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**Unformatted text preview: **k =−� The required condition, then,would be:
�
� �
� �
� |ak |2 , where R=0.9 (⇒) k =−� Then, let’s calculate the Fourier series coeﬃcients of the output, bk :
⎪
⎪
ak , |k�0 | ← W
ak , |k | ← W/�0
� bk = ak H (jk�0 ) � bk =
=
0, |k�0 | > W
0, |k | > W/�0
1 And ﬁnally, we plug the expressions of ak and bk in the required condition and simplify. By matching the the expression of x(t) with the synthesis equation, we can conclude that �0 = � and ak = �|k| 4
�
� Px = 2 |ak | = k =−� Py = 2 |bk | = k =−� = N
� k =−N |k | 2 |� | = 2 k =−� N
� |ak |2 �
� �2k − 1 (� � is real) k =0 2
− 1 (� 0 < |�| < 1)
1 − �2 = �
� �
� , where N is the largest integer, such that N ← W/�0 k =−N |�|k||2 = 2 N
� �2k − 1 (� � is real) k =0 2 N +1 1 − (� )
=2
1 − �2 −1 � M
� 1 − � M +1
�
, for any complex � ≤= 0
�n =
1−�
n=0 � Plugging in (⇒):
N +1 1...

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