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Using z-transformation find the solution y(k) of the difference equation y(k+2)-y(k+1)+ 0.25y(k)= u(k+2) where the initial conditions are y(0) = 1 and y(1)= 2. The input is u(k) is the discrete-time unit step u(k)= 1 for k= 0, 1, 2, 3, 4... and u(k)=0 otherwise, which means that there are initial conditions for the input.

Using z-transformation find the solution y(k) of the difference equation y(k+2)-y(k+1)+ 0.25y(k)= u(k+2) where the initial conditions are y(0) = 1 and y(1)= 2. The input is u(k) is the discrete-time unit step u(k)= 1 for k= 0, 1, 2, 3, 4... and u(k)=0 otherwise, which means that there are initial conditions for the input.

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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Using z-transformation find the solution y(k) of the difference equation y(k+2)-y(k+1)+ 0.25y(k)= u(k+2) where the initial conditions are y(0) = 1 and y(1)= 2. The input is u(k) is the discrete-time unit step u(k)= 1 for k= 0, 1, 2, 3, 4... and u(k)=0 otherwise, which means that there are initial conditions for the input.
Transcribed Image Text:Using z-transformation find the solution y(k) of the difference equation y(k+2)-y(k+1)+ 0.25y(k)= u(k+2) where the initial conditions are y(0) = 1 and y(1)= 2. The input is u(k) is the discrete-time unit step u(k)= 1 for k= 0, 1, 2, 3, 4... and u(k)=0 otherwise, which means that there are initial conditions for the input.
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