2nd order sigma delta modulator

Posted by Krishna Pillai

April 8, 2007

In a previous post, the variance of the in-band quantization noise for a first order sigma delta modulator was derived. Taking it one step furhter, let us find the variance of the quantization noise filtered by a second order filter.

With a first order filter, the quantization noise passes through a system with transfer function $H_n(z) = 1 - z^{-1}$ (Refer Eq. 9.2.17 in [1]). The frequency response of $H_n(z)$ is

$H_n(F) = 2\left|sin(\frac{\pi F}{F_s})\right|$ (Refer Eq. 9.2.18 in [1]).

For a sigma delta modulator with a second order filter, transfer function of the noise shaping filter is $H_n(z) = [1 - z^{-1}]^2$ .

The frequency response of $H_n(z)$ is

$H_n(F) = 4sin^2(\frac{\pi F}{F_s})$ .

The quantization noise in the desired signal bandwith $B$ can be computed as,

$\begin{eqnarray}\sigma_n^{2} & = & \frac{\sigma_e^2}{F_s}\int_{-B}^{+B}|H_n(F)|^2 \\ & = & \frac{4\sigma_e^2}{F_s}\int_{-B}^{+B} \left[1 - cos(\frac{2\pi F}{F_s})\right]^2 \\ & = & \frac{4\sigma_e^2}{F_s}\int_{-B}^{+B} \left[1 - 2cos(\frac{2\pi F}{F_s}) + \frac{1}{2} + \frac{1}{2}cos(\frac{4\pi F}{F_s})\right]\end{eqnarray}$ .

Applying limits of the integration,

$\begin{eqnarray}\sigma_n^{2}& = & \frac{4\sigma_e^2}{F_s} \left[3B - \frac{2F_s}{\pi}sin(\frac{2\pi F}{F_s}) + \frac{F_s}{4\pi}sin(\frac{4\pi F}{F_s})\right]\\& = & \frac{4\sigma_e^2}{\pi} \left[\frac{3B \pi}{F_s} - 2sin(\frac{2\pi B}{F_s}) + \frac{1}{4}sin(\frac{4\pi B}{F_s})\right]\end{eqnarray}$ .

Assuming $F_s>>B$ , the three term Taylor series expansion for sine wave is $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120}$ .

Expanding,

$\begin{eqnarray}\sigma_n^{2} & = & \frac{4\sigma_e^2}{\pi} \left[\frac{3B \pi}{F_s} - 2\left(\frac{2\pi B}{F_s} - \frac{8}{6}\frac{\pi^3 B^3}{F_s^3} + \frac{32}{120}\frac{\pi^5 B^5}{F_s^5} \right) + \frac{1}{4}\left(\frac{4\pi B}{F_s} - \frac{64}{6}\frac{\pi^3 B^3}{F_s^3} + \frac{1024}{120}\frac{\pi^5 B^5}{F_s^5} \right)\right]\\ & = & \frac{4\sigma_e^2}{\pi} \left( -\frac{64}{120}\frac{\pi^5 B^5}{F_s^5} + \frac{256}{120}\frac{\pi^5 B^5}{F_s^5}\right)\\& = & \frac{4\sigma_e^2}{\pi} \left( \frac{8}{5}\frac{\pi^5 B^5}{F_s^5}\right) \\& = & \frac{\pi^4\sigma_e^2}{5} \left(\frac{2 B}{F_s}\right)^5\end{eqnarray}$ .

Note that, the above equation is different from the answer provided in Problem 9.11(c) of [1]. Per my understanding factor of $\pi^3$ is missing in Problem 9.11(c).