Frequency Response

The filter's frequency response is acting as a complex multiply, which means a gain scaling and phase shift on the input signal.
The complex filter gain in polar form is given by

$\displaystyle H(e^{j\omega T}) = G(\omega)e^{j\Theta(\omega)}$
where the amplitude and phase response are given by

$\displaystyle G(\omega) \triangleq \vert H(e^{j\omega T})\vert$ and $\displaystyle \quad \Theta(\omega) \triangleq \angle H(e^{j\omega T}).$
To solve, we may ``balance the exponents'' to obtain

$\displaystyle H(e^{j\omega T})$ $\displaystyle =$ $\displaystyle (1+e^{-j\omega T})$

$\displaystyle =$ $\displaystyle (e^{j\omega T/2} + e^{-j\omega T/2})e^{-j\omega T/2}$

$\displaystyle =$ $\displaystyle 2\cos(\omega T/2)e^{-j\omega T/2},$

where the amplitude response is then given by

$\displaystyle G(\omega)$ $\displaystyle =$ $\displaystyle \left\vert 2\cos(\omega T/2)e^{-j\omega T/2}\right\vert$

$\displaystyle =$ $\displaystyle 2\cos(\pi fT), \qquad \vert f\vert\le f_s/2,$

and the phase response is given by

$\displaystyle \Theta(\omega) = -\frac{\omega T}{2} = -\pi\frac{f}{f_s}, \qquad \vert f\vert\le f_s/2.$

``Mus 270a: Introduction to Digital Filters'' by Tamara Smyth, Department of Music, University of California, San Diego.

$\displaystyle H(e^{j\omega T})$	$\displaystyle =$	$\displaystyle (1+e^{-j\omega T})$
	$\displaystyle =$	$\displaystyle (e^{j\omega T/2} + e^{-j\omega T/2})e^{-j\omega T/2}$
	$\displaystyle =$	$\displaystyle 2\cos(\omega T/2)e^{-j\omega T/2},$

$\displaystyle G(\omega)$	$\displaystyle =$	$\displaystyle \left\vert 2\cos(\omega T/2)e^{-j\omega T/2}\right\vert$
	$\displaystyle =$	$\displaystyle 2\cos(\pi fT), \qquad \vert f\vert\le f_s/2,$