0
\$\begingroup\$

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!

\$\endgroup\$
2
\$\begingroup\$

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)\$.

|improve this answer|||||
\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.