Inverse Z-Transform Methods: Partial Fractions Analysis
Learn about finding the inverse of Z-transform using partial fraction method for complex functions with multiple order poles. Understand the process step-by-step with practical examples and solutions provided.
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3rd Class : Engineering Analysis Assist Prof. Dr. Fadhil S. Hasan Imaginary ZT-5: The Inverse of Z-transform ? ??(?) :- outer real ? ? 1. ???(?) = 1 ? ?= ???(?) ? ? ,??? ? > ? a Imaginary inner ? ? 2. ???( ? 1) = 1 = ???( ? 1) ? ? ,??? ? < ? ? ? real b ? ? 1 3. ?( ?) = ? 1 ,??? ? < 1 = ?( ? 1) ? 1 ? ? 1 4. ?(?) = ? 1 ,??? ? > 1 = ?(?) ? 1
11 = ?(n) 5. ?(?) = 1 ,??? ??? ? ? ? (? 1)2 ROC: ? > 1 1 (? 1)2=n u(n) 6. ??(?) = There are many methods to find the inverse of Z-transform 1. Partial fraction method i. Distinct poles X ? ? ?1 ?2 ?? = + + + ? ?1 ? ?2 ? ?? X ? ? ??= lim ? ??? ?? 1 Example : Find ? ? for X ? = 1 1.5? 1+0.5? 2 for the following ROC's: ?) ? > 1 , b) ? < 0.5 and c) 0.5 < ? < 1
Solution: ?2 ?2 1.5? + 0.5 X ? ? ?1 ?2 X ? = = (? 1)(? 0.5)= ? 1+ ? ? 0.5 ? ? ?1= lim = 2, ?2= lim = 1 ? 0.5 ? 1 ? 1 ? 0.5 X ? ? 2 1 ? ? = ? 1 ? 0.5, X ? = 2 ? 1 ? 0.5 a. a. ??? ? > ? ? ? = 2? ? 0.5??(?)
?.??? ? < 0.5 ? ? = 2? ? 1 + 0.5??( ? 1 ) ?.??? 0.5 < ? < 1 ? ? = 2? ? 1 0.5??(? )
ii. Multiple order poles 1 Example: Find ? ? ) for X ? = (1+? 1)(1 ? 1)2 ?3 1 Solution: X ? = (1+? 1)(1 ? 1)2= (?+1)(? 1)2 ?2 ) X(? ? ?12 ? 12+ ?11 ? 1+ ?3 = ? + 1) ? 12= ? + 1 ?2 =1 ?12= lim 2, ? + 1 ? 1 2? ? + 1 ?2 ? + 12 =3 ?11= lim 4 ? 1 ?2 ? 12=1 ?3= lim 4 ? 1
1 2? 3 4? ? 1+ 1 4? ? + 1 X(?) = ? 12+ ?(?) =1 2??(?) +3 4?(?) +1 4 1??(?) Notes: ???(?)is a right hand sequence (causal or positive time sequence). ???( ? 1)is a left hand sequence (anti-causal or negative time sequence). Example : Find the inverse Z-transform of ?2+ ? (? 0.5)3(? 0.25) ??? ? ? ??? ? >1 X ? = 2 Solution: ?13 ?12 ?11 ?4 X ? ? ? + 1 = ? 0.53+ ? 0.52+ + = ? 0.53(? 0.25 ? 0.5 ? 0.25 )
? + 1 ? 0.25 ? 0.25 ? 1 ? 0.252 ?13= lim = 6, ?12= lim = 20 ? 0.5 ? 0.5 ?2 ??2 ?11=1 ? + 1 ? 0.25 =1 ? ?? 1.25 ? 0.252 2!lim ? 0.5 2.5 ? 0.25 ? 0.254= 80 2!lim ? 0.5 =1 2lim ? 0.5 ? + 1 ? 0.53= 80 ?4= lim ? 0.25 6? 20? 80? ? 0.5 80? X ? = ? 0.53 ? 0.52+ ? 0.25 ? 2 ? 1 ? ? x ? = 6? ? 1 1 2 1 2 1 2 1 4 ? ? 20? ? ? + 80 ? ? 80 ?(? 2
Notes: At ?? ??? ? > ? ? ? 1 ? ?= ???(?). ? ? 1 (? ?)2= ??? 1?(?) ? ?(? 1) ? 1 ?? 2?(?) (? ?)3= 2! ? ? ? 1 ? 2 (? ?+2) In general ? 1 ?? ?+1?(?) (? ?)?= (? 1)! HW: Using the partial fraction method to find ?(?) for ?(?2 4?+5) (? 3)(? 1)(? 2) ??? ??? ? (?) 2 < ? < 3,(?) ? > 3,(?) ? < 1 1. X ? = ?3+2?2 ? 0.52(?+3 4)??? ??? ? >3 2. X ? = 4
