hw01-f06-sol - EECS 20N Structure and Interpretation of...

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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 π 6 z 2 = e i 5 π 6 = cos 5 π 6 + i sin 5 π 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 π 6 e i 5 π 6 = e = - 1 (c) (i) z 2 1 = ( e i π 6 ) 2 = e i π 3 (ii) z 3 1 = ( e i π 6 ) 3 = e i π 2 = i (iii) z 6 1 = ( e i π 6 ) 6 = e = - 1 (iv) z 6 2 = ( e i 5 π 6 ) 6 = e i 5 π = - 1 HW 1.2 (a) How do we evaluate i i withour a calculator? We know e i π 2 = i, i i = e - π 2 . From the graph of e x , we know e - π 2 is a positive value < 1 . In fact, i i = 0 . 2079 Rewriting, t R , f ( t ) = 0 t < 0 e - π 2 t t 0 ⇒ | f ( t ) | = 0 t < 0 e - π 2 t t 0 f ( t ) = 0 t < 0 0 t 0 (b) G ( ω ) = cos( ω 2 ) e i ( - ω 2 ) . | G ( ω ) | = | cos( ω 2 ) | . G ( ω ) = cos( ω/ 2) · (cos( - ω/ 2) + i sin( - ω/ 2)) = | G ( ω ) | e i G ( ω ) . When cos( ω/ 2) > 0 , G ( ω ) = - ω/ 2 . When cos( ω/ 2) < 0 , G ( ω ) = π - ω/ 2 . As can be seen from Figure 1, the plots repeat every 2 π increment. So, it is sufficient to sketch the graphs by only plotting an interval of length 2 π . HW 1.3 1. Using the given hint we write e i 2 θ = ( e ) 2 , so cos(2 θ ) + i sin(2 θ ) = cos( θ ) 2 - sin( θ ) 2 + i 2 cos( θ ) sin( θ ) 1
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0 1 - 1 1 (a) | f ( t ) | . 0
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