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HW02s09_soln - ECE-2025 Homework#2 Solutions Problem 2.1(a...

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ECE-2025 Homework #2 Solutions Spring-2009 Problem 2.1: (a) ) 6 . 0 999 cos( 5 ) 2 . 0 999 cos( 3 ) ( 1 π π + + = t t t x . First define t j j t j j j t j j t j j e Ae e e e e e e e t z 999 999 6 . 0 2 . 0 999 6 . 0 999 2 . 0 1 ) 5 3 ( 5 3 ) ( φ π π π π = + = + = , and )] ( Re[ ) ( 1 1 t z t x = , then we can get the amplitude and phase with a calculator, or get the amplitude via the following mathematical operations. The amplitude is 1192 . 3 )] 2 . 0 6 . 0 cos( 30 25 9 [ ] ) 6 . 0 sin 5 2 . 0 sin 3 ( ) 6 . 0 cos 5 2 . 0 cos 3 [( 2 / 1 2 / 1 2 2 = + + + = + + + = π π π π π π A ; and phase π π π π π φ 4088 . 0 )] 6 . 0 cos 5 2 . 0 cos 3 /( ) 6 . 0 sin 5 2 . 0 sin 3 [( tan 1 = + + = . Finally, we get ) 4088 . 0 999 cos( 1192 . 3 ) 999 cos( ) ( 1 π φ + = + = t t A t x . (b) ) 8 / 3 cos( 8 ) 8 / 9 cos( 8 ) 8 / 5 cos( 7 ) 8 / cos( 7 ) ( 2 π π π π π π π π + + + + = t t t t t x ; First define t j j t j j t i j j j t j j j e e e e e e e e e e e t z π π π π π π π π π π π 8 / 5 8 / 3 4 / 4 / 8 / 3 8 / 5 8 / 4 2 7 2 7 ) ( 7 ] [ 7 ) ( = = + = + = and t j j t i j j j t j j j e e e e e e e e e t z π π π π π π π π π 8 / 5 4 / 4 / 8 / 5 8 / 3 8 / 9 5 2 8 ) ( 8 ] [ 8 ) ( = + = + = . Then define t j j t j j e e A e e t z t z t z π φ π π 2 2 8 / 5 5 4 2 2 ) ( ) ( ) ( = = + = Therefore, we have ) 8 / 5 cos( 2 )] ( Re[ ) ( 2 2 π π + = = t t z t x .
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