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Consider the sequence in figure below, Assuming that fs=100 Hz, compute the amplitude spectrum, phase spectrum, and power spectrum. X(2)=-2 Y(3)=-2-j2 x(n) 4 3+ 2+ 1+ 0 2 1 TNT Similarly, for k= 3, f = Solution:- Since N=4, and using the DFT shown in example 4.1. we find the DFT coefficients to be X(0)=10 X(1)=-2+j2 for k= 1, f = 1 x 100/4 = 25 Hz, Solution: Since N= 4, and using the DFT shown in Example 4.1, we find the DFT coefficients to be X(0) 10 X(1) = -2+12 X(2) = -2 X(3)=-2-12 The amplitude spectrum, phase spectrum, and power density spectrum are computed as follows: for k= 0, f = kfs/N = 0 x 100/4 = 0 Hz, A = 1X (0)| = 2.5, %0 = tan-¹ for k= 2, f= 2 x 100/4= 50 Hz, A₁ = X(1)| = 0.7071, ₁ = tan-1 A2 = X(2)| = 0.5, 02 = tan 3 3 x 100/4 = 75 Hz, A3 = 1X (3)| = 0.7071, 43 Imag[X(0)] Real(X(0)] 5 an -1/Imag[X(2)] eal X(2) n -1 (Imag{X (1)]) Real | = 0, Po = AXIO) = 6.25 (Imag[X(3)]) Real X(3) = 1350, P₁= X(1)² = 0.5 = 0.5000 = 180°, P₂=X(2) ²0 0.2500 = -1350 P3=2X(3) 1² = 0.5000.

Consider the sequence in figure below, Assuming that fs=100 Hz, compute the amplitude spectrum, phase spectrum, and power spectrum. X(2)=-2 Y(3)=-2-j2 x(n) 4 3+ 2+ 1+ 0 2 1 TNT Similarly, for k= 3, f = Solution:- Since N=4, and using the DFT shown in example 4.1. we find the DFT coefficients to be X(0)=10 X(1)=-2+j2 for k= 1, f = 1 x 100/4 = 25 Hz, Solution: Since N= 4, and using the DFT shown in Example 4.1, we find the DFT coefficients to be X(0) 10 X(1) = -2+12 X(2) = -2 X(3)=-2-12 The amplitude spectrum, phase spectrum, and power density spectrum are computed as follows: for k= 0, f = kfs/N = 0 x 100/4 = 0 Hz, A = 1X (0)| = 2.5, %0 = tan-¹ for k= 2, f= 2 x 100/4= 50 Hz, A₁ = X(1)| = 0.7071, ₁ = tan-1 A2 = X(2)| = 0.5, 02 = tan 3 3 x 100/4 = 75 Hz, A3 = 1X (3)| = 0.7071, 43 Imag[X(0)] Real(X(0)] 5 an -1/Imag[X(2)] eal X(2) n -1 (Imag{X (1)]) Real | = 0, Po = AXIO) = 6.25 (Imag[X(3)]) Real X(3) = 1350, P₁= X(1)² = 0.5 = 0.5000 = 180°, P₂=X(2) ²0 0.2500 = -1350 P3=2X(3) 1² = 0.5000.

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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Consider the sequence in figure below, Assuming that fs=100 Hz, compute the amplitude spectrum, phase spectrum, and power spectrum. x(n) 3 2 X(0) = 10 X(1) = -2+12 X(2) = -2 X(3)=-2-12 1 0 T = NT Solution:- Since N=4, and using the DFT shown in example 4.1. we find the DFT coefficients to be X(0)=10 X(1)=-2+j2 X(2)=-2 Y(3)=-2-j2 A₂ = X(2)| = 0.5, 02 = tan 1 Solution: Since N 4, and using the DFT shown in Example 4.1, we find the DFT coefficients to be Similarly, for k= 3, f= 3 x 100/4 = 75 Hz, 4 A3 = X(3)| = 0.7071, 43 H The amplitude spectrum, phase spectrum, and power density spectrum are computed as follows: for k= 0, f = kfs/N = 0 x 100/4 = 0 Hz, A = 1X (0)| = 2.5, %= tan-1 Imag[X(0)] Real([X(0) an for k= 1, f 1 x 100/4 = 25 Hz, A₁ = X(1)| = 0.7071, ₁ = tan-1 (Imag(X(1)]) = 135⁰, P₁= X(1)² = 0.5000 Real X(1)]) for k= 2, f = 2 x 100/4 = 50 Hz, -1 5 Imag[X(2)]) eal X(2) n | = 0, Po = AX(0)2 = 6.25 Imag[X(3)] Real X(3)] = 180°, P₂ = x(2)² = 0 0.2500 =-1350 P3 = X(3)1²=0.5000. "dy
Transcribed Image Text:Consider the sequence in figure below, Assuming that fs=100 Hz, compute the amplitude spectrum, phase spectrum, and power spectrum. x(n) 3 2 X(0) = 10 X(1) = -2+12 X(2) = -2 X(3)=-2-12 1 0 T = NT Solution:- Since N=4, and using the DFT shown in example 4.1. we find the DFT coefficients to be X(0)=10 X(1)=-2+j2 X(2)=-2 Y(3)=-2-j2 A₂ = X(2)| = 0.5, 02 = tan 1 Solution: Since N 4, and using the DFT shown in Example 4.1, we find the DFT coefficients to be Similarly, for k= 3, f= 3 x 100/4 = 75 Hz, 4 A3 = X(3)| = 0.7071, 43 H The amplitude spectrum, phase spectrum, and power density spectrum are computed as follows: for k= 0, f = kfs/N = 0 x 100/4 = 0 Hz, A = 1X (0)| = 2.5, %= tan-1 Imag[X(0)] Real([X(0) an for k= 1, f 1 x 100/4 = 25 Hz, A₁ = X(1)| = 0.7071, ₁ = tan-1 (Imag(X(1)]) = 135⁰, P₁= X(1)² = 0.5000 Real X(1)]) for k= 2, f = 2 x 100/4 = 50 Hz, -1 5 Imag[X(2)]) eal X(2) n | = 0, Po = AX(0)2 = 6.25 Imag[X(3)] Real X(3)] = 180°, P₂ = x(2)² = 0 0.2500 =-1350 P3 = X(3)1²=0.5000. "dy
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