Problem 2Each of the following figures shows the impulse response of a LTI system h[n]. For eachfigure in (a), (b), (c) and (d):(i) Firstly write an expression of h[n] and then find an expression of the frequencyresponse of the system:h[n]e-inw(ii) say whether H(e) as a function of w is even or odd or neither(iii) and sketch/plot the magnitude of the frequency response, i.e., H(ew)|, over -27 <w 2π (you may use Matlab).(a)(b)In each figure, the value indicated for each impulse means the amplitude of that impulse.For example, if the value of an impulse at n = 3 is indicated to be 2j, it means 2jd[n – 3].The value of h[n] is zero for all n outside the range shown in the figure.h[n](c)H(ejw) =h[n]1 1 1-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6h[n]-6 -5 -4 -3 -2 -1 0 1-6-5-4-3-2-1 01n=-∞111 1 123 4562 3 4 5 6nnh[n] is real-valued.h[n] is real-valued.h[n] is real-valued.

Given: Three different Impulse response sequences. To find: For each of impulse response, we have to…

(ii) say whether H(e) as a function of w is even or odd or neither (iii) and sketch/plot the magnitude of the frequency response, i.e., |H(e³w)|, over −2ñ < w ≤ 2π (you may use Matlab). (a) response of the system: In each figure, the value indicated for each impulse means the amplitude of that impulse. For example, if the value of an impulse at n = 3 is indicated to be 2j, it means 2jd[n − 3]. The value of h[n] is zero for all n outside the range shown in the figure. h[n] (b) (c) H(ejw) = Σ h[n]e¯¡nw, n=-x h[n] il -6-5-4-3-2-1 0 1 2 3 4 5 6 1 1 1 h[n] -6-5-4-3-2-1 0 1 2 3 456 -6 -5 -4 -3 -2 -1 0 −1 1 1 1 1 1 2 3 4 5 6 n h[n] is real-valued. h[n] is real-valued. h[n] is real-valued.

(ii) say whether H(e) as a function of w is even or odd or neither (iii) and sketch/plot the magnitude of the frequency response, i.e., |H(e³w)|, over −2ñ < w ≤ 2π (you may use Matlab). (a) response of the system: In each figure, the value indicated for each impulse means the amplitude of that impulse. For example, if the value of an impulse at n = 3 is indicated to be 2j, it means 2jd[n − 3]. The value of h[n] is zero for all n outside the range shown in the figure. h[n] (b) (c) H(ejw) = Σ h[n]e¯¡nw, n=-x h[n] il -6-5-4-3-2-1 0 1 2 3 4 5 6 1 1 1 h[n] -6-5-4-3-2-1 0 1 2 3 456 -6 -5 -4 -3 -2 -1 0 −1 1 1 1 1 1 2 3 4 5 6 n h[n] is real-valued. h[n] is real-valued. h[n] is real-valued.

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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