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A. Given sequences, x₁ (n) = 56(n) - 26(n − 2) X2(n) = 36(n − 3) a) X₁(z) = 52z-², X₂(z) = 3z-³, X(z) = X₁(z)X₂(z), ⇒ X(z) = (5 – 2z-²) (3z-³), ⇒ X(z) = (15z−³ – 6z-5), - b) Taking inverse z-transform of X(z) obtained in step (a), x[n] = 158[n - 3] - 68[n - 5] Explanation According to convolution theorem, convolution in time domain is equivalent to the multiplication in frequency domain.

A. Given sequences, x₁ (n) = 56(n) - 26(n − 2) X2(n) = 36(n − 3) a) X₁(z) = 52z-², X₂(z) = 3z-³, X(z) = X₁(z)X₂(z), ⇒ X(z) = (5 – 2z-²) (3z-³), ⇒ X(z) = (15z−³ – 6z-5), - b) Taking inverse z-transform of X(z) obtained in step (a), x[n] = 158[n - 3] - 68[n - 5] Explanation According to convolution theorem, convolution in time domain is equivalent to the multiplication in frequency domain.

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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A. Given sequences, X₁ (n) = 56(n) — 26(n − 2) x2 (n) = 38(n − 3) a) X₁(z) = 52z-2, X₂(z) = 3z-³, X(z) = X₁(z)X₂(z), ⇒ X(z) = (5 – 2z-²) (3z−³), - ⇒ X(z) = (15z−³ – 6z−5), b) Taking inverse z-transform of X(z) obtained in step (a), x[n] = 156[n – 3] — 68[n – 5] - - Explanation According to convolution theorem, convolution in time domain is equivalent to the multiplication in frequency domain.
Transcribed Image Text:A. Given sequences, X₁ (n) = 56(n) — 26(n − 2) x2 (n) = 38(n − 3) a) X₁(z) = 52z-2, X₂(z) = 3z-³, X(z) = X₁(z)X₂(z), ⇒ X(z) = (5 – 2z-²) (3z−³), - ⇒ X(z) = (15z−³ – 6z−5), b) Taking inverse z-transform of X(z) obtained in step (a), x[n] = 156[n – 3] — 68[n – 5] - - Explanation According to convolution theorem, convolution in time domain is equivalent to the multiplication in frequency domain.
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