162 112 96z 16 + 5(2z-1) %3D - (2+1) (4z-1) (2z-1) 85(2+1) 85(2+1) Using above in (7), we get 96z 112 85(2+1) 16 32 85(z+1) + 5(27-1) 17(4z-1) 112z 96z 16z 32z %3D - 85(-211 85(-2+1) 5(27-1) 17(4z-1) 112z 96z 8z 87 85(2+1) 85(2+1) 5(z-1/2) 17(z-1/8) Since we have = sin (). z-A kn = COS Z-I 2+1 and Z- = a*. %3D Taking inverse z-transform of (8), we get using above formula
162 112 96z 16 + 5(2z-1) %3D - (2+1) (4z-1) (2z-1) 85(2+1) 85(2+1) Using above in (7), we get 96z 112 85(2+1) 16 32 85(z+1) + 5(27-1) 17(4z-1) 112z 96z 16z 32z %3D - 85(-211 85(-2+1) 5(27-1) 17(4z-1) 112z 96z 8z 87 85(2+1) 85(2+1) 5(z-1/2) 17(z-1/8) Since we have = sin (). z-A kn = COS Z-I 2+1 and Z- = a*. %3D Taking inverse z-transform of (8), we get using above formula
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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![63.l ll
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보 CHAT
VX MATH SOLV
7
How did he get the
value inside the line
167
112
96z
16
(z+1)(4z-1)(2z-1)
85(z+1)
85(2+1)
+
5(2z-1)
Using above in (7), we get
112
967
16
32
17(4z-1)
85(2+1)
85(22+1)
5(27-1)
112z
96z
16z
32z
85(-211)
85(2+1)
5(27-1)
17(4z-1)
112z
962
87
87
Z.
85(2+1)
85(2+1)
5(z-1/2)
17(z-1/8)
Since we have
z-=sin (블), Z-
and Z- ] = a*.
= COS
-2+1
-2+
Taking inverse z-transform of (8), we get using
above formula
(1) = z-
112z
96z
= Z-1
- z-!
85(7+1)
+Z-
85(+1)
y(k)3D sin (뜰)-cos (뜰) +](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ef9c729-562c-4f81-8ef4-617818240d6c%2Ffb96a78e-f15b-4362-8720-5e954f42c8da%2F6rzv4w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:63.l ll
www.bartleby.com/questions-ar
Q SEARCH & ASK
보 CHAT
VX MATH SOLV
7
How did he get the
value inside the line
167
112
96z
16
(z+1)(4z-1)(2z-1)
85(z+1)
85(2+1)
+
5(2z-1)
Using above in (7), we get
112
967
16
32
17(4z-1)
85(2+1)
85(22+1)
5(27-1)
112z
96z
16z
32z
85(-211)
85(2+1)
5(27-1)
17(4z-1)
112z
962
87
87
Z.
85(2+1)
85(2+1)
5(z-1/2)
17(z-1/8)
Since we have
z-=sin (블), Z-
and Z- ] = a*.
= COS
-2+1
-2+
Taking inverse z-transform of (8), we get using
above formula
(1) = z-
112z
96z
= Z-1
- z-!
85(7+1)
+Z-
85(+1)
y(k)3D sin (뜰)-cos (뜰) +
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