(*) Y(z) 16z (2²+1)(4z-1)(2z-1) Now partial fraction is 16z 112 96z 16 (z²+1)(4z-1)(2z-1) 85(+1) 85(z2+1) + 5(2z-1) Using above in (7), we get Y(z) 112 96z 16 + 5(27-1) 32 - - 85(z+1) 85(z+1) 17(4z-1) II

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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How did he get the
value inside the line
LV
Y(2) -Y(z
+Y(z
2+1
r(:) [1 - & + ÷) =
2z
8z2
z2+1
82-6z+1
8z2
2z
→ Y( z
%3D
2+1
(4z-1)(2z-1)
27
» Y( z
%3D
8z2
z2+1
Y(z)
16z?
(z²+1)(4z-1)(2z-1)
Now partial fraction is
16z
112
967
16
(22+1)(4z-1)(2z-1)
85(2+1)
85(+1)
5(27-1)
Using above in (7), we get
Y(z)
112
967
16
32
85(z²+1)
85(z+1)
5(27-1)
17(4z-1)
r(:)
96z2
+
85(2+1)
112z
16z
32z
%3D
85 (z2+1)
5(27-1)
17(4z-1
112z
96z2
8z
8z
+
85(z+1)
85(2+1)
5(z-1/2)
17(z-1/
Transcribed Image Text:How did he get the value inside the line LV Y(2) -Y(z +Y(z 2+1 r(:) [1 - & + ÷) = 2z 8z2 z2+1 82-6z+1 8z2 2z → Y( z %3D 2+1 (4z-1)(2z-1) 27 » Y( z %3D 8z2 z2+1 Y(z) 16z? (z²+1)(4z-1)(2z-1) Now partial fraction is 16z 112 967 16 (22+1)(4z-1)(2z-1) 85(2+1) 85(+1) 5(27-1) Using above in (7), we get Y(z) 112 967 16 32 85(z²+1) 85(z+1) 5(27-1) 17(4z-1) r(:) 96z2 + 85(2+1) 112z 16z 32z %3D 85 (z2+1) 5(27-1) 17(4z-1 112z 96z2 8z 8z + 85(z+1) 85(2+1) 5(z-1/2) 17(z-1/
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