Homework1_Solutions (1).pdf - Homework 1 Solutions ECE 333 Professor Amir Arbabi Problem 1 a(1 4j(\u22122 3j = \u22122 3j \u2212 8j \u2212 12 = \u221214 \u2212 5j To

# Homework1_Solutions (1).pdf - Homework 1 Solutions ECE 333...

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Solutions Homework 1 ECE 333 Professor Amir Arbabi Problem 1 a. (1 + 4 j )( - 2 + 3 j ) = - 2 + 3 j - 8 j - 12 = - 14 - 5 j To convert to Polar Form use equation (1.41) in text: | z | = p ( - 14) 2 + ( - 5) 2 = 14 . 87 Before calculating phase, note from rectangular form that the point lies in the 3rd quadrant of the complex plane. arctan will return a value between - π 2 and π 2 . Therefor any point that lies in quadrants II or III needs to be shifted by π . So our modified equation for phase is θ = π + arctan y x , and thus θ = π + arctan - 5 - 14 = π + . 34 = 3 . 48 Rectangular: - 14 - 5 j Polar: | z | = 14 . 87, θ = 3 . 48 b. 2 + j 1 - 2 j = 2 + j 1 - 2 j · 1 + 2 j 1 + 2 j = 2 + j + 4 j - 2 1 + 4 = 5 j 5 = j | z | = p (0) 2 + (1) 2 = 1 1
Solutions Homework 1 ECE 333 Professor Amir Arbabi θ = arctan 1 0 = π 2 Rectangular: j Polar: | z | = 1, θ = π 2 c. e 2 - j = e 2 · e - j = 7 . 39 e - j using equations (1.40) from text we can solve for rectangular form : x = | z | cos θ = 7 . 39 cos - 1 = 3 . 99 y = | z | sin θ = 7 . 39 sin - 1 = - 6 . 22 Rectangular: 3 . 99 - 6 . 22 j Polar: | z | = 7 . 39, θ = - 1 d. (4 - 3 j ) - 1 = 1 4 - 3 j · 4 + 3 j 4 + 3 j = 4 + 3 j 25 = . 16 + . 12 j | z | = p ( . 16) 2 + ( . 12) 2 = . 2 2
Solutions Homework 1 ECE 333

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