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x(п) %3D п а" и(-n) x (п) %3D п (п + 1) и(п)

x(п) %3D п а" и(-n) x (п) %3D п (п + 1) и(п)

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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Determine the z transform for 5 and 6 please

4. x(п) — п а" и(п) 1 Z{a"u(n)} ROC: |z| > [a| %3D 1 - az-1 And Z{n x(n)} : = -Z dz * X(z) , differentiationin z – domain property d = -Z 1 Z{n a"u(n)} here x(n) = a"u(n) %3D dz 1 – az-1 d d (1 – az-1)1– · dz dz(1– az-1) = -z · (1 – az-1)2 0 + a(-1)z-2 = -Z · (1 – az-1)² a z-1 ROC : |z| > [a| %3| (1 – az-1)² az-1 Thus па" и(п) ROC: |z| > [a| (1-az-1)² x(п) %3D п а" и(-n) x(п) %3D п (п + 1) и(п) 5. 6.
Transcribed Image Text:4. x(п) — п а" и(п) 1 Z{a"u(n)} ROC: |z| > [a| %3D 1 - az-1 And Z{n x(n)} : = -Z dz * X(z) , differentiationin z – domain property d = -Z 1 Z{n a"u(n)} here x(n) = a"u(n) %3D dz 1 – az-1 d d (1 – az-1)1– · dz dz(1– az-1) = -z · (1 – az-1)2 0 + a(-1)z-2 = -Z · (1 – az-1)² a z-1 ROC : |z| > [a| %3| (1 – az-1)² az-1 Thus па" и(п) ROC: |z| > [a| (1-az-1)² x(п) %3D п а" и(-n) x(п) %3D п (п + 1) и(п) 5. 6.
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