Homework2_Solutions.pdf - Homework 2 Solutions ECE 333 Professor Amir Arbabi Problem 1 2 f(x 1.5 1 0.5 \u22122 2 4 x(meters 6 a Direction of travel and

# Homework2_Solutions.pdf - Homework 2 Solutions ECE 333...

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Solutions Homework 2 ECE 333 Professor Amir Arbabi Problem 1 - 2 2 4 6 0 . 5 1 1 . 5 2 x (meters) f ( x ) a. Direction of travel and velocity: As t increases, y ( x, t ) moves in the + x direction. For every increment of t (1 second), y ( x, t ) moves 1 meter along x -axis, thus the velocity is 1 m s b. - 2 2 4 6 0 . 5 1 1 . 5 2 x (meters) s 1 s 2 1
Solutions Homework 2 ECE 333 Professor Amir Arbabi c. h ( - 2) = f (1 - ( - 2)) = f (3) = 0 h ( - 1) = f (1 - ( - 1)) = f (2) = 0 h (0) = f (1 - 0) = f (1) = 2 h (1) = f (1 - 1) = f (0) = 1 h (2) = f (1 - 2) = f ( - 1) = 0 h (3) = f (1 - 3) = f ( - 2) = 0 - 2 - 1 1 2 3 0 . 5 1 1 . 5 2 t (sec) h ( t ) 2
Solutions Homework 2 ECE 333 Professor Amir Arbabi Problem 2 a. f 1 ( t ) = - 5 sin( π 4 - ωt ) f 1 ( t ) = 5 sin( ωt - π 4 ) = 5 cos( ωt - π 4 - π 2 ) = 5 cos( ωt - 3 π 4 ) = < [5 e j ( ωt - 3 π 4 ) ] = < [5 e j - 3 π 4 e jωt ] thus, ˜ F 1 = 5 e j - 3 π 4 b. f 2 ( t ) = cos( ωt + π 4 ) - 2 sin( ωt + π 3 ) f 2 ( t ) = cos( ωt + π 4 ) - 2 sin( ωt + π 3 ) = cos( ωt + π 4 ) - 2 cos( ωt - π 6 ) = < [ e j ( ωt + π 4 ) - 2 e j ( ωt - π 6 ) ] = < [( e j π 4 - 2 e - j π 6 ) e jωt ] thus, ˜ F 2 = e j π 4 - 2 e - j π 6 = cos( π 4 ) + j sin( π 4 ) - 2[cos( π 6 ) - j sin( π 6 )] = . 71 + . 71 j - 2[ . 87 - . 5 j ] = - 1 . 03 + 1 . 71 j 3
Solutions Homework 2 ECE 333 Professor Amir Arbabi c.

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