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solution0 - ECE 101 Linear Systems Fall 2009 Problem Set#0...

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ECE 101 – Linear Systems, Fall 2009 Problem Set #0 Solutions September 24, 2009 (send comments/questions to [email protected]) 1.48 Using Euler’s relation, we have: z 0 = r 0 e 0 = r 0 cos θ 0 + jr 0 sin θ 0 = x 0 + jy 0 Then z 1 through z 5 are: (a) z 1 = x 0 - jy 0 (b) z 2 = q x 2 0 + y 2 0 (c) z 3 = - x 0 - jy 0 = - z 0 (d) z 4 = - x 0 + jy 0 (e) z 5 = x 0 + jy 0 = z 0 (recall, e is periodic with period 2 π ) Plots: 1
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1.49 Once in polar form re , the magnitude is given by r and angle given by θ . (a) 2 e jπ/ 3 (b) 5 e (c) 5 2 e j 5 π/ 4 (d) 5 e jtan - 1 (4 / 3) = 5 e j 53 . 13 o (e) 8 e - (f) 4 2 e j 5 π/ 4 (g) 2 2 e - j 5 π/ 12 (h) e - j 2 π/ 3 (i) e jπ/ 6 Plot: 1.51 By Euler’s relation, we have (i) e = cos θ + j sin θ and (ii) e - = cos θ - j sin θ (a) Summing (i) + (ii) yields cos θ = 1 2 ( e + e - ). (b) Subtracting (i) - (ii) yields sin θ = 1 2 j ( e - e - ). 1.52 (c) z + z * = x + jy + x - jy = 2 x = 2 Re { z } (d) z - z * = x + jy - x + jy = 2 jy = 2 jIm { z } 2
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1.53 (a) ( e z ) * = ( e x + jy ) * = ( e x e jy ) * = e x e - jy = e x - jy = e z * (c) | z | = re = | r | = re - = | z * | In words, taking the conjugate means flipping across the real axis; this negates the angle, but does not affect the magnitude.
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