1 00:00:00,090 --> 00:00:02,490 The following content is provided under a Creative 2 00:00:02,490 --> 00:00:04,030 Commons license. 3 00:00:04,030 --> 00:00:06,330 Your support will help MIT OpenCourseWare 4 00:00:06,330 --> 00:00:10,720 continue to offer high quality educational resources for free. 5 00:00:10,720 --> 00:00:13,320 To make a donation or view additional materials 6 00:00:13,320 --> 00:00:17,280 from hundreds of MIT courses, visit MIT OpenCourseWare 7 00:00:17,280 --> 00:00:18,450 at ocw.mit.edu. 8 00:00:52,530 --> 00:00:54,240 ALAN OPPENHEIM: Well, in the last lecture 9 00:00:54,240 --> 00:00:58,130 we introduced the z transform. 10 00:00:58,130 --> 00:01:00,860 In this lecture, I'd like to go backwards. 11 00:01:00,860 --> 00:01:05,630 That is, I'd like to introduce the inverse z transform 12 00:01:05,630 --> 00:01:11,510 and demonstrate some of its properties with a few examples. 13 00:01:11,510 --> 00:01:17,210 To begin with, let me remind you of the z transform relationship 14 00:01:17,210 --> 00:01:20,970 as we talked about it in the last lecture. 15 00:01:20,970 --> 00:01:25,970 The z transform X of z of a sequence x of n 16 00:01:25,970 --> 00:01:33,130 is given by the sum of x of n times z to the minus n. 17 00:01:33,130 --> 00:01:35,710 The inverse z transform, of course, 18 00:01:35,710 --> 00:01:40,120 is the relationship, or the set of rules, 19 00:01:40,120 --> 00:01:45,970 that allow us to obtain x of n the original sequence 20 00:01:45,970 --> 00:01:50,850 from its z transform, x of z. 21 00:01:50,850 --> 00:01:54,570 There are a variety of methods that 22 00:01:54,570 --> 00:01:59,130 can be used for implementing the inverse z transform. 23 00:01:59,130 --> 00:02:02,610 Some of them are somewhat informal methods. 24 00:02:02,610 --> 00:02:06,660 And others, one in particular which we'll talk about, 25 00:02:06,660 --> 00:02:10,229 is a somewhat formal method. 26 00:02:10,229 --> 00:02:13,350 To begin with, an informal method 27 00:02:13,350 --> 00:02:16,020 and a method that, in fact, most of us 28 00:02:16,020 --> 00:02:19,290 tend to use the most often is a method 29 00:02:19,290 --> 00:02:25,350 that we could refer to as the inspection method for computing 30 00:02:25,350 --> 00:02:28,350 the inverse z transform. 31 00:02:28,350 --> 00:02:31,770 Basically, the inspection method says 32 00:02:31,770 --> 00:02:37,230 if we've computed a z transform from a sequence, 33 00:02:37,230 --> 00:02:39,630 then by looking at that z transform, 34 00:02:39,630 --> 00:02:44,130 we can recognize more or less by inspection the sequence 35 00:02:44,130 --> 00:02:47,400 that that z transform came from. 36 00:02:47,400 --> 00:02:53,360 For example, we have worked several times now 37 00:02:53,360 --> 00:02:56,300 the computation of the z transform 38 00:02:56,300 --> 00:02:58,640 of an exponential sequence, which 39 00:02:58,640 --> 00:03:03,290 is a right sided sequence, and gotten is the answer, 40 00:03:03,290 --> 00:03:08,550 or the z transform, 1 over 1 minus a z to the minus one, 41 00:03:08,550 --> 00:03:11,870 with a region of convergence for the magnitude of z greater 42 00:03:11,870 --> 00:03:14,790 than the magnitude of a. 43 00:03:14,790 --> 00:03:18,910 Consequently, having worked that example, 44 00:03:18,910 --> 00:03:23,550 if I now present you with this z transform and ask you 45 00:03:23,550 --> 00:03:26,280 what sequence generated it. 46 00:03:26,280 --> 00:03:31,470 One way of doing that, obviously, is by inspection. 47 00:03:31,470 --> 00:03:35,850 By recognizing that this sequence generates 48 00:03:35,850 --> 00:03:39,230 that z transform. 49 00:03:39,230 --> 00:03:42,050 Another example that we've worked several times 50 00:03:42,050 --> 00:03:47,630 now is the exponential sequence, which is left sided. 51 00:03:47,630 --> 00:03:50,630 That is 0 for n greater than zero, 52 00:03:50,630 --> 00:03:53,990 generating the same ratio of polynomials 53 00:03:53,990 --> 00:03:57,170 as in the right sided exponential, 54 00:03:57,170 --> 00:03:59,390 but with a region of convergence which 55 00:03:59,390 --> 00:04:02,330 is the interior of a circle rather than 56 00:04:02,330 --> 00:04:04,310 in this case where the region of convergence 57 00:04:04,310 --> 00:04:07,430 is the exterior of a circle. 58 00:04:07,430 --> 00:04:12,050 And in fact, it was the comparison between these two z 59 00:04:12,050 --> 00:04:14,870 transforms that caused us to focus 60 00:04:14,870 --> 00:04:17,000 in the last lecture on the importance 61 00:04:17,000 --> 00:04:19,200 of the region of convergence. 62 00:04:19,200 --> 00:04:24,660 So this is one method, which I emphasize again is in fact, 63 00:04:24,660 --> 00:04:26,480 one of the most common methods. 64 00:04:26,480 --> 00:04:30,500 That is, just simply recognizing the z transform as something 65 00:04:30,500 --> 00:04:33,560 that we've computed previously from a sequence. 66 00:04:33,560 --> 00:04:37,640 And that obviously also as you work more examples, 67 00:04:37,640 --> 00:04:43,400 you build up a larger list of sequences and z transforms 68 00:04:43,400 --> 00:04:47,140 which pairs what you can recognize. 69 00:04:47,140 --> 00:04:55,520 The second method for computing the inverse z transform 70 00:04:55,520 --> 00:05:03,200 is based on the fact that the z transform relationship is 71 00:05:03,200 --> 00:05:07,340 basically a power series in z. 72 00:05:07,340 --> 00:05:13,710 That is, if we look back here at the z transform expression, 73 00:05:13,710 --> 00:05:18,470 this is essentially a power series relationship which 74 00:05:18,470 --> 00:05:22,880 expands x of z in powers of z. 75 00:05:22,880 --> 00:05:26,990 Consequently, given a z transform, 76 00:05:26,990 --> 00:05:31,250 if we can expand it in terms of powers of z, 77 00:05:31,250 --> 00:05:35,330 then we can pick off the coefficients in that expansion 78 00:05:35,330 --> 00:05:39,530 as values of the sequence x of n. 79 00:05:39,530 --> 00:05:43,470 For example, let's take the sequence, 80 00:05:43,470 --> 00:05:48,622 sorry the z transform x of z, which is 1 over 1 minus a z 81 00:05:48,622 --> 00:05:50,830 to the minus 1. 82 00:05:50,830 --> 00:05:54,700 Or we can rewrite it as z over z minus a. 83 00:05:54,700 --> 00:06:00,400 Now this of course is a z transform 84 00:06:00,400 --> 00:06:04,690 whose inverse we know how to compute by inspection. 85 00:06:04,690 --> 00:06:07,470 However, we can also compute the inverse z 86 00:06:07,470 --> 00:06:14,040 transform by expanding this expression in a power series. 87 00:06:14,040 --> 00:06:19,500 Let's divide 1 by 1 minus a z to the minus 1, 88 00:06:19,500 --> 00:06:21,390 and then recognize the coefficients 89 00:06:21,390 --> 00:06:26,530 in that expansion as values of the sequence X of n. 90 00:06:26,530 --> 00:06:28,540 So carrying that out. 91 00:06:28,540 --> 00:06:33,180 We have 1 divided into 1 is 1. 92 00:06:33,180 --> 00:06:38,730 Multiplying back we have 1 minus a z to the minus 1. 93 00:06:38,730 --> 00:06:43,780 Subtracting we then have a z to the minus 1. 94 00:06:43,780 --> 00:06:50,122 Dividing by 1 we then have a times z to the minus 1. 95 00:06:50,122 --> 00:06:55,380 Multiplying down again, we have a z to the minus 1 96 00:06:55,380 --> 00:07:00,130 minus a squared z to the minus 2. 97 00:07:00,130 --> 00:07:05,740 Subtracting we have a squared z to the minus 2. 98 00:07:05,740 --> 00:07:11,650 Dividing by 1 we have a squared z to the minus 2. 99 00:07:11,650 --> 00:07:16,360 And if we continue this, we'll generate terms 1, a times 100 00:07:16,360 --> 00:07:21,500 z to the minus 1, a squared times z to the minus 2, 101 00:07:21,500 --> 00:07:28,670 a cubed times z to the minus 3, et cetera. 102 00:07:28,670 --> 00:07:32,800 This series will continue with an infinite number of terms. 103 00:07:32,800 --> 00:07:40,180 And we can recognize then this as the expansion of x of z 104 00:07:40,180 --> 00:07:44,770 in the form the sum of x n times z to the minus n. 105 00:07:44,770 --> 00:07:51,550 Consequently, we can recognize that apparently x of n 106 00:07:51,550 --> 00:07:56,290 is 1 at n equal zero, a at n equals 1, a squared at n 107 00:07:56,290 --> 00:08:01,070 equals 2, a cubed at n equals 3, etc. 108 00:08:01,070 --> 00:08:04,730 Of course, there are lots of ways 109 00:08:04,730 --> 00:08:10,070 of expanding a function like this in a power series. 110 00:08:10,070 --> 00:08:15,200 For example, if I simply reverse the order of these two terms 111 00:08:15,200 --> 00:08:19,850 and carried out the division, this isn't the series I'd get. 112 00:08:19,850 --> 00:08:24,800 What I'd get instead is the series, which is minus a 113 00:08:24,800 --> 00:08:28,970 to the minus 1 times z, minus a to the minus 2 times 114 00:08:28,970 --> 00:08:34,710 z squared, minus a to the minus 3 times z cubed, etc. 115 00:08:34,710 --> 00:08:38,640 In that case, if this was the expansion, 116 00:08:38,640 --> 00:08:41,640 then what I would have to recognize as the sequence 117 00:08:41,640 --> 00:08:46,790 corresponding to this is the sequence which is 0 at n 118 00:08:46,790 --> 00:08:48,560 equals zero. 119 00:08:48,560 --> 00:08:53,240 It has no terms of the form z to the minus a positive power. 120 00:08:53,240 --> 00:08:57,890 So apparently, it's all 0 for n greater than zero. 121 00:08:57,890 --> 00:09:02,810 And it's minus a to the minus 1 at n equals minus 1, minus a 122 00:09:02,810 --> 00:09:06,260 to the minus 2 at n equals minus 2, etc. 123 00:09:06,260 --> 00:09:12,140 So in fact, for this example, I could expand this either 124 00:09:12,140 --> 00:09:18,550 in this power series, or I could expand it in this power series. 125 00:09:18,550 --> 00:09:23,410 Well, which power series is the right power series? 126 00:09:23,410 --> 00:09:27,620 We know in fact that given just this ratio of polynomials, 127 00:09:27,620 --> 00:09:31,330 there are several choices for the sequence that it 128 00:09:31,330 --> 00:09:35,550 corresponds to depending on the region of convergence. 129 00:09:35,550 --> 00:09:40,620 If I associate as the region of convergence 130 00:09:40,620 --> 00:09:45,240 the magnitude of z greater than the magnitude of a, 131 00:09:45,240 --> 00:09:49,940 then it's this power series that converges. 132 00:09:49,940 --> 00:09:52,740 And this power series diverges. 133 00:09:52,740 --> 00:09:54,620 So for that region of convergence, 134 00:09:54,620 --> 00:09:58,220 that is with the magnitude of z greater 135 00:09:58,220 --> 00:10:01,730 than the magnitude of a as the region of convergence, 136 00:10:01,730 --> 00:10:03,590 this is the power series. 137 00:10:03,590 --> 00:10:08,600 And consequently, the answer is a to the n times 138 00:10:08,600 --> 00:10:11,840 u of n, which of course is consistent with what 139 00:10:11,840 --> 00:10:14,900 we know already or equivalently consistent with the inspection 140 00:10:14,900 --> 00:10:16,380 method. 141 00:10:16,380 --> 00:10:19,380 On the other hand, if the region of convergence 142 00:10:19,380 --> 00:10:24,810 was the magnitude of z less than the magnitude of a, 143 00:10:24,810 --> 00:10:28,620 then it's this power series that converges. 144 00:10:28,620 --> 00:10:30,570 And consequently in that case, we 145 00:10:30,570 --> 00:10:35,040 get a left sided sequence, which again is consistent with what 146 00:10:35,040 --> 00:10:37,350 we know either from the inspection method 147 00:10:37,350 --> 00:10:39,180 or equivalently from the examples 148 00:10:39,180 --> 00:10:42,150 that we've worked before. 149 00:10:42,150 --> 00:10:46,680 One obvious drawback to the power series 150 00:10:46,680 --> 00:10:49,440 is the fact that when we generate 151 00:10:49,440 --> 00:10:52,590 the inverse z transform this way, 152 00:10:52,590 --> 00:10:57,360 we get the values in the sequence as individual terms. 153 00:10:57,360 --> 00:11:03,210 That is, we generate x of n not in closed form. 154 00:11:03,210 --> 00:11:05,450 We generate it as a sequence. 155 00:11:05,450 --> 00:11:07,160 For simple sequences of course, we 156 00:11:07,160 --> 00:11:10,610 can recognize a closed form form for that. 157 00:11:10,610 --> 00:11:12,880 But in more complicated cases, we can't. 158 00:11:16,700 --> 00:11:17,200 All right. 159 00:11:17,200 --> 00:11:21,260 A third method for generating the inverse z 160 00:11:21,260 --> 00:11:26,090 transform is basically an extension 161 00:11:26,090 --> 00:11:28,970 of the inspection method, which allows 162 00:11:28,970 --> 00:11:34,700 us to take a more complicated z transform expression and expand 163 00:11:34,700 --> 00:11:42,390 it out in terms of terms that we can recognize by inspection. 164 00:11:42,390 --> 00:11:46,490 And this method I'll refer to as the method 165 00:11:46,490 --> 00:11:50,700 based on the partial fraction expansion. 166 00:11:50,700 --> 00:11:53,910 I assume that some of you are familiar 167 00:11:53,910 --> 00:11:57,480 with the partial fraction expansion in general 168 00:11:57,480 --> 00:11:59,700 for a rational function. 169 00:11:59,700 --> 00:12:02,970 But in case you're not or in case you're rusty on it, 170 00:12:02,970 --> 00:12:06,690 let me just quickly review what the partial fraction 171 00:12:06,690 --> 00:12:07,960 expansion consists of. 172 00:12:10,670 --> 00:12:17,330 Let's consider the general function of a variable, 173 00:12:17,330 --> 00:12:20,540 I'll take as a complex variable x, which 174 00:12:20,540 --> 00:12:24,200 is a rational function in x. 175 00:12:24,200 --> 00:12:28,280 It is one polynomial in the numerator divided 176 00:12:28,280 --> 00:12:31,370 by a denominator polynomial. 177 00:12:31,370 --> 00:12:35,210 And let me remark incidentally that you shouldn't confuse 178 00:12:35,210 --> 00:12:38,810 this x with the sequence values x of n 179 00:12:38,810 --> 00:12:40,520 that we've been referring to. 180 00:12:40,520 --> 00:12:45,490 This is simply some arbitrary complex number, 181 00:12:45,490 --> 00:12:48,570 or complex variable. 182 00:12:48,570 --> 00:12:55,650 Now I can in general expand the rational function 183 00:12:55,650 --> 00:13:03,000 in terms of an expansion of a form which, 184 00:13:03,000 --> 00:13:06,490 at least under certain simple conditions, 185 00:13:06,490 --> 00:13:12,520 is a sum of terms of the form of a constant, 186 00:13:12,520 --> 00:13:21,380 Rk divided by x minus xk, where the x minus xk 187 00:13:21,380 --> 00:13:26,690 are the roots of the denominator polynomial q of x. 188 00:13:26,690 --> 00:13:32,860 And these coefficients are referred to as the residues. 189 00:13:32,860 --> 00:13:40,290 Now to expand this rational function in this form 190 00:13:40,290 --> 00:13:45,420 without additional terms, I need to apply the restriction, which 191 00:13:45,420 --> 00:13:48,910 I'll do for the purposes of this lecture, 192 00:13:48,910 --> 00:13:52,570 that the order of the numerator polynomial 193 00:13:52,570 --> 00:13:56,330 is less than the order of the denominator polynomial, 194 00:13:56,330 --> 00:14:00,490 and that in addition, there are no multiple order 195 00:14:00,490 --> 00:14:02,890 roots of the denominator polynomial. 196 00:14:02,890 --> 00:14:07,510 In other words, the denominator polynomial is n-th order, 197 00:14:07,510 --> 00:14:11,290 then the n roots of the denominator polynomial 198 00:14:11,290 --> 00:14:13,570 are distinct. 199 00:14:13,570 --> 00:14:17,980 In the text, there is the more general discussion 200 00:14:17,980 --> 00:14:20,380 of the partial fraction expansion 201 00:14:20,380 --> 00:14:22,810 when the order of the numerator is 202 00:14:22,810 --> 00:14:25,720 greater than or equal to the order of the denominator 203 00:14:25,720 --> 00:14:28,000 or if there are multiple order roots. 204 00:14:28,000 --> 00:14:30,530 The essential ideas of course don't change. 205 00:14:30,530 --> 00:14:34,480 It's just the mechanics of implementing the expansion 206 00:14:34,480 --> 00:14:36,100 in the form of the expansion. 207 00:14:36,100 --> 00:14:39,580 So for this lecture we'll assume that the expansion simply 208 00:14:39,580 --> 00:14:41,340 has terms of this form. 209 00:14:44,370 --> 00:14:47,430 And this then is referred to as the partial fraction 210 00:14:47,430 --> 00:14:51,560 expansion of f of x. 211 00:14:51,560 --> 00:14:55,640 Now how do we get the coefficients Rk 212 00:14:55,640 --> 00:14:57,950 in the expansion? 213 00:14:57,950 --> 00:15:03,450 Well essentially by recognizing that if we 214 00:15:03,450 --> 00:15:10,680 multiply f of x by one of the denominator factors, 215 00:15:10,680 --> 00:15:19,870 say x minus xr, and evaluate that product of x equals xr. 216 00:15:19,870 --> 00:15:23,890 Then relating that to this sum, as I've 217 00:15:23,890 --> 00:15:30,430 indicated here, when k is equal to r, 218 00:15:30,430 --> 00:15:34,460 these two factors cancel out. 219 00:15:34,460 --> 00:15:37,730 When k is not equal to r, then when 220 00:15:37,730 --> 00:15:45,060 I substitute x equals xr in here, then this term vanishes. 221 00:15:45,060 --> 00:15:50,620 So basically then, what happens inside this sum is 222 00:15:50,620 --> 00:15:56,720 that the term inside this sum is equal to 0 223 00:15:56,720 --> 00:16:03,950 for k not equal to r, and it's equal to capital Rr, that 224 00:16:03,950 --> 00:16:14,090 is the residue at the pole, x equals xr when k is equal to r. 225 00:16:14,090 --> 00:16:17,120 So consequently, there's only one term 226 00:16:17,120 --> 00:16:20,960 left in this sum, which is capital Rr. 227 00:16:20,960 --> 00:16:25,910 And therefore the residues, that is capital Rr, 228 00:16:25,910 --> 00:16:34,900 can be obtained by multiplying f of x by x minus xr, 229 00:16:34,900 --> 00:16:38,960 evaluating the result at x equals xr. 230 00:16:38,960 --> 00:16:41,480 And the resulting number is the residue 231 00:16:41,480 --> 00:16:46,550 of f of x at the pole x equals xr. 232 00:16:46,550 --> 00:16:51,020 That's for the case of simple order 233 00:16:51,020 --> 00:16:55,430 poles, that is first order roots of the denominator polynomial. 234 00:16:55,430 --> 00:16:58,040 And also for the case when the order of the numerator 235 00:16:58,040 --> 00:17:01,820 polynomial is less than or equal to the order of the denominator 236 00:17:01,820 --> 00:17:03,470 polynomial. 237 00:17:03,470 --> 00:17:07,520 And just store away for later, because I want to refer back 238 00:17:07,520 --> 00:17:10,260 to this point, that in particular, 239 00:17:10,260 --> 00:17:12,440 when there are multiple order poles, 240 00:17:12,440 --> 00:17:16,010 the evaluation of the residue is somewhat more complicated. 241 00:17:16,010 --> 00:17:17,930 In particular it involves computing 242 00:17:17,930 --> 00:17:19,760 some derivatives, et cetera. 243 00:17:19,760 --> 00:17:23,069 And the more the higher the multiplicity of the pole, 244 00:17:23,069 --> 00:17:25,260 the more complicated it is. 245 00:17:25,260 --> 00:17:28,099 And that's a point that will sneak back at us 246 00:17:28,099 --> 00:17:30,590 toward the end of the lecture. 247 00:17:30,590 --> 00:17:35,640 Now this of course is for a general rational function. 248 00:17:35,640 --> 00:17:40,890 Let's apply this now to the inverse z transform. 249 00:17:40,890 --> 00:17:45,370 Well, we know that the z transform-- 250 00:17:45,370 --> 00:17:47,940 well let's restrict ourselves to the case 251 00:17:47,940 --> 00:17:52,260 where the z transform is a rational function of z 252 00:17:52,260 --> 00:17:55,030 or z to the minus 1. 253 00:17:55,030 --> 00:18:00,130 And consequently we can, referring back 254 00:18:00,130 --> 00:18:04,390 to the general notion of the partial fraction expansion, 255 00:18:04,390 --> 00:18:08,330 either think of the variable z as what 256 00:18:08,330 --> 00:18:12,100 we're calling x over there, in which case 257 00:18:12,100 --> 00:18:17,170 we have an expansion of the z transform in the form 258 00:18:17,170 --> 00:18:24,130 of the sum on k of ak divided by z minus little ak, 259 00:18:24,130 --> 00:18:27,520 where the little ak's are the poles, and the capital 260 00:18:27,520 --> 00:18:30,550 Ak's are the residues. 261 00:18:30,550 --> 00:18:37,360 Alternatively, if we interpret z to the minus 1 as x, then 262 00:18:37,360 --> 00:18:43,070 we can expand x of z in the form of some other residues 263 00:18:43,070 --> 00:18:52,050 Bk divided by z to the minus 1 minus some coefficient, little 264 00:18:52,050 --> 00:18:52,950 bk. 265 00:18:52,950 --> 00:18:57,000 Or we can rewrite that as capital Bk divided 266 00:18:57,000 --> 00:19:00,940 by 1 minus ak z to the minus 1. 267 00:19:00,940 --> 00:19:05,280 So here we're expanding in terms of-- 268 00:19:05,280 --> 00:19:08,880 we're treating the z transform as a rational function in z. 269 00:19:08,880 --> 00:19:10,590 Here we're treating the z transform 270 00:19:10,590 --> 00:19:15,090 as a rational function in z to the minus 1. 271 00:19:15,090 --> 00:19:18,150 So what does the method consist of? 272 00:19:18,150 --> 00:19:22,920 Well the method consists of expanding x of z 273 00:19:22,920 --> 00:19:26,190 in terms of simple terms like this, 274 00:19:26,190 --> 00:19:28,650 and then using the inspection method 275 00:19:28,650 --> 00:19:33,600 to recognize what the inverse transform is for each 276 00:19:33,600 --> 00:19:35,670 of the individual terms. 277 00:19:35,670 --> 00:19:38,730 To get the final sequence then, we add those up. 278 00:19:38,730 --> 00:19:41,040 That is, we've expressed the z transform 279 00:19:41,040 --> 00:19:43,990 as a sum of simple factors. 280 00:19:43,990 --> 00:19:46,910 So the sequence, the original sequence, 281 00:19:46,910 --> 00:19:50,220 we can express as the sum of the inverse z 282 00:19:50,220 --> 00:19:53,850 transform obtained by the expect inspection method 283 00:19:53,850 --> 00:19:56,600 of each of the individual factors. 284 00:19:56,600 --> 00:19:57,100 All right. 285 00:19:57,100 --> 00:20:00,330 Let's just work a simple example to show you 286 00:20:00,330 --> 00:20:03,960 what the mechanics are of this and how it works out. 287 00:20:03,960 --> 00:20:10,980 Let's take an example where x of z is 1 over 1 minus 1/2 288 00:20:10,980 --> 00:20:16,470 z to the minus 1 times 1 minus 1/4 z to the minus 1 289 00:20:16,470 --> 00:20:19,050 with a region of convergence which 290 00:20:19,050 --> 00:20:23,990 I'll take as the magnitude of z greater than a half. 291 00:20:23,990 --> 00:20:25,610 This has two poles of course. 292 00:20:25,610 --> 00:20:29,360 One at z equals 1/2, the other at z equals 1/4. 293 00:20:29,360 --> 00:20:32,940 And the region of convergence is outside the outermost pole. 294 00:20:32,940 --> 00:20:36,590 So what we expect to get is a right sided sequence. 295 00:20:36,590 --> 00:20:40,350 And let's work this one by treating 296 00:20:40,350 --> 00:20:46,110 the variable x in our previous discussion as z to the minus 1. 297 00:20:46,110 --> 00:20:48,900 And obviously, it doesn't matter which one we use. 298 00:20:48,900 --> 00:20:52,110 Although it turns out actually that depending 299 00:20:52,110 --> 00:20:54,450 on how you pick it, you either end up 300 00:20:54,450 --> 00:20:56,790 with a numerator polynomial whose order 301 00:20:56,790 --> 00:20:59,360 is greater than the denominator polynomial, or the other way 302 00:20:59,360 --> 00:20:59,860 around. 303 00:21:02,651 --> 00:21:03,150 OK. 304 00:21:03,150 --> 00:21:09,880 Well, so we want then to expand this z transform 305 00:21:09,880 --> 00:21:14,700 in terms of a sum of factors. 306 00:21:14,700 --> 00:21:18,510 And let's just rewrite this to make things 307 00:21:18,510 --> 00:21:26,930 a little clearer by multiplying denominator and numerator by 2, 308 00:21:26,930 --> 00:21:30,830 by minus 2 rather, and then also by minus 4. 309 00:21:30,830 --> 00:21:36,260 We can simply rewrite this as 8 divided by z to minus 1 minus 2 310 00:21:36,260 --> 00:21:39,110 times z to the minus 1 minus 4. 311 00:21:39,110 --> 00:21:44,240 So that it's more in the form of x minus a constant times 312 00:21:44,240 --> 00:21:47,960 x minus another constant. 313 00:21:47,960 --> 00:21:48,560 All right. 314 00:21:48,560 --> 00:21:53,480 Well, so here then we have the z transform 315 00:21:53,480 --> 00:21:55,640 that we want to expand in a partial fraction 316 00:21:55,640 --> 00:22:01,310 expansion in terms of treating z to the minus 1 317 00:22:01,310 --> 00:22:06,210 as the complex variable x in the expansion. 318 00:22:06,210 --> 00:22:10,460 We have two poles, two roots of the denominator polynomial, 319 00:22:10,460 --> 00:22:13,680 one at z to the minus 1 equals 2, the other 320 00:22:13,680 --> 00:22:16,260 at z the minus 1 equals 4. 321 00:22:16,260 --> 00:22:18,660 So we need to calculate the residues 322 00:22:18,660 --> 00:22:20,930 at each of those roots. 323 00:22:20,930 --> 00:22:26,510 To calculate the residue at z to the minus 1 equals 2, 324 00:22:26,510 --> 00:22:27,650 what are the steps? 325 00:22:27,650 --> 00:22:34,810 Well, we multiply this by z to the minus 1 minus 2. 326 00:22:34,810 --> 00:22:38,890 That is, x minus 2 where x we're interpreting as z 327 00:22:38,890 --> 00:22:40,960 to the minus 1. 328 00:22:40,960 --> 00:22:49,700 And then evaluate the result at z to the minus 1 equals 2. 329 00:22:49,700 --> 00:22:57,550 Well obviously of course, this term cancels out that term. 330 00:22:57,550 --> 00:23:00,490 And we want to do that before we substitute z to the minus 1 331 00:23:00,490 --> 00:23:02,260 equals 2. 332 00:23:02,260 --> 00:23:04,360 And then substituting z to the minus 1 333 00:23:04,360 --> 00:23:10,390 equals 2, we get minus 2 left over in the denominator divided 334 00:23:10,390 --> 00:23:11,500 into eight. 335 00:23:11,500 --> 00:23:17,480 And that's equal to minus 4. 336 00:23:17,480 --> 00:23:21,320 So the residue then at z to the minus 1 337 00:23:21,320 --> 00:23:26,650 equals 2 is equal to minus 4. 338 00:23:26,650 --> 00:23:30,320 Similarly, to calculate the residue at the root 339 00:23:30,320 --> 00:23:35,020 z to the minus 1 equals 4, we multiply by z 340 00:23:35,020 --> 00:23:37,470 to the minus 1 minus 4. 341 00:23:37,470 --> 00:23:41,440 Then these two terms will cancel out. 342 00:23:41,440 --> 00:23:44,410 Evaluate the resulting expression at z to the minus 1 343 00:23:44,410 --> 00:23:46,060 equals 4. 344 00:23:46,060 --> 00:23:50,350 So we have 4 minus 2 is 2 and to 8 is 4. 345 00:23:50,350 --> 00:23:53,800 And the residue then at z to the minus 1 equals 4, 346 00:23:53,800 --> 00:23:57,010 the residue is equal to 4. 347 00:23:57,010 --> 00:24:02,080 Consequently, the partial fraction expansion for x of z, 348 00:24:02,080 --> 00:24:09,850 for this x of z, is minus 4, minus 4, over z 349 00:24:09,850 --> 00:24:18,500 to the minus 1 minus 2 plus 4 over z to the minus 1 minus 4. 350 00:24:18,500 --> 00:24:20,710 Well finally, let's just put this back 351 00:24:20,710 --> 00:24:25,630 in a form that is more the form that we're used to looking 352 00:24:25,630 --> 00:24:28,106 at the z transform in. 353 00:24:28,106 --> 00:24:32,450 In particular multiplying this piece, numerator 354 00:24:32,450 --> 00:24:37,430 and denominator, by minus 1/2, we get 2 over 1 minus 1/2 z 355 00:24:37,430 --> 00:24:39,960 to the minus 1. 356 00:24:39,960 --> 00:24:44,280 Here, multiplying numerator and denominator by minus 1/4, 357 00:24:44,280 --> 00:24:50,320 we get minus 1 over 1 minus 1/4 z to the minus 1. 358 00:24:50,320 --> 00:24:54,540 And now we've broken x of z into a form that allows us 359 00:24:54,540 --> 00:24:57,420 to use the inspection method. 360 00:24:57,420 --> 00:25:00,510 That is, we can recognize from this-- 361 00:25:00,510 --> 00:25:03,360 remember that the region of convergence 362 00:25:03,360 --> 00:25:06,250 was outside a circle. 363 00:25:06,250 --> 00:25:09,090 So we want to interpret the region of convergence 364 00:25:09,090 --> 00:25:11,430 to be the same here. 365 00:25:11,430 --> 00:25:19,830 So consequently for this piece we end up with 2 times 1/2 366 00:25:19,830 --> 00:25:24,810 to the n times u of n. 367 00:25:24,810 --> 00:25:27,760 That's an n. 368 00:25:27,760 --> 00:25:36,370 And for this piece we end up with minus 1 times a quarter 369 00:25:36,370 --> 00:25:40,490 to the n times u of n. 370 00:25:40,490 --> 00:25:48,380 So by breaking x of z into a sum of components, simple terms, 371 00:25:48,380 --> 00:25:51,910 we were then able to apply the inspection method 372 00:25:51,910 --> 00:25:58,090 to each one of those to get the resulting z transform 373 00:25:58,090 --> 00:26:02,110 in a closed form expression, or at least the sum 374 00:26:02,110 --> 00:26:03,975 of some closed form expression. 375 00:26:06,481 --> 00:26:06,980 All right. 376 00:26:06,980 --> 00:26:10,910 So that then is basically an extension 377 00:26:10,910 --> 00:26:13,010 of the inspection method. 378 00:26:13,010 --> 00:26:17,260 When the inspection method by itself doesn't work, 379 00:26:17,260 --> 00:26:21,340 it's usually the partial fraction expansion method 380 00:26:21,340 --> 00:26:25,780 that one uses, perhaps also in conjunction 381 00:26:25,780 --> 00:26:28,960 with some properties of the z transform, which 382 00:26:28,960 --> 00:26:34,640 is something that will be going into more in the next lecture. 383 00:26:34,640 --> 00:26:40,340 Now finally, there is a somewhat more formal way 384 00:26:40,340 --> 00:26:44,360 of evaluating the inverse z transform. 385 00:26:44,360 --> 00:26:48,410 And in some cases, it's important to actually use 386 00:26:48,410 --> 00:26:52,520 this to explicitly evaluate the inverse z transform. 387 00:26:52,520 --> 00:26:57,110 In other cases, it's important to make reference to the more 388 00:26:57,110 --> 00:27:00,920 formal inverse transform relationship 389 00:27:00,920 --> 00:27:04,400 to discuss properties of the inverse z transform. 390 00:27:04,400 --> 00:27:08,570 So sometimes the formal inverse transform 391 00:27:08,570 --> 00:27:11,270 is computationally useful. 392 00:27:11,270 --> 00:27:14,840 In other cases, it's useful in proving 393 00:27:14,840 --> 00:27:16,970 theorems or developing properties 394 00:27:16,970 --> 00:27:19,360 or something of the sort. 395 00:27:19,360 --> 00:27:22,990 Well this method, the final method and the most 396 00:27:22,990 --> 00:27:28,360 formal method, is the method that I refer to here 397 00:27:28,360 --> 00:27:32,330 as contour integration. 398 00:27:32,330 --> 00:27:36,170 And let me basically hand it to you. 399 00:27:36,170 --> 00:27:39,830 That is, rather than go through a derivation, 400 00:27:39,830 --> 00:27:44,180 let me just simply state what the inverse z transform 401 00:27:44,180 --> 00:27:46,190 relationship is. 402 00:27:46,190 --> 00:27:50,640 In the text there is a derivation of this. 403 00:27:50,640 --> 00:27:54,460 And so for the purposes of the lecture itself, 404 00:27:54,460 --> 00:27:56,040 let me just presented. 405 00:27:56,040 --> 00:27:57,580 Any objections? 406 00:27:57,580 --> 00:27:59,480 No. 407 00:27:59,480 --> 00:28:06,500 So and then also work through at least one example to show you 408 00:28:06,500 --> 00:28:09,199 what the mechanics are. 409 00:28:09,199 --> 00:28:11,240 It's another nice thing about videotape lectures, 410 00:28:11,240 --> 00:28:12,734 there are never any objection. 411 00:28:15,400 --> 00:28:15,900 OK. 412 00:28:15,900 --> 00:28:17,160 Well, here it is. 413 00:28:17,160 --> 00:28:19,020 This is the inverse z transform. 414 00:28:19,020 --> 00:28:19,950 What does it say? 415 00:28:19,950 --> 00:28:28,540 It says the sequence x of n is 1 over 2 pi j times a contour 416 00:28:28,540 --> 00:28:33,460 integral, which I'll return to in a second, of x of z times 417 00:28:33,460 --> 00:28:37,530 z to the n minus 1 dz. 418 00:28:37,530 --> 00:28:40,260 If you're not used to looking at an inverse transform 419 00:28:40,260 --> 00:28:42,180 relationship like this, let me just 420 00:28:42,180 --> 00:28:47,700 indicate that it's surprising how easy it 421 00:28:47,700 --> 00:28:50,400 is to verify that, in fact, this is the inverse z 422 00:28:50,400 --> 00:28:54,180 transform relationship, basically using the Cauchy 423 00:28:54,180 --> 00:28:56,270 residue theorem. 424 00:28:56,270 --> 00:28:58,620 But this is it. 425 00:28:58,620 --> 00:29:02,170 This is, of course, a complex function. 426 00:29:02,170 --> 00:29:04,540 This is a complex integral. 427 00:29:04,540 --> 00:29:08,290 And the important thing to straighten out 428 00:29:08,290 --> 00:29:11,680 is this contour c. 429 00:29:11,680 --> 00:29:19,210 The contours c is a closed contour in the z plane 430 00:29:19,210 --> 00:29:21,970 which encircles the origin. 431 00:29:21,970 --> 00:29:23,700 So it's a closed contour. 432 00:29:23,700 --> 00:29:24,780 It encircles the origin. 433 00:29:24,780 --> 00:29:28,840 It's like a circle that goes around the origin, 434 00:29:28,840 --> 00:29:31,180 with another very important condition on it. 435 00:29:31,180 --> 00:29:35,050 Which is that it lies inside the region of convergence 436 00:29:35,050 --> 00:29:36,520 of x of z. 437 00:29:36,520 --> 00:29:39,490 Well, that's not surprising, because the region 438 00:29:39,490 --> 00:29:43,480 of convergence for x of z is the only place where we're allowed 439 00:29:43,480 --> 00:29:45,670 to stick in values for z. 440 00:29:45,670 --> 00:29:49,980 So obviously quote, we're not allowed 441 00:29:49,980 --> 00:29:54,930 to look at values of this integrand except where x of z 442 00:29:54,930 --> 00:29:57,970 makes sense, that is except in the region of convergence. 443 00:29:57,970 --> 00:30:01,020 So obviously then, we have to evaluate this integral 444 00:30:01,020 --> 00:30:03,750 considering only values of z inside the region 445 00:30:03,750 --> 00:30:06,660 of convergence. 446 00:30:06,660 --> 00:30:10,490 So contour c is a closed contour in the z plane, 447 00:30:10,490 --> 00:30:14,060 encircles the origin, and it lies inside the region 448 00:30:14,060 --> 00:30:16,360 of convergence. 449 00:30:16,360 --> 00:30:20,440 Now for x of z, a rational function of z, 450 00:30:20,440 --> 00:30:23,610 which are the cases we've been talking about, 451 00:30:23,610 --> 00:30:28,400 this entire integrand is a rational function. 452 00:30:28,400 --> 00:30:34,010 And consequently, a contour integral of this type 453 00:30:34,010 --> 00:30:41,830 can be simply evaluated using the notion of residues. 454 00:30:41,830 --> 00:30:45,380 I'll state, and hopefully this is a property 455 00:30:45,380 --> 00:30:49,400 of complex contour integrals that you're familiar with, 456 00:30:49,400 --> 00:30:56,130 that the integral of a rational function of that type, 457 00:30:56,130 --> 00:30:59,880 the integral itself is 2 pi j times the sum of the residues 458 00:30:59,880 --> 00:31:02,900 inside the contour. 459 00:31:02,900 --> 00:31:07,640 And that 2 pi j cancels out as 1 over 2 pi j. 460 00:31:07,640 --> 00:31:10,250 And consequently, x of n is simply 461 00:31:10,250 --> 00:31:15,740 the sum of the residues of x of z times z to the n minus 1 462 00:31:15,740 --> 00:31:19,950 at the poles inside the contour c. 463 00:31:19,950 --> 00:31:21,540 The poles of what? 464 00:31:21,540 --> 00:31:26,610 The poles not of x and z only, but the poles 465 00:31:26,610 --> 00:31:28,680 of the entire integrand. 466 00:31:28,680 --> 00:31:32,400 That is, x of z multiplied by z to the n minus 1. 467 00:31:32,400 --> 00:31:37,100 Important point, which it's important to keep in mind. 468 00:31:37,100 --> 00:31:39,950 The integral or the residues that we're looking at 469 00:31:39,950 --> 00:31:43,970 are the residues, not just of x of z, but x of z 470 00:31:43,970 --> 00:31:46,280 multiplied by z to the n minus 1. 471 00:31:49,160 --> 00:31:52,920 Another statement, just tuck it away, 472 00:31:52,920 --> 00:31:58,850 is that this is valid for both n greater than zero 473 00:31:58,850 --> 00:32:02,860 and equal to zero and n less than 0. 474 00:32:02,860 --> 00:32:06,460 We'll see in a minute that for n greater than zero, 475 00:32:06,460 --> 00:32:09,490 it might be easier to evaluate than for n less than 0, 476 00:32:09,490 --> 00:32:11,290 depending on the example. 477 00:32:11,290 --> 00:32:14,710 But tuck away the fact that this expression is 478 00:32:14,710 --> 00:32:17,590 a valid expression, no matter whether n 479 00:32:17,590 --> 00:32:21,980 is positive or negative or of course zero. 480 00:32:21,980 --> 00:32:22,480 All right. 481 00:32:22,480 --> 00:32:26,170 Let's look at the mechanics of evaluating the inverse z 482 00:32:26,170 --> 00:32:28,360 transform this way. 483 00:32:28,360 --> 00:32:35,410 Let's return to the example x of z equals 1 over 1 minus 1/2 z 484 00:32:35,410 --> 00:32:39,160 to the minus 1 with a region of convergence 485 00:32:39,160 --> 00:32:42,810 the magnitude of z greater than 1/2. 486 00:32:42,810 --> 00:32:46,680 And by the way, by now if there's 487 00:32:46,680 --> 00:32:49,260 one thing you should have gotten out of these lectures, 488 00:32:49,260 --> 00:32:51,990 it certainly should be to be able to recognize 489 00:32:51,990 --> 00:32:54,540 what the inverse z transform of this is. 490 00:32:54,540 --> 00:32:59,310 It's an example that I guess we bounced around so many times. 491 00:32:59,310 --> 00:32:59,880 All right. 492 00:32:59,880 --> 00:33:05,240 Let's look at this z transform. 493 00:33:05,240 --> 00:33:11,280 First, let's just look at x of z in the z plane. 494 00:33:11,280 --> 00:33:13,280 Here's the z plane. 495 00:33:13,280 --> 00:33:18,750 We have a pole at z equals 1/2. 496 00:33:18,750 --> 00:33:23,700 And we have a 0 at z equals zero. 497 00:33:23,700 --> 00:33:25,380 Here's the unit circle. 498 00:33:25,380 --> 00:33:27,840 And in green, what I've indicated 499 00:33:27,840 --> 00:33:31,430 is the region of convergence. 500 00:33:31,430 --> 00:33:33,340 Now to do the contour integration, 501 00:33:33,340 --> 00:33:35,710 it's not just x of z that we're looking at. 502 00:33:35,710 --> 00:33:40,390 It's x of z times z to the n minus 1 equals 503 00:33:40,390 --> 00:33:45,070 z to the n over z minus 1/2. 504 00:33:45,070 --> 00:33:49,070 Incidentally, what is the contour 505 00:33:49,070 --> 00:33:54,770 of integration look like then the z plane for this example? 506 00:33:54,770 --> 00:33:58,820 It's a closed contour inside the region of convergence 507 00:33:58,820 --> 00:34:00,590 and encircling the origin. 508 00:34:00,590 --> 00:34:08,989 So for example, it's a contour that looks like this. 509 00:34:08,989 --> 00:34:12,679 And incidentally, it is and I don't 510 00:34:12,679 --> 00:34:18,020 think I stress this before, it's a counter-clockwise contour. 511 00:34:18,020 --> 00:34:21,600 So this is the contour c. 512 00:34:21,600 --> 00:34:24,480 It's a counterclockwise contour. 513 00:34:24,480 --> 00:34:27,900 As I've drawn it, I didn't draw a very good circle, 514 00:34:27,900 --> 00:34:29,730 partly because I can't. 515 00:34:29,730 --> 00:34:33,030 But in fact, it also emphasizes the fact 516 00:34:33,030 --> 00:34:34,530 that it doesn't have to be a circle. 517 00:34:34,530 --> 00:34:37,889 It just has to be closed, inside the region of convergence, 518 00:34:37,889 --> 00:34:39,630 and circling the origin. 519 00:34:39,630 --> 00:34:43,150 And a counterclockwise contour. 520 00:34:43,150 --> 00:34:43,750 All right. 521 00:34:43,750 --> 00:34:48,940 How many poles of x of z times z to the n minus 1 are there? 522 00:34:48,940 --> 00:34:50,889 That is, how many places are there 523 00:34:50,889 --> 00:34:53,590 that we have to evaluate residues? 524 00:34:53,590 --> 00:34:55,750 Well let's look, first of all, at n 525 00:34:55,750 --> 00:34:59,550 greater than or equal to 0. 526 00:34:59,550 --> 00:35:02,070 This doesn't introduce any poles. 527 00:35:02,070 --> 00:35:06,750 This introduces one pole equals 1/2. 528 00:35:06,750 --> 00:35:11,410 So we only need to evaluate the residue of this at this one 529 00:35:11,410 --> 00:35:15,270 pole that is at z equals 1/2. 530 00:35:15,270 --> 00:35:20,010 So we want to compute the residue then of x of z times 531 00:35:20,010 --> 00:35:25,630 z to the n minus 1 at z equals 1/2. 532 00:35:25,630 --> 00:35:29,500 So that z to the n over z minus 1/2 533 00:35:29,500 --> 00:35:35,310 multiplied by z minus 1/2 evaluated and z equals 1/2. 534 00:35:35,310 --> 00:35:38,690 These two cancel out. 535 00:35:38,690 --> 00:35:44,110 And we end up with simply 1/2 to the n. 536 00:35:44,110 --> 00:35:49,420 So for n greater than or equal to 0, the sequence x of n 537 00:35:49,420 --> 00:35:53,600 is just a half to the n. 538 00:35:53,600 --> 00:35:57,470 So that's part of the result. Now we 539 00:35:57,470 --> 00:36:04,030 have to get the inverse z transform for n less than zero. 540 00:36:04,030 --> 00:36:07,960 Well for n less than 0, again of course we have 1 pole 541 00:36:07,960 --> 00:36:10,830 and z equal to 1/2. 542 00:36:10,830 --> 00:36:15,550 But now, remember that we had this factor, which 543 00:36:15,550 --> 00:36:18,080 was z to the n. 544 00:36:18,080 --> 00:36:23,660 And for n negative that introduces poles at the origin. 545 00:36:23,660 --> 00:36:27,120 And in fact it, introduces n poles. 546 00:36:27,120 --> 00:36:32,600 So for n less than 0, we have not only this pole, 547 00:36:32,600 --> 00:36:36,530 but we have n poles at z equals 0. 548 00:36:36,530 --> 00:36:38,000 That's a problem. 549 00:36:38,000 --> 00:36:41,270 Because now we're going to have to evaluate the residues 550 00:36:41,270 --> 00:36:44,420 and each of these poles. 551 00:36:44,420 --> 00:36:48,650 And the more negative value of n we're trying to get x of n for, 552 00:36:48,650 --> 00:36:51,950 the worse it gets, because the higher the multiplicity 553 00:36:51,950 --> 00:36:54,100 of the pole. 554 00:36:54,100 --> 00:36:54,600 All right. 555 00:36:54,600 --> 00:36:58,030 But look, we've seen this example before. 556 00:36:58,030 --> 00:37:03,190 We know that x of n is 1/2 to the n for n 557 00:37:03,190 --> 00:37:06,540 greater than or equal to 0 and it's 0 for n less than 0. 558 00:37:06,540 --> 00:37:10,000 As we knew that from other times that we've worked this example. 559 00:37:10,000 --> 00:37:13,000 So obviously, there must be some easy way 560 00:37:13,000 --> 00:37:17,260 to get the fact that x of n, in fact for this example, 561 00:37:17,260 --> 00:37:20,391 is 0 for n less than 0. 562 00:37:20,391 --> 00:37:20,890 All right. 563 00:37:20,890 --> 00:37:24,960 Let's take a look at a way, an easy way. 564 00:37:24,960 --> 00:37:28,830 Here we have, again, the inverse z transform. x of n 565 00:37:28,830 --> 00:37:33,330 is 1 over 2 pi j times this contour integral. 566 00:37:33,330 --> 00:37:37,620 And let's simply make a substitution of variables. 567 00:37:37,620 --> 00:37:40,980 Let's make the substitution of variables z 568 00:37:40,980 --> 00:37:46,590 equal to p to the minus 1, in which case z to the n minus 1 569 00:37:46,590 --> 00:37:51,800 is equal to p to the minus n plus 1. 570 00:37:51,800 --> 00:37:57,225 Which means incidentally that with z equal to r e 571 00:37:57,225 --> 00:38:03,890 to of the j theta, p, this new variable, is 1 over r times 572 00:38:03,890 --> 00:38:07,170 e to the minus j theta. 573 00:38:07,170 --> 00:38:10,070 So now we want to make the substitution of variables 574 00:38:10,070 --> 00:38:13,080 in this contour integral. 575 00:38:13,080 --> 00:38:18,280 Well, let's see-- dz is minus p to the minus 2, 576 00:38:18,280 --> 00:38:21,120 just differentiating this. 577 00:38:21,120 --> 00:38:23,940 Well, I'll let you go through the algebra on your own. 578 00:38:23,940 --> 00:38:27,090 But if you take this substitution of variables, 579 00:38:27,090 --> 00:38:31,860 stuff it into that contour integral, then the result 580 00:38:31,860 --> 00:38:38,120 that we end up with is that x of n is equal to minus 1 581 00:38:38,120 --> 00:38:41,450 over 2 pi j, the minus coming from the fact 582 00:38:41,450 --> 00:38:46,250 that dz is minus p to the minus 2 times dp. 583 00:38:46,250 --> 00:38:48,200 That's that minus sign. 584 00:38:48,200 --> 00:38:53,930 x of 1 over p times p to the minus n minus 1 dp. 585 00:38:53,930 --> 00:38:56,420 And what happens to the contour? 586 00:38:56,420 --> 00:39:02,400 Well the contour previously was, let's say r e to the j theta. 587 00:39:02,400 --> 00:39:04,700 Let's take it as a circular contour. 588 00:39:04,700 --> 00:39:10,190 So the contour now is 1 over r e to the minus j theta. 589 00:39:10,190 --> 00:39:13,070 That means as one consequence that 590 00:39:13,070 --> 00:39:14,960 our old counter-clockwise contour 591 00:39:14,960 --> 00:39:17,900 turns into a clockwise contour. 592 00:39:17,900 --> 00:39:20,630 So that's what I've indicated here. 593 00:39:20,630 --> 00:39:23,720 But we also have this minus sign. 594 00:39:23,720 --> 00:39:27,470 And we can use that minus sign to change 595 00:39:27,470 --> 00:39:29,490 the direction of the contour. 596 00:39:29,490 --> 00:39:34,610 So if we change the contour back to a counterclockwise contour, 597 00:39:34,610 --> 00:39:36,860 get rid of the minus sign. 598 00:39:36,860 --> 00:39:39,740 Then finally we have the expression x of n 599 00:39:39,740 --> 00:39:44,330 is 1 over 2 pi j times the contour integral of capital X 600 00:39:44,330 --> 00:39:49,070 of 1 over p p to the minus n minus 1 dp. 601 00:39:49,070 --> 00:39:51,680 And what is this contour now? 602 00:39:51,680 --> 00:39:57,560 Well if the old contour C was circular with a radius of r, 603 00:39:57,560 --> 00:40:04,350 this contour c prime is a circle with a radius of 1 over r. 604 00:40:04,350 --> 00:40:08,595 Well let's just look at what this substitution of variable z 605 00:40:08,595 --> 00:40:13,280 to 1 over p did as a matter of fact. 606 00:40:13,280 --> 00:40:17,450 Here I've illustrated the z plane. 607 00:40:17,450 --> 00:40:21,726 Here I've illustrated the p plane. 608 00:40:21,726 --> 00:40:27,150 The unit circle in the z plane, when I replace z by 1 over p, 609 00:40:27,150 --> 00:40:30,840 turns again into the unit circle in the p plane, 610 00:40:30,840 --> 00:40:34,610 because the unit circle is where the magnitude of z is 1. 611 00:40:34,610 --> 00:40:38,760 And if I take the reciprocal of that, the magnitude is still 1. 612 00:40:38,760 --> 00:40:41,970 The important thing is what happened to this contour. 613 00:40:41,970 --> 00:40:46,290 Well this contour got converted if it was outside the unit 614 00:40:46,290 --> 00:40:51,720 circle into a new contour which is inside the unit circle. 615 00:40:51,720 --> 00:40:56,400 And in fact, in general, stuff that was in here, 616 00:40:56,400 --> 00:40:58,800 the inside of the unit circle, is 617 00:40:58,800 --> 00:41:03,960 going to end up outside the unit circle in the p plane. 618 00:41:03,960 --> 00:41:09,750 And things that were outside the unit circle in the z plane 619 00:41:09,750 --> 00:41:14,310 are going to end up inside the unit circle in the p plane. 620 00:41:14,310 --> 00:41:18,270 This transformation simply takes stuff outside the unit circle 621 00:41:18,270 --> 00:41:21,000 and folds it inside the unit circle, 622 00:41:21,000 --> 00:41:22,890 takes things inside the unit circle 623 00:41:22,890 --> 00:41:25,080 and folds them outside the unit circle. 624 00:41:25,080 --> 00:41:29,011 And it's just simply a substitution of variables. 625 00:41:29,011 --> 00:41:29,510 All right. 626 00:41:29,510 --> 00:41:31,660 Now with this substitution of variables, 627 00:41:31,660 --> 00:41:37,077 let's return to the example that we've been working. 628 00:41:40,350 --> 00:41:41,365 We have our example. 629 00:41:45,220 --> 00:41:49,430 x of z is 1 over 1 minus 1/2 z to the minus 1, 630 00:41:49,430 --> 00:41:52,950 the magnitude of z greater than 1/2. 631 00:41:52,950 --> 00:41:55,890 We want to replace z by 1 over p. 632 00:41:55,890 --> 00:42:00,020 So we want x of 1 over p. 633 00:42:00,020 --> 00:42:03,590 x of 1 over p, then simply replacing z to the minus 1 634 00:42:03,590 --> 00:42:09,040 by p, is 1 over 1 minus 1/2 times p. 635 00:42:09,040 --> 00:42:10,930 If this was true for the magnitude 636 00:42:10,930 --> 00:42:14,230 of z greater than 1/2, then this is 637 00:42:14,230 --> 00:42:17,530 true for the magnitude of p less than 1/2. 638 00:42:17,530 --> 00:42:23,090 So we get minus 2 divided by p minus 2. 639 00:42:23,090 --> 00:42:27,640 If we look at this in the p plane, 640 00:42:27,640 --> 00:42:31,690 we have then a pole in the p plane. 641 00:42:31,690 --> 00:42:34,660 The pole in the z plane was at z equals 1/2. 642 00:42:34,660 --> 00:42:40,470 So the pole in the p plane is at p equals 2. 643 00:42:40,470 --> 00:42:43,610 That's now outside the unit circle. 644 00:42:43,610 --> 00:42:46,520 The region of convergence, now, is inside 645 00:42:46,520 --> 00:42:49,220 the circle of radius 2. 646 00:42:49,220 --> 00:42:52,550 And where is the contour of integration c prime? 647 00:42:52,550 --> 00:42:56,960 Well that now is inside the circle of radius 2. 648 00:42:56,960 --> 00:43:00,440 For example, let's draw it here. 649 00:43:00,440 --> 00:43:04,900 So this is now the contour c prime. 650 00:43:04,900 --> 00:43:06,220 All right. 651 00:43:06,220 --> 00:43:11,710 We want finally to evaluate the inverse z transform, 652 00:43:11,710 --> 00:43:17,340 or the inverse p transform, by looking 653 00:43:17,340 --> 00:43:24,091 at the residues of x of 1 over p times p to the minus n minus 1. 654 00:43:24,091 --> 00:43:30,900 The residues of that at its roots, at its poles, 655 00:43:30,900 --> 00:43:36,610 that are inside the contour of integration c prime. 656 00:43:36,610 --> 00:43:43,770 Well, for n less than 0, we have one pole at p equals 2. 657 00:43:43,770 --> 00:43:49,470 And for n negative, this will contribute no poles. 658 00:43:49,470 --> 00:43:56,610 Therefore, there are no roots of this inside this contour. 659 00:43:56,610 --> 00:44:01,970 And consequently, the contour integral is obviously zero. 660 00:44:01,970 --> 00:44:07,090 So x of n is equal to zero for n less than 0. 661 00:44:07,090 --> 00:44:11,080 Now for greater than or equal to 0, 662 00:44:11,080 --> 00:44:15,500 we have one pole at p equals 2. 663 00:44:15,500 --> 00:44:20,240 But then because of this factor, p to the minus n minus 1, 664 00:44:20,240 --> 00:44:25,680 we have n plus 1 poles at p equals 0. 665 00:44:25,680 --> 00:44:29,210 So now it seems like for n greater than or equal to 0, 666 00:44:29,210 --> 00:44:30,710 we're back where we started from. 667 00:44:30,710 --> 00:44:34,662 Because now we have these multiple order poles at p 668 00:44:34,662 --> 00:44:35,980 equals 0. 669 00:44:35,980 --> 00:44:37,917 What do we do about that? 670 00:44:37,917 --> 00:44:39,750 Well, we don't have to do anything about it. 671 00:44:39,750 --> 00:44:44,010 Because we already considered the case for n greater than 672 00:44:44,010 --> 00:44:46,250 or equal to 0. 673 00:44:46,250 --> 00:44:49,550 In that case, we didn't run into multiple order poles 674 00:44:49,550 --> 00:44:50,690 at the origin. 675 00:44:50,690 --> 00:44:52,700 We used the substitution of variables, 676 00:44:52,700 --> 00:44:56,600 z equals 1 over p to basically avoid that problem 677 00:44:56,600 --> 00:44:59,870 and to get the answer for n less than 0. 678 00:44:59,870 --> 00:45:02,700 So consequently, the answer that we get, 679 00:45:02,700 --> 00:45:07,580 which we finally know to be the right answer in fact, 680 00:45:07,580 --> 00:45:12,890 is that x of n is equal to 0 for n less than zero. 681 00:45:12,890 --> 00:45:17,330 And as we worked out before, x of n is equal to 1/2 682 00:45:17,330 --> 00:45:21,260 to the n for n greater than or equal to 0. 683 00:45:21,260 --> 00:45:33,300 Or finally, x of n is equal to 1/2 to the n times u of n. 684 00:45:33,300 --> 00:45:33,800 All right. 685 00:45:33,800 --> 00:45:36,370 That's an example we've worked out now three times. 686 00:45:36,370 --> 00:45:40,581 And three times we got the same answer, so it must be right. 687 00:45:40,581 --> 00:45:41,080 All right. 688 00:45:41,080 --> 00:45:45,720 Well this roughs out the inverse z 689 00:45:45,720 --> 00:45:49,830 transform, several methods for getting the inverse z 690 00:45:49,830 --> 00:45:51,730 transform. 691 00:45:51,730 --> 00:45:55,900 Obviously, to become fluent with the inverse z transform 692 00:45:55,900 --> 00:45:59,410 requires working a lot of examples. 693 00:45:59,410 --> 00:46:02,230 And you know where you going to have 694 00:46:02,230 --> 00:46:03,790 all those examples to work. 695 00:46:03,790 --> 00:46:06,100 That is, in the study guide. 696 00:46:06,100 --> 00:46:09,940 And you'll find as I stressed previously, 697 00:46:09,940 --> 00:46:14,530 that as you work more and more examples, what you find 698 00:46:14,530 --> 00:46:16,690 in computing inverse z transforms 699 00:46:16,690 --> 00:46:19,330 is that more and more frequently, 700 00:46:19,330 --> 00:46:22,970 you're able to recognize the answer. 701 00:46:22,970 --> 00:46:27,920 Well, in the next lecture, we'll complete the discussion 702 00:46:27,920 --> 00:46:29,630 of the z transform. 703 00:46:29,630 --> 00:46:32,270 In particular, we'll among other things 704 00:46:32,270 --> 00:46:35,240 present some properties of the z transform, which 705 00:46:35,240 --> 00:46:40,010 also by the way, help us in computing inverse z transforms. 706 00:46:40,010 --> 00:46:42,620 And so in the next lecture, we'll 707 00:46:42,620 --> 00:46:46,280 be continuing the topic of the z transform, 708 00:46:46,280 --> 00:46:47,960 and in fact concluding the topic. 709 00:46:47,960 --> 00:46:49,810 Thank you.