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The length of an FIR filter is 7. If this filter has a linear phase, show that $ \sum_{n=0}^{N-1} h(n) \sin (\alpha-n) \omega=0 $ is satisfied.
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Solution:

The length of the filter is 7. Therefore, for the linear phase,

$ \alpha=\frac{N-1}{2}=\frac{7-1}{2}=3\\ $

$ \begin{aligned}\\ \sum_{n=0}^{N-1} h(n) \sin (\alpha-n) \omega= & \sum_{n=0}^6 h(n) \sin (3-n) \omega \\\\ = & h(0) \sin 3 \omega+h(1) \sin 2 \omega+h(2) \sin \omega+h(3) \sin 0+h(4) \sin (-\omega) \\\\ & +h(5) \sin (-2 \omega)+h(6) \sin (-3 \omega) \\\\ = & 0\\ \end{aligned}\\ $

Hence, the equation is satisfied.

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