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Impulse response of a discrete time system h[n] and an input x[n] has been given for a linear shift invariant (LSI) system as follows: h[n]u[n+2]-u[n-1] x[n] = 8[n]-8[n 1] (a) Plot the system function h[n] and the input x[n]. (b) Specify and plot response of the system (n)) for the input x[n] using convolution sum (analytical method). (e) Specify whether the system is causal or not by indicating a reason.

Impulse response of a discrete time system h[n] and an input x[n] has been given for a linear shift invariant (LSI) system as follows: h[n]u[n+2]-u[n-1] x[n] = 8[n]-8[n 1] (a) Plot the system function h[n] and the input x[n]. (b) Specify and plot response of the system (n)) for the input x[n] using convolution sum (analytical method). (e) Specify whether the system is causal or not by indicating a reason.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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DISCRETE SYSTEM PROBLEM( NEED NEAT HANDWRITTEN SOLUTION ONLY OTHERWISE DOWNVOTE).

**Discrete Time System Analysis** _Impulse response of a discrete-time system \( h[n] \) and an input \( x[n] \) has been given for a linear shift-invariant (LSI) system as follows:_ \[ h[n] = u[n+2] - u[n-1] \] \[ x[n] = \delta[n] - \delta[n-1] \] ### Problem Statement: **(a)** Plot the system function \( h[n] \) and the input \( x[n] \). **(b)** Specify and plot the response of the system \( y[n] \) for the input \( x[n] \) using convolution sum (analytical method). **(c)** Specify whether the system is causal or not by indicating a reason. --- **Explanation of Symbols and Functions:** - \( u[n] \): Unit step function. \( u[n] = 0 \) for \( n < 0 \) and \( u[n] = 1 \) for \( n \geq 0 \). - \( \delta[n] \): Unit impulse function. \( \delta[n] = 1 \) for \( n = 0 \) and \( \delta[n] = 0 \) for \( n \neq 0 \). ### Detailed Steps to Solution: **(a) Plots:** - **System Function \( h[n] \):** The unit step function \( u[n+k] = 1 \) for \( n \geq -k \). Therefore: - \( u[n+2] \): This function shifts the unit step function to the left by 2 units. - \( -u[n-1] \): This function shifts the unit step function to the right by 1 unit and inverts it. The combination \( u[n+2] - u[n-1] \) results in a function that equals 1 between the shifted locations of the unit step functions (i.e., from \( n = -2 \) to \( n = 0 \)). - **Input \( x[n] \):** The input \( x[n] = \delta[n] - \delta[n-1] \) is a combination of two impulse functions. It consists of an impulse at \( n = 0 \) and an inverted impulse at \( n = 1 \). **(b)
Transcribed Image Text:**Discrete Time System Analysis** _Impulse response of a discrete-time system \( h[n] \) and an input \( x[n] \) has been given for a linear shift-invariant (LSI) system as follows:_ \[ h[n] = u[n+2] - u[n-1] \] \[ x[n] = \delta[n] - \delta[n-1] \] ### Problem Statement: **(a)** Plot the system function \( h[n] \) and the input \( x[n] \). **(b)** Specify and plot the response of the system \( y[n] \) for the input \( x[n] \) using convolution sum (analytical method). **(c)** Specify whether the system is causal or not by indicating a reason. --- **Explanation of Symbols and Functions:** - \( u[n] \): Unit step function. \( u[n] = 0 \) for \( n < 0 \) and \( u[n] = 1 \) for \( n \geq 0 \). - \( \delta[n] \): Unit impulse function. \( \delta[n] = 1 \) for \( n = 0 \) and \( \delta[n] = 0 \) for \( n \neq 0 \). ### Detailed Steps to Solution: **(a) Plots:** - **System Function \( h[n] \):** The unit step function \( u[n+k] = 1 \) for \( n \geq -k \). Therefore: - \( u[n+2] \): This function shifts the unit step function to the left by 2 units. - \( -u[n-1] \): This function shifts the unit step function to the right by 1 unit and inverts it. The combination \( u[n+2] - u[n-1] \) results in a function that equals 1 between the shifted locations of the unit step functions (i.e., from \( n = -2 \) to \( n = 0 \)). - **Input \( x[n] \):** The input \( x[n] = \delta[n] - \delta[n-1] \) is a combination of two impulse functions. It consists of an impulse at \( n = 0 \) and an inverted impulse at \( n = 1 \). **(b)
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