete domain;
222
00:44:10,990 --> 00:44:16,230
as you have done in the continuous domain
Laplace transform, you take and then then
223
00:44:16,230 --> 00:44:20,380
take the ratio of the output by input. So,
this is the transfer function the discrete
224
00:44:20,380 --> 00:44:25,039
domain.
225
00:44:25,039 --> 00:44:31,460
Given a transfer function we can find out
its impulse response; just take the Laplace
226
00:44:31,460 --> 00:44:38,460
the Z inverse of this and you have seen for
calculating Z inverse there are three methods;
227
00:44:39,490 --> 00:44:46,490
you can use any one of them, so you make factors,
find out the factors factorizing this and
228
00:44:49,619 --> 00:44:54,190
then making partial fractions either calculate
the residue or by the partial fraction method,
229
00:44:54,190 --> 00:45:00,309
you can find out the solution. Otherwise,
by long division you can calculate H K. So,
230
00:45:00,309 --> 00:45:07,309
this is normally written as H z or G z.
231
00:45:12,180 --> 00:45:19,180
Now, let us come back to discrete domain functions
once again; before we go further because we
232
00:45:24,710 --> 00:45:31,710
will be requiring the discrete domain descriptions,
especially in the frequency domain and. Like
233
00:45:40,569 --> 00:45:47,569
the the properties that we just now discussed
in the Z domain; you have also in the frequency
234
00:45:49,609 --> 00:45:56,609
domain similar properties that is some of
them probably discussed earlier.
235
00:45:57,650 --> 00:46:04,650
Linearity, the same principle a times X 1
n plus b times X 2 n; will give you a times
236
00:46:12,710 --> 00:46:19,710
X 1 e to the power j omega plus b times X
2 e to the power j omega, probably we mentioned
237
00:46:20,700 --> 00:46:27,700
it earlier, delay or shift as very similar
to Z transforms. Then your convolution that
238
00:46:34,500 --> 00:46:41,500
will also be similar; now for convolution
for convolution, suppose we have output Y
239
00:46:54,700 --> 00:47:01,700
is convolve convolved with X and H, so it
will be in the frequency domain, X e to the
240
00:47:07,730 --> 00:47:14,730
power j omega, H e to the power j omega. If
you can find out the values, it is okay, if
241
00:47:15,930 --> 00:47:22,930
you can get an analytical analytical expression
and if you can take the inverse.
242
00:47:25,579 --> 00:47:32,579
So, we calculate x e to the power j omega
and H e to the power j omega, all right. So,
243
00:47:46,670 --> 00:47:53,670
if you are given a sequence x n as x 0, x
1 etcetera; so calculate x 0 plus x 1 e to
244
00:47:57,200 --> 00:48:03,230
the power minus j omega and so on. If it is
a mathematical description; that is given
245
00:48:03,230 --> 00:48:09,450
and if you can find the convergence, if you
can reduce it to a close form, say like if
246
00:48:09,450 --> 00:48:16,450
the sequence is 2 to the power, say 0.2 to
the power n u n then you know the sequence
247
00:48:18,460 --> 00:48:25,460
is 1, 0.2, 0.04 and so on. So, you can write
the expression since it is a convergence series;
248
00:48:28,980 --> 00:48:35,980
we will be getting the solution very identical
to Z transform, X e to the power j omega will
249
00:48:37,359 --> 00:48:44,359
be 1 by 1 minus 0.2 e to the power minus j
omega, all right.
250
00:48:45,569 --> 00:48:52,569
This kind of a series, if you take the discrete
discrete time Fourier transform all right,
251
00:48:52,960 --> 00:48:59,960
so the frequency domain description of this
sequence will be this, is a g p series I have
252
00:49:02,220 --> 00:49:03,299
reduced it.
253
00:49:03,299 --> 00:49:08,490
Since e to the power j omega is always having
a magnitude 1, so 0.2 times this will not
254
00:49:08,490 --> 00:49:15,490
blow up, it will be a convergence series okay
because the magnitude is reducing, all right.
255
00:49:17,210 --> 00:49:24,210
So, from here if you similarly get e to the
power H, e to the power j omega, in such a
256
00:49:26,039 --> 00:49:33,039
form close form and take the product and then
take the inverse; how would you compute the
257
00:49:33,109 --> 00:49:35,470
inverse, how would you compute the inverse?
258
00:49:35,470 --> 00:49:42,470
Many of classical Fourier series you have
seen; a periodic function, any periodic function
259
00:49:44,119 --> 00:49:51,119
say this a periodic function in the time domain,
can be expressed in the frequency domain,
260
00:49:55,500 --> 00:50:02,500
can be expressed in the frequency domain as
a 0, I will write in a complex form C n e
261
00:50:05,740 --> 00:50:12,740
to the power j omega summation, all right,
okay plus, in varying from minus infinite
262
00:50:17,630 --> 00:50:22,640
to plus infinite. If you combine together
and then write in the complex form, it will
263
00:50:22,640 --> 00:50:25,640
be like this, okay.
264
00:50:25,640 --> 00:50:32,640
So, C n e to the power j omega and we get
lines, these are called harmonics, okay. I
265
00:50:38,069 --> 00:50:45,069
can just number them n harmonic number 1,
2, 3, 4 and so on, these are the frequencies
266
00:50:45,880 --> 00:50:52,880
present. Now what is the frequency transform
of the sequence x 0, x 1, x 2? What is the
267
00:50:57,730 --> 00:51:04,730
frequency transform? x 0 plus x 1 and this
we saw earlier; this is a periodic function
268
00:51:05,520 --> 00:51:12,520
all right of period two p i, is it not? We
discussed earlier, this is having a periodicity
269
00:51:13,190 --> 00:51:17,680
of two p i.
270
00:51:17,680 --> 00:51:24,680
So, if I now take omega and X e to the power
j omega, u take it magnitude as well as phase;
271
00:51:29,140 --> 00:51:36,140
either way it is periodic function, may be
like this and this is 2 p i minus to plus
272
00:51:45,640 --> 00:51:52,640
p i, 2 p i and 4 p i and so on, so it is a
periodic function. In the time domain, if
273
00:51:54,760 --> 00:52:01,760
it is a periodic function then in the frequency
domain; it is a discrete discrete line, is
274
00:52:05,930 --> 00:52:10,710
it not.
So, you just stretch your imagination in the
275
00:52:10,710 --> 00:52:17,710
frequency domain, if it is periodic then what
is it in the time domain? It will be lines
276
00:52:18,680 --> 00:52:25,680
and precisely that is what we started with,
it was x n, is it not, all right. How did
277
00:52:31,970 --> 00:52:38,970
you compute these magnitudes in the normal
Fourier series? You integrate say a 0’s
278
00:52:40,940 --> 00:52:46,630
and sorry a i’s and b i’s, how did you
compute or complex C n?
279
00:52:46,630 --> 00:52:53,630
Integration over one period then the function
f t then e to the power plus j omega n t,
280
00:52:57,440 --> 00:53:04,440
d t was it plus or minus, minus okay. So,
in the time domain if it is periodic; the
281
00:53:12,039 --> 00:53:19,039
corresponding frequency domain constants are
evaluated from here, divided by T, okay. In
282
00:53:21,440 --> 00:53:28,440
this case by the same argument, I can evaluate
x n, what is the periodicity? 2 p i 1 by 2
283
00:53:32,559 --> 00:53:39,559
p i minus p i to plus p i, okay; X e to the
power j omega, e to the power I will take
284
00:53:42,799 --> 00:53:49,740
plus j omega and d omega okay, that will give
X n.
285
00:53:49,740 --> 00:53:56,740
So, once you have got X, sorry you have got
X and H and if they are convolved in the time
286
00:54:01,770 --> 00:54:07,500
domain; so in the frequency domain you have
to just take their products to calculate output
287
00:54:07,500 --> 00:54:12,619
and then take inverse of that product, and
for taking the inverse will referred to these
288
00:54:12,619 --> 00:54:19,619
expression. So, our Y n will be 1 by 2 p i
minus p i to plus p i product of X into H,
289
00:54:25,819 --> 00:54:31,799
all right.
290
00:54:31,799 --> 00:54:38,799
So given two sequences, take the frequency
domain transform, take the product then take
291
00:54:39,460 --> 00:54:45,690
the inverse; that is integration over the
period 2 p i, after multiplying by e to the
292
00:54:45,690 --> 00:54:52,690
power j omega, okay. So, we will stop here
for today and we will continue further in
293
00:54:53,190 --> 00:54:57,010
the next class. Thank you very much.