ansT4 - = n t jn n e c t y , we have = , ( ) n j c n n 2 8...

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EE3118 Linear Systems and Signal Analysis KST SemA 2008/09 Answer of Tutorial 4 Q1. (a) () [] () ( ) ω 5 sinc 100 10 sinc 200 + = t x F (b) () [] () ω 5 . 2 sinc 100 5 sinc 400 2 2 = t x F Q2. () ω j x F 2 sgn = Q3. (a) (b) () () () π 5 . 0 5 . 0 sin ) 2 1 ( 3 . 0 5 . 0 5 . 0 cos 6 . 0 3 . 0 2 2 + + + = t t t y Q4. The relationship: 0 0 2 0 2 3 0 3 6 5 v dt dv dt v d dt v d v i + + + = , hence () () () () () 1 6 5 1 2 3 + + + = ω j j j H V V i o (a) () + + = 2075 . 0 6 2 cos 515 . 0 π t t v o (The complete output response is plotted in the figure. The result obtained is the steady state output response) (b) () = + = 1 0 3 cos n n t n a a t π where 3 1 0 = a and + + + = 27 2 9 5 1 3 2 cos 3 3 2 sin 2 3 3 3 2 2 2 2 2 π n n j n n n n n n a n Q5. (a) () t j t j e j e j t π 2 2 4 8 1 4 8 1 + + = (b) () −∞ =
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Unformatted text preview: = n t jn n e c t y , we have = , ( ) n j c n n 2 8 1 1 + − − = ′ Q6. (a) ( ) 1 = H and ( ) 2 2 1 2 tan − − = ∠ − a a H (b) ( ) ( ) ( ) t t u ae t y t a δ − = − 2 (c) ( ) ( ) t u e b a a e b a b a t y t a t b − − − + = − − 2 0.8 0.8 π ( ) H-0.8 π 0.8 π ( ) H ∠ ω ω...
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