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.