Plot the amplitude and phase spectrum (single and double sided) for the following signal: 1. f(t)=5+3 sin(2* 10³t + 30°) - 4 cos(4л * 10³t - 20°) 2. f(t)=2+ 10 sin(2* 105t -20°) - 14 cos(6л * 105t + 30°) + 7 sin(4π 105t - 50°) 3. f(t)=1+4 cos(10000t + 30°) Fit) = 1 + 4 Cos(10000++38) F(t) = 1 Cos()) + 4 Cos (1 +30°) fit) W2 Wi 2x = 0 Wz =100007-02x+2=10000/-07₂ = 500 Single Amp. 1 4 Phas 30 5310 √ble 12=2 Amp. ½ 22 Phase 5000
Plot the amplitude and phase spectrum (single and double sided) for the following signal: 1. f(t)=5+3 sin(2* 10³t + 30°) - 4 cos(4л * 10³t - 20°) 2. f(t)=2+ 10 sin(2* 105t -20°) - 14 cos(6л * 105t + 30°) + 7 sin(4π 105t - 50°) 3. f(t)=1+4 cos(10000t + 30°) Fit) = 1 + 4 Cos(10000++38) F(t) = 1 Cos()) + 4 Cos (1 +30°) fit) W2 Wi 2x = 0 Wz =100007-02x+2=10000/-07₂ = 500 Single Amp. 1 4 Phas 30 5310 √ble 12=2 Amp. ½ 22 Phase 5000
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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Transcribed Image Text:Plot the amplitude and phase spectrum (single and double sided) for the following
signal:
1. f(t)=5+3 sin(2* 10³t + 30°) - 4 cos(4л * 10³t - 20°)
2. f(t)=2+ 10 sin(2* 105t -20°) - 14 cos(6л * 105t + 30°) +
7 sin(4π 105t - 50°)
3. f(t)=1+4 cos(10000t + 30°)

Transcribed Image Text:Fit) = 1 + 4 Cos(10000++38)
F(t) = 1 Cos()) + 4 Cos (1 +30°)
fit)
W2
Wi
2x = 0
Wz
=100007-02x+2=10000/-07₂ = 500
Single Amp.
1
4
Phas
30
5310
√ble
12=2 Amp.
½ 22
Phase
5000
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