How did he get the value inside the line -dz X [n] = 2ri P z-1)(z-2) X Im] = f 의ada z'dz (z-1)(z-2) 2ni Residue at z = 1 Res (z – 1) × |(e - 1)x E- 1 (z-1)(z-2) z=1 Res due at z = 2 Res (2 – 2) × = 2 | (z-1)(z-2) Hence, x (k) = z-" - -1) x (k) %3D 2'и (k) — и (k) | x (k) = (2k – 1)u (k)
How did he get the value inside the line -dz X [n] = 2ri P z-1)(z-2) X Im] = f 의ada z'dz (z-1)(z-2) 2ni Residue at z = 1 Res (z – 1) × |(e - 1)x E- 1 (z-1)(z-2) z=1 Res due at z = 2 Res (2 – 2) × = 2 | (z-1)(z-2) Hence, x (k) = z-" - -1) x (k) %3D 2'и (k) — и (k) | x (k) = (2k – 1)u (k)
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
2i P 7-1)(2-2)
-dz
(z-1)(z-2)
Residue at z =
1
|(e -
Res (z – 1) ×
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) = =-" -
-1)
x (k) = 2*u (k) – u (k)
|
x (k) = (2* – 1)u (k)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdcca79f6-5174-417c-9283-484df30db65c%2F3157ecad-8c42-478b-b9f3-42b5a4f040c6%2Fmr6mq19_processed.jpeg&w=3840&q=75)
Transcribed Image Text:How did he get the
value inside the line
X [n] = $
X [n] = $ "dz
2i P 7-1)(2-2)
-dz
(z-1)(z-2)
Residue at z =
1
|(e -
Res (z – 1) ×
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) = =-" -
-1)
x (k) = 2*u (k) – u (k)
|
x (k) = (2* – 1)u (k)
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