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= The impulse response of the causal system with transfer function H(z) = O 1. None of the answers O 2. h(n)=E=o 0.5”+1u(n+1) 3. O 4. O 5. - h(n) = ((0.5)n+1 − (0.5)n+¹)u(n) Bet h(n) = ((0.5)” — (0.5)”)u(n) h(n) = ((0.5) + (0.5)") (Z-0.5) (Z+0.5) is:

= The impulse response of the causal system with transfer function H(z) = O 1. None of the answers O 2. h(n)=E=o 0.5”+1u(n+1) 3. O 4. O 5. - h(n) = ((0.5)n+1 − (0.5)n+¹)u(n) Bet h(n) = ((0.5)” — (0.5)”)u(n) h(n) = ((0.5) + (0.5)") (Z-0.5) (Z+0.5) is:

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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Signalse and system
The impulse response of the causal system with transfer function H(z) = (Z-0.5) (Z+0.5) z² is: O 1. None of the answers 0 2. 3. 4. O 5. h(n) = ((0.5)n + (0.5)") h(n) = -0 0.5n+¹u(n+1) n=0 h(n) = ((0.5) n+1- (0.5)n+¹)u(n) h(n) = ((0.5) - (0.5)")u(n)
Transcribed Image Text:The impulse response of the causal system with transfer function H(z) = (Z-0.5) (Z+0.5) z² is: O 1. None of the answers 0 2. 3. 4. O 5. h(n) = ((0.5)n + (0.5)") h(n) = -0 0.5n+¹u(n+1) n=0 h(n) = ((0.5) n+1- (0.5)n+¹)u(n) h(n) = ((0.5) - (0.5)")u(n)
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