Introducing Fractional-Order Dynamics to Sigma–Delta Modulators

Abstract

The aim of the present paper is to investigate the performance of a fractional-order sigma–delta modulator wherein the integer-order integrator is replaced by a fractional integrator of order $\alpha \,(1 <\alpha < 2)$. A generalized approach to both linear frequency domain and non-linear time domain modeling and characterization of fractional-order sigma–delta modulator has been discussed. The performance of such modulator has been studied and compared with the corresponding integer-order modulators through simulation.

Appendix

For fractional-order modulators, $H_n(e^{j\omega })$ is given by replacing z by $e^{j\omega }$ in Eq. (10), i.e.

$$\begin{aligned} H_n(e^{j\omega }) = \frac{(1-e^{-j\omega })^\alpha }{e^{-j\omega }+(1-e^{-j\omega })^\alpha }. \end{aligned}$$

Using the fact that if a stationary random process with power spectral density $P(e^{j\omega })$ is input to a linear filter with transfer function $H(e^{j\omega })$, the power spectral density of the output process is $P(e^{j\omega })|H(e^{j\omega })|^2$, the power spectal density of quantization noise at modulator output is given by: $P_{ny}(e^{j\omega }) = P_n(e^{j\omega })|H_n(e^{j\omega })|^2$. Assuming white noise process, in-band noise power $\sigma _{ny}^2$ at the output of A/D is:

$$\begin{aligned} \sigma _{ny}^2 = \frac{1}{2\pi }\int _{-\omega _B}^{\omega _B}P_n(e^{j\omega })|H_n(e^{j\omega })|^2 \hbox {d}\omega = \frac{\sigma _n^2}{2\pi }\int _{-\omega _B}^{\omega _B}|H_n(e^{j\omega })|^2 \hbox {d}\omega . \end{aligned}$$

Unlike integer-order modulators, neither it is easy to perform the integration directly to get the in-band noise power nor to arrive at a tidy and closed form equation for in-band noise power as a function of oversampling ratio for fractional-order case which leaves only way of numeric integration to perform the task. Despite availability of several techniques, trapezoidal rule has been utilized in this work to perform the integration for its simplicity. After this, putting the values in the definition of SNR we found

$$\begin{aligned} SNR = 10\log \bigg (\frac{\sigma _x^2}{\sigma _{ny}^2}\bigg ) = 10\log \bigg (\frac{\sigma _x^2}{\sigma _e^2}\bigg )-10\log \bigg (\frac{1}{2\pi }\int _{-\omega _B}^{\omega _B}|H_n(e^{j\omega })|^2 \hbox {d}\omega \bigg ). \end{aligned}$$