1 choosing any cuto frequencies which satisfy 21

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Unformatted text preview: � 1 1 = 1 − ej (3)�0 t − ej (−3)�0 t = bk ejk�0 t 2 2 k = 1− � ⎧ k=0 �1, 1 � bk − 2 , k = ±3 ⎧ � 0, otherwise � bk = ak H (jk�0 ) � H (jk�0 ) = bk ak , for ak ≤= 0. y (t) has only three non-zero components, therefore H (j� ) has a non-zero value at those three components, corresponding to A1 , A2 and A3 as follows: 1 b0 = = 2 = A2 . a0 1/2 1 sin(3 · ω /2) −j (3)�/2 b3 e H (j (3)�0 ) = =− ÷2 a3 2 j (3 · ω )2 3� j (9)ω 2 ej 2 =− 4 sin(3ω/2) j 9ω 2 9 =− (−j ) = ω 2 = A3 . 4(−1) 4 1 sin(−3 · ω /2) −j (−3)�/2 b−3 =− ÷2 e H (j (−3)�0 ) = a−3 2 j (−3 · ω )2 −3� j (9)ω 2 ej 2 =− 4 sin(−3ω/2) j 9ω 2 9 =− (j ) = ω 2 = A1 . 4(1) 4 H (j (0)�0 ) = H (j� ) also needs to eliminate the other frequency components of x(t) which do not exist in the output y (t). � H (jk�0 ) = 0, for k = 0, ±3 ≤ To meet the above conditions, the cutoff frequencies �1,2,3 must be chosen to pass the desired components and reject the undesired components. The following inequalities meet that 6 requirement: 0 < �1 < (1)�0 , (2)�0 < �2 < (3)�0 , (3)�0 < �3 < (4)�0 (2.1) Choosing any cut...
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