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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 1 SOLUTIONS HW 1.1 z 1 = 3 2 + 1 2 i = s ( 3 2 ) 2 + ( 1 2 ) 2 e i (arctan 1 3 ) = e i 6 z 2 = e i 5 6 = cos 5 6 + i sin 5 6 = 3 2 + 1 2 i (a) z 1 + z 2 = ( 3 2 + 1 2 i ) + ( 3 2 + 1 2 i ) = i (b) z 1 z 2 = e i 6 e i 5 6 = e i = 1 (c) (i) z 2 1 = ( e i 6 ) 2 = e i 3 (ii) z 3 1 = ( e i 6 ) 3 = e i 2 = i (iii) z 6 1 = ( e i 6 ) 6 = e i = 1 (iv) z 6 2 = ( e i 5 6 ) 6 = e i 5 = 1 HW 1.2 (a) How do we evaluate i i withour a calculator? We know e i 2 = i, i i = e 2 . From the graph of e x , we know e 2 is a positive value < 1 . In fact, i i = 0 . 2079 Rewriting, t R , f ( t ) = t < e 2 t t  f ( t )  = t < e 2 t t f ( t ) = t < t (b) G ( ) = cos( 2 ) e i ( 2 ) .  G ( )  =  cos( 2 )  . G ( ) = cos( / 2) (cos( / 2) + i sin( / 2)) =  G ( )  e i G ( ) . When cos( / 2) > , G ( ) = / 2 . When cos( / 2) < , G ( ) =  / 2 . As can be seen from Figure 1, the plots repeat every 2 increment. So, it is sufficient to sketch the graphs by only plotting an interval of length 2 ....
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This note was uploaded on 04/02/2008 for the course EE 20N taught by Professor Ayazifar during the Fall '08 term at University of California, Berkeley.
 Fall '08
 Ayazifar
 Electrical Engineering

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