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:


(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 $\color{red}{e^{-j2 \pi n}}$?


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

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}

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.