PS1Solns

# PS1Solns - UCSB Fall 2007 ECE 130A Solutions to Problem Set...

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UCSB Fall 2007 ECE 130A: Solutions to Problem Set 1 Problem 1 1 . 48 ( b ) z 2 = x + jy = r 0 = x = r 0 , y = 0 . 1 . 48 ( c ) z 3 = x + jy = r 0 e j ( θ 0 + π ) = r 0 e 0 e x + jy = r 0 e 0 ( cos ( π ) + jsin ( π )) = - r 0 e 0 = - r 0 ( cos ( θ 0 ) + jsin ( θ 0 )) x = - r 0 cosθ 0 , y = - r 0 sinθ 0 1 . 48 ( d ) z 4 = x + jy = r 0 e j ( - θ 0 + π ) = r 0 e - 0 e x + jy = r 0 e - 0 ( - 1) = - r 0 ( cos ( θ 0 ) - jsin ( θ 0 )) x = - r 0 cosθ 0 , y = r 0 sinθ 0 Figure 1: r 0 = 2 , θ 0 = π 4 Figure 2: r 0 = 2 , θ 0 = π 2 Problem 2 1 . 49 ( b ) z 1 = - 5 = 5 e r 0 = 5 , θ 0 = π. 1 . 49 ( d ) z 2 = 3 + j 4; r 0 = p 3 2 + 4 2 = 5 θ 0 = arctan( 4 3 ) = 0 . 9273 radians = 53 . 13 degrees

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z 2 = 5 e j 0 . 9273 1 . 49 ( f ) z 3 = (1 + j ) 5 , 1 + j = ( p 1 2 + 1 2 ) e j (arctan( 1 1 )) = 2 e j π 4 z 3 = ( 2 e j π 4 ) 5 = 4 2 e j 5 π 4 r 0 = 4 2 , θ 0 = 5 π 4 = 225 degrees Figure 3: Complex numbers in polar form (Problem 2) Problem 3 1 . 3 ( a ) x 1 ( t ) = e - 2 t u ( t ) E = R -∞ ( e - 2 t u ( t )) 2 dt = R 0 e - 4 t dt = 1 4 P = lim T →∞ E 2 T = 0, since E is finite. 1 . 3 ( b ) x 2 ( t ) = e j (2 t + π 4 ) | x 2 ( t ) | = 1. Hence, E = R -∞ | x 2 ( t ) | 2 dt = . Also, x 2 ( t ) is periodic. It must have infinite en- ergy and finite average power.
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