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i have a symmetric impulse response: $h[n]=-h[M-n]$ , $0\leq n \leq M$

Frequency response is: $H(e^{j\omega})=je^{-j\omega N}R(e^{-j\omega})$

$R(e^{-j\omega})$ is a real-valued function. I want to derive the function $R(e^{-j\omega})$ for $h[n]$ when M is even.

So what i have till now is that: $$H(e^{j\omega})= \sum_{n=0}^M h[n]e^{-j\omega n}$$

With the symmetry condition: $$H(e^{j\omega})= e^{-j\omega \frac{M}{2}} (\sum_{k=0}^{\frac{M}{2}} a[k] \cos(\omega k))$$

But the function $R(e^{-j\omega})$ has to be expressed as a sum of weighted sin-terms. How can i get from the cos-terms to sin-terms ? Do i have to add something like +90° or so ?

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up vote 1 down vote accepted

You can use the identity $\cos x=\sin (x-\frac{\pi}{2})$ if you want to change the cosines to sines.

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