hw1solns.220.06

# hw1solns.220.06 - 1 ECE 220 HOMEWORK 1 SOLUTION I P ROBLEM...

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1 ECE 220 HOMEWORK 1 SOLUTION I. P ROBLEM 1 (a) See Fig1. Re(z) Im(z) z=0+j2 z=-1+j z=-j Fig. 1. Plots for Problem 1 (b) z = 0 + j 2 = 2 π/ 2 = 2 - 3 π/ 2 z = ( - 1 , 1) = 2 3 π/ 4 = 2 - 5 π/ 4 z = (0 , - 1) = 1 3 π/ 2 = 1 - π/ 2 . (c) z = 0 + j 2 z = 2 π/ 4 or 2 - 3 π/ 4 z = ( - 1 , 1) z = 2 1 / 4 3 π/ 8 or 2 1 / 4 - 5 π/ 8 z = (0 , - 1) z = 1 3 π/ 4 or 1 - π/ 4 . II. P ROBLEM 2 (a) See Fig2. (b) z = 2 e j 3 π 4 = 2( - 2 / 2 + j 2 / 2) = - 1 + j z = 3 e - j π 2 = 3(0 - j ) = - 3 j . III. P ROBLEM 3 (a) j 3 = j 2 j = - j = (0 , 1) (b) e j ( π +2 πm ) = cos( π + 2 πm ) + j sin( π + 2 πm ) = cos π + j sin π = - 1 = ( - 1 , 0) for all m . (c) j 2 n = ( j 2 ) n = ( - 1) n = ( - 1 , 0) if n is odd ; (1 , 0) if n is even . (d) j 1 / 2 = ( e j π 2 ) 1 / 2 = e j π 4 or e - j 3 π 4 = ( 2 / 2 , 2 / 2) or ( - 2 / 2 , - 2 / 2) .

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2 Re(z) Im(z) z = 2 e j (3 π/ 4) z = 3 e - j ( π/ 2) Fig. 2. Plots for Problem 2 IV. P ROBLEM 4 Let z 0 = e j 2 π N . Since z 0 6 = 0 for any N , this problem is equivalent to prove z N - 1 0 z 0 = 1 i.e. z N 0 = 1 . Clearly, z N 0 = ( e j 2 π N ) N = e j 2 π N N = e j 2 π = 1 , which ends the proof.
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• Spring '05
• JOHNSON
• 2m, 2k, Euler, 3j, 0 - J

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