x(п) %3D п а" и(-n) 5. =
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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Question
please solve the z-transform for 5, 6 in similar way to the solved one above if that possible
![4.
х(п) %3D п а" и(п)
1
Z{a"u(n)} =
ROC : |z| > |a|
1 - az-1
d
And Z{n x(n)} = -z
dz
X(z) , differentiationin z – domain property
d
1
Zin a"u(п)}
here x(n) = a"u(n)
= -Z
dz 1- az-1
|
d
dz
(1 – az-1)2
d
(1 – az"
= -z
-1) 음1-1.
(1 – az-1)
|
dz
0 + a(-1)z-2
= -z
(1 – az-1)2
a z-1
ROC : |z| > |a|
(1 – az-1)2 '
a z-1
Thus
па" и(п)
ROC : |z| > ]a|
|(1-az-1)2
x (п) %3D п а" и(-п)
x(п) %3D п (п + 1) и (п)
5.
6.
||](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6fe45da-ee62-4004-8275-058a02bbba24%2F00ac1319-c13d-4e7e-9b0b-10647b0d231f%2Fjvm6wl7_processed.png&w=3840&q=75)
Transcribed Image Text:4.
х(п) %3D п а" и(п)
1
Z{a"u(n)} =
ROC : |z| > |a|
1 - az-1
d
And Z{n x(n)} = -z
dz
X(z) , differentiationin z – domain property
d
1
Zin a"u(п)}
here x(n) = a"u(n)
= -Z
dz 1- az-1
|
d
dz
(1 – az-1)2
d
(1 – az"
= -z
-1) 음1-1.
(1 – az-1)
|
dz
0 + a(-1)z-2
= -z
(1 – az-1)2
a z-1
ROC : |z| > |a|
(1 – az-1)2 '
a z-1
Thus
па" и(п)
ROC : |z| > ]a|
|(1-az-1)2
x (п) %3D п а" и(-п)
x(п) %3D п (п + 1) и (п)
5.
6.
||
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