Attenuation of ``Harmonics''

On each pass through the delay-line loop, a partial at frequency $f$ is subject to an attenuation equal to the loop amplitude response $\vert H(\omega T)\vert$.

The frequency response $H(\omega T)$ of the simple lowpass filter may be found by testing with a complex sinusoid $x(n) = e^{\omega nT}$:

$$y(n) = x(n) + x(n-1)$$

$$= e^{j\omega nT} + e^{j\omega (n-1)T}$$

$$= e^{j\omega nT} + e^{j\omega nT}e^{-j\omega T}$$

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

$$= (1 + e^{-j\omega T})x(n),$$

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

The gain of the filter is given by

$$G(\omega) = \vert H(e^{j\omega T})\vert$$

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

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

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

$$= 2\cos(\omega T/2)$$