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**Unformatted text preview: **1 ,
⎧
⎧ 2j
⎧
⎧
�2,
� ak = 1
⎧ 2j ,
⎧
⎧
⎧
⎧−1,
⎧
⎧
⎧
�
0, k
k
k
k
k
k = −2
= −1
=0
=1
=2
= 3, 4, 5 And ﬁnally, we ﬁnd the Fourier series coeﬃcients, bk , of the output y [n]:
� bk = ak H (e
= ak · jk�0 1 + 2 e−jk�04
1 + 2 e−jk 4 4
) = ak ·
= ak ·
�
1
1 − 4 e−jk�0
1 − 1 e−jk 4
4 1 + 2 e−jk�
�
1 − 1 e−jk 4
4 We have only few non-zero coeﬃcients, so we can go ahead and evaluate them. In doing so,
it is sometimes useful while computing the value of the complex exponential, to visualize the
the complex vector ejkω going around the unit circle. As the integer k increases by one, the
complex vector’s angle increases by an angle of β .
�
⎩
1 + 2 e−j (0)�
1+2
6
b0 = a0 H ej (0)�0 = (2)
= (2)
1=3
1 −j (0) �
4
1− 4
1− 4e
4
=8
��
��
� j (1)�0 ⎩
1
1 + 2(−1)
1 1 + 2 e−j (1)�
=
b1 = a 1 H e
=
1
1
1 −j (1) �
4
2j 1 − 4 e
2j 1 − 1 ( �2 − j �2 )
4
= 0.1247 + j 0.5806
��
��
� j...

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