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In Digital Filter Design by Parks and Burrus, p. 19.


The transfer function of an FIR filter is given by the $\mathcal Z$-transform of $h(n)$ as:

$$H(z)=\sum_{n=0}^{N-1}h(n)z^{-n}$$

(where $h$ is the filter)

The frequency response of a filter is defined as

$$H(\omega)=\sum_{n=0}^{N-1} h(n)e^{-j\omega n}$$

where $\omega$ is frequency in $\textrm{rad/sec}$.

Then the text proceeds to show that $H(\omega)$ is periodic with period $2\pi$:

\begin{align} H(\omega + 2 \pi)&= \sum_{n=0}^{N-1} h(n) e^{-j(\omega+2\pi)n}\\ &= \sum_{n=0}^{N-1} h(n) e^{- j \omega n} \color{red}{e^{-j2\pi n}}\\ &=H(\omega) \end{align} Could someone clarify how is that equal to $H(\omega)$ when there's the extra term $\color{red}{e^{-j2 \pi n}}$?

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$$\sum_{n=0}^{N-1}h(n)e^{-j\omega n}e^{-j2\pi n}=\sum_{n=0}^{N-1}h(n)e^{-j\omega n}\cdot1$$

Since \begin{align} e^{-j2\pi n}&=\cos(-2\pi n) + j\sin(-2\pi n)\\ &=\cos(2\pi n) - j\sin(2\pi n)\\ &=1-0\\ &=1 \end{align}

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  • $\begingroup$ And why is that? $\endgroup$ – mavavilj May 29 '16 at 21:17
  • $\begingroup$ @mavavilj please see update. $\endgroup$ – Gilles May 29 '16 at 21:21
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Use Euler's formula, which allows you to write the extra term as,

$$ e^{-j2\pi n}=\cos(2\pi n)-j\sin(2\pi n). $$

Since $n$ is an integer you can simplify this expression to,

$$ e^{-j2\pi n}=1. $$

Since multiplying by one does not change anything, then the expression with the extra term has to be the same as without it.

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