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Q6) a)Match the signals below with the candidate sketches below (i.e., you are supposed to choose which graph below corresponds to the signal below). Note: u(t) is the unit step function; i.e., u(t) = 1 for t>=0s and u(t) = 0 for t<0s. X₁(t) = u(t+1)-u(t-3) X₂(t) = u(t-3)-u(t-4) X3(t) = u(t+1)-u(t-1) d(t) = x₁(-t) e(t) = 2 x₁(t-1.5) f(t) = x₁(2t + 1.5) g(t) = x4(2t) h(t) = x3(t/2) i(t) = 2 x3(t) j(t) = -0.5 x₂(t) k(t) = x₁(1.5 t + 2.5) y(t) = 2 x₁(-t/2+1.5) 432 1 0 -1 4321OT 0 -1 -6-4-20 2 4 6 (1) 4321OT 0 -1 -6-4-20 246 (5) n -6-4-20 2 4 6 (9) 43210 3 -1 4321OT -1 43210 -6-4-2 0 2 4 6 (2) -1 -6-4-20246 (6) 2 -6-4-20246 (10) 4321OT 0 -6-4-2 0 2 4 6 (3) 4321OT 43210 -6-4-2 0 2 4 6 (7) -6-4-20 246 (11) 4321OT 0 -1 -6-4-20 2 4 6 4321OT 43210 -6-4-20 2 4 6 (8) 3 2 0 -6-4-20 246 (12)

Q6) a)Match the signals below with the candidate sketches below (i.e., you are supposed to choose which graph below corresponds to the signal below). Note: u(t) is the unit step function; i.e., u(t) = 1 for t>=0s and u(t) = 0 for t<0s. X₁(t) = u(t+1)-u(t-3) X₂(t) = u(t-3)-u(t-4) X3(t) = u(t+1)-u(t-1) d(t) = x₁(-t) e(t) = 2 x₁(t-1.5) f(t) = x₁(2t + 1.5) g(t) = x4(2t) h(t) = x3(t/2) i(t) = 2 x3(t) j(t) = -0.5 x₂(t) k(t) = x₁(1.5 t + 2.5) y(t) = 2 x₁(-t/2+1.5) 432 1 0 -1 4321OT 0 -1 -6-4-20 2 4 6 (1) 4321OT 0 -1 -6-4-20 246 (5) n -6-4-20 2 4 6 (9) 43210 3 -1 4321OT -1 43210 -6-4-2 0 2 4 6 (2) -1 -6-4-20246 (6) 2 -6-4-20246 (10) 4321OT 0 -6-4-2 0 2 4 6 (3) 4321OT 43210 -6-4-2 0 2 4 6 (7) -6-4-20 246 (11) 4321OT 0 -1 -6-4-20 2 4 6 4321OT 43210 -6-4-20 2 4 6 (8) 3 2 0 -6-4-20 246 (12)

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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Q6) a)Match the signals below with the candidate sketches below (i.e., you are supposed to choose which graph below corresponds to the signal below). Note: u(t) is the unit step function; i.e., u(t) = 1 for t>=0s and u(t) = 0 for t<0s. X₁(t) = u(t+1)-u(t-3) X₂(t) = u(t-3)-u(t-4) X3(t) = u(t+1)-u(t-1) d(t) = x₁(-t) e(t) = 2 x₁(t-1.5) f(t) = x₁(2t + 1.5) g(t) = x4(2t) h(t) = x3(t/2) i(t) = 2 x3(t) j(t) = -0.5 x₂(t) k(t) = x₁(1.5 t + 2.5) y(t) = 2 x₁(-t/2+1.5) 432 1 0 -1 4321OT 0 -1 -6-4-20 2 4 6 (1) 4321OT 0 -1 -6-4-20 246 (5) n -6-4-20 2 4 6 (9) 43210 3 -1 4321OT -1 43210 -6-4-20246 (2) -1 -6-4-20246 (6) 2 -6-4-20246 (10) 4321OT 0 4321OT -6-4-2 0 2 4 6 (3) -6-4-2 0 2 4 6 (7) 4321O -6-4-20246 (11) 4321OT 0 -1 -6-4-20 2 4 6 (4) 4321OT -6-4-20 2 4 6 (8) 43210 3 2 0 -6-4-20 246 (12)
Transcribed Image Text:Q6) a)Match the signals below with the candidate sketches below (i.e., you are supposed to choose which graph below corresponds to the signal below). Note: u(t) is the unit step function; i.e., u(t) = 1 for t>=0s and u(t) = 0 for t<0s. X₁(t) = u(t+1)-u(t-3) X₂(t) = u(t-3)-u(t-4) X3(t) = u(t+1)-u(t-1) d(t) = x₁(-t) e(t) = 2 x₁(t-1.5) f(t) = x₁(2t + 1.5) g(t) = x4(2t) h(t) = x3(t/2) i(t) = 2 x3(t) j(t) = -0.5 x₂(t) k(t) = x₁(1.5 t + 2.5) y(t) = 2 x₁(-t/2+1.5) 432 1 0 -1 4321OT 0 -1 -6-4-20 2 4 6 (1) 4321OT 0 -1 -6-4-20 246 (5) n -6-4-20 2 4 6 (9) 43210 3 -1 4321OT -1 43210 -6-4-20246 (2) -1 -6-4-20246 (6) 2 -6-4-20246 (10) 4321OT 0 4321OT -6-4-2 0 2 4 6 (3) -6-4-2 0 2 4 6 (7) 4321O -6-4-20246 (11) 4321OT 0 -1 -6-4-20 2 4 6 (4) 4321OT -6-4-20 2 4 6 (8) 43210 3 2 0 -6-4-20 246 (12)
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