12. Difference Equations - Inverse Z transforms - Using Residue method - 21MAB201T — Transcript

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1. 0:00dear students welcome to the problem
2. 0:01solving session on difference equations
3. 0:04using is transform this is part two
4. 0:07video in part one we learned how to
5. 0:10solve differen equations using is
6. 0:13transform by partial fraction method in
7. 0:16this session we are going to solve the
8. 0:18same problem but we are going to use
9. 0:21residual method to solve the problem so
10. 0:24I'll be little quick uh in doing the
11. 0:27first part of the problems where in the
12. 0:29second part in of partial fractions we
13. 0:31are going to use residue method already
14. 0:33we discussed a lot about residue method
15. 0:36you can go and check the playlist and
16. 0:38watch the residue method and come back
17. 0:39here so the advantage of residue method
18. 0:41is you don't need to remember any of the
19. 0:44is transform formula either is a
20. 0:46transform or inverse is a transform just
21. 0:48if you know two formula how to find the
22. 0:51residue for simple pole and how to find
23. 0:54the residue for pole of order M when M
24. 0:56greater than 1 and do some simple
25. 0:58calculation we get the answer okay let
26. 1:00us go into the session I'm taking the
27. 1:03same problem solve un n + 2 - 5 unu n +
28. 1:071 + 6 unu n = to 4^ n as I said whatever
29. 1:11the problem is this convert into my
30. 1:14convenient format y of n so convert this
31. 1:17into y of n + 2 - 5 y of n + 1 + 6 y of
32. 1:23n is equal to 4^ n the given conditions
33. 1:27y of 0 is 0 y of 1 1 next we have to
34. 1:31write the formula for Z of Y of n that
35. 1:34is f of Z so I'm going to find Z inverse
36. 1:38of capital F of Z that is going to be my
37. 1:41answer y of n so Z of Y of n is equal to
38. 1:44F of z z into F of zus Z into y now we
39. 1:49know how to find Z of y n + 2 just
40. 1:53multiply is on the above equation we'll
41. 1:56be getting z s capital F of Z minus z s
42. 1:59y - Z into y of 1 now take this as
43. 2:04equation 1 and taking Z transform on
44. 2:07both sides and applying this formulas
45. 2:10there we can get a equation in terms of
46. 2:13f of Z let us do this now we know y not
47. 2:16is 0 I'm just applying here and
48. 2:19assembling the remaining data z s - 5 Z
49. 2:23+ 6 I'm taking this F of Z as common so
50. 2:27we will be getting z s - 5 + 6 into
51. 2:31capital F of Z - z y of 1 is 1 is = z by
52. 2:36Z - 4 now we have to simplify this we
53. 2:39know that z s - 5 Z + 6 can be written
54. 2:43as Z - 2 Z - 3 either use your
55. 2:46calculator or manually you can get this
56. 2:49because this is simply a quadratic
57. 2:51equation next we have to simplify this
58. 2:54taking LCM on the right hand side we'll
59. 2:56be getting this now z s - 3 3 Z / by Z -
60. 3:014 then taking Z in common we get Z into
61. 3:05Z - 3 / Z -4 bringing this to the other
62. 3:10side zus 3 zus 3 will get cancel
63. 3:13therefore my final answer is f of Z is
64. 3:16equal to Z / Z - 2 Z - 4 in partial
65. 3:20fraction do you remember students the
66. 3:23code is simply F of Z by Z then only I
67. 3:26can do partial fraction method in
68. 3:29residual method we have to do z^ nus1
69. 3:33into capital F of Z for this Final
70. 3:35Answer F of Z we have to multiply z^ nus
71. 3:391 on both sides then we can get the
72. 3:42required form which can be used for
73. 3:45finding residues now make the
74. 3:47denominator 0o Z - 2 into Z - 4 is = 0
75. 3:51we have two Roots 2A 4 and see the power
76. 3:54here it is one therefore zal 2 is a
77. 3:58simple pole as well as Z equal 4 is also
78. 4:01a simple pole the first two steps over
79. 4:04next we have to find the residues r of 2
80. 4:07and R of 4 we know that general formula
81. 4:11R of a is limit Z TS to a zus a into z^
82. 4:15nus1 capital F of Z because the poll is
83. 4:19simple this is the formula for simple
84. 4:22Pole now the value of a is 2 so we get R
85. 4:25of 2 is limit Z TS to 2 Z - 2 into z^
86. 4:29nus 1 capital F of Z from step one we
87. 4:33know this value just substituting this I
88. 4:35can cancel Z - 2 and Z - 2 now when I
89. 4:39apply the limit I'll be getting 2^ n / 2
90. 4:43- 4 that is 2^ n by -2 therefore R of 2
91. 4:48is - 2^ n by 2 see the beauty of this
92. 4:52residual method students I don't use any
93. 4:54Z transform formula I get the solution
94. 4:56for R of2 you can see the typed version
95. 4:59similarly I can find R of 4 now writing
96. 5:03the value of z^ nus1 into F of Z I can
97. 5:07cancel Z - 4 Z - 4 applying the limit
98. 5:10we'll be getting 4^ n / 4 - 2 so 4^ n /
99. 5:15by 2 you can see the typed version and
100. 5:18finally we have to do the substitution Z
101. 5:20inverse of Z by zus 2 into Z - 4 we take
102. 5:24this as Z inverse of f of Z inverse of f
103. 5:28of Z is nothing but sum of residues of Z
104. 5:31power nus 1 into capital F of z r of 2 +
105. 5:36r of 4 just substitute the values we get
106. 5:39the solution and we know that Z inverse
107. 5:42of f of Z is nothing but y of n that is
108. 5:45my required solution so we found y of
109. 5:49n in the similar way we are going to
110. 5:52solve the second problem also as usual
111. 5:55convert into our required format write
112. 5:57the formula Z of YN Z of YN + 1 Z of YN
113. 6:01+ 2 here both the initial conditions are
114. 6:04zero f of z f of z f of Z so I'll be
115. 6:08getting z s + 6 Z + 9 into F of Z this
116. 6:12can be simplified as Z +
117. 6:153² now I'm writing this therefore my
118. 6:18capital F of Z is z by Z - 2 into Z + 3²
119. 6:23as I said for residual method very
120. 6:26important thing we have to multiply by Z
121. 6:29n -1 into capital F of Z that is going
122. 6:33to be my step one now making the
123. 6:35denominator zero I'll be getting Roots Z
124. 6:39= 2 and Z = -3 tce so I can say Z = 2 is
125. 6:45a simple pole Z = to minus 3 is a pole
126. 6:49of order 2 because it is repeated twice
127. 6:51next we have to find the residue for
128. 6:54simple pole we know the formula R of a
129. 6:56is equal to limit Z TS to a z a z^ nus 1
130. 7:01into capital F of Z when you substitute
131. 7:03a = 2 and substituting the value from
132. 7:06step one we can cancel this and this and
133. 7:10applying the limits we'll be getting 2^
134. 7:12n / 2 + 3² so my R of 2 is 2^ n / 25
135. 7:20next R of -3 this is a PO of order 2 so
136. 7:24I have to write the formula limit Z TS
137. 7:27to a 1 by m - 1 factorial d^ M - 1 / d
138. 7:33z^ m- 1 Z - a^ M z^ nus 1 into capital F
139. 7:40of Z now we can see a value is min -3 so
140. 7:44we have to substitute and we have to
141. 7:46find the solution and also m is going to
142. 7:49be two because pole of order two so M
143. 7:52minus 1 is 1 so we are getting d by d z
144. 7:56now first apply the values of Z n -1
145. 8:00into capital F of Z after final
146. 8:02simplification we can go for
147. 8:04differentiation now Z + 3² z + 3 S get
148. 8:09cancel and we will be getting z^ n / zus
149. 8:132 we have to do U by V rule denominator
150. 8:16v² numerator V duu minus U DV now doing
151. 8:21the differentiation we'll be getting Z -
152. 8:242 whole squ in the denominator in the
153. 8:26numerator we have Z - 2 and
154. 8:29differentiation of z^ n is n into z^
155. 8:32nus1 minus now keeping z^ n
156. 8:35differentiation of Z - 2 is 1 just apply
157. 8:39the limit when you apply the limit
158. 8:40you'll be getting - 3 - 2 N - 3^ n - 1 /
159. 8:46- 3^ n in the denominator we'll be
160. 8:49getting -3 - 2 s you can see the typed
161. 8:53version now simplifying this my
162. 8:55numerator will become -5 n into -3 ^ n -
163. 9:001 - - 3^ n / 25 I want the answers in
164. 9:06powers of N and now simplifying this - 5
165. 9:10N - 3 power N - 3^ - 1 will become -3
166. 9:16into 25 and splitting the second term
167. 9:19minus of - 3^ n by 25 now I can cancel
168. 9:23this minus and minus and I can cancel
169. 9:25this 5 and 25 therefore my final answer
170. 9:29is n into -3 n / 15 - -3 n / 25 see the
171. 9:38beauty of this problem students if you
172. 9:40go for partial fraction you have to
173. 9:42remember what is the formula for Z
174. 9:44inverse of Z by z- a what is the formula
175. 9:47for Z inverse of Z by Z - 1 s that is n
176. 9:52this is a power n and Z inverse of Z by
177. 9:56Z + a² etc etc but the residue method
178. 10:01will give the exact answer without
179. 10:03knowing any formula so in the case of
180. 10:06difference equation I did two problems
181. 10:08for part one video as well as part two
182. 10:11video but in two different method in
183. 10:13part one video we see using partial
184. 10:17fraction method in part two we solve the
185. 10:20same problem using residual method I
186. 10:24hope you understand now you'll be having
187. 10:26confident to do any problems in
188. 10:28differenti equation
189. 10:29now finally we have to write the answer
190. 10:32z inverse of f of Z is sum of the
191. 10:34residues of z^ n minus1 into capital F
192. 10:38of Z this is nothing but R of 2 and R
193. 10:41ofus 3 and we know that Z inverse of f
194. 10:44of Z is y of n this is my required
195. 10:50answer thanks for watching subscribe to
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197. 10:55see you in the next video bye-bye