Following is the problem:

A sinusoidal signal $$A_c\cos(2πf_ct + φ)$$, where φ is uniformly distributed over $$[−π, π]$$, is quantized by a 1-bit quantizer. Assume that the signal utilizes all the representation levels of the quantizer. What will be the signal-to-quantization noise ratio?

I have done some basic problems on quantization but I am not sure how to proceed when a probability distribution is given.

Any help is appreciated. Thanks!


without loss of generality, it doesn't matter what \$\varphi\$ is and it can be assumed \$ A_c \ge 0 \$.

since this is a 1-bit quantizer, the sole quantizing threshold is at 0. then the error will depend on the ratio of the output level of the quantizer and \$ A_c \$. let the 1-bit quantizer be defined as:

$$ \hat{x}(t) \triangleq \frac{\Delta}{2}\operatorname{sgn}\{ x(t) \} $$

where \$ x(t) = A_c \sin(2 \pi f_c t + \varphi) \$ and the stepsize \$ \Delta > 0 \$.

and the quantization error is

$$ \epsilon_x(t) \triangleq \hat{x}(t) - x(t) $$

if the output level of the 1-bit quantizer is \$\pm 1\$ (which means \$\Delta=2\$) and \$ A_c \$ is a million, you can expect an awful lotta quantization error. but this 1-bit quantizer will behave exactly the same as if \$ A_c = 2 \$ in which the quantization error would be far less. somehow, a 1-bit quantizer has an inherent gain factor in it and that has to be modeled from some knowledge of the nature of the amplitude of \$x(t)\$.

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  • \$\begingroup\$ Thanks for your comment but I still do not see how to solve the problem at hand. How do I go about finding the noise here? Sorry if I miss something obvious here. Thanks again! \$\endgroup\$ – Pranav Arora Nov 26 '16 at 8:29
  • \$\begingroup\$ well, when \$x(t)>0\$, then \$\hat{x}(t)=+\frac{\Delta}{2}\$ and when \$x(t)<0\$, then \$\hat{x}(t)=-\frac{\Delta}{2}\$. see if you can come up with an expression for \$\epsilon_x(t)\$. then square it and find the average of the square. \$\endgroup\$ – robert bristow-johnson Nov 26 '16 at 18:36
  • \$\begingroup\$ To find the average, I need to integrate it over something but what is the integration variable here? How do I factor in the distribution of $\varphi$ here? Can you please help me set up the integral for this? \$\endgroup\$ – Pranav Arora Nov 26 '16 at 18:43
  • \$\begingroup\$ turns out that it doesn't matter what \$ \varphi \$ is. you can set it to any number you want. the averaging of \$ |\epsilon_x(t)|^2 \$ is over time and you can do that averaging over one period. \$\endgroup\$ – robert bristow-johnson Nov 26 '16 at 18:45
  • \$\begingroup\$ Ok, let \$ \varphi \$ be zero. Then the integral is: \$ \frac{1}{T}\int_0^{T} (\hat{x}(t)-A_c\sin(2\pi f_c t))^2dt \$. Solving this is not too hard but I fail to see why the distribution doesn't matter here. :/ \$\endgroup\$ – Pranav Arora Nov 26 '16 at 18:49

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