**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...

View
Full
Document