An LTI system is described by the difference equation y[n] = r[n] – 2x[n – 1] – x[n – 3]. What is the output y[n] due to the input a[n] = d[n] + 28[n – 1] – 8[n – 2]? Select one: y[r] = d[n] – 58[n – 2] + S[n – 3] – 25[n – 4] + S[n – 5] y[n] = 8[n] – 38[n – 2] – 38[n – 3] – 20[n – 4] – 8[n – 5] y[n] = 8[n] – 65[n – 2] + 8[n – 4] y/n] = 8{n] – 50[n – 1] + 8[n – 2] – 25[n – 3] + 8[n – 4] y(n] = 8[n] – 50[n – 2] + 8[n – 3] – 28[1n – 4] – 8[n – 5]
An LTI system is described by the difference equation y[n] = r[n] – 2x[n – 1] – x[n – 3]. What is the output y[n] due to the input a[n] = d[n] + 28[n – 1] – 8[n – 2]? Select one: y[r] = d[n] – 58[n – 2] + S[n – 3] – 25[n – 4] + S[n – 5] y[n] = 8[n] – 38[n – 2] – 38[n – 3] – 20[n – 4] – 8[n – 5] y[n] = 8[n] – 65[n – 2] + 8[n – 4] y/n] = 8{n] – 50[n – 1] + 8[n – 2] – 25[n – 3] + 8[n – 4] y(n] = 8[n] – 50[n – 2] + 8[n – 3] – 28[1n – 4] – 8[n – 5]
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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![An LTI system is described by the difference equation y[n] = x[n] – 2x[n - 1] – [n – 3]. What is the output y[n] due to the input
a[n] = d[n] + 25[n – 1] – S[n – 2)?
Select one:
y[r] = S[n] – 58[n – 2] + S[n – 3] – 25[n – 4] + S[n – 5]
y[n] = 8[n] – 35[n – 2] – 38[n – 3] – 20[n – 4] – 8[n – 5]
y/n] = 8[n] – 68[n – 2] + 8[n – 4]
y/n] = 8{n] – 50[n – 1] + 8[n – 2] – 25[n – 3] + 8[n – 4]
y(n] = 8[n] – 50[n – 2] + 8[n – 3] – 28[1n – 4] – 8[n – 5]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F245da8b8-c018-48cc-b5f1-380e26f41ef7%2F300c7ca3-dad8-467e-974d-6b4901547fa3%2Fiti9be9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:An LTI system is described by the difference equation y[n] = x[n] – 2x[n - 1] – [n – 3]. What is the output y[n] due to the input
a[n] = d[n] + 25[n – 1] – S[n – 2)?
Select one:
y[r] = S[n] – 58[n – 2] + S[n – 3] – 25[n – 4] + S[n – 5]
y[n] = 8[n] – 35[n – 2] – 38[n – 3] – 20[n – 4] – 8[n – 5]
y/n] = 8[n] – 68[n – 2] + 8[n – 4]
y/n] = 8{n] – 50[n – 1] + 8[n – 2] – 25[n – 3] + 8[n – 4]
y(n] = 8[n] – 50[n – 2] + 8[n – 3] – 28[1n – 4] – 8[n – 5]
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