ee20-hw01-s08-sol

ee20-hw01-s08-sol - EECS 20N: Structure and Interpretation...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences UNIVERSITY OF CALIFORNIA BERKELEY Problem Set 1 SOLUTIONS HW 1.1 (a) z 1 = 1 + i 3 = 2( 1 2 + i 3 2 ) = 2 exp( i 3 ) z 2 = exp( i 2 3 ) = cos( 2 3 ) + i sin( 2 3 ) = 1 2 + i 3 2 See Fig. 1 2 1 1/2 z1 z1* 1/z1* 1/z1 z2 = 1/z2* z2* = 1/z2 Real Axis Imaginary Axis-1/2-1-2 Figure 1: Problem 1.1a (b) z 1 + 2 z 2 = 1 1 + i ( 3 + 3) = i 2 3 z 2 1 + z 2 = (2 exp( i 3 )) 2 + exp( i 2 3 ) = 4 exp( i 2 3 ) + exp( i 2 3 ) = 5 exp( i 2 3 ) 1 2 z 1 + z * 2 = 1 2 (1 + i 3) + ( 1 2 + i 3 2 ) * = 1 2 + i 3 2 1 2 i 3 2 = 0 (c) | z 1 | = | 2 exp( i 3 ) | = 2 | z 2 | = | exp( i 2 3 ) | = 1 | z 1 z 2 | = | z 1 || z 2 | = 2 | z 1 z * 2 | = | z 1 || z * 2 | = | z 1 || z 2 | = 2 1 | z 1 /z 2 | = | z 1 | / | z 2 | = 2 | z 2 /z 1 | = | z 2 || z 1 | = 1 / 2 (d) (i) z 2 1 = (2 exp( i 3 )) 2 = 4 exp( i 2 3 ) (ii) z 3 1 = (2 exp( i 3 )) 3 = 8 exp( i 3 3 ) = 8 (iii) z 6 1 = (2 exp( i 3 )) 6 = 64 exp( i 6 3 ) = 64 exp( i 2 ) = 64 (iv) z 4 2 = (exp( i 2 3 )) 4 = exp( i 8 3 ) = exp( i (2 + 2 3 )) = exp( i 2 3 ) = z 2 (e) z 1 / 4 2 = exp( i (2 n + 2 3 )) 1 / 4 n { , 1 , 2 , 3 } z 1 / 4 2 = exp( i ( 2 3 )) 1 / 4 , exp( i (2 + 2 3 )) 1 / 4 , exp( i (4 + 2 3 )) 1 / 4 , exp( i (6 + 2 3 )) 1 / 4 z 1 / 4 2 = exp( i 6 ) , exp( i ( / 2 + 6 )) , exp( i ( + 6 )) , exp( i (3 / 2 + 6 )) z 1 / 4 2 = exp( i 6 ) , exp( i ( 2 3 )) , exp( i ( 7 6 )) , exp( i ( 10 6 )) See Figure 2. The circles are the roots and the cross is z 2 ....
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ee20-hw01-s08-sol - EECS 20N: Structure and Interpretation...

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