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)\$.
