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lab01_sol_su09 - ECE220 Problem Lab#1 Solutions I...

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ECE220 Problem Lab #1 Solutions I. Operations on functions Problem 1.16 (pg 40) f (0) = 1 - cos (0) 2(0) = 1 - 1 2(0) = 0 0 (1) Since we cannot define what happens when 0 is divided by 0, we apply LHopitals Rule: f (0) = lim t 0 r ( t ) s ( t ) = r 0 (0) s 0 (0) = sin (0) 2 = 0 2 = 0 (2) Problem 1.35 (pg 46) 1. ) -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Plot of f(t) = sin(2 π Ft), F=1 y(t) t 1
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2. ) -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 Plot of f(t) = sin(2 π Ft), F=-1 y(t) t 3. ) f(t) = sin(2 πFt ) = 1, when the argument to the sin function is π 2 or any multiple of 2 π away from π 2 2 πFt = π 2 + 2 πk (3) t = π 2 + 2 πk 2 πF = 1 + 4 k 4 F (4) where k is any integer 4. ) f(t) = sin(2 πFt ) = 0, when the argument to the sin function is 0 or any multiple of π away from 0 2 πFt = πk (5) t = πk 2 πF = k 2 F (6) where k is any integer 5. ) df ( t ) dt = d sin(2 πFt ) dt = 2 πF cos(2 πFt ) (7) 2
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6. ) Z f ( t ) dt = Z sin(2 πFt ) dt = - cos(2 πFt ) 2 πF (8) 7. ) Taylor series expansion of sin(2 πFt ) is sin(2 πFt ) = X k =0 ( - 1) k (2 πFt ) 2 k +1 (2 k + 1)! (9) f (0) = sin(2 πF 0) = X k =0 ( - 1) k (2 πF 0) 2 k +1 (2 k + 1)! = 0 (10) Problem 1.38 (pg 47) Point Quadrant Angle (1,1) I arctan(1/1) = 45 o (1,2) I arctan(2/1) = 63 . 4 o (1,-1) IV - 45 o + 360 o = 315 o (1,-2) IV - 63 . 4
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