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B. Find the unit impulse response for each of the following linear system У(п) %3D —0.8 у (п— 1) +x(п — 1) for n 2 0, x(-1) %3D о, у(-1) 3D о The right answer is h(m) %3D1.258(m) -1.25(-0.8)"; п 20 Don't use Z-transform Use Digital Convolution How to find it/what is the solving method to find this answer? Helpful info: Note: let x(n) = 8(n) y(n) = h(k)x(n – k) k=-00 =...+ h( – 1)x(n+1)+h(0)x(n) + h(1)x(n – 1) + h(2)x(n – 2) + ... Using a conventional notation, we express the digital convolution as y(n) = h(n)*x(n). Note that for a causal system, which implies its impulse response h(n) = 0 for n < 0, the lower limit of the convolution sum begins at 0 instead of o, that is y(n) = h(k)x(n – k) = x(k)h(n – k). k=0 k=0

B. Find the unit impulse response for each of the following linear system У(п) %3D —0.8 у (п— 1) +x(п — 1) for n 2 0, x(-1) %3D о, у(-1) 3D о The right answer is h(m) %3D1.258(m) -1.25(-0.8)"; п 20 Don't use Z-transform Use Digital Convolution How to find it/what is the solving method to find this answer? Helpful info: Note: let x(n) = 8(n) y(n) = h(k)x(n – k) k=-00 =...+ h( – 1)x(n+1)+h(0)x(n) + h(1)x(n – 1) + h(2)x(n – 2) + ... Using a conventional notation, we express the digital convolution as y(n) = h(n)*x(n). Note that for a causal system, which implies its impulse response h(n) = 0 for n < 0, the lower limit of the convolution sum begins at 0 instead of o, that is y(n) = h(k)x(n – k) = x(k)h(n – k). k=0 k=0

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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B. Find the unit impulse response for each of the following linear system y(n) = -0.8 y (n – 1) + x(n – 1) for n 2 0, x(-1) = 0, y(-1) = o The right answer is h(n) %3D1.256(п)-1.25(-0.8)"; п20 Don't use Z-transform Use Digital Convolution How to find it/what is the solving method to find this answer? Helpful info: Note: let x(n) = 8(n) y(n) = h(k)x(n – k) k=-00 = ...+h( – 1)x(n+ 1) + h(0)x(n) + h(1)x(n – 1) + h(2)x(n – 2) + ... Using a conventional notation, we express the digital convolution as y(n) = h(n)*x(n). Note that for a causal system, which implies its impulse response h(n) = 0 for n < 0, the lower limit of the convolution sum begins at 0 instead of oo, that is y(n) = h(k)x(n – k) = x(k)h(n – k). %3D k=0 k=0
Transcribed Image Text:B. Find the unit impulse response for each of the following linear system y(n) = -0.8 y (n – 1) + x(n – 1) for n 2 0, x(-1) = 0, y(-1) = o The right answer is h(n) %3D1.256(п)-1.25(-0.8)"; п20 Don't use Z-transform Use Digital Convolution How to find it/what is the solving method to find this answer? Helpful info: Note: let x(n) = 8(n) y(n) = h(k)x(n – k) k=-00 = ...+h( – 1)x(n+ 1) + h(0)x(n) + h(1)x(n – 1) + h(2)x(n – 2) + ... Using a conventional notation, we express the digital convolution as y(n) = h(n)*x(n). Note that for a causal system, which implies its impulse response h(n) = 0 for n < 0, the lower limit of the convolution sum begins at 0 instead of oo, that is y(n) = h(k)x(n – k) = x(k)h(n – k). %3D k=0 k=0
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