I have a gaussian signal $x$ with zero mean and unit variance. I am quantizing it by using a uniform quantizer of step size $q$ calculated by

$$q = \frac{x_\mathrm{max} - x_\mathrm{min}}{2^{M_\mathrm{ENOB}}},$$

where $M_\mathrm{ENOB}$ is the effective number of bits. Is this the correct way of quantizing a Gaussian signal?

  • $\begingroup$ Your question is unanswerable since you do not provide any explanation of what your symbols mean. A Gaussian random variable takes on values from $-\infty$ to $\infty$. So what does maxRange or minRange mean? For a one-bit quantizer and a three-bit quantizer, read a more reasonable statement of the problem in this problem set and its answer in the solution set. $\endgroup$ Nov 3, 2015 at 15:13
  • $\begingroup$ @DilipSarwate, a long time ago, in the context of audio, i derived an expression essentially for $x_\text{max}$ (and consequently $x_\text{min} = -x_\text{max}$) for a given symmetrical p.d.f. (with zero mean) and a given number of uniformly-spaced quantization levels. there is an optimal value of the quantization spacing so that the total quantization error variance, including the error of the tails of the p.d.f. extending beyond the outside steps of quantization, is minimum. i haven't thought about that problem for decades. i wonder if this is about what zsha is concerned. $\endgroup$ Nov 5, 2015 at 4:20
  • $\begingroup$ @DilipSarwate, i think the result i had was the same or very similar to the Bernhard paper cited by Deve. $\endgroup$ Nov 5, 2015 at 4:24

1 Answer 1


The problem with quantizing Gaussian distributed signals (like the real/imaginary part of an OFDM signal) is that they can take any value in theory. It is thus necessary to clip such signals at threshold $C$ prior to quantization.

Low $C$ increases the distortion noise in this process, while large $C$ will lead to strong quantization noise. It can be shown, that there is an optimum value for $C$ that depends on the variance $\sigma^2$ of the signal and the quantizer resolution $M$. For details see [1,2]. After $C$ has been determined, the step size $q$ is calculated by $$ q=\frac{2C}{2^M} $$

By the way, I don't think it makes sense to use the ENOB (of a DAC) for calculating the step size. The ENOB is a kind of signal-to-noise ratio combining quantization noise and other effects like frequency selectivity and spurious frequencies of the DAC device. The number of input bits is still the nominal number of bits $M$.

Example: let $\sigma^2=1$ and $M=8$. Then with eq. (7) from [2] we can calculate the SNR $\gamma$ after clipping and quantization as a function of the clipping ratio $\zeta=C^2/\sigma^2$. The SNR should be maximized. I don't know a closed-form expression for the optimum $\zeta_\mathrm{opt}$ yielding $\gamma_\mathrm{max}$ but we can determine it numerically. For $M=8$ and with Fig. 2 from [2] we have $\zeta_\mathrm{opt} \approx 11.87\,\text{dB} \approx 15.38$. Furthermore, $C_\mathrm{opt}=\sqrt{\zeta_\mathrm{opt}\sigma^2}$ and with $\sigma^2=1$ we finally get $C_\mathrm{opt}\approx3.92$.

Remark: as you tagged this question you might be dealing with complex signals. In that case be careful with the definition of $\sigma^2$. In my answer it is the variance of a real-valued, Gaussian distributed signal. If the signal in question is complex-valued with independent and identically Gaussian distributed real and imaginary part, it has variance $2\sigma^2$, where $\sigma^2$ is the variance of the real or imaginary part.


[1] D. J. G. Mestdagh, P. Spruyt, and B. Biran, “Analysis of clipping effect in DMT-based ADSL systems,” in Communications, 1994. ICC’94, SUPERCOMM/ICC'94, Conference Record,'Serving Humanity Through Communications.'IEEE International Conference on, 1994, pp. 293–300.

[2] M. Bernhard, D. Rörich, T. Handte, and J. Speidel, “Analytical and numerical studies of quantization effects in coherent optical OFDM transmission with 100 Gbit/s and beyond,” in ITG Symposium on Photonic Networks, 2012, pp. 34–40. Online

  • $\begingroup$ thank you for your suggestion. Would it be reasonable to use the maximum dynamic range of the quantizer as the clipping threshold C? $\endgroup$
    – zsha
    Nov 3, 2015 at 16:05
  • $\begingroup$ This strongly depends on $\sigma^2$. If $\sigma^2=1$ it still depends on $M$. $\endgroup$
    – Deve
    Nov 3, 2015 at 17:00
  • $\begingroup$ okay, but is there some formula that relates M to C? In my case the variance of the signal is 1 and M is equal to 8. And in the two references u have mentioned, I couldn't see any formula that related M to C. $\endgroup$
    – zsha
    Nov 4, 2015 at 8:35
  • $\begingroup$ As per this paper clipping ratio is equal to the unclipped amplitude of the OFDM signal if its variance equals 1(page 26). So I think it would be reasonable to use the amplitude of the signal as the clipping ratio! $\endgroup$
    – zsha
    Nov 4, 2015 at 9:39
  • $\begingroup$ @zsha Regarding your previous comment: I updated my answer. Regarding your most recent comment: the paper just defines the clipping ratio. However, it doesn't say how we should choose the clipping ratio. Additionally, from a quick look I think they do not take quantization effects into account. $\endgroup$
    – Deve
    Nov 4, 2015 at 9:56

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