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How did he get the value inside the line 2ni (z-1)(z-2) X [n] = $ 1 %3D 1 2ni (z–1)(z–2) "dz Residue at z = 1 Res (2 – 1) × - Z. = -1 (z-1)(z-2) Res due at z = 2 Res[(z – 2) × = 2 Z. | (z-1)(z-2) z=2

How did he get the value inside the line 2ni (z-1)(z-2) X [n] = $ 1 %3D 1 2ni (z–1)(z–2) "dz Residue at z = 1 Res (2 – 1) × - Z. = -1 (z-1)(z-2) Res due at z = 2 Res[(z – 2) × = 2 Z. | (z-1)(z-2) z=2

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How did he get the value inside the line X [n] = $ X [n] = $ "dz 2ri P Tz-1)(z-2) -dz 2ni z'dz (z-1)(z-2) Residue at z = 1 Res (z – 1) × |(e 1 (z–1)(z–2) z=1 Res due at z = 2 Res (z – 2) × - 2) x = 2 | (z-1)(z-2) Hence, x (k) = z- (z–2) -1) x (k) = 2*u (k) – u (k) | x (k) = (2* – 1)u (k)
Transcribed Image Text:How did he get the value inside the line X [n] = $ X [n] = $ "dz 2ri P Tz-1)(z-2) -dz 2ni z'dz (z-1)(z-2) Residue at z = 1 Res (z – 1) × |(e 1 (z–1)(z–2) z=1 Res due at z = 2 Res (z – 2) × - 2) x = 2 | (z-1)(z-2) Hence, x (k) = z- (z–2) -1) x (k) = 2*u (k) – u (k) | x (k) = (2* – 1)u (k)
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